vector
geometry
algebra
polynomial
geometry proposed

[quote="Amir.S"]we have a regular $ p \minus{} \text{gon}$ $ A_1A_2\dots A_p$ and $ O$ be its center.we have vectors $ \overrightarrow \alpha_1,\dots,\overrightarrow \alpha_n$ that vector $ \overrightarrow \alpha_i$ is parallel to $ OA_i$.  if $ \overrightarrow \alpha_1 \plus{} \dots \plus{} \overrightarrow \alpha_p \equal{} 0$
then prove :$ |\overrightarrow \alpha_1| \equal{} \dots \equal{} |\overrightarrow \alpha_p|$ where $ p$ is a prime number.[/quote]

I think you want to add the condition that the lengths of $ \alpha_i$ all are rational... And then again, it is not a geometry problem, but a consequence of the fact that cyclotomic polynomials are irreducible in $ \mathbb{Q}\left[X\right]$.

  darij


determine the mathematical word whose letters have been rearranged:
1. bdeiiiilnsv
2. dehlnoopry
3. aacglnoot
4. ceegimort
5. ceimoopst
6. acdiiimnnrst
7. accdiinnoortt
8. accenost
9. ddeeennpt
10. aeeilnnoptx
11. aaeeltuv
12. ddehinor
13.yhrynageunir.    :P  HAVE FUN!

[quote="eryaman"]errr...didn't you also post this in fun and games?[/quote]
YES, i just want to get more replies................  :lol:

i know none of them...
perhaps i won't recognize them even if you put them in right orders...

We are told that this is from the CPSC.  Thread locked.


Five couples were at a party. If each person shook hands exactly once with everyone else except his/her spouse, how many handshakes were exchanged? (Note: One obviously doesn't shake hands with oneself.)

There were 10 people in total, who shook hands with 8 other people, to make a total of $ 10 \times 8 \equal{} 80$. However, we're overcounting - that is, we counted Person A shaking hands with Person B and Person B shaking hands with Person A twice - so we have to divide by two to get a total answer of $ \boxed{40}$.

Another way to do it:
The total number of handshakes was 10 choose 2 which is 45.

But the couples did not shake hands.

Since there are 5 couples, the answer is $ \boxed{40} $


A quadrilateral has the following vertices: (2,0), (4,5), (5,2), (1,4).
How many square units are in its area?

[geogebra]939681c0825b248237a7d3f479b6ec5d2ade7b58[/geogebra]

Answer: 20-3-2-1.5-1.5=12

Many ways: shoelace, area chasing, Pick's theorem, etc.

Unless my Shoelace Theorem fails, isn't this:

$ \frac{1}{2}  | (2 \times 5 + 4 \times 2 + 5 \times 4 + 1 \times 0 ) - (0 \times 4 + 5 \times 2 + 5 \times 4 + 1 \times 2 )  | = \frac{1}{2}  | 38-32  |=\boxed{3}$...

But that doesn't seem right, because just graphing it looks bigger...

You have to put the first point twice...

So:

$ (2,0)$
$ (1,4)$
$ (4,5)$
$ (5,2)$
$ (2,0)$

Which yields an answer of $ 12$.


[quote="Temperal"]I noticed the "Featured Article" box is gone from the front page. Did WotW replace it?[/quote]

I guess so: [url]http://www.artofproblemsolving.com/Wiki/index.php?title=Main_Page&diff=prev&oldid=15807[/url]

I think it should be returned. WotW is an entirely different concept from featured articles, almost the opposite, really.


A grocer bought 15 dozen oranges at $1.00 a dozen. She threw away 20 rotten oranges,
and then sold the rest at 8 oranges for 85 cents. How much profit did the grocer make,
in dollars and cents?

let's do it without triggering LATEX...
[hide]she brought everything for 15 dollars. after throwing 20 away, she had 160 oranges. so she had 20 8-orange packs, selling for 17 dollars.
She make a 2 dollar profit. [/hide]

15*12=180=15d
15\12=125 cents
so an orange is 125 cents
180-20=160
she had left oranges worth 2 dollars
20*85c= 1700c=17d
she gained 17 dollars by selling them  but lost 15 dollars by buying them
so she gained a total of 2 dollars and zero cents


[color=darkred]$1.\blacktriangleright$ Se dau doua drepte concurente neperpendiculare $d_{1}$ , $d_{2}$ si constantele reale pozitive $k_{1}$ , $k_{2}$ , $k\ .$ Sa se determine locul geometric al punctelor $L$ din planul $\pi=(d_{1},d_{2})$ pentru care $k_{1}\cdot \delta_{d_{1}}(L)+k_{2}\cdot \delta_{d_{2}}(L)=k\ .$ Sa se stabileasca relatia intre $k_{1}$ , $k_{2}$ , $k$ si masura unghiului ascutit dintre cele doua drepte ca locul geometric sa fie nevid ([u]relatia de compatibilitate[/u]).
[b]Aplicatie.[/b] Sa se arate ca intr-un patrulater circumscriptibil  mijloacele diagonalelor si centrul cercului inscris sunt coliniare ([u]dreapta lui Newton[/u]).[hide="Raspuns."]Un paralelogram.[/hide]
$2.\blacktriangleright$ Se dau doua puncte fixe $A$ , $B$ si constantele reale nenule $k_{1}$ , $k_{2}$ , $k$. Sa se determine locul geometric al punctelor $L$ pentru care $k_{1}\cdot LA^{2}+k_{2}\cdot LB^{2}=k$. Sa se stabileasca relatia de compatibilitate. Discutie.
[b]Aplicatie.[/b] Se considera patrulaterul convex $ABCD$. Sa se determine locul geometric al punctelor $L$ pentru care $LA^{2}+2\cdot LB^{2}+3\cdot LC^{2}=6\cdot LD^{2}+k\cdot [ABCD]$. [hide="Raspuns."] $k_{1}+k_{2}=0\Longrightarrow$ O dreapta perpendiculara pe $AB$;
in caz contrar $\Longrightarrow$ un cerc cu centrul pe dreapta $AB$.[/hide]
$3.\blacktriangleright$ Se da un triunghi $ABC$, o dreapta (un cerc) $d\ (w)$, un punct fix $F$ si un punct mobil $M\in d\ (w)$. Sa se determine locul geometric al punctului $L$ pentru care $\triangle FML\sim\triangle ABC$.
[b]Aplicatie.[/b] Fiind date doua drepte $d_{1}$ , $d_{2}$ si un cerc $w$, sa se construiasca cel putin un triunghi $XYZ$ asemenea cu un triunghi dat (in particular $XYZ$- triunghi echilateral) avand $X\in d_{1}$ , $Y\in d_{2}$ si $Z\in w$.[hide="Raspuns."]Reuniunea a doua drepte (a doua cercuri).[/hide]
[b]Notatii[/b].
$\delta_{d}(X)$ - distanta punctului $X$ la dreapta $d$;
$[XYZT]$- aria patrulaterului convex $XYZT$.[/color]

1.Notam distanta de la $L$ la dreapta $d_{1}$ cu $LL_{1}$ si distanta de la $L$ la dreapta $d_{2}$ cu $LL_{2}$.
Daca $LL_{2}=0$ rezulta $L\equiv M\in d_{2}\Rightarrow LL_{1}\equiv MM_{1}=\frac{K}{K_{1}}$
Analog
Daca $LL_{1}=0$ rezulta $L\equiv N\in d_{1}\Rightarrow LL_{2}\equiv NN_{2}=\frac{K}{K_{2}}$
Vom arata ca segmentul $MN$ apartine locului geometric.
Fie $L\in MN$. Rezulta usor $\frac{LL_{1}}{MM_{1}}=\frac{NL}{MN}\Rightarrow LL_{1}=\frac{NL}{NM}\cdot \frac{K}{K_{1}}\ (1)$
Analog
$\frac{LL_{2}}{NN_{2}}=\frac{ML}{MN}\Rightarrow LL_{2}=\frac{ML}{NM}\cdot \frac{K}{K_{2}}\ (2)$
Din $(1)$ si $(2)$ avem $K_{1}\cdot LL_{1}+K_{2}\cdot LL_{2}=\frac{K}{NM}(ML+NL)=\frac{K}{NM}\cdot NM=K$
Notam $M'$ si $N'$ simetricele lui $M$ si $N$ fata de intersectia $O$ a dreptelor $d_{1}$ si $d_{2}$. Evident ca $M'N'$ apartine locului geometric.
Luam $L'\in MN'$. Ducem analog perpendicularele pe $d_{1}$ si $d_{2}$ si exact ca mai inainte aratam ca si punctele segmentului $MN'$ apartin locului geometric si evident ca si punctele segmentului $M'N$ apartin locului geometric. Asadar locul geometric este constituit din laturile paralelogramului $MN'M'N$.
Solutia este valabila pentru orice $K_{1},\ K_{2},\ K$ pozitive si nu depinde nici de unghiul ascutit dintre cele doua drepte. Daca $K_{1}=K_{2}$ paralelogramul devine dreptunghi.

2.[u]Cazul I[/u] $k_{1}k_{2}>0$ si pentru compatibilitate $k$ are semnul lui $k_{1}$ si $k_{2}$.
Se reduce la $k_{1},\ k_{2},\ k$ pozitive.
Impartim segmentul fix $AB$ in $k_{1}+k_{2}$ parti de lungime
$p$. Fie $D\in AB$ astfel incat $AD=k_{2}p$ si $DB=k_{1}p$ Conform
teoremei lui Stewart, pentru orice $L$ avem:
$pk_{1}LA^{2}+pk_{2}LB^{2}=pLD^{2}(k_{1}+k_{2})+k_{1}k_{2}(k_{1}+k_{2})p^{3}$
Deci
$k_{1}LA^{2}+k_{2}LB^{2}=LD^{2}(k_{1}+k_{2})+k_{1}k_{2}(k_{1}+k_{2})p^{2}$
Rezulta
$LD=\sqrt{\frac{k-k_{1}k_{2}(k_{1}+k_{2})p^{2}}{k_{1}+k_{2}}}$ $D$ este fix, iar $LD$ este constant, rezulta locul geometric este
cercul de centru $D$ si raza
$\sqrt{\frac{k-k_{1}k_{2}(k_{1}+k_{2})p^{2}}{k_{1}+k_{2}}}$

[u]Cazul II[/u]  $k_{1}k_{2}<0$
Presupunem $k_{1}>0$ si notam $k_{2}=-|k_{2}|$
De asemenea presupunem $|k_{2}|>k_{1}$. Generalitatea nu se restrange pentru ca daca conditia nu este indeplinita atunci relatia din enunt se inmulteste cu $(-1)$.
Se imparte segmentul AB in $|k_{2}|-k_{1}$ parti, o parte notandu-se cu $p$. Se ia $D\in AB$ astfel incat $B\in [AD]$ si $BD=k_{1}p$
Se scrie relatia lui Stewart in triunghiul $ADL$ si se obtine $D$ fix si $DL$ constant, deci locul geometric este cercul de centru $D$ si raza $DL$.

Pentru $k_{1}=|k_{2}|$ relatia din enunt devine $LA^{2}-LB^{2}=\frac{k}{k_{1}}$, caz pe care il rezolvam cu teorema lui Pitagora, obtinand ca loc geometric o perpendiculara pe $AB$ la stanga sau la dreapta mijlocului, dupa cum $\frac{k}{k_{1}}$ este negativ sau pozitiv. Pentru $k=0$ se afla mediatoarea segmentului.

[u]Aplicatia de la 1.[/u]
Fie $ABCD$ patrulaterul circumscriptibil, $M$ si $N$ mijloacele diagonalelor $BD$ si $AC$, iar $O$ centrul cercului inscris.
Vom demonstra ca $M,\ N,\ O$ apartin locului geometric de la 1., dreptele $d_{1}$ si $d_{2}$ fiind $AD$, respectiv $BC$.
$k_{1}=AD$, $k_{2}=BC$ si $k=S_{ABCD}$
Pentru $M$ avem:
$AD\cdot d(M,AD)+BC\cdot d(M,BC)=S_{ABD}+S_{BCD}=S_{ABCD}$
Pentru $N$ avem:
$AD\cdot d(N,AD)+BC\cdot d(N,BC)=S_{ADC}+S_{ABC}=S_{ABCD}$
Pentru $O$ avem:
$AD\cdot d(O,AD)+BC\cdot d(O,BC)=r(AD+BC)=r(AD+BC+AB+CD)\cdot \frac{1}{2}=S_{ABCD}$

[u]Aplicatia de la 2.[/u]
Fie $D_{1}\in [AB]$ astfel incat $\frac{AD_{1}}{AB}=\frac{2}{3}$
Din Stewart in triunghiul $LAB$, dupa simplificarea cu $\frac{AB}{3}$ avem:
$LA^{2}+2LB^{2}=3LD_{1}^{2}+\frac{2}{3}AB^{2}$
Inlocuim in relatie si avem:
$3LD_{1}^{2}+3LC^{2}-6LD^{2}=k\cdot [ABCD]-\frac{2}{3}AB^{2}$
Luam $D_{2}$ mijlocul lui $CD_{1}$ si cu teorema medianei avem:
$3(\frac{4LD_{2}^{2}+CD_{1}^{2}}{2})-6LD^{2}=k\cdot [ABCD]-\frac{2}{3}AB^{2}$
Deci
$LD_{2}^{2}-LD^{2}=\frac{k\cdot [ABCD]-\frac{3}{2}CD_{1}^{2}-\frac{2}{3}AB^{2}}{6}$
Locul geometric este deci o dreapta perpendiculara pe $DD_{2}$.

[quote="Virgil Nicula"][color=darkred]$2.\blacktriangleright$ Se dau doua puncte fixe $A$ , $B$ si constantele reale nenule $k_{1}$ , $k_{2}$ , $k$. Sa se determine locul geometric al punctelor $L$ pentru care $k_{1}\cdot LA^{2}+k_{2}\cdot LB^{2}=k\ \ (*)$. Sa se stabileasca relatia de compatibilitate. Discutie.
[b]Aplicatie.[/b] Se considera patrulaterul convex $ABCD$. Sa se determine locul geometric al punctelor $L$ pentru care $LA^{2}+2\cdot LB^{2}+3\cdot LC^{2}=6\cdot LD^{2}+k\cdot [ABCD]$. [hide="Raspuns."] $k_{1}+k_{2}=0\Longrightarrow$ O dreapta perpendiculara pe $AB$;
in caz contrar $\Longrightarrow$ un cerc cu centrul pe dreapta $AB$.[/hide][/color][/quote]
[color=darkblue][b]Demonstratie.[/b] Pentru $k_{1}+k_{2}=0$ relatia $(*)$ devine "diferenta $LA^{2}-LB^{2}$ este constanta", ceea ce inseamna ca locul geometric cautat este o dreapta perpendiculara pe dreapta $AB$. Presupunem $k_{1}+k_{2}\ne 0$ si notam punctul $C\in AB$ pentru care $k_{1}\cdot \overline{CA}+k_{2}\cdot\overline{CB}=0\ \ (1)$. Se observa usor ca relatia $(1)$ inseamna $(2)\ \{\begin{array}{c}(k_{1}+k_{2})\cdot\overline{CA}=-k_{2}\cdot\overline{AB}\\\ (k_{1}+k_{2})\cdot\overline{CB}=-k_{1}\cdot \overline{AB}\end{array}$.
In triunghiul $ALB$ aplicam relatia Stewart cevienei $[LC$ :
$LA^{2}\cdot\overline{BC}+LB^{2}\cdot\overline{CA}+LC^{2}\cdot\overline{AB}+\overline{BC}\cdot\overline{CA}\cdot\overline{AB}=0$.
Folosind relatiile $(2)$ obtinem :
$(k_{1}+k_{2})\cdot (k_{1}\cdot LA^{2}+k_{2}\cdot LB^{2})=(k_{1}+k_{2})\cdot LC^{2}+k_{1}k_{2}\cdot AB^{2}$, adica lungimea segmentului $LC$ este constanta si punctul $C$ este fix, ceea ce inseamna ca locul geometric cautat in acest caz este un cerc cu centrul pe dreapta $AB$. Acest cerc exista daca si numai daca se verifica urmatoarea relatie de compatibilitate : $\boxed{\ k(k_{1}+k_{2})\ge k_{1}k_{2}\cdot AB^{2}\ }$.[/color]

3.Fie $\frac{FL}{FM}=\frac{AC}{AB}=K$, constant.
Avand pe FM luam un L' pe FM astfel incat $FL'=FM\cdot K$ si apoi rotim pe FL' cu unghiul A. Astfel L' se suprapune peste L.
Observam ca L' descrie prin omotetie o dreapta d', paralela cu d, sau un cerc C', omotetic cu C.
Roting figura respectiva cu unghiul A, obtinem locul geometric cautat.
In cazul dreptei locul geometric va fi o dreapta care face unghiul A cu dreapta data.
In cazul cercului locul geometric va fi un cerc cu raza de k ori mai mare, iar unghiul format de dreptele care unesc centrele cercurilor cu punctul F va fi de asemenea A.

[u]Aplicatia 3[/u]
Luam varful $Z'$ pe cerc, punct fix.
Daca $Y'$ descrie dreapta $d_{1}$ atunci locul geometric al lui $X'$ astfel incat $X'Y'Z'\sim XYZ$ este, conform problemei 3, o dreapta ce face unghiul $XZY$ cu $d_{1}$ si care se gaseste conform solutiei respective. Atunci $X'$ se va gasi la intersectia dintre dreapta respectiva si $d_{2}$
Avand $X'$ si $Z'$ gasim pe $Y'\in d_{1}$


[quote="Seo-Yeon, Sohn"]$ \frac{ab}{c(c\plus{}a)} \plus{} \frac{bc}{a(a\plus{}b)} \plus{} \frac{ca}{b(b\plus{}c)}\ge \frac{a}{a\plus{}c} \plus{} \frac{b}{b\plus{}c} \plus{} \frac{c}{c\plus{}b}$
[/quote]

you are right, ElChapin...

how about you to try to prove it??

i tried it ,, but i can't

Is there a typo...like...$ \frac{b}{b\plus{}c} \rightarrow \frac{b}{a\plus{}b}$?

[quote="Zellex"]Is there a typo...like...$ \frac {b}{b \plus{} c} \rightarrow \frac {b}{a \plus{} b}$?[/quote]

 sry, it is typo...


[quote="keta"]Hi, are there any awards or scholarships that are currently awarded to the top scorers on the two tests?[/quote]

There were scholarships in 2001 and 2002, but they lost the funding. None before or since, although I suppose it could happen again if they got some funding for it.

There are substantial prizes for high performance on the USAMO. The Clay Math Award carries a prize of $\$5000 (given for most elegant proof), while the third, second, and first prizes are somewhere around $\$10,000, $\$15,000, and $\$20,000(maybe 25,000 - not sure).


A tiling is assigned a product number based on the number of triangles and quadrilaterals that are visible in the tiling. A tiling's product number is equal to three times the number of triangles plus four times the number of quadrilaterals. What is the product number assigned to the pictured tiling?
[asy]import olympiad; size(100); import geometry_dev; import graph; defaultpen(linewidth(0.8));
draw(origin--(1.5,0)--(1.5,1)--(0,1)--cycle); draw(origin--(0.75,1)--(1.5,0));[/asy]

There are three triangles and three quadrilaterals (a trapezoid on either side plus the rectangle). 3(3) +3(4)=9+12= 21.


$ f$ is a real-valued function on the reals such that $ | f(x) |\geq 1$ and $ f(x \plus{} \frac {13}{42}) \plus{} f(x) \equal{} f(x \plus{} \frac{1}{6}) \plus{} f(x \plus{} \frac {1}{7})$ for all $ x$. Show that there is a real number $ c > 0$ such that $ f(x \plus{} c) \equal{} f(x)$ for all $ x$.

[quote="tdl"]$ f$ is a real-valued function on the reals such that $ | f(x) |\geq 1$ and $ f(x \plus{} \frac {13}{42}) \plus{} f(x) \equal{} f(x \plus{} \frac {1}{6}) \plus{} f(x \plus{} \frac {1}{7})$ for all $ x$. Show that there is a real number $ c > 0$ such that $ f(x \plus{} c) \equal{} f(x)$ for all $ x$.[/quote]
Must be $ |f(x)|\leq 1$


limit
floor function
inequalities
calculus
calculus computations

Denote with $ f(k)$ greatest odd divisor (divider) of $ k\in\mathbb{N}$. Calculate
(a) $ L: \equal{}\lim_{n}\frac{1}{n}\sum_{k\equal{}1}^{n}\frac{f(k)}{k}$
(b)$ \lim_{N\rightarrow\infty}\sum_{n\equal{}1}^{N}(\sum_{k\equal{}1}^{n}\frac{f(k)}{k}\minus{}nL)$

For $ 2^{m}\leq n <2^{m+1}$ let $ \lfloor x\rfloor$ denote the largest integer smaller or equal to $ x$ and set $ T_{n}^{\pm}=\frac{2}{3}n\left(1-4^{-(1+m)}\right)\pm\left(1-2^{-(1+m)}\right)$ and calculate:

$ S_{n}: =\sum_{k=1}^{n}\frac{f(k)}{k}=\sum_{i=0}^{m}\sum_{j=0}^{\lfloor\frac{n-2^{i}}{2^{i+1}}\rfloor}\frac{f(2^{i}(2j+1))}{2^{i}(2j+1)}=\sum_{i=0}^{m}\sum_{j=0}^{\lfloor\frac{n-2^{i}}{2^{i+1}}\rfloor}2^{-i}=\sum_{i=0}^{m}2^{-i}\lfloor\frac{n+2^{i}}{2^{i+1}}\rfloor$

i.e. from the inequalities $ \frac{n-2^{i}}{2^{i+1}}<\lfloor\frac{n+2^{i}}{2^{i+1}}\rfloor\leq\frac{n+2^{i}}{2^{i+1}}$ we can deduce $ T_{n}^{-}<S_{n}\leq T_{n}^{+}$

This immediately gives $ L=\frac{2}{3}$ and for the second part let $ n=\sum_{k=0}^{m_{n}}\epsilon_{k,n}2^{k}$ with $ \epsilon_{k,n}\in\{0,1\},\epsilon_{m,n}=1$ (i.e. we have updated the $ m$-notation to $ m_{n}$).

This gives $ \lfloor\frac{n+2^{i}}{2^{i+1}}\rfloor=\epsilon_{i,n}+\sum_{k=i+1}^{m_{n}}\epsilon_{k,n}2^{k-1-i}$ and substitution into $ S_{n}$ equates to $ S_{n}=\frac{2}{3}n+\frac{1}{3}\sum_{k=0}^{m_{n}}\epsilon_{k,n}2^{-k}$. In particular, we observe $ S_{n}>nL$ and $ R_{N}=\sum_{n=1}^{N}(S_{n}-nL)$ is an increasing sequence.

For the special case of $ N=2^{M}$ we compute

$ R_{N}=\frac{1}{3}\sum_{n=1}^{2^{M}}\sum_{k=0}^{m_{n}}\epsilon_{k,n}2^{-k}=\sum_{k=0}^{M}2^{-k}\left|\left\{n=1,\ldots,2^{M}|k\leq m_{n}\right\}\right|=\\ \;\;\;\;=\sum_{k=0}^{M}2^{-k}(2^{M}-2^{k}+1)\end{array}$

which converges to infinity

P.S. (hopefully no typos)

I spotted a typo, although it does not change the conclusions; the last lines should correctly read

$ R_{N}\equal{}\frac{1}{3}\sum_{n\equal{}1}^{2^{M}}\sum_{k\equal{}0}^{m_{n}}\epsilon_{k,n}2^{\minus{}k}\equal{}\\ \;\;\;\equal{}\frac{1}{3}\sum_{k\equal{}0}^{M}2^{\minus{}k}\left|\left\{n\equal{}1,\ldots, 2^{M}|k\leq m_{n},\epsilon_{k,n}\equal{}1\right\}\right|\equal{}\\ \;\;\;\equal{}\frac{1}{3}\sum_{k\equal{}0}^{M}2^{\minus{}k}\left|\left\{n\equal{}1,\ldots,2^{M}|\epsilon_{k,n}\equal{}1\right\}\right|\equal{}\\ \;\;\;\equal{}\frac{1}{3}\left[2^{\minus{}M}\plus{}\sum_{k\equal{}0}^{M\minus{}1}2^{\minus{}k}2^{M\minus{}1}\right]\equal{}\frac{1}{3}[2^{M}\plus{}2^{\minus{}M}\minus{}1]$

which converges to infinity, too


geometry circumcircle inradius inequalities AMC

Consider acute $\triangle ABC$. Let $R$, $r$, and $h_c$ denote the circumradius, inradius, and height to side $AB$ from point $C$ respectively.

Prove $R+r \leq h_c$

Is this really true? Let $C = \frac{\pi}{2} - 2\epsilon$, $A = B = \frac{\pi}{4} + \epsilon$ and $\epsilon \rightarrow 0$. Then we have $h_c = \frac{c}{2} = R$, but $r = \frac{a+b-c}{2} = \frac{c\sqrt{2} - c}{2}$ so that $R + r = \frac{\sqrt{2}}{2}c > \frac{c}{2} = h_c$.


algebra unsolved
algebra
2-variable inequality
Diophantine equation

Determine all pairs of positive numbers $a,b$ such that

\[9(a+b) +\frac{1}{a}+\frac{1}{b} \geq 10 +\frac{a}{b} + \frac{b}{a}\]

[i]unknown source[/i]

Dear Manlio,

Why did you send this one to Number theory? Maybe you forgot something here?

Namdung

well I'll move it to algebra, it seems like an algebra problem, if anything was forgotten by manlio I'll move it back! :D :D

I dont see the meaning of this problem. But I have do solve it.

1) a = b=1/3;
2) b>=1 or b<=1/9 and a >= positive root of
      (9b-1)x^2 + (9b^2-10b+1)x + b-b^2 = 0

Namdung


Find, as a proper base ten fraction, the sum of the following infinite series of repeating decimals, all written in base eight:

S = .07* + .007* + .0007* + ...   (where * indicates a repeating digit)

I did it slightly different:
8S=1+S (.7* in base 8 is 1, just like .9* in base 10 is 1)

Solvins for S easily gives you 1/7.


Find the minimum value of $ \sum_{k \equal{} 1}^{2n}|x \minus{} k|$, where $ x\in\mathbb{R}$.

[hide="yay"]
Let $ E(n) \equal{} \min\sum_{k\equal{}1}^{n} |x\minus{}k|$.

For odd $ n \equal{} 2k\plus{}1$ we have $ E(n) \equal{} \frac{n^2\minus{}1}{4}$.

For even $ n \equal{} 2k$ we have $ E(n) \equal{} \frac{n^2}{4}$.

Proof: For odd $ n$, we pair up the integers in the interval $ [1,n]$ as $ (1,n), (2,n\minus{}1), (3,n\minus{}2), \cdots, (k,k\plus{}2)$ with $ k\plus{}1$ left over.  Consider the $ k\minus{}1$ intervals $ [1,n]$, $ [2,n\minus{}1]$, $ \cdots$.  Obviously if $ x$ is in one of these intervals $ [a,b]$, then the sum $ |x\minus{}a|\plus{}|x\minus{}b|$ is constant and less than the sum when $ x \notin [a,b]$.  Hence the sum attains a minimum when $ x$ is in all of the intervals - that is, when $ x \in [k,k\plus{}2]$.  But we have the additional $ |x\minus{}(k\plus{}1)|$ term.  Hence the minimum occurs when $ x \equal{} k\plus{}1$.  The sum is relatively easily computed as $ E(n) \equal{} 2\sum_{i\equal{}1}^k i \equal{} k^2\plus{}k \equal{} \frac{n^2\minus{}1}{4}$.

For even $ n$, we have the intervals $ [1,n], [2,n\minus{}1], \cdots, [k,k\plus{}1]$.  By similar reasoning the minimum occurs when $ x \in [k,k\plus{}1]$.  Obviously $ E(n) \equal{} \sum_{i\equal{}1}^k 2i\minus{}1 \equal{} k^2 \equal{} \frac{n^2}{4}$.
[/hide]


Let $ A,B,C,D,E$ be points on a circle such that $ \angle ABC\equal{}\angle BCD\equal{}\angle CDE\equal{}45^{\circ}$. Prove that the region above the zig-zag line $ ABCDE$ and the region below it have equal areas.

Let $ X \equiv CB \cap AE , Y \equal{} CD \cap AE.$ Then $ \triangle CAE$ is right with $ CA \equal{} CE \equal{} L.$ Now, we shall prove that the area of $ \triangle CXY$ equals the sum of  the areas of $ \triangle ABX$ and $ \triangle EDY.$ Since these triangles are similar, because of $ AB \parallel CY$ and $ ED \parallel CX,$ we shall show $ XY^2 \equal{} AX^2 \plus{} EY^2.$ $ \triangle CXY \sim \triangle AYC \sim \triangle EXC$ yields

$ \frac {AY}{L} \equal{} \frac {L}{EX} \Longrightarrow AY \cdot EX \equal{} L^2 \equal{} \frac {_1}{^2}AE^2$

$2(AX \plus{} XY)(EY \plus{} XY) \equal{} (XY \plus{} AX \plus{} EY)^2  \Longrightarrow XY^2 \equal{} AX^2 \plus{} EY^2.$


Whats yours?

I started playing FICS chess today and i got my rating to 1372.

I have been playing on yahoo chess, but recently it has not working....in the middle of games, i get disconnected and my rating drops.  :(  With FICS theres a lot less glitches.

Is this blitz or standard? (blitz is deflated and standard is inflated on FICS..)

you need to specify what rating?

yahoo-1400
FICS- umm 1400?
USCF- 1100 lol

USCF-1223 :D
FICS-1300-ish

What's sad is that my little brother's rating is higher than mine  :rotfl:

lol I'm 1493 USCF

1832 Yahoo

1980 Blitz Fics

yahoo-1400ish, yeah I know I suck cuz I gave up in chess :(

I once had a USCF but haven't kept it up
I do play actively on Gameknot.com (good well reputed chess website) with rating around 1700, although I haven't played very many games (it will be going up... I should be around 1900)

Cool, what's your handle on gameknot?  We can play sometimes.  My is chess55555, and my rating is 1880 right now.   :)

my national is 1765 before the update...

my online is 2200

my gameknot.com is 1781

i drew a 2150 otb today :D

For someone new at playing chess competitively, (I've played several times with friends/family) what is the suggested website?

By ......playing rating tmt of course! :D
Are you new in chess?

I am new to this rating thing.. you  mind explaining. ??

You havent played in competitions then??
Anyway i will try to explain.Rating is just used to describe the stength of a player.For example,a player of rating 2000 is (considered)stronger than the one having 1200.To get the rating you have to score at least some points( i guess min2.5 of 9,I dont know about recent rules) against rated players in a rating tmt(most tmt are rating tmt though).Then you earn rating based on your performance.From then you can increase your rating by beating other rated players.The more your opponents rating the more you gain.If you lose against a rated then you lose some rating depending upon his rating.To become an international master(IM) you need more than 2400,and more than 2600 to become Grand master(GM).At the point of writing, the top rated player in the world is Magnus carlsen with rating above 2800.
So that is all about rating.

I was reading this thread, and saw this:
[quote=BOGTRO]USCF - 1815

ICC Standard - about 1825

ICC Blitz - about 1750 (Yeah I know, crashed)

Draw a GM OTB on Monday =)

btw Alex Lenderman is on ICC as 'manest'

My account is 'Cryptochess'[/quote]
I burst out laughing....and then looked at the timestamp...

[quote=Bob_Smith]I have 384 on chess.com.

I hate timed chess.[/quote]

I was 600 when I stopped. I think I can maybe get to 800, now.
I also hate timed chess, but I hate it even more when you have to give up because your opponent is taking 3 days between each move, and then your rating goes down. :(


Let $ a$, $ b$, $ c$, $ d$ be four positive reals such that $ d\equal{}a\plus{}b\plus{}c\plus{}2\sqrt{ab\plus{}bc\plus{}ca}$. Prove that $ a\equal{}b\plus{}c\plus{}d\minus{}2\sqrt{bc\plus{}cd\plus{}db}$.

\begin{eqnarray*} d & = & a + b + c + 2\sqrt {ab + bc + ca} \\
d - a - b - c & = & 2\sqrt {ab + bc + ca} \\
a^2 + b^2 + c^2 + d^2 + 2ab + 2bc + 2ca - 2ad - 2bd - 2cd & = & 4ab + 4bc + 4ca \\
a^2 + b^2 + c^2 + d^2 - 2ab + 2bc - 2ca - 2ad + 2bd + 2cd & = & 4bc + 4bd + 4cd \\
\left(a - b - c - d\right)^2 & = & \left(2\sqrt {bc + bd + cd}\right)^2 \\
d > a + b + c > a - b - c & \Rightarrow & a - b - c - d < 0 \\
a - b - c - d & = & - 2\sqrt {bc + bd + cd} \\
a & = & b + c + d - 2\sqrt {bc + bd + cd} \end{eqnarray*}
qed.


algorithm
number theory
relatively prime

Prove that $(f_n, f_{n-1})=1$, where $f_n$ is the nth Fibonacci number.

I'm stuck :(.

Hint. .....If two numbers have a common factor, their difference also is divisible by that factor  ... [hide]Think if  $f_{100}$ and $f_{99}$  had common factor (say $d$ ) won't it also divide $f_{98}$ .. How about $f_{97}$ ? ... and  remember $f_1$ has no divisors except 1  :)[/hide]

[quote="jensen"]we have $f_n=f_{n-1}+f{n-2}$ and $f_0=f_1=1$ and we have one more thing $f_n= \frac{(x_1)^n-(x_2)^n}{\sqrt{5}}\leq 1$ and $x_1=\frac{1+\sqrt{5}}{2}$ and ${x_2=\frac{1- \sqrt{5}}{x}}$.
and i think with these we can prove it easily. :D[/quote]

This won't help. Look at Gyan's hint.

Other  similar, well known , and  "easy" to prove  problems are of the type:

  1.   $f_n$ will divide $f_m$ if  $n$ divides $m$.
  2.    If $f_n$ is prime, then $n$ is prime.  ([b]Well , except for the case n=4)[/b]  (Reverse of course  need not be true.)
  3.   For n>4, if $n$ is prime, $f_{n+1}$ is never a prime.

And  one not so easy to prove ('cause if you show this you'd be famous, .. I don't think it has been proven yet)  :)

4.    There are infinite fibonacci primes.

Heres a pretty cool theorem you could use, (if you can prove it!)

$\gcd{(F_n,F_m)}=F_{\gcd{(n,m)}}$

if you use the reverse of the euclidean algorithm, and add one each time, you generate the fibbonacci sequence

[hide="Solution"] We can use induction.  We know that 1 and 1, the first two Fibonacci numbers, are relatively prime.  Now we show that if $F_{n},F_{n+1}$ are relatively prime, then $F_{n+1},F_{n+2}$ are also relatively prime.

Let $m$ be any divisor of $F_{n+1}$.  We know that $F_{n+1}\equiv 0\mod{m}$, and $F_{n}\equiv a_{m}\mod{m}$, for some $a\not\equiv 0\mod{m}$.  Adding the two to generate $F_{n+2}$ in mod $m$, we see that $F_{n+2}\equiv a_{m}\mod{m}$, so $m$ is not a divisor of $F_{n+2}$.  Since this works for any divisor of $F_{n+1}$, we know that $F_{n+1},F_{n+2}$ have no common divisors and are therfore relatively prime.

Q.E.D.[/hide]


that's a nice solution. if you do it the long way you run into some combinatorics that could lead to an error

[quote="drunner2007"]Simplify the sum
$i+2i^2+3i^2+\ldots2007i^{2007}$[/quote]

The solution by [b]Altheman[/b] is great! I would like to suggest yet another way of computing the given complex sum that uses the idea of differentiation. It might look a bit cumbersome at first glance but knowing this method will help you tackle a bunch of similar problems in a more general way. Anyway, here is the method and the solution.

We know
$1 + x + x^2 + x^3 + ... + x^n = \frac{x^{n+1} - 1}{x-1}$

Now differentiating both sides of the equation w.r.t. $x$, we obtain
$1 + 2x + 3x^2 + ... + nx^{n-1} = \frac{(x-1)(n+1)x^n - (x^{n+1}-1)}{(x-1)^2}$

Multiplying both sides of the equation with $x$, we obtain
$x + 2x^2 + 3x^3 + ... + nx^n = \frac{x}{(x-1)^2}.[(n+1)x^{n+1}-(n+1)x^n-x^{n+1}+1]$

$\therefore x + 2x^2 + 3x^3 + ... + nx^n = \frac{x}{(x-1)^2}.[nx^{n+1}-(n+1)x^n+1]$

Now put $x=i$ and $n=2007$ in the above equation to get
$i + 2i^2 + 3i^3 + ... + 2007i^{2007} = \frac{i}{(i-1)^2}.[2007i^{2008}-2008i^{2007}+1]$
$= -\frac12.[2007+2008i+1] = -1004(1+i)$


find the remainder when 2000! is divided 2003?

Hint: By Willson's theorem we have $2002!\equiv-1(\mod 2003)$.

[quote="Xaero"]I hope this is right...

[hide="Solution"]By Wilson's theorem, we have $2002!\equiv-1 (\text{mod 2003})$. This equals $2001!\equiv 1(\text{mod 2003})$.
This can be transformed into $3\cdot 23\cdot 29\cdot 2000!\equiv 2004(\text{mod 2003})=23\cdot 29\cdot 2000!\equiv 668(\text{mod 2003})$
$\Longrightarrow 23\cdot 29\cdot 2000!\equiv-668334(\text{mod 2003})\Longrightarrow 2000!\equiv 1001(\text{mod 2003})$. Our answer is then $\boxed{1001}$. [/hide][/quote]

[hide] After you get $2001! \equiv 1 \pmod{2003}$, we can simply divide by $-2$ to get $2000! \equiv \frac{2004}{-2}\equiv-1002 \equiv 1001 \pmod{2003}$.[/hide]

Yes that is correct.

How would you do it if it said the remainder when 1000! is divided by 2003?

Well, for $1000! \pmod{2003}$, we'll still consider wilson's theorem such that $2002!\equiv-1 \pmod{2003}$. Since we're are in $\mod 2003$, $2002!\equiv 1001! \cdot-(1001!) \equiv-(1001!)^{2}$. This will give two answers $1000!\pmod{2003}=2001$ or $1000!\pmod{2003}=2$ because it is a quadratic.


geometry
parallelogram
geometric transformation
vector
rotation
angle bisector
geometry solved

Are you sure this is the statement? I don't think it's a fixed value.. :?

I saw this problem in a magazine, I think it should be right !

But think about it: we can fix $A,D$, and move $B,C$ on the circle so that $\angle BPC+\angle APD=\pi$. The triangle $PBC$ won't have all its angles fixed, in general.

[quote="payman_pm"]let $ABCD$ be a cyclic quadrilateral and $P$ be a point such that
$\angle APD+\angle BPC=180$ find
$\angle PAD+\angle PCB$[/quote]

Indeed, < PAD + < PCB is not a fixed value. But the following modified version of the problem is correct:

[color=blue]Let ABCD be a parallelogram and P a point inside it such that < APD + < BPC = 180°. Prove that < PAD + < PCB = < BAD.[/color]

The [i]solution[/i] is canonical: Since the quadrilateral ABCD is a parallelogram, we have the vectorial equality $\overrightarrow{AB}=\overrightarrow{DC}$. Now, let Q be a point such that $\overrightarrow{PQ}=\overrightarrow{AB}=\overrightarrow{DC}$ (this point Q can be constructed as the image of the point P under the translation along the vector $\overrightarrow{AB}=\overrightarrow{DC}$). Since $\overrightarrow{PQ}=\overrightarrow{AB}$, the quadrilateral APQB is a parallelogram, and thus QB || PA and < PQB = < PAB.

Since $\overrightarrow{PQ}=\overrightarrow{AB}=\overrightarrow{DC}$, the points Q, B and C are the images of the points P, A and D under a given translation. Since translations preserve angles, we thus have < BQC = < APD. Thus, the equation < APD + < BPC = 180° becomes < BQC + < BPC = 180°. Hence, the quadrilateral BPCQ is cyclic, and it follows that < PQB = < PCB. Comparing this with < PQB = < PAB, we get < PCB = < PAB, and thus < PAD + < PCB = < PAD + < PAB = < BAD. Problem solved.

  Darij

No,you are actually wrong...the problem statement is very corect...and here is the sollution.
So, if we name M the intersection between [BC] and [AP] and Q1 the intersection of [BD] and [AM] and Q2= the int between [CE] and [AM] and also <(BAM)=x,<(MAC)=y,<(ABP)= :alpha: <(PBM)= :beta: <(PCM)= :gamma: ,<(ACP)= :lambda  and <(PMB)= :theta: , it is obvious that <(ABQ)=:alpha/2 and <(APQ2)= :lambda:/2.Thus,we shall prove that Q1==Q2.
AQ1/AP=sin(:alpha:/2)/sin:alpha:
BQ2/BP=sin(:lambda:/2)/sin: lambda:
AQ1/Q1M=AQ1/Q1M and also AQ2/Q2M=AQ2/Q2M
so AQ1/sin( : alpha :/2)=AB/sin(:alpha:/2+APB)
Q1M/sin(:alpha:/2+:beta:)=BM/sin(180-APB)=BQ/sin :theta:
but it is kmowm that sin(180-APB)=sin APB, so
AQ1/Q1m=AB/BM*sin(:alpha:/2)/sin(:alpha:/2+APB)*sin(:alpha:/2+:beta:)/sin(APB) and so on,because at this point the sloveent is obvious...still thinikng of a 100% sollution of my own still.confused

Hmm... sounds like nonsense...  :?

  Darij

[quote="pestich"]This problem is from  Polish olympiad of 1971 problem #4,
 and your saw the official solution  with lambdas and what not.
 Only 2% of participants solved it.

   For APD + BPC =180    P is on AC so that AC is angle bisector of BPD.
   Diagonal BD  makes angle X with diagonal AC  and this angle is
   exactly  PAD + PCB  as half sum of arcs AB and CD.[/quote]

Please explain why you think that P must lie on AC (actually, my sketch proves this wrong)...

  Darij

When P is on AC   then APB+BPC = 180
        If AC is angle bisector of BPD  then APB = APD   and   APD+BPC=180
       as required.

         Maj. Pestich

    P.S.
    Are you pulling my leg (and everything that goes with it) , Darij ?
    To repay me for all the contraversial nonsense I put up here?

[quote="pestich"]For APD + BPC =180    P is on AC so that AC is angle bisector of BPD.
   Diagonal BD  makes angle X with diagonal AC  and this angle is
   exactly  PAD + PCB  as half sum of arcs AB and CD.

   Maj.  Pestich[/quote]

I do not understand why $P$ must lie on the angle bisector $\overline{AC}$ of $\angle{BPD}$. Obviously we can rotate rays $PB$ and $PC$ about the point $P$, which would move $P$ off of the diagonal.

[quote="pestich"]When P is on AC   then APB+BPC = 180
        If AC is angle bisector of BPD  then APB = APD   and   APD+BPC=180
       as required.[/quote]

But, please, Pestich, you showed that if P is a point on the diagonal AC such that AC bisects < BPD, then < APD + < BPC = 180°, and now you claim the converse, namely that if < APD + < BPC = 180°, then P lies on AC and AC bisects < BPD. That's just wrong! You could easily see that there are infinitely many points P inside the quadrilateral ABCD satisfying < APD + < BPC = 180°, but only one point P on the diagonal AC such that AC bisects < BPD (at least, in the non-degenerate case). Hence your argument cannot be true.

  Darij

Calm down guys. Pestich a small request from me would be, if possible, to give a little bit more detailed proofs. I admit that I have tried to read some of your geo proofs, but it is really hard. You leap over huge steps and sometimes we can't make the connections. ;)


How many even four-digit counting numbers can be formed by
choosing digits from the set $ \{1, 2, 3, 4, 5\}$ if digits can be repeated?

The first three-digits can be any one of the $ 5$ digits, as they will not break restrictions.  The last digit, however, must be either $ 2$ or $ 4$ because the number must be even.


In how many ways can we give 6 computers to 3 schools such that each school receives at least a computer??
Consider the cases:1)the computers are distinct( example of arrangement: computer 1 to school 3,computer 2 to school 2,...,computer 6 to school 1)
2)the computers are not distinct(example of arrangement:2 computers at school 1,2 computers at school 2,2 computers at school 3)

Generalization.

[quote="xpmath"][hide="hint"]Say each school gets 1, then the problem reduces to how many ways can you give out 3 computers to 3 school.[/hide][/quote]

Then would it be

[hide]
If the computers are distinct;
Lets call the 3 computers 1, 2 and 3 respectively.
Then all the possible ways to arrange them between the schools would be:

1, 1, 1
0, 1, 2
0, 2, 1
1, 2, 0,
1, 0, 2
2, 1, 0,
2, 0, 1
3, 0, 0
0, 3, 0
0, 0, 3

Adding all them up gives us  10 possibilities.

[/hide]

Faster way-
[hide]Let the computers be $ a_1,a_2,$ and $ a_3$, so $ a_1\plus{}a_2\plus{}a_3\equal{}3$, and the number of solutions by balls and urns is 10.[/hide]

Or, if you think about, we have the five spots in between the 6 computers, and we need to choose 2 of them to insert dividers, so $ \binom{5}{2}\equal{}10$.

Example: Let x=computer, |=divider

xx|x|xxx


I'm new to computer programming, and I was wondering about the very first USACO problem in the training pages: Your Ride Is Here

[size=100][code] It is a well-known fact that behind every good comet is a UFO. These UFOs often come to collect loyal supporters from here on Earth. Unfortunately, they only have room to pick up one group of followers on each trip. They do, however, let the groups know ahead of time which will be picked up for each comet by a clever scheme: they pick a name for the comet which, along with the name of the group, can be used to determine if it is a particular group's turn to go (who do you think names the comets?). The details of the matching scheme are given below; your job is to write a program which takes the names of a group and a comet and then determines whether the group should go with the UFO behind that comet.

 Both the name of the group and the name of the comet are converted into a number in the following manner: the final number is just the product of all the letters in the name, where "A" is 1 and "Z" is 26. For instance, the group "USACO" would be 21 * 19 * 1 * 3 * 15 = 17955. If the group's number mod 47 is the same as the comet's number mod 47, then you need to tell the group to get ready! (Remember that "a mod b" is the remainder left over after dividing a by b; 34 mod 10 is 4.)

 Write a program which reads in the name of the comet and the name of the group and figures out whether according to the above scheme the names are a match, printing "GO" if they match and "STAY" if not. The names of the groups and the comets will be a string of capital letters with no spaces or punctuation, up to 6 characters long.[/code][/size]

How do you read each letter one by one? And for converting each letter to the corresponding integer, I was going to cast each one as an int and do the necessary subraction/addition stuff, but was wondering if there was a shorter way (a method that does that?). I've looked online at the Sun webpage at the InputStreamReader class, but I couldn't find anything  :maybe: [/code]

well... i dont know what do u use c++ or java.

but actually i advise u to take the word as char input. thn loop through the letters, convert them to integer by using their ascii value. then multiply and mod them... thats all.

Hello.

What you can do is this:
[code]
int main(){
char word[5];
ofstream fout ("filename.out");
ifstream fin ("filename.in");

fin >> word[0] >> word[1] >> word[2] >> word[3] >> word[4];

}

[/code]
This should work. You could also do:
[code]
#include <iostream>
#include <string.h>
#include <fstream>

using namespace std;

int main(){
String word;
ofstream fout ("filename.out");
ifstream fin ("filename.in");

fin >> word;

}
[/code]
and from here use the string length function that will give you the characters from n to k (specified in the argument). If you do n to n+1, that will return that character. I don't remember the exact syntax, but I hope this will help.
EDIT: This is c++.

With C++, you'd use cin.get(char& c).

So, as an example:
[hide="C++ Solution"]
[code]
#include <iostream>
using namespace std;

int word_total()
{
  char in;
  int k, total=1;
  cin.get(in);
  while(!cin.fail() && in!='\n')
  {
    k = (int) in - (int) 'A' + 1;
    total *= k%47;
    total = total%47;
    cin.get(in);
  }
  return total;
}

int main()
{
  int word1, word2;
  word1 = word_total();
  word2 = word_total();
  if(word1 == word2)
    cout << "GO";
  else
    cout << "STAY";
  return 0;
}
[/code][/hide]

But you mentioned Sun, so I assume you're using Java. I haven't used Java in a while, but I think what you're looking for is InputStreamReader::read (http://java.sun.com/j2se/1.5.0/docs/api/java/io/InputStreamReader.html#read()). The algorithm will be the exact same as in the C++ example (read a character, add it to the total, total=total%47, and repeat until the character read is a newline).

metehan says  [quote="metehan"][quote]But you mentioned Sun, so I assume you're using Java. I haven't used Java in a while, but I think what you're looking for is InputStreamReader::read (http://java.sun.com/j2se/1.5.0/docs/api/java/io/InputStreamReader.html#read()). The algorithm will be the exact same as in the C++ example (read a character, add it to the total, total=total%47, and repeat until the character read is a newline).
[/quote]

so it is impossible
 :P[/quote]

but i don't understand why it is impossible.. :!:

Hashus, if you can't understand this, please don't write anything to this forum.  :mad:

metehan, I don't see why it's impossible either. Please edify us with why "it" is impossible...

I would recommend reading the "Submitting Solutions" text in the training pages if you haven't already. However, to answer your immediate question for JAVA:

[code]
/*
ID: your_id_here
LANG: JAVA
TASK: ride
*/
import java.io.*;
import java.util.*;

public class ride {
  public static void main (String [] args) throws IOException {
// Use BufferedReader rather than RandomAccessFile; it's much faster
    BufferedReader f = new BufferedReader(new FileReader("ride.in"));    // input file name goes here
    PrintWriter out = new PrintWriter(new BufferedWriter(new FileWriter("ride.out")));

    String s = f.readLine();   // gets entire line
    char[] c = s.toCharArray();
    int[] arr = new int[c.length];
    for(int i=0;i<c.length;i++)
       arr[i]=c[i]-'A'+1;
    //do stuff with your array
    out.close();       // close the output file
    System.exit(0);  // don't omit this!
  }
}
[/code]

You can use out.print() to print to your file, and out.println() to print with a newline at the end. Also, if you want to split up space-separated integers or something (for later problems), I would recommend looking up the StringTokenizer class.


Euler
calculus
number theory proposed
number theory

Can anyone make use of the Euler Product formula to prove Dirichlet's theorem?

PS: Try to find an elementary proof (ie don't use calculus techniques from analytic number theory).

Dirichlet's Theorem is a very difficult result, and I [b]seriously[/b] doubt that an elementary proof exists.  The Euler Product formula is used (on Dirichlet $ L$-series) in the standard proof, but the meat of the proof lies elsewhere.


Given that
f(x+y)=[(f(x)+f(y)]/[1-f(x)f(y)]
prove f(0)=0

firstly, set $f(x)= \tan{g(x)}$

This yields $\tan{g(x+y)}= \displaystyle \frac{\tan{g(x)}+\tan{g(y)}}{1-\tan{g(x)} \tan{g(y)}}$

$\rightarrow \tan{g(x+y)}=\tan{(g(x)+g(y))}$

$\rightarrow  g(x+y)=g(x)+g(y)+n \pi$ for $n$ an element of the integers.

$x=y=0 \rightarrow g(0)=2g(0)+n \pi$
$\rightarrow g(0)=-n \pi$
$\rightarrow \tan{g(0)}=f(0)=0$

Qed

Im setting both of them equal to zero because it gives me loads of $g(0)$'s which is good for what i need. OK?

I assume the function is R to R.

Put f(0) = c.  Try y=0.

Now $f(x) = [f(x) + c]/[1 - cf(x)]$.

It leads to $c[ 1 + f(x)^2 ] = 0$, an identity for all x.  But f(x) is real, so c = 0.

Nice one! I should have tried that, i ust saw tan and ran with it!

but how do you know that it is the tangent function?, i realize that is the expansion identity for the tan function, but that does not mean there are not other functions that satisfy those conditons

I didnt. I simply used tan to simplify it... see? I said $f(x)=\tan{g(x)}$
As opposed to $f(x)= \tan{x}$

Once you define $f(x)=\tan g(x)$, you cannot recover $g$ completely, so that step doesn't really work.

I could put other limits on it to make it work....
Tan not one to one?

The property you use is not being one to one, but being a bijection between an open interval $]-\pi/2,\pi/2[$ and the real line.
For every real $x$, there is plenty of $y$ (surjectivity) such that $\tan (y)=f(x)$ and if we ask $-\pi<2y<\pi$, this $y$ is unique (one-to-oneness); so we can define a function $g$, by putting $g(x)=y$.


The odds of an event happening successfully are expressed as $ a: b$, where $ a: b$ is the ratio of the probability of a successful event to the probability of an unsuccessful event.  The odds of rolling a 5 on an unfair die are 1:3.  What is the probability of rolling a 5 with this die?  Express your answer as a percent.

So isn't it just $ 1/4$ because $ 1: 3$: The 3 is the probability of the 5, and the 3 is the probability of everything else. So it's $ 1/4\equal{}75$%!!

A slight correction jonathanchou711...

$ \frac{1}{4} \neq 75\%$
$ \frac{1}{4} \equal{} 25\%$

But everything else is right :)


AMC USA(J)MO USAMO geometry Stanford college MATHCOUNTS

Below is a list of those high schools with the most USAMO qualifiers over the last 10 years (1996 to 2005).  The schools are listed in descending order.

To conserve space, if a school had 10 or more qualifiers in a year, I represented the number of qualifiers with the code A = 10, B = 11, C = 12, D = 13, E = 14, F = 15, G = 16, and H = 17.

Let me know if you see any errors in the list.

[code]
                     YEAR
              9 9 9 9 0 0 0 0 0 0
SCHOOL  STATE 6 7 8 9 0 1 2 3 4 5 TOTAL
T. Jeff.   VA 5 4 3 4 B E H E E B  97
IMSA	    IL 6 5 3 3 5 8 9 6 6 5  56
Stuyvesant NY 7 5 1 4 5 9 8 8 4 3  54
AAST       NJ 5 2 F 2 2 3 8 5 6 5  53
Homeschool -- 1 3 4 4 9 6 7 5 7 5  51
Blair      MD 2 3 2 0 5 2 9 8 5 7  43
Exeter     NH 1 2 2 1 1 1 4 9 B 8  40
Lexington  MA 2 2 1 1 5 6 4 4 3 3  31
Gunn       CA 4 3 2 2 2 3 3 3 5 2  29
Lynbrook   CA 1 2 1 0 1 4 3 4 3 5  24
Naperv. N. IL 1 0 1 3 2 6 2 3 3 1  22
Mon. Vista CA 1 1 1 2 5 4 3 2 0 2  21
Andover    MA 2 2 1 0 2 3 4 4 0 1  19
Hunter     NY 0 1 2 1 3 4 4 2 2 0  19
W. Windsor NJ 0 0 1 1 4 2 2 4 2 2  18
J. Madison WI 0 1 1 2 3 2 2 2 2 2  17
Madison W. WI 0 0 2 2 4 3 2 2 1 0  16
Saratoga   CA 0 0 0 0 3 2 2 3 2 2  14
NCSSM      NC 3 3 0 2 1 1 1 1 1 1  14
St. Mark's TX 0 0 1 2 2 4 3 1 1 0  14
LB Johnson TX 4 2 1 1 0 1 0 1 1 2  13
Rox. Latin MA 0 0 0 2 1 3 2 3 0 1  12
[/code]

Very Well Done!

Looking at the chart, I wonder if the noticeable rise at Exeter in recent years coincides with Zuming Feng's tenure.

Cool! Interesting list. 17 in one year for TJ is amazing. (and Go Homeschoolers!)

My school (Monta Vista) has had well more than 12 qualifiers.  Note that Stephen Guo, listed in the archives as from "Olympia Institute," attended Monta Vista.

Since 1999, going by the official archive, I know we've had: 2 in 1999, 5 in 2000, 4 in 2001, 3 in 2002, 2 in 2003, 0 in 2004, 2 in 2004, adding up to 18 already.  Whether or not you count Stephen, we should be on your list, so I believe your data is incomplete.

Thanks.  I added Monta Vista (CA) to the list.  I counted Stephen Guo, as you recommended.  Monta Vista also had 1 USAMO qualifier (L. Chen) in 1996 and 1997.

Does anyone have a list of 1998 qualifiers?  I don't have one, so I treated that year as a 0 for Monta Vista, pending further info.

By the way, what is the Olympia Institute?

[quote="vRusczyk"]We're gonna have to add the AoPS school to that list :).[/quote]Ravi, you can probably devise a 3-deminensional representation, as nice and as concise as your 2-dimensional worksheet, of what she wants.  :)

(assuming USAMO qualifiers reveal their double-identities of "School X and AoPS")

Olympia Institute is a place that offered both AMC's rather than just one.  Monta Vista has only offered one of the two AMC's for every year until 2005.

It's likely that both Shuang You and Oaz Nir qualified for USAMO in 1998 as well, but I don't have the 1998 AMC records, so I don't know for sure.

Hopefully Choate will get on that list pretty soon!

entirely irrelevant, but i knew i'd seen your name somewhere.  you have a perfect score on the AMC 10 A and a picture on the amc site!

[quote]It's likely that both Shuang You and Oaz Nir qualified for USAMO in 1998 as well[/quote]
Oaz Nir lived in Mississippi in 1998. He qualified for the USAMO only by the one-per-state rule, as the Mississippi representative. Then he did well enough on the USAMO to be invited to MOP. He moved to California the next year.

You can't count him in 1998 for Monta Vista.

Shuang You was a 9th grader in 1998; I'll have to see if I have any lists or books from that year to see whether she qualified.

Uh, I know LBJ (Johnson) high school had a qualifier in 2003 (me).

[quote="orl"][quote="vRusczyk"]We're gonna have to add the AoPS school to that list :).[/quote]

I am very interested in it. I guess it must be better by far than T. Jeff. School.  :)

So what are the stats ?  ;)[/quote]I think this warrants a poll! :)

Hmm, this is getting more serious :-) I think some middle schoolers were counted as high schoolers (Hint: Blair and Lexington)

This is the list of 8th or 7th graders from 2004 and 2005 USAMO qualifiers list

2004:

Caroline Suen	8	Miller Middle School	San Jose	CA
Jiale Shu	8	Mission San Jose High School	Fremont	CA
Joseph Chu	8	Miller Middle School	San Jose	CA
Peter Chien	8	Redwood Middle School	Saratoga	CA
Shotaro Makisumi	8			CA
Stephen Ge	8	EPGY-Stanford University	Stanford	CA
Harrison Brown	8	Milton High School	Alpharetta	GA
Alex Zhai	8	University Laboratory High School	Urbana	IL
Gregory Gauthier	8	Wheaton North High School	Wheaton	IL
David Benjamin	7	William Henry Harrison High School	West Lafayette	IN
Sway Chen	8	Lexington High School	Lexington	MA
Shui Hu	8	Irondale Senior High School	New Brighton	MN
Samuel Balinghasay	8	Jamestown High School	Jamestown	ND
Kevin Kowalski	8	The Meadows School	Las Vegas	NV
Danny Zhu	8	Hunter College High School	New York	NY
Colin Sandon	8	Essex High School	Essex Junction	VT

2005:

David Benjamin	8	Honey Creek Middle School	Terre Haute	IN
Sergei Bernstein	8	Belmont High School	Belmont	MA
Akimitsu Hogge	8	Montgomery Blair High School	Silver Spring	MD
David Rush	8	Scarsdale Senior High School	Scarsdale	NY
Eric Larson	8	Theo Roosevelt Middle School	Eugene	OR
Seungjun Kim	8	Shahala	Vancouver	WA
Chen Sun	7	London Central Secondary School	London	ON

Yeah, you did an amazing job!---9 quals per year is unmatched by any school except TJ.  And btw, qvc, where are u from?

I think having the rank 1, 2, and 3 scorers on the 2004 USAMO, as well as 3 of the 6 IMO team members last year is a more impressive achievement and better barometer of the Exeter machine than counting qualifiers (because these aren't just qualifiers, they're GOOD qualifiers).

Just wondering, which of those top schools are public, and which are private. I know Lexington is public, because that's where I took the test. Others?

I believe Exeter, Andover, St. Marks, and Roxbury Latin are private.  I believe the rest of the schools on the list are public, but I am not 100% sure.  The list contains many public magnet schools.

So I'm guessing Lexington is the highest ranked public non-magnet school? And yes, all four of the schools you mentioned above are private

darktreb, IMSA shouldn't count as public!  There's selective admission :P.

Then no school with any form of magnet program should be considered public.

Basically, I'm going by "public = state funded, not privately funded", which IMSA falls under (as evidenced by the variety of changes year to year due to various state Senate demands, etc.)

Yeah, I think selective admission should count as private.

There should be a separate category for selective magnet schools then. They don't take just any person in the area, but they're still publicly funded. You can't really consider then a private school.

Ok, then we'll have three seperate categories, magnet(selective admission, no cost), private(selective admission, cost), public(no selective admission, no cost)

I knew someone was going to say that :lol:

[quote="tetrahedr0n"]What about no selection admission, cost ? :)[/quote]

The selective qualification there would be that you need to be wealthy enough  :D .

But seriously, how about no academically selective admissions?


A permutation $ \sigma: [2n]\rightarrow [2n]$ is called graceful if $ \{ |\sigma(i)\minus{}\sigma(i\plus{}1)|\}_{i\equal{}1}^{2n\minus{}1}\equal{}[2n\minus{}1]$.  Given that $ \sigma$ is graceful, prove that $ \{\sigma(2k)\}_{k\equal{}1}^n\equal{}[n]$ if and only if $ \sigma(1)\minus{}\sigma(2n)\equal{}n$.

Sorry, I don't understand your notation. Please explain.. :oops:

We call a  permutation gracefull when $ \sigma: \{1, 2,\cdots ,2n\}\rightarrow \{1, 2, \cdots , 2n\n\}$ that $ \{ |\sigma(1) - \sigma(2)|, |\sigma(2) - \sigma(3)|, \cdots , |\sigma(2n - 1) - \sigma(2n)|\} = \{1, 2, \cdots 2n - 1\}$.

Show that $ \sigma$ is gracefull then $ \{\sigma (2), \sigma (4), \cdots ,\sigma(2n)\} = \{1, 2, \cdots , n\} \Leftrightarrow |\sigma(1) - \sigma(2n)| = n$ (I assume that it must have modulus...  :wink: )

Preaty hard isn't it? :D :D

I guess that we can sum all the elements of $ \{1, 2, \cdots ,2n - 1\}$ and notice that is the same sum of $ \{ |\sigma(1) - \sigma(2)|, |\sigma(2) - \sigma(3)|, \cdots , |\sigma(2n - 1) - \sigma(2n)|\}$ by definition, and must be $ 2n^2 - n$, and we can take that the parity of  $ |\sigma(1) - \sigma(2n)|$ is the same of $ n$ allways.

Btw: notice the notation-»$ [n] = \{1, 2, \cdots , n\}$ and $ \{f(i)\}_{i = a}^b = \{f(a), f(a + 1), \cdots , f(b)\}$ as the same of the sumatory :D

I think the modulus makes the problem wrong. For example, consider the permutation $ \{3,4,2,5,1,6\}$. This satisfies $ |\sigma(1) \minus{} \sigma(2n)| \equal{} n$, but not $ \{\sigma (2), \sigma (4), \cdots ,\sigma(2n)\} \equal{} \{1, 2, \cdots , n\}$.

The only if part first:

Since $ \sigma$ is difficult and too long to type, I simply write $ g$. Assume $ \{g(2),g(4),\ldots,g(2n)\}\equal{}\{1,2,\ldots,n\}$. Let $ f(k)\equal{}|g(k)\minus{}g(k\plus{}1)|$. Hence $ f(k)\equal{}g(k)\minus{}g(k\plus{}1)$ for odd $ k$ and $ f(k)\equal{}\minus{}g(k)\plus{}g(k\plus{}1)$ for even $ k$.

\[ f(1)\plus{}f(2)\plus{}\ldots\plus{}f(2n\minus{}1)\\\equal{}(g(1)\plus{}2g(3)\plus{}\ldots\plus{}2g(n\minus{}1))\minus{}(2g(2)\minus{}2g(4)\minus{}\ldots\minus{}2g(2n\minus{}2)\minus{}g(2n))\\\equal{}2(g(1)\plus{}g(3)\plus{}\ldots\plus{}g(2n\minus{}1))\minus{}2(g(2)\plus{}g(4)\plus{}\ldots\plus{}g(2n))\plus{}g(2n)\minus{}g(1)
\]

Note that $ f(1)\plus{}f(2)\plus{}\ldots\plus{}f(2n\minus{}1)\equal{}1\plus{}2\plus{}\ldots\plus{}2n\minus{}1\equal{}2n^2\minus{}n$, $ g(1)\plus{}g(3)\plus{}\ldots\plus{}g(2n\minus{}1)\equal{}(n\plus{}1)\plus{}(n\plus{}2)\plus{}\ldots\plus{}2n\equal{}3/2n^2\plus{}n/2$, $ g(2)\plus{}g(4)\plus{}\ldots\plus{}g(2n)\equal{}1\plus{}2\plus{}\ldots\plus{}n\equal{}n(n\plus{}1)/2$. From here it is easy to see that $ g(1)\minus{}g(2n)\equal{}n$.


calculus
derivative
analytic geometry
graphing lines
slope
limit
logarithms

Hello. Today, I was curious how the derivatives were so, as a result, I proved it :).  Now, I would like to know how you would approach the method of proof.  My proof is hidden below.

[b][i]Proof of the Derivatives[/i][/b]

It is quite clear that the proof is mainly focused about the equation $y=ax^n$ with degree $n$.
We need to prove that the derivative of $y=ax^n$ is $y'=anx^{n-1}$.
Let us say that we have a fixed point $\text{A}(x,y)=\text{A}(x,ax^n)$ and a point $\text{P}(x+h,y+k)=\text{P}(x+h,a(x+h)^n)$.
The slope of the secant $AP$ is
$\\m_{AP}=\frac{a(x+h)^n-ax^n}{x+h-x}=\frac{a(x+h)^n-ax^n}{h}\\=\frac{a(x^n+nx^{n-1}h+\frac{n(n-1)}{2}x^{n-2}h^2+\frac{n(n-1)(n-2)}{6}x^{n-3}h^3+...+h^{n-1}+h^n)-ax^n}{h}\\= \frac{a(nx^{n-1}h+\frac{n(n-1)}{2}x^{n-2}h^2+\frac{n(n-1)(n-2)}{6}x^{n-3}h^3+...+h^{n-1}+h^n)}{h}\\ =anx^{n-1}+\frac{n(n-1)}{2}x^{n-2}h+\frac{n(n-1)(n-2)}{6}x^{n-3}h^2+...+h^{n-2}+h^{n-1}$
Since we wish to find the slope of the tangent line passing through point $\text{A}$, $\text{P}$ has to approach point $\text{A}$, hence $h\rightarrow 0$.  In order to do this, we evaluate
$\\f'(x)=\lim_{h\rightarrow 0}m_{AP}\\ =\lim_{h\rightarrow 0} anx^{n-1}+\frac{n(n-1)}{2}x^{n-2}h+\frac{n(n-1)(n-2)}{6}x^{n-3}h^2+...+h^{n-2}+h^{n-1}\\=anx^{n-1}$.

Since we know that
$D_x(ax^{n_1}+bx^{n_2}+cx^{n_3}+...+d)\\=aD_xx^{n_1}+bD_xx^{n_2}+cD_xx^{n_3}+...+D_xd\\=an_1x^{n_1-1}+bn_2x^{n_2-1}+cn_3x^{n_3-1}+...+0$,
it is clear that the pattern works for any equation having any number of terms.

$\mathbb{QED}$

EDIT: I'm not quite sure if this is more calculus or algebra.  Since the proof is mainly algebraic, I have placed it here.

Masoud Zargar

You wanted to know why $\frac{d}{dx}(x^n)=nx^{n-1}$?

Well, we have

$f'(x)=\lim_{t\to x}\frac{f(t)-f(x)}{t-x}$ by definition of derivative.  Substituting in, we get

$f'(x)=\lim_{t\to x}\frac{t^n-x^n}{t-x}=\lim_{t\to x} (t^{n-1}+xt^{n-2}+\ldots+x^{n-1})=nx^{n-1}$

$f(x) = x^n = e^{n\ln x}$, so $f'(x) = \frac{n}{x} e^{n\ln x} = nx^{n-1}$. This is only true for x > 0, though, so to show that it is true for all x for integer values of n you need to do something similar to what you did and then differentiate $x^{-n}, n >0$ using that.

Indeed.  Logarithmic differentiation also works.


What is $ 781\equiv 1 \pmod{71}$? I am wanting to show that $ 781\equal{}11\cdot 71$ is a 17-pseudoprime. Also, i think i am using the chinese remainder theorem here?

Thanks!

$ 781 \equiv 0 \pmod{71}$

The definition of congruence is $ a \equiv b \pmod{n}$ iff $ n|(a \minus{} b)$. Since $ 781 \equal{} 11 \times 71$, $ 71|781 \minus{} 0$

I believe what you need to show is that $ 17^{780} \equiv 1 \pmod {11}$ and that $ 17^{780} \equiv 1 \pmod {71}$, which, since $ \gcd(17,71) \equal{} 1$ means that $ 17^{780} \equiv 1 \pmod {781}$ by the CRT. Which means that 781 is a  17-pseudoprime (if I understood your question correctly?)

The easiest way I know of calculating the two congruences above is using the square and multiply algorithm.[url]http://en.wikipedia.org/wiki/Exponentiation_by_squaring[/url] (look under further applications).

EDIT: Hmmm... the wikipedia article isn't the best for communicating the simple use of the algorithm that I was trying to describe...

edited twice for english  :P

Thanks, max! yes, i had made an error and you were on point.

Los

Actually, having just re-read this, using FLT would be a quicker way to finish up those congruences...


geometry
trigonometry
incenter
analytic geometry
function
parallelogram
projective geometry

Let $ABCD$ be an arbitrary quadrilateral. The bisectors of external angles $A$ and $C$ of the quadrilateral intersect at $P$; the bisectors of external angles $B$ and $D$ intersect at $Q$. The lines $AB$ and $CD$ intersect at $E$, and the lines $BC$ and $DA$ intersect at $F$. Now we have two new angles: $E$ (this is the angle $\angle{AED}$) and $F$ (this is the angle $\angle{BFA}$). We also consider a point $R$ of intersection of the external bisectors of these angles. Prove that the points $P$, $Q$ and $R$ are collinear.

Well....what comes to my mind immediately is I prefer solving it by co-geom/complex number rather than drawing the diagram accurately, no matter by hand or geometry software :D

We denote centers of excircles of different triangles as $I_1, I_2, I_3, I_4, J_1, J_2, J_3, J_4$
(from sketch)
It is not hard to prove that $I_1I_2I_3I_4$ is concyclic by angle chasing.

$< I_4I_1I_3 = < J_4 J_2J_3$ from concyclity of $J_2EDI_1$

$< I_4I_1I_3 = < I_4I_2I_3 = < FI_2B = < BJ_1F = < J_4J_1J_3 $ (from concyclity $J_1FBI_2$)

Hence $ < J_4 J_2J_3 = < J_4J_1J_3 $ and $J_1J_2J_3J_4$ is concyclic.

Let $U$ be the intersection of $ I_1I_4$  and $ I_2I_3$, similarly $V$ - $ J_1J_4$  and $J_2J_3$.

We will demonstrate that $P,U,V$ are collinear.

$\frac {\sin UPI_2}{\sin UPI_3}= \frac {UI_2\sin I_1I_2I_3}{UI_3\sin I_4I_3I_2}= \frac{\sin I_2I_3I_1}{\sin I_4I_1I_3}$

$\frac {\sin VPJ_2}{\sin VPJ_3}= \frac{\sin J_2J_3J_1}{\sin J_4J_1J_3}$

As we showed $< I_4I_1I_3 = < J_4J_1J_3$

And we can see that $< I_2I_3I_1 = 180- < DI_3I_2 = < DJ_3F = 180 - J_2J_3J_1$

So we can say fractions are equal hence $P,U,V$ are collinear.

Last step is from projective geometry:
$PU$ is polar to $Q$ through angle $I_3PI_1$ so it divides it in harmonic ratio.
So lines $PI_1, PU, PI_3, PQ$ are in fascicol.
Analogously for - $PJ_1, PV, PJ_3, PR$ are in fascicol.
Three lines are identical so $P,Q,R$ are collinear.

I think I solved a little bit different problem about intersection
of internal bisectors of $A,C$ and $B,D$.

Continuing solution:
Note $T_1,T_2,T_3,T_4$ intersections of external bisectors of angles $A,B,C,D$

Let $W$ be the intersection of external bisector of $A$ and $C$.
From problem we need to prove that $W,R,V$ are collinear.
(sorry, but in sketch they are not)

Again same idea we easily can prove that $T_1,T_2,T_3,T_4$ is concyclic.
It’s $V,T_1,T_2,T_3,T_4,W$ is complete quadrilateral and it’s diagonal intersection is $U$

It means $VW,VT_3,VU,VT_1$ is fascilcol armonic.
From first post we had $VR,VJ_3,VU,VJ_1$ is fascilcol armonic.

Three lines coinside - $V,R,W$ are collinear.

I hate monologs. Can anybody accept my solution ?

prowler, do you mind editing your diagram slightly to use different colours for the sides of the quadrilateral, the external angle bisectors at $A, B, C, D$ and the external angle bisectors at $E$ and $F$ (something like what Darij does for his diagrams)?  It would be much easier to compare your solution with the diagram in that case.  Right now, all I see is a whole lot of lines... :?

Would you please state exactly which triangles $I_1, I_2, I_3, I_4, J_1, J_2, J_3, J_4$ are excentres of?  It's really hard to tell from your diagram at present.  Perhaps if the diagram was coloured, it might be easier.  Thanks!

$I_1$ - excenter $AED$
$I_2$ - excenter $AFB$
$I_3$ - incenter $CFD$
$I_4$ - incenter $BEC$

$J_1$ - excenter $ABF$
$J_2$ - excenter $ADE$
$J_3$ - excenter $DCF$
$J_4$ - incenter $BCE$

$T_1$ - excenter $BEC$
$T_2$ - incenter $ABF$
$T_3$ - excenter $DCF$
$T_4$ - incenter $AED$

My advice to print sketch.
P.S Which program Darij uses to draw sketches with colors, or dinamic one ?

Prowler, your solution is correct and nice, but it can be simplified. In fact, you proved the following three assertions (where I am using your notations, not the notations of the original problem):

- The point of intersection P of the internal angle bisectors of the angles A and C, the point of intersection V of the external angle bisectors of the angles B and D, and the point of intersection U of the internal angle bisectors of the angles E and F are collinear. (This is the collinearity of the points P, U, V, proven by you in post #3.)

- The point of intersection P of the internal angle bisectors of the angles A and C, the point of intersection Q of the internal angle bisectors of the angles B and D, and the point of intersection R of the external angle bisectors of the angles E and F are collinear. (This is the collinearity of the points P, Q, R, proven by you in post #3. This is also problem 6 of the 239MO 1997, classes 10-11.)

- The point of intersection W of the external angle bisectors of the angles A and C, the point of intersection V of the external angle bisectors of the angles B and D, and the point of intersection R of the external angle bisectors of the angles E and F are collinear. (This is the collinearity of the points V, R, W proven by you in post #5, and this is also the original problem of this thread.)

But actually, from the more convenient viewpoint of directed lines, these three assertions are all equivalent to each other. They all can be formulated in one theorem:

[color=blue][b]Theorem 1.[/b] Let x, y, z, w be four directed lines. Then, the points $\left[x;\;y\right]\cap\left[z;\;w\right]$, $\left[y;\;z\right]\cap\left[w;\;x\right]$ and $\left[x;\;z\right]\cap\left[y;\;w\right]$ are collinear.[/color]

Here, for any two directed lines g and h, the abbreviation [g; h] means the exterior angle bisector between the lines g and h. This angle bisector is defined as the locus of all points P such that d(P; g) = - d(P; h) (this locus is one line, not two lines, since we are using directed distances!). Here, the notation d(U; t) stands for the directed distance from a point U to a directed line t. Note that, if two directed lines g and h are parallel and have the same direction, then their exterior angle bisector [g; h] is the line parallel to them and lying in the middle between them; if two directed lines g and h are parallel and have opposite directions, then their exterior angle bisector [g; h] doesn't exist, strictly speaking; we can say that this exterior angle bisector is the line at infinity.

I have explained the necessary fundamentals about directed lines and directed distances in http://www.mathlinks.ro/Forum/viewtopic.php?t=6559 post #5. You can now take x = DA, y = AB, z = BC and w = CD, and you will get all three assertions you showed as consequences of Theorem 1 for different directions of the lines x, y, z, w. So it is enough to prove Theorem 1.

And here goes a particularly beautiful proof of Theorem 1, which, if I remember correctly, was not my idea, but I don't remember anymore where I learnt it:

[i]Proof of Theorem 1.[/i] Let $\lambda$ be the locus of all points P such that d(P; x) + d(P; y) + d(P; z) + d(P; w) = 0. This locus $\lambda$ is a linear locus, i. e. it is either the empty set, or a straight line, or the whole plane. This is because, by the formula for the directed distance from a point to a line in a cartesian coordinate system, the directed distances d(P; x), d(P; y), d(P; z) and d(P; w) can be written as linear functions of the cartesian coordinates of the point P (in some fixed cartesian coordinate system), and thus, the equation d(P; x) + d(P; y) + d(P; z) + d(P; w) = 0 is a linear equation in the coordinates of P and therefore represents a linear locus. [You can also show this without the use of Cartesian coordinates.]

So, as we said, the locus $\lambda$ is either the empty set, or a straight line, or the whole plane. Now, we can easily see that the case when the locus $\lambda$ is the whole plane can be left out of consideration - for an explanation why, open the following hide tag:

[hide="Why we can assume that the locus is not the whole plane"]

Why can we assume that the locus $\lambda$ is not the whole plane? Well, assume that the locus $\lambda$ is the whole plane.

The four lines x, y, z, w can be grouped into 6 pairs: (x; y), (y; z), (z; w), (w; x), (x; z) and (y; w). If, at least, two of these pairs consist of parallel lines (i. e., if, for instance, x || y and w || x, or if x || y and z || w), then Theorem 1 becomes rather easy (we either have three among the lines x, y, z, w being parallel, or the four lines x, y, z, w form a parallelogram), so that we can assume that at most one of the 6 pairs consists of parallel lines. WLOG, let this pair be (y; z). So, if two among the four lines x, y, z, w are parallel at all (of course, in the "usual" case, no two of the lines are parallel), then these are the lines y and z.

Thus, the lines x and y are not parallel, so that they have a finite point of intersection A, and the lines z and w are not parallel, so that they have a finite point of intersection C.

Since the locus $\lambda$ is the whole plane, the point A also lies on the locus $\lambda$. Hence, d(A; x) + d(A; y) + d(A; z) + d(A; w) = 0. Since the point A lies on the lines x and y, we have d(A; x) = 0 and d(A; y) = 0, so that we get d(A; z) + d(A; w) = 0. Thus, d(A; z) = - d(A; w), so that the point A lies on the line [z; w]. Of course, the point of intersection C of the lines z and w also trivially lies on the line [z; w]. Thus, the line [z; w] is the line AC. Similarly, the line [x; y] is the line AC. Hence, the lines [x; y] and [z; w] coincide, so that their point of intersection $\left[x;\;y\right]\cap\left[z;\;w\right]$ cannot be reasonably defined, and Theorem 1 loses its sense.

Hence, we can we assume that the locus $\lambda$ is not the whole plane.[/hide]

Hence, it remains to consider the case when the locus $\lambda$ is either the empty set, or a straight line.

In this case, consider the point $T=\left[x;\;y\right]\cap\left[z;\;w\right]$. Since this point T lies on the exterior angle bisector [x; y], we have d(T; x) = - d(T; y). Since this point T lies on the exterior angle bisector [z; w], we have d(T; z) = - d(T; w). Thus, d(T; x) + d(T; y) + d(T; z) + d(T; w) = (- d(T; y)) + d(T; y) + (- d(T; w)) + d(T; w) = 0. Hence, the point T lies on the locus $\lambda$. Hence, the locus $\lambda$ is a line (in fact, we know that the locus $\lambda$ is either the empty set, or a line; but it cannot be an empty set, since the point T lies on this locus, so it must be a line), and the point $T=\left[x;\;y\right]\cap\left[z;\;w\right]$ lies on this line. Similarly, the points $\left[y;\;z\right]\cap\left[w;\;x\right]$ and $\left[x;\;z\right]\cap\left[y;\;w\right]$ lie on the same line $\lambda$. Thus, the points $\left[x;\;y\right]\cap\left[z;\;w\right]$, $\left[y;\;z\right]\cap\left[w;\;x\right]$ and $\left[x;\;z\right]\cap\left[y;\;w\right]$ lie on one line, and Theorem 1 is proven.

[quote="prowler"]P.S Which program Darij uses to draw sketches with colors, or dinamic one ?[/quote]

I use [url=http://dynageo.de]Euklid DynaGeo[/url], a very nice shareware dynamic geometry program. Alas, it is avaliable in German only.

  Darij

[quote="prowler"]
P.S Which program Darij uses to draw sketches with colors, or dinamic one ?[/quote]
Prowler, try [url=http://www.mit.edu/~ibaran/kseg-0.401.zip]this program[/url] it's colorful, dynamic, in English and freeware :)


algebra
polynomial
linear algebra
matrix
linear algebra unsolved

Let $K$ be an infinite field and consider $f \in K[X], \, \deg f = n$.

Is it true that we can find infinitely many $n \times n$ matrices $A$ such that $p_{A}= f$? What about $m_{A}= f$?
($p_{A}$ = characteristic polynomial; $m_{A}$ = minimal polynomial.)

PS: I know that we can find at least one matrix such that $m_{A}= f$ (Kent Merryfield posted this quite recently).
PPS: A particular case of this appears in one of the possible subjects for the (Romanian) Baccalaureat exam. I have a book in which [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=18653]this guy[/url] gives a solution, but I don't think it can be generalized.

Of course- just apply a change of basis to that companion matrix. If the coefficients of the polynomial are in the bottom row, we can apply a similarity that makes the top row into anything but a multiple of $(1,0,\dots,0)$. That similarity does not change the minimal polynomial.


Compute the integral( an easy one)
\[\displaystyle\int{\frac{dx}{xlnxln(lnx)ln(ln(lnx))}}\]
where $lnx=\log_e{x}$

[hide="answer"]$=\ln\ln\ln\ln{x}+C$ (assuming $x$ is positive)[/hide]

By the way, I don't read the pre-olympiad forum much; is calculus allowed here?

[quote="AntonioMainenti"]By the way, I don't read the pre-olympiad forum much; is calculus allowed here?[/quote]

It's probably better suited for one of the Calculus forums  :?

[quote="seamusoboyle"]AS a matter of interest how did you solve it?[/quote]
let $u=\ln{\ln{x}}$ then $du=\frac{1}{x\ln{x}}dx$ thus
$\int{\frac{dx}{x\ln{x}\ln{(\ln{x})}\ln{(\ln{(\ln{x})})}}=\int{\frac{1}{u\ln{u}}du}=\ln{\ln{u}}=\ln\ln\ln\ln{x}+C}$

Isn't this one of Kunny's "Today's Calculation of Integral" problems without the limits?

[quote="amirhtlusa"]not any of the recent one...i guess :|[/quote]

I could be wrong, I just know I've seen this before.


Maths is not my strong subject so forgive my ignorance.

I have been trying to match shell images (2 nautilus and one ammonite fossil) to the Fibonacci sequence but I couldn't. So instead I just sized a picture of a shell to 377 pixels drew the rectangles on freehand and then measured the sides. The following image is the result. The numbers in red is the ratio of the two sides and averages app 1.36.

What am I doing wrong?

[img]http://www.robfrost.co.uk/temp/notfibonacci.jpg[/img]

This is the Flash I've been working on:

[url=http://www.robfrost.co.uk/temp/goldenRectangle.html]www.robfrost.co.uk/temp/goldenRectangle.html[/url]

Many thanks

http://goldennumber.net/spirals.htm
check this link it might shed some light... ps I love phi!

That's a wonderful connection between the real world and math. I assume this pattern is a result of a certain growing pattern , i.e. according to some sort of differential eqn etc...

Thanks for the posts har0bed1813 and Rushil. I'm afaraid they have not brought me any closer to solving my problem with the ratio averaging 1.36 as opposed to 1.65.

So still looking for a good maths brain to find where I'm going wrong.

i'm sorry but I'm zero in Biology. Although I do think I might be able to help you with your math problems. Can you post your equations (after applying the biology) so that I can try my hand at them!!

The equations are in the image above. I drew the squares freehand onto the shell and then measured them. Maybe it isn't a good maths brain I need just someone who can explain the correlation between the two, as the actual shape of a nautilus and amonite are much taller than the actual layout in the linked flash model I built.

This page might explain better what I mean:
[url]http://www.robfrost.co.uk/temp/goldenRectangle.html[/url]

Hold on, lemme try to dig out my book on golden ratios, I think there was a picture of what you're looking for in there.

Off the top tho, aren't the "rectangles" supposed to be squares?

They are squares in my example of the actual Fibonacci sequence. See link:
[url]http://www.robfrost.co.uk/temp/goldenRectangle.html[/url]

The fact they are not when drawn freehand on the nautilus is precisely my point.

The ratios of the terms in the Fibonacci series are NOT all the golden ration, phi, which is about 1.618. The ratios of the successive terms APPROACH phi as you get larger.

1,1,2,3,5,8,13, 21…

2/1 = 2

3/2 = 1.5

5/3 = 1.666666

8/5=1.6

21/13 =1.61538

So first of all, your first few ratios you should expect to jump around a bit, which they seem to do before settling down a bit more on your diagram.

The next thing is, from what I can tell, the nautilus spirals are similar to Fibonacci spirals in that they are both logarithmic spirals, (meaning they decrease by a constant factor each turn as you spiral in), but they don’t necessarily have the *same* factor. I think the idea that they’re both phi is kind of a romantic misnomer.

So, the Fibonacci spiral increases by a factor of phi every quarter turn, (which means every full turn it will increase by a factor of phi^4, or about 6.85). In comparison, your shell increases by 1.3 every quarter turn, or about 2.85 every full turn, which matches another nautilus picture I measured. (http://www.allposters.com/-sp/Nautilus-Posters_i380011_.htm).

Finally, if the nautilus isn’t cut through very evenly down the center, you can get weird calculations.

Many thanks for your hard work and for explaining it to me. I had almost given up on getting a reply.

It is a bit of a shame that it is only a "romantic misnomer" but that is what I expected. I notice the site with the nautilus picture also has a sunflower which is supposed to have the same ratio in the seeds, but is probably just another logarithmic spiral.

Spirals are everywhere in nature so still quite cool even though they are not some magic secret discovered by Leonardo da Vinci.

Hello,

I believe that is a picture of a goniatite, not sure if it makes much difference if you use a goniatite, ammonite or nautilus.

I find it all very interesting though.

Regards
Oscar

...you say goniatite I say ammonite let's call the whole thing off..... :wink:

Thanks for the correction Oscar.


Hi everybody I have question that I can't solve.

Let $ \mathbb{R}^{\mathbb{R}}$ be the set of real functions on $ \mathbb{R}$ with the product topology = pointwise convergence.

Let $ E \equal{} \{ f \in \mathbb{R}^{\mathbb{R}} : f(x) \equal{} 2 \text{ except finitely often} \}$. Show that the zero function is in the closure of $ E$.
I was thinking about constructing a net in $ E$ with limit the zero function... But then we have to construct $ \mathbb{R}$ from finite sets... Any idea ?

Well, what is a neighborhood of the zero function?

Under the product topology, a basis of open sets are products of open sets where all but finitely many are $ \mathbb{R}$. So, show that every basic open set containing the zero function contains an element of $ E$ (look at those open sets which are not equal to $ \mathbb{R}$ -- what can you do in those coordinates?)

Okay maybe I get it... So yeah if we take an nbhd $ U \equal{} \Pi_{x\in\{x_1,x_2,...,x_n\}} U_x \times \Pi_{x\neq x_i} \mathbb{R}$, the function which maps all the $ x_i$ to $ 0$ and the rest to $ 2$ is in $ U$ and in $ E$ and I construct the net by indexing with the nbhd ?d

You don't need to construct a net. An element $ g$ is in the closure of $ E$ iff every open neighborhood of $ g$ intersects $ E$ nontrivially. It suffices to check this condition on basic open sets.

Yeah okay, sorry forgot to tell, the next question is "construct a net in $ E$ with limit to the zero function"...

The full answer is here if someone need it http://www.mathlinks.ro/viewtopic.php?t=303374


Could someone help me with the following questions please. I'm not sure how i go about them - thanks.

http://members.optusnet.com.au/billy06/maths.JPG

I heard someone said that calculus problems should be posted here, i.e. Calculus and real analysis.
http://www.mathlinks.ro/Forum/index.php?f=218

Another member of the forum has posted the same question with a link to the same webpage.  So, either you (1) are the same person being really obnoxious or (2) you are both trying to cheat on the same assignment.  In either case, this thread should be locked.


Prove: $ a,b,c,d\ge 0\Rightarrow a^3\plus{}b^3\plus{}c^3\plus{}d^3\ge abc\plus{}bcd\plus{}cda\plus{}dab$

It's obvious by Muihard.

Different solution :

By AM-GM :

$ abc\le\frac{a^3\plus{}b^3\plus{}c^3}{3}$

Adding three similar inequalities we have :

$ abc\plus{}bcd\plus{}cda\plus{}dab\le a^3\plus{}b^3\plus{}c^3\plus{}d^3$


In triangle ABC, we have AB = 1 and AC = 2. Side BC and the median from A to BC have the same length. What is BC?

Even though this problem is a bit too hard for Getting Started, I will solve it:
[hide="The solution..."]This is an easy application of Stewart's Theorem. Let $d$ = the length of the median, $a$ the length of $BC$, $b$ the length of $AC$, $c$ the length of $AB$, and $m$ the length of $\frac{1}{2}$ of $BC$ (in other words, $d$ bisects $a$ into 2 equal pieces of length $m$). Applying Stewart's Theorem,

$dad+mam = cmc+bmb$
$d^{2}(d)+m^{2}(2m) = m+4m = 5m = \frac{5}{2}a = \frac{5}{2}d$

$d^{3}+2m^{3}= \frac{5d}{2}$
$d^{3}+2(\frac{d}{2})^{3}= \frac{5d}{2}$
$\frac{5d^{3}}{4}= \frac{5d}{2}$

$\frac{d^{2}}{4}= \frac{1}{2}$

$d^{2}= 2$

$d=\sqrt{2}$

$d=a=\sqrt{2}$

The length of $BC$ is $\sqrt{2}$.[/hide]

Hint.

Try to make a triangle into a parallelogram and note that the sum of the squares of the sides equals the sum of the squares of the diagonals.

[hide]Extend the median from B to D such that ABDC is a parallelogram. Since the sum of the squares of the diagonals of a parallelolgram equals the sum of the squares of the four sides,
$(2x)^{2}+x^{2}=1+1+2^{2}+2^{2}=10$
$5x^{2}=10$
$x^{2}=2$
$x=\sqrt{2}$.[/hide]

[quote="mathgeniuse^ln(x)"][quote="vishalarul"]Even though this problem is a bit too hard for Getting Started, I will solve it:
[/quote]

This isn't getting started...[/quote]

A mod moved it from the Getting Started forum to here I think.

[hide="more elementry solution"]
not as elegant, but doing this analytically isn't too bad.

Let $A$ be at (0,0).  Let $B$ be at (1,0).  Let $C$ be at $(x,y)$.

We know that $x^{2}+y^{2}= 4$ and $BC = \sqrt{(x-1)^{2}+y^{2}}= \sqrt{x^{2}+y^{2}-2x+1}= \sqrt{5-2x}$

Let $D$ be the midpoint of $BC$.  D is at $(\frac{x+1}2,\frac{y}2)$.

Since $AD=BC$,

$\sqrt{(\frac{x+1}2)^{2}+(\frac{y}2)^{2}}= \sqrt{(x-1)^{2}+y^{2}}$

Squaring each side, canceling stuff out, and substituting $x^{2}+y^{2}= 4$, we get $x=3/2$.  Plugging this into the equation above, $\sqrt{5-2x}= \sqrt{2}$.

Not too messy for an analytical solution.

[/hide]

This WAS Getting Started after all :D

[hide="Very elementary"]Let $D$ be the midpoint of $BC$. If $BC=a$, then $DC={a\over 2}$ and $AD=a$. Hence triangles $ABC$ and $DAC$ have one angle in common and the same proportion of two sides - $2: 1=a:{a\over 2}$ - therefore they are similar. So we have $DA: AC=AB: BC\implies a: 1=2: a\implies a=\sqrt{2}$[/hide]

well, you can also manipulate Stewart's Theorem.

Assume $BC=x$, then the length of the median is also x.

$x*(\frac{x}2*\frac{x}2+x^{2})=1^{2}*\frac{x}2+2^{2}*\frac{x}2$
$\frac{5x^{3}}4=\frac{5x}2$
$x^{3}=2x$
$x^{3}-2x=0$
$x(x^{2}-2)=0$
$x(x-\sqrt2)(x+\sqrt2)=0$
$x=0$, $\sqrt2$, or $-\sqrt2$.
however, we have to reject $0$ and $-\sqrt2$

Therefore, $BC=\boxed{\sqrt2}$.

Edit:  :oops_sign: , sorry vishalarul, didn't see your solution, so I posted the exact copy of yours. sorry


Prove the following statement.
If a number is deficient by one, then it is a power of 2.

Hmm... a proof. What should I do with this... ?  :?

With all the controversy over proofs in this section, I'll let Silverfalcon make a decision on it. It can stay till now.  ;)

By "Number is deficient by 1", I belive he   means,  it is  almost a perfect number, In other words,   sum of all its factors  (not counting the number it self) is 1 less than the number.

The proof for this, i belive, is not suitable for this forum, May be in Olympiad section, as it is, to put it mildly,  NOT an easy (I spent some time looking at it) . In fact,  I belive that it is an  open  problem and  [b]no one [/b]  as yet has any proof.

This does not mean that proof does not exist, it just that,, if you do provide a proof here, you would be famout, so good luck. :)

There is also a probablity, I think,  that  the statment may not be true (If so the number has to be VERY large - ) In many ways this looks like it is related to the problem of finding an odd perfect number.

BTW: The  proof in "other direction" is fairly easy

Hope the mod  moves the topic to other forum...


complex numbers
advanced fields
advanced fields theorems

I am looking for a proof of the complex Descartes' Circle Theorem

[b]Theorem[/b] (the complex Descartes' Circle Theorem):
Given any four mutually tangent circles with curvatures $ k_1,k_2,k_3,k_4$ and centered at the complex numbers $ z_1,z_2,z_3,z_4$ respectively,
\[ (k_1z_1\plus{}k_2z_2\plus{}k_3z_3\plus{}k_4z_4)^2\equal{}2(k_1^2z_1^2\plus{}k_2^2z_2^2\plus{}k_3^2z_3^2\plus{}k_4^2z_4^2).\]


function
AoPSwiki
AMC
USA(J)MO
USAMO
algebra unsolved
algebra

A sequence of real functions $ \mathop f\nolimits_n$ is defined as follow
 $ \mathop f\nolimits_1 (x)  \equal{}\sqrt {x^2 \plus{} 48}$, $ \mathop f\nolimits_{n \plus{} 1}(x)  \equal{}\sqrt {x^2 \plus{} 6\mathop f\nolimits_n (x)}$ for $ n>0$
For each positive integer n, find all real solutions of the equation $ \mathop f\nolimits_1 (x)\equal{}2x$

AoPSWiki: [[1990 USAMO Problems/Problem 2]]


[b]p[/b] it's an odd, prime number from the set $ \mathbb{N}$ . we have the numbers [b]a[/b] and [b]b[/b] from the set $ \mathbb{Z}$, thus $ a^p^ + ^1 + b^p^ + ^1$. Knowing that $ p^3$ devide $ a{}^2{}^p + b{}^2{}^p$ than we have to  prove that $ p^3$ divide $ a{}^2{}^P + b{}^2{}^P$.

what? p is an even prime? it's 2 maybe??

Um it would be nice if the problem made sense.


Let be $\ n$ a natural number.Find $\ n$ so that the number $\ n^4+\ 4$ is prime.

I think it's a classic. Google Sophie Germain.

[quote="mathmanman"]It is prime for $n=1$ only.
For $n > 1$, Sophie Germain's identity allows you to prove that it's composite.

This is really well-known, I'm pretty sure it's already on the forum.[/quote]
    What is $\ Sophie\ Germain\ Identity?$I don't know it.

[quote="kunny"]$n^4+4=(n^2+2)^2-(2n)^2=(n^2+2n+2)(n^2-2n+2).$[/quote]
     Thank you Kunny. :lol: This is the easiest way to solve this problem.I have also another question:can you generalise it?

[quote="ZetaX"]It's also in the Formulary :)[/quote]
    Thank you ZetaX!It's beautiful what you did there!All the theorems! :lol:


Woah!

I’ve stumbled across a problem at work and mathematics isn’t my strong side.

I have a series of interconnected nodes (see attached picture) and I need to find some sort of algorithm to find:

a) each 'unique' networks and
b) what nodes are in each unique network.

(I refer to a 'unique network' as a group of connected nodes, again, see attached picture)

I figure it must be some sort of iterative loop checking each node, but I’m at a loss as to how to do it.

Does any body have any suggestions or ideas about how this problem can be solved?

Greatly appreciative for any advise!
Ben

This is probably better suited for "Other Problem Solving Topics" or one of the Olympiad Combinatorics forums.  (Hopefully a moderator will move it there.)  I'm somewhat unclear on what exactly your task is.  Is one of the following correct?

You have some arbitrary network composed of a number of nodes which are connected to each other.  You are interested in counting the number of sub-networks of this network which have the same "shape" (we might call this "graph isomorphism classes") as one of the three networks you listed.  Or, you are interested in finding all different subgraphs?

If you could clarify exactly what you are doing, it would be helpful.  From what it sounds like, you may be dealing with an NP-or-worse problem (one for which no efficient algorithm is known).


[quote="LeFromage"]Just test $ x=0,1,2,3,4$.[/quote]

This will not work as there are infinitely many solutions.

$ x^{2}-2x+2\equiv0\;(mod5)$
$ x(x-2)\equiv-2\;(mod5)$

Hence the only two possible solutions are:

$ x\equiv3\;(mod5)$
$ x\equiv4\;(mod5)$

what is mod ? what is mod5?
what is 0(mod5)?
can someone explain this in simple words please.

It is a modulus.
Almost like division with the remainder coming before the $ \bmod$ and the divisor being $ n$ in $ \bmod n$.

For example
$ 11 \equiv 2 \bmod 3$ or $ 2 \equiv-1 \bmod 3$.
When using $ \bmod$ the equivalence sign is used instead of the equal sign.
Then are also a couple properties of $ \bmod$ such as
$ 11^{2}\equiv 2^{2}\bmod 3$ Simplify $ 121 \equiv 4 \bmod 3 \equiv 1 \bmod 3$
and
$ 2^{2}\equiv \left(-1 \right)^{2}\bmod 3$ Simplify $ 4 \equiv 1 \bmod 3$.

modulus is this $ \parallel$ isnt it?
Xample $ |x|$=modulus of $ x$?
isnt that modulo ?
still i dont get it

That is a different type of modulus.  In the States that is called absolute value.

Here are some links that will probably explain it better.
[url=http://en.wikipedia.org/wiki/Modulo]Wikipedia Modulo[/url].
[url=http://en.wikipedia.org/wiki/Modulo]Wikipedia Modulus[/url].
[url=http://en.wikipedia.org/wiki/Modular_arithmetic]Wikipedia Modular Arithmetic[/url].

[quote="IrinaSharova"][quote="LeFromage"]Just test $ x=0,1,2,3,4$.[/quote]

This will not work as there are infinitely many solutions.

$ x^{2}-2x+2\equiv0\;(mod5)$
$ x(x-2)\equiv-2\;(mod5)$

Hence the only two possible solutions are:

$ x\equiv3\;(mod5)$
$ x\equiv4;(mod5)$[/quote]
It does work, because all other solutions will be congruent. For example $ x=3$ and $ x=4$ work. So the solutions are $ x\equiv 3$ and $ x\equiv 4$ mod 5.


in Amir's game, you pick a number from 1 to 100.  Any time your number is not a square, it is increased by 5, and anytime your number is a square, you lose.  you win by getting your number over 100,
how many numbers win?

[quote="Meghrabi"][hide]hmm

96,97,98 and 99 ?

so 4 numbers win[/hide][/quote]

nope! each time you dont have a perfect square you add five

for example 2 and 3 win too ;)

[hide="Brute force since I am bored"]Well, the ways of losing are by arriving at 1, 4, 9, 16, 25, 36, 49, 64, 81, or 100.
This gives the following winners:
Any number that isn't a multiple of 5 which is above 81
82-84, 86-89, 91-94, 96-99
Any number that isn't one more than a multiple of 5 or a multiple of 5 above 64
67-69, 72-74, 77-79
Not 0/1/4 more than a multiple of 5 (above 0)
2, 3, 7, 8, 12, 13, 17, 18, 22, 23, 27, 28, 32, 33, 37, 38, 42, 43, 47, 48, 52, 53, 57, 58, 62, 63

Sooo.... this gives the set of winning numbers
2, 3, 7, 8, 12, 13, 17, 18, 22, 23, 27, 28, 32, 33, 37, 38, 42, 43, 47, 48, 52, 53, 57, 58, 62, 63, 67, 68, 69, 72, 73, 74, 77, 78, 79, 82, 83, 84, 86, 87, 89, 91, 92, 93, 94, 96, 97, 99
Which is 48 numbers.
Wow, what a waste of brute force. There is a much nicer way of doing it.[/hide]

[hide]
The perfect squares are 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100. Now, group them by the remainder (if any) they have when dividing by 5 and find the highest number in each group. Giving you: 64, 81, and 100
Now, this means these 13 numbers don't work: $64, 59, 54...4$
These 17 numbers also don't work: $81, 76, 71...1$
Lastly, these 20 numbers don't work: $100, 95, 90...5$

That means that $50$ numbers work.[/hide]

Haha I wasted my time brute forcing and also got it wrong. What a waste  :P
Nice solution [b]ignite168[/b].


How many natural numbers require 3 digits when wriiten in base 12, but require 4 digits when written in base 9?

[hide="Answer"][b]Answer:[/b] $ 999$ [/hide]

I didn't quite get the solution in the book..

[hide]
Smallest three digit number in base 12 is 144, and the largest is 1727.  The largest 4 digit base 9 number is 6561, smallest is 729.  We are looking for all numbers $ x$ so that
$ x\in \mathbb{N}$
$ 144\leq x\leq 1727$
$ 729\leq x\leq 6561$
Clearly the 2 inequalities mean that $ 729\leq x\leq 1727$, and there are $ \boxed{999}$ numbers that satisfy that and the first fact.
[/hide]


vector
linear algebra
matrix
geometry
geometric transformation
rotation
analytic geometry

Let $ O$ be the center of the regular polygon $ A_1A_2\cdots A_n$. Prove that $ \overrightarrow{OA_1} \plus{} \overrightarrow{OA_2} \plus{} \cdots \plus{} \overrightarrow{OA_n} \equal{} \mathbf0$.

Also, how to type the vector sign in a better way?

You can use matrix of rotation or complex plane of rotation, de Moivre's theorem.

[code]\overrightarrow{OA}[/code]

$ \overrightarrow{OA}$

Thank you!
Okey, now I assume that we don't know matrix and linear algebra.

Wait, doesnt the arrow stand for ray?
and since ray's have infinite length, how would it be 0?
Unless the arrow sign means something else.

It stands for a vector, although it should be obvious by the title

Place the center $ O$ at the origin of the complex plane $ \Pi$ with $ A_1$ on the postive $ x$ axis.  Then you may represent each the vectors $ \overrightarrow{OA_1},...,\overrightarrow{OA_n}$ with the complex numbers $ r,r\varepsilon,r\varepsilon^2,..., r\varepsilon^{n \minus{} 1},\ \varepsilon \equal{} e^{i\frac {2\pi}{n}}$, respectively.  Then we have $ \sum \overrightarrow{OA_i} \equal{} \bold{0}$ since $ \sum_{k \equal{} 0}^{n \minus{} 1}\varepsilon^k \equal{} 0$ ($ \varepsilon^n \equal{} 1$).

How to do it in a geometric way? (not using matrix, complex number, coordinate, etc.)

Suppose that the sum is not 0 and rotate everything by $ \frac{2\pi}{n}$ along the origin. The conclusion follows.

Physical interpretation gives the sum of $ n'$ s forces is zero vector when these forces are balanced.


Does anyone know of any websites that have problems?

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=133189[/url]

Browse or bump that topic please.


Please find out the range of the following function and show me the solution:

$ f(x) \equal{} \frac{1}{\sqrt{x^{2}\minus{}1}}$

Thank you.

Is the entire denominator supposed to be the radicand? I.e., should it look like this: $ f(x) \equal{} \frac{1}{\sqrt{x^2 \minus{} 1}}$?

Assume you meant the function to be $ f(x) \equal{} \frac {1} {\sqrt {x^2 \minus{} 1}}$. Clearly $ f: \mathbb{R} \setminus [ \minus{} 1, 1] \to \mathbb{R}_{ \plus{} }$. Now take $ 0 \leq y \equal{} \frac {1} {\sqrt {x^2 \minus{} 1}}$ and get $ x \equal{} \pm \sqrt {1 \plus{} \frac {1} {y^2}}$, with only restriction $ y \neq 0$, hence the range of $ f$ is $ (0, \infty)$. Of course, the same result follows from $ \lim_{x \to \pm 1} f(x) \equal{} \infty$, while $ \lim_{x \to \pm \infty} f(x) \equal{} 0$, so $ f$ being continuous on its domain takes all intermediate values.
If, otherwise, the function would be $ f(x) \equal{} \frac {1} {|x| \minus{} 1}$, it would have as domain  $ \mathbb{R} \setminus \{ \minus{} 1, 1\}$, while its range would be $ ( \minus{} \infty, \minus{} 1] \cup (0, \infty)$.

EDIT: Sorry, I kept having both formulae simultaneously present in my mind.

[quote="mavropnevma"]If, otherwise, the function would be $ f(x) \equal{} \frac {1} {|x| \minus{} 1}$, it would have the same domain,[/quote]The domain would be different because for this one its only $ |x|\ne1$ and the other one is $ |x|>1$ ;)


Does anyone else notice that Kiran Kedlaya's Geometry Lecture Notes on the AMC site are missing diagrams? That is, the notes say "DIAGRAM" wherever a diagram should be. Does anyone happen to have a version of the lecture notes with diagrams?

I think they're unfinished; he meant to go back and put in diagrams. For example, in the inequality package, there's a big space near the beginning that says "INSERT EXAMPLES HERE" or something.

his notes are an unfinished manuscript. there are no figures at all. fierytycoon, how much of this have you done? i started working with it recently. did you check out his guide to inequalities? i think it was pretty good.

I am not working on them right now, I was just poking around a bit.

The Sunday math class that I'm currently attending is using his math notes.  We've been going through it quickly because this class is primarily intended for only 3 of the students (who already did these hard olympiad problems last year) as a review.  Oh well, I can still learn quite a bit.

I have also noticed that there are diagrams missing...Hopefully he'll update it.

-interesting_move

Some of Tom Davis' Math Circle notes have diagrams missing, too.  These are at [url]http://www.geometer.org/mathcircles/[/url], in case you haven't seen them...

Enjoy!

mathfanatic


In a store display there are 60 shirts, each shirt either red or blue. There are 10 more blue shirts than red shirts. Each blue shirt has five stars on it, and each red shirt has one star. When $ \frac45$ of the blue shirts and none of the red shirts have been sold, how many stars are left in the display?

There are 35 blue shirts and 25 blue shirts (easy part)

After selling, there are 7 blue shirts and 25 red shirts.
There are 7(5)+25(1)=60 stars left.


Ring Theory
abstract algebra
superior algebra
superior algebra solved

Let $R$ be a PID, and $I$ is a nonzero ideal of $R$. Prove that $\frac{R}{I}$ is noetherian and artinian.

there's a monotonic bijection

{ideals of $R$ containing $I$} $\to$ {ideals of $R/I$}

given by $J \to J/I$. this implies $R$ noetherian $\Rightarrow R/I$ noetherian. since every PID is noetherian, $R/I$ is noetherian. further, every every descending chain in $R/I$ comes from a descending chain in $R$ of the form $J_{1}\supseteq J_{2}\supseteq J_{3}\supseteq ... ... \supseteq I$. write $J_{i}= R a_{i}$ and $I = R b$ where $b \notin R^{x}\cup \{0\}$. then we have an ascending chain of divisors $a_{1}| a_{2}| a_{3}| ... | b$, which has to terminate since $R$ is also UFD. thus, $R/I$ is artinian.


If a,b,c lengths of the side of a triangle proove that
\[ a^4+b^4+c^4+abc(a+b+c)\geq 2(a^2b^2+b^2c^2+c^2a^2)  \]

We have $a^4+b^4+c^4\geq a^2b^2+b^2c^2+c^2a^2$  and  $abc(a+b+c)\geq 0$

Add these and you will take the desired ..

Please donot put such inequalities here .Post them in the Intermiadiate section  :mad:

Maybe , but even if the inequality in very easy for Olympiad section

[quote="luke999"]If a,b,c lengths of the side of a triangle proove that
\[ a^4+b^4+c^4+abc(a+b+c)\geq a^2b^2+b^2c^2+c^2a^2  \][/quote]
   Hey, Luca, it's trivial!Don't post this kind of problems... :P

[quote="eLeM"]I think inequality should be:
$a^4+b^4+c^4+abc(a+b+c)\geq 2(a^2b^2+b^2c^2+c^2a^2)$, because $a^4+b^4+c^4 \geq a^2b^2+b^2c^2+c^2a^2$ by AM-GM.[/quote]

Can someone post a solution or give a link to this inequality?

[quote="stancioiu sorin"][quote="eLeM"][quote="luke999"]If a,b,c lengths of the side of a triangle proove that
\[ a^4+b^4+c^4+abc(a+b+c)\geq a^2b^2+b^2c^2+c^2a^2  \][/quote]

I think inequality should be:
$a^4+b^4+c^4+abc(a+b+c)\geq 2(a^2b^2+b^2c^2+c^2a^2)$, because $a^4+b^4+c^4 \geq a^2b^2+b^2c^2+c^2a^2$ by AM-GM.[/quote]
    Yes, you are right, Elem.The first inequality is from AM-GM, and the second is Muirhead.That's trivial![/quote]

But using Muirhead second inequality is reversed. $\sum_{sym}a^2bc \leq \sum_{sym}a^2b^2$ since $(2,2,0) \succ (2;1;1)$. Or I understand it incorrectly?

[quote="eLeM"][/quote][quote="stancioiu sorin"][quote="eLeM"][quote="luke999"]If a,b,c lengths of the side of a triangle proove that
\[ a^4+b^4+c^4+abc(a+b+c)\geq a^2b^2+b^2c^2+c^2a^2  \][/quote]

I think inequality should be:
$a^4+b^4+c^4+abc(a+b+c)\geq 2(a^2b^2+b^2c^2+c^2a^2)$, because $a^4+b^4+c^4 \geq a^2b^2+b^2c^2+c^2a^2$ by AM-GM.[/quote]
    Yes, you are right, Elem.The first inequality is from AM-GM, and the second is Muirhead.That's trivial![/quote][quote="eLeM"]

[/quote]

It is trivial.
$a^4+b^4+c^4+abc(a+b+c)- 2(a^2b^2+b^2c^2+c^2a^2)$$= \sum a^2(a-b)(a-c)+ac \sum (a-c)^2 \geq 0$ :D

[quote="Gibbenergy"]
It is trivial.
$a^4+b^4+c^4+abc(a+b+c)- 2(a^2b^2+b^2c^2+c^2a^2)$$= \sum a^2(a-b)(a-c)+ac \sum (a-c)^2 \geq 0$ :D[/quote]

Thanks. It is not so trivial for me but still nice :)

[quote="spix"]It is quite obvious you don't  know what Muirhead is.[/quote]
   Why do you say that?You don't know what I know!! :mad: Because I am only seventh grade you think that I don't know this famous inequality?Please, don't write such messages...

Don`t get me wrong. I have nothing with you. But , you have a very annoying manner of posting. and you spam.(p.s.: your enthusiasm is remarkable .. keep math-ing)

OK, spix.In the future I won't [color=red]spam[/color], if you think that.Anywhay, thank you for advertisment. :)

By Schur and then Muirhead

$a^4+b^4+c^4+abc(a+b+c)\geq \sum a^3b \geq 2\sum a^2bc$

(note that the first sum contains six terms, whereas the second only contains three)

Nowhere did we need to use that $a,b,c$ are sides of a triangle; all we need is $a,b,c\geq 0$.