A family of four dines at a restaurant where the cost of the buffet is $13.25 per person, plus 4.5% tax.  How many dollars is the total bill, to the nearest cent, if tax is added after the bill is totalled?

[hide]
If one member costs $\$$13.25,
4 people cost $\$$13.25*4=53
after 4.5% tax=53*(100%+4.5%)
                   =53*104.5%
=55.385
                   The answer is $\$$55.39 when rounded to the nearest cent[/hide]

Right.  That's what the answer key says.  I'm not sure why $55.38 is incorrect though.  Aren't $55.38 and $55.39 equally near to $55.385?

[quote="slamchess"]when it is equal you always round up
(unless it is a problem where you always round down like #of people)[/quote]

that's not called rounding down, its called flooring
the def. of rounding is to go to the nearest, up or down
the def. of flooring is the greastest int. < or = to the number.

[quote="Ravi B"]Aren't 55.38 and 55.39 equally near to 55.385?[/quote]

Yes, but by convention $0.5$ is rounded up.  (Edit:  This is so that if you pick an arbitrary digit and round it to the nearest ten, you are equally likely to round either up or down since $0, 1, 2, 3, 4$ round down but $5, 6, 7, 8, 9$ round up.)

[quote="davidlizeng"]that's not called rounding down, its called flooring[/quote]

...?  "Rounding down" is a perfectly sensible way to describe flooring.

Where does the problem say "round"?  I just see the word "nearest".  Since 55.38 and 55.39 are equally near, I conclude the problem is ambiguous.  MathCounts can't redefine the terms "near" and "nearest", can it?  Otherwise, we would be in the awkward position of saying that 55.38 and 55.39 are equally near, but 55.39 is nearest.

P.S.  When I was a kid, we were taught a different rounding convention, round to even, so that 55.375 and 55.385 would both round to 55.38.  The idea was that such roundings tend to cancel each other out.

Good point. Mathcounts ought to create a term dictionary, especially that this problem causes controversy about rounding 55.385 to $55.39 rather than$55.38. I wonder how this would be handled in real life.

4*13.25 = 53 * 1.045. um... multiply that out and round. :)

Also, Mathcounts Rules probably says something about problems that require you (imply) to round, in which case you would use the 5 and up rule.

[quote="Ravi B"]Where does the problem say "round"?  I just see the word "nearest".  [/quote]

In the contexts in which I've seen it, "to the nearest" implies "rounded to the nearest."


function
inequalities
algebra open
algebra

Find  $ f: \mathbb{N}^* \to\mathbb{ N}^*$ such that:
i. $ f(2) = 2$
ii.$ f(mn)= f(m)\cdot f(n)$
iii.$ f(m) < f(n)$ with $ m < n$.

Obviously, we only have to define it for the primes, because of prime decomposition, and we know that $ f(1)\equal{}1$ since we know that $ 1 \le f(1)<f(2)\equal{}2$, which is consistent with the second restriction. All we have to do now is show that the third condition will only hold with the identity function when we define it on the primes.

If $ f(p_i)<p_1$ then consider the inequality $ 2^k \le p_i^n$ <--> $ k/n \le log_2p_i$ Now, let $ k/n >log_2f(p_i)$. Such k,n exists, since between every 2 distinct real numbers are infinately many rational numbers.

$ f(2^k) \le f(p_i^n)$ <--> $ 2^k \le f(p_i)^n$ <--> $ k/n \le log_2f(p_i)$ Contradiction.

Similarly for the other way.

Therefore, $ f(x)\equal{}x$ is the only solution.

[hide="Alternate way of clamping the value at primes"]By strong induction. To show $ f(p) \equal{} p$ for a prime $ p$ given that $ f(q) \equal{} q$ for all primes $ q < p$, note that the prime factorizations of $ p\plus{}1, p\minus{}1$ are composed of primes less than $ p$, all of which we know map to themselves. So we must have $ f(p\plus{}1) \equal{} p\plus{}1, f(p\minus{}1) \equal{} p\minus{}1$ by the inductive hypothesis. Then use condition (iii).[/hide]


Two 1.9 kg balls are attached to the ends of a thin rod of negligible mass, 54 cm in length. The rod is free to rotate in a vertical plane about a horizontal axis through its center. With the rod initially horizontal as shown, a 53 gm wad of wet putty drops onto one of the balls with a speed of 3.5 m/sec and sticks to it. a) What is the angular speed of the system just after the putty wad hits?

Just before the wad hits, all of the energy in the system is in the form of kinetic energy in the wad.  After the wad hits, it is all rotational kinetic energy.  Also remember conservation of angular momentum.


pigeonhole principle
algebra unsolved
algebra

Let A=(a1,a2,a3...a2000) be a sequence of integers each lying in the interval[-1000,1000]Suppose that the entries in sum up to 1. Show that some nonempty subsequence of A sums to zero.

I believe that it's a hard and very beautiful problem  ;)

[i]Solution:[/i]

We can suppose that non of the subsequence of $A$ is not equal to zero.Now we construct the sequence $A$ with constructing the sequence $B$ s.t. $b_1>0$ and $b_2<0$ and for every $i=1,2,...,2000$ we know that there exist $b_i$ s.t. the sign of the $b_i$ be the opposite of the sign of the sum  $S_{i-1}=b_1+b_2+...+b_{i-1}$.(suppose that for every $S_{i-1}$ is not equal to zero)  and we know $a_1+a_2+...+a_{2000}=1$ and it show that one of the subsequece of $A$ is equal to zero or has an opposite sign to the sign of the $S_{i-1}$.
So that every of this $S_1,S_2,...,S_{2000}$ is one of the $1999$ number ,so that according to the pigeonhole principle there exist $i,j$ s.t. $S_i=S_j$ where $1\leq j\leq i \leq 2000$ so that $b_{j+1}+b_{j+2}+...+b_i=0$ and the proof is complete  :D


[quote="Hustla25"]put up the figure

can't see it[/quote]

It's there, I made it with asymptote.

Sorry for reviving but i thought this should have an answer.

I got 6, which is E

Any better methods than brute force spatially visualization?

There are only 9 things to check, and it really shouldn't be hard to visualize once you see how #4-9 can be put up (since 6 is the max answer).

Tho might be a dumb question but don't you need 6 congruent squares to form the cube? For this problem I only see 5

[quote="worthawholebean"]How many of the nine resulting polygons can be folded to form a cube with [b][color=#FF0000]one face missing[/color][/b]?[/quote]


geometry
geometric transformation
rotation
complex numbers

Let $ a,b,c$ be points in the complex plane. Prove that the triangle determined by the three complex numbers is equilateral if and only if $ a^{2}+b^{2}+c^{2}=ab+bc+ca$.

[hide="Quick and dirty"]  $ a^{2}+b^{2}+c^{2}= ab+bc+ca \Leftrightarrow (a-b)^{2}+(b-c)^{2}+(c-a)^{2}= 0$

Therefore WLOG $ a = 0$.  The original equation becomes $ b^{3}+c^{3}= 0$, so $ b$ is an appropriate rotation of $ c$.  The conclusion follows. [/hide]

mkay what about $ a^{2}+b^{2}+c^{2}-ab-bc-ca=(a+\omega b+\omega^{2}c)(a+\omega^{2}b+\omega c)=0$ where $ \omega^{2}+\omega+1=0$


superior algebra
superior algebra unsolved

Suppose an extension $ E/K$ of degree $ 2n$ is of the form $ E\equal{}K(\sqrt{a_1}, \sqrt{a_2}, \dots , \sqrt{a_n})$ with $ a_i \in K$. Prove that if $ char K \not\equal{} 2$ then $ \alpha : \equal{} \sqrt{a_1}\plus{} \sqrt{a_2}\plus{} \dots\plus{} \sqrt{a_n}$ is a primitive element of $ E/K$

If so, [url=http://www.mathlinks.ro/viewtopic.php?t=244104]here[/url] is a solution, posted 3 days ago...


Let's discuss this olympiad and its problems.What is your first day results?Of course not official,at least rumours,or your own opinion.
Is here someone who participated at this olympiad,actually,at this time at first day(second day will be tomorrow)?
P.S:
I think that this olympiad problems are much harder than it was at Zimo 2007.What is your opinion?
I hope tomorrow we will continue this topic(of course if i'll not be very busy). :)
I'll post all problems tomorrow. :wink:

[quote="cefer"]What about you Erken? I hope you solved all of them

Anyway,what about problems? Can anybody post them?  :)[/quote]
I don't know official results yet,but i'll post all of them,as soon as possible.

[quote="AdrianDD"]Please, can you post the results? Thank you![/quote]
There is a list of results at my school.I'll come back from there soon,and post the results.
Good bye.

1st Day:
[url=http://www.mathlinks.ro/viewtopic.php?p=1013352#1013352]Problem 1[/url]
[url=http://www.mathlinks.ro/viewtopic.php?p=1013354#1013354]Problem 2[/url]
[url=http://www.mathlinks.ro/viewtopic.php?p=1013356#1013356]Problem 3[/url]

2nd Day:
[url=http://www.mathlinks.ro/viewtopic.php?p=1013358#1013358]Problem 4[/url]
[url=http://www.mathlinks.ro/viewtopic.php?p=1013359#1013359]Problem 5[/url]
[url=http://www.mathlinks.ro/viewtopic.php?search_id=2087670879&t=183916]Problem 6[/url]

You are MONSTERS :lol:
2 gold medals of 4 participants is a very good rezult)))

ps OPUBLIKUYTE REZULTATY plz!!!

[quote="A_L_E_N"]You are MONSTERS :lol:
2 gold medals of 4 participants is a very good rezult)))
 [/quote]
HI!
Bulgarians have 3 gold meals!!!

Gold Medals:
1 Воробьев Илья  Россия, Москва,СУНЦ МГУ
2 Клочков Егор Казахстан, Павлодар, сборная
3 Далиев Асет Казахстан, Алматы, РСФМСШИ
4 Борислав Атанасов Вълков Болгария, София, Софийская математическая гимназия
5 Александър Кирилов Даскалов Болгария, София, Софийская математическая гимназия
6 Тусупбеков Еркен Казахстан, Алматы, РСФМСШИ
7 Оразбаев Санжар Казахстан, Актобе, сборная
8 Temuge Enkhbaatar  Монголия, Улан-Батор, Школа №1
9 Любослав Николаев Панчев Болгария, София, Софийская математическая гимназия
10 Deneanu Andrei-Florin Румыния, Бухарест, International Computer Highschool of Bucharest

Silver medals:

1 Bozgan Francisc Teodor Румыния, Бухарест, International Computer Highschool of Bucharest
2 Sarkhan Badirli  Азербайджан, Баку, Baku Turkish High School
3 Семенова Вера Украина, Киев, Украинский ФМЛ КНУ им. Т.Шевченко
4 Мулдагалиев Биржан Казахстан, Алматы, РСФМСШИ
5 Калин Иванов Ланджев Болгария, София, Софийская математическая гимназия
6 Смирнов Дмитрий Россия, Санкт-Петербург, ФТШ при ФТИ им. А.Ф. Иоффе
7 Сатылханов Канат Казахстан, Талдыкорган, сш №20
8 Bumbacea Radu       Румыния, Бухарест, TUDOR VIANU National High School of computer science
9 Коробейников Игорь  Таджикистан, Душанбе, Таджикско-Турецкий лицей
10 Byambajav Namsraijav  Монголия, Улан-Батор, Школа №1
11 Кабасов Багдат Казахстан, Актобе, сборная
12 Хаджимуратов Нурсултан Казахстан, Талдыкорган, сш №20
13 Касенов Ескендир Казахстан, Астана, сборная
14 Атласова Мария  Республика Саха, Якутск, Республиканский лицей-интернат
15 Myagmarjav Bold  Монголия, Улан-Батор, Школа №1
16 Калдыбаев Нуразем Кыргызстан,БишкекКыргызско-Турецкий лицей "Себат"

Bronze medals:

1 Лакирбая Лаша Грузия, Тбилиси, Школа № 199 "Комаров"
2 Олимджони Пирмахмад Таджикистан, Душанбе, Таджикско-Российская гимназия-интернат
3 Никифоров Дьулустан  Республика Саха, Якутск, Республиканский лицей-интернат
4 Ruslan Muslimov  Азербайджан, Баку, Baku Turkish High School
5 Равчеев Никита  Россия, Красноярск, сборная
6 Кхомерики Георг Грузия, Тбилиси, Школа № 199 "Комаров"
7 Досымбаев Олжас Казахстан, Алматы, РСФМСШИ
8 Гуроян Ваган Армения, Ереван, ФМШ при ЕрГУ
9 Zorigoo Ochirhuyag  Монголия, Улан-Батор, Школа №1
10 Tsogbayar Idertsogt  Монголия, Улан-Батор, Школа №1
11 Хайдаров Фарход Узбекистан, Ташкент, Школа "Иктидор"
12 Шульга Олесандр Украина, Киев, Украинский ФМЛ КНУ им. Т.Шевченко
13 Moraru Laurentiu        Румыния, Бухарест, TUDOR VIANU National High School of computer science
14 Войделевич Алексей Беларусь, Минск, Средняя школа № 51
15 Ферадзе Лаша Грузия, Тбилиси, Школа № 199 "Комаров"
16 Нукежанов Ернар Казахстан, Алматы, РСФМСШИ
17 Кенжетаев Азамат Казахстан, Павлодар, сборная
18 Жолдас Мерей Казахстан, Усть-Каменогорск, КТЛ для одаренных детей
19 Georgescu Flavian Румыния, Бухарест, International Computer Highschool of Bucharest
20 Ерпылев Алексей  Россия, Москва,СУНЦ МГУ
21 Мирзаев Иномжон  Таджикистан, Душанбе, Таджикско-Турецкий лицей
22 Нурсултанов Медет Казахстан, Астана, сборная
23 Присивко Вячеслав Россия, Санкт-Петербург, ФТШ при ФТИ им. А.Ф. Иоффе
24 Camburu Oana-Maria Румыния, Бухарест, TUDOR VIANU National High School of computer science
Sorry for russian language,cause it is too long to translate all the names into english. :roll:

Клочков Егор??? Hmmm... 4th participation in ZIMO)))
Can you post post points nof gold medalists?

I can post the points of all medalists :) :
1 Воробьев Илья  математика 1RUS 7 7 0 7 7 7 35 GOLD
2 Клочков Егор математика 3KAZ 7 3 0 7 7 7 31 GOLD
3 Далиев Асет математика 9KAZ 5 4 0 7 7 7 30 GOLD
4 Борислав Атанасов Вълков математика BGR 5 3 0 7 7 7 29 GOLD
5 Александър Кирилов Даскалов математика BGR 7 7 0 6 7 1 28 GOLD
6 Тусупбеков Еркен математика 9KAZ 7 3 0 7 7 3 27 GOLD
7 Оразбаев Санжар математика 2KAZ 7 7 0 6 7 0 27 GOLD
8 Temuge Enkhbaatar  математика MNG 0 7 3 7 7 1 25 GOLD
9 Любослав Николаев Панчев математика BGR 7 4 0 7 7 0 25 GOLD
10 Deneanu Andrei-Florin математика 2ROM 7 3 2 6 7 0 25 GOLD
11 Bozgan Francisc Teodor математика 2ROM 7 3 0 6 7 1 24 SILVER
12 Sarkhan Badirli  математика 1AZE 7 3 0 7 0 7 24 SILVER
13 Семенова Вера математика UKR 7 2 0 7 7 0 23 SILVER
14 Мулдагалиев Биржан математика 10KAZ 7 0 0 7 7 1 22 SILVER
15 Калин Иванов Ланджев математика BGR 7 3 0 5 7 0 22 SILVER
16 Смирнов Дмитрий математика 3RUS 7 6 2 7 0 0 22 SILVER
17 Сатылханов Канат математика 4KAZ 0 3 0 3 7 7 20 SILVER
18 Bumbacea Radu       математика 1ROM 7 3 0 3 7 0 20 SILVER
19 Коробейников Игорь  математика 2TJK 7 0 0 5 7 0 19 SILVER
20 Byambajav Namsraijav  математика MNG 7 1 3 7 1 0 19 SILVER
21 Кабасов Багдат математика 2KAZ 7 0 0 5 7 0 19 SILVER
22 Хаджимуратов Нурсултан математика 4KAZ 7 3 0 2 7 0 19 SILVER
23 Касенов Ескендир математика 1KAZ 7 3 0 7 2 0 19 SILVER
24 Атласова Мария  математика YKT 7 3 0 2 7 0 19 SILVER
25 Myagmarjav Bold  математика MNG 7 4 0 5 2 0 18 SILVER
26 Калдыбаев Нуразем математика 3KGZ 7 3 0 2 5 1 18 SILVER
27 Лакирбая Лаша математика GEO 7 3 0 7 0 0 17 BRONZE
28 Олимджони Пирмахмад математика 1TJK 7 0 0 3 7 0 17 BRONZE
29 Никифоров Дьулустан  математика YKT 7 3 0 0 7 0 17 BRONZE
30 Ruslan Muslimov  математика 1AZE 7 3 0 7 0 0 17 BRONZE
31 Равчеев Никита  математика 5RUS 7 0 0 2 7 0 16 BRONZE
32 Кхомерики Георг математика GEO 7 0 0 2 7 0 16 BRONZE
33 Досымбаев Олжас математика 9KAZ 4 3 0 1 7 1 16 BRONZE
34 Гуроян Ваган математика ARM 7 3 0 6 0 0 16 BRONZE
35 Zorigoo Ochirhuyag  математика MNG 7 0 0 7 2 0 16 BRONZE
36 Tsogbayar Idertsogt  математика MNG 7 2 0 7 0 0 16 BRONZE
37 Хайдаров Фарход математика UZB 7 0 0 2 7 0 16 BRONZE
38 Шульга Олесандр математика UKR 7 0 0 2 0 7 16 BRONZE
39 Moraru Laurentiu        математика 1ROM 0 3 0 6 7 0 16 BRONZE
40 Войделевич Алексей математика 2BLR 7 0 0 1 7 0 15 BRONZE
41 Ферадзе Лаша математика GEO 7 2 0 5 1 0 15 BRONZE
42 Нукежанов Ернар математика 11KAZ 0 0 0 7 5 3 15 BRONZE
43 Кенжетаев Азамат математика 3KAZ 7 1 0 7 0 0 15 BRONZE
44 Жолдас Мерей математика 5KAZ 7 0 0 1 7 0 15 BRONZE
45 Georgescu Flavian математика 2ROM 0 0 0 2 6 7 15 BRONZE
46 Ерпылев Алексей  математика 1RUS 7 2 0 5 0 0 14 BRONZE
47 Мирзаев Иномжон  математика 2TJK 7 0 0 7 0 0 14 BRONZE
48 Нурсултанов Медет математика 1KAZ 0 3 0 7 3 1 14 BRONZE
49 Присивко Вячеслав математика 3RUS 7 0 0 7 0 0 14 BRONZE
50 Camburu Oana-Maria математика 1ROM 4 0 0 3 7 0 14 BRONZE


quadratics
modular arithmetic
number theory proposed
number theory

for all $ x,y,n,m,k \in N$ prove that $ 4kxy\minus{}1 \nmid x^m\plus{}y^n$

[quote="shoki"]for all $ x,y,n,m,k \in N$ prove that $ 4kxy \minus{} 1 \nmid x^m \plus{} y^n$[/quote]
Proof.
First, we have a lemma:
Lemma:
Assume that $ m, x \in \mathbb{N}$ such that: $ 4x\mid t\plus{}1$. Prove that: $ \minus{}x$ isn't quadratic residue $ \pmod t$.

Back to the problem.

Assume that $ 4kxy \minus{} 1 \mid x^m \plus{} y^n$

With $ t\equal{}4kxy\minus{}1$, we have $ \minus{}x, \minus{}y$ isn't quadratic residue $ \pmod t$.

If $ m, n$ is even number, we have: $ 4kxy\minus{}1\mid (x^{m'})^2\plus{}(y^{n'})^2$ inferred $ \gcd(x^{m'}, 4kxy\minus{}1)>1$, this is not true.

Assume that $ n$ is odd number. Call $ p$ is divisor number of $ n$ such that $ p\equal{}4i\plus{}3$. We have
$ \binom{x}{p}\equal{}\binom{\minus{}x}{p}\cdot\binom{\minus{}1}{p}\equal{}1$. Thus $ x$ is quadratic residue $ \pmod t$
Moreover $ p\mid x^m\plus{}y^n$ so $ x^m\equiv (\minus{}y)^n\pmod p$. (1)

Because $ \minus{}y$ isn't quadratic residue $ \pmod t$, $ n$ odd, $ x$ is quadratic residue $ \pmod t$, so (1) not true.

So $ 4kxy \minus{} 1 \nmid x^m \plus{} y^n$ for all $ x,y,n,m,k \in \mathbb{N}$


geometry
topology
real analysis
real analysis theorems

Hello,

yet again I would like to ask a question concerning seminorms, an area in which my experience is next to nothing.

Let us agree we write $p_1 \leq \leq p_2$ for seminorms if there is a constant$C>0$ such that $p_1\leq C p_2$

Suppose you have a space V with a countable set of seminorms P.

prove that this system of seminorms is equivalent with an increasing sequence of seminorms :

$p_1\leq\leq p_2 \leq \leq p_3 \cdots$

my professor and costudents all tell me this is so obvious , but unfortunately I don't think it is.  My main difficulty is that I think it requires infinite rearranging of the seminorms, and thus not a really construction process of the sequence of seminorms.

All comments are welcome and much appreciated!

I think it would be enough to prove that for any two seminorms $a,b$ of $V$ there is a $C$ such that:

$a \leq \leq b$

how would that help?  And usually that is not true?   I really don't understand you radusorici I am sorry  :?

Let the original norms be $s_1,s_2,\dots$. I'm pretty sure just taking the sums $p_1=s_1,p_2=s_1+s_2,p_3=s_1+s_2+s_3,\dots$ would work. Since $p_k(x_n)\to0$ for every $k$ if and only if $s_k(x_n)\to0$ for every $k$, the two families of seminorms generate the same topology.

hmm, but someone else told me that inducing the same topology is not the same as equivalence of systems of seminorms, it is post somewhere by the way

boy, am I bad at this :?


Just to keep the game going, I'm going to make a Game 3.
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1734033730&t=149694]Instructions, Game 1[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1734033730&t=151415]Game 2[/url]

Lingomaniac88 is modding.

I'll mod this one.

We need about 10-15 people.  For those not familiar with the rules:

In each round, everyone PM's me an integer between 0 and 100.  Once everyone sends me a number or when the submission deadline passes, I will post half the average and what everyone sent.

If anyone sent me a number below half the average, that person is out.  Otherwise, that person receives the difference between the number submitted and half the average.  So if you sent me 36 and half the average was 25, you'd earn 11 points.  If you earn over 100 points, you're out.

Note: if you don't submit a number, I will assume you submit 100.

We continue until there's only one person left.  Last one standing wins.

[b]PLAYERS:[/b]
perfect628
xpmath
hunter34
mustafa
mnmath
cincodemayo5590

We have 7 people so far.  I'll wait 24 hours for more people.  Those who joined can submit your numbers for round 1 right now if you want.

Anyone else want to play?  You can still sign up.

I guess we're just playing with seven people.  Round 1 is due in 48 hours. (Sunday at 10:00PM EST)

I'm going to have to drop out, because I'm going to camp for a week. Sorry

[quote="lingomaniac88"]You know what?  Forget it.  The game is off.  We don't have a lot of players.[/quote]

You should edit your signature then...


geometry
combinatorics unsolved
combinatorics

[quote="Zorro"]Prove that in a convex hexagon whith area $1$ there exists diagonal which cuts a triangle with area not greater than $\frac{1}{6}$[/quote]

But this is not a combinatorics problem, and I think is rather classical. Draw ALL three diagonals and reason with areas. You must be able to prove that, in case that all the triangles you refer have areas greater than 1/6, every big diagonal of the hexagon cuts it in two halves, each one of them having an area greater than 1/2

Daniel

[quote="Zorro"]sorry but I still don't know how to do it[/quote]

Ok, sorry about that. Draw the three big diagonals of the hexagon, assume first the case in which they are concurrent, say at point $O$. If the points in the hexagon are labelled $A_1A_2A_3A_4A_5A_6$, this defines six triangles, $OA_iA_{i+1}$.  You can prove that, if all triangles $A_iA_{i+1}A_{i+2}$ have area greater than 1/6, then the area of each $OA_iA_{i+1}$ triangle is greater than 1/6 and therefore the area of the hexagon must be greater than 1 thus contradicting the assumption. The reason for which the area of $OA_iA_{i+1}$ is greater than 1/6 is that its value must lie between the the area of trangle $A_{i-1}A_iA_{i+1}$ and area of triangle $A_{i}A_{i+1}A_{i+2}$ because they share base and $O$ lies between $A_{i-1}$ and $A_{i+2}$, and we assume that both of the latter have areas greater than 1/6.
similarly one can analyze the case in which the three diagonals do not intersect at a point. Hope now it's understandable

Daniel

well the case when they intersect I understand, but I don't know how to do the other case


On the AMC 12, each correct answer is worth $ 6$ points, each incorrect answer is worth $ 0$ points, and each problem left unanswered is worth $ 2.5$ points. If Charlyn leaves $ 8$ of the $ 25$ problems unanswered, how many of the remaining problems must she answer correctly in order to score at least $ 100$?

$ \textbf{(A)}\ 11 \qquad \textbf{(B)}\ 13 \qquad \textbf{(C)}\ 14 \qquad \textbf{(D)}\ 16 \qquad \textbf{(E)}\ 17$

[hide]

From the unanswered problems, Charlyn has already $ 8\cdot 2.5 \equal{} 20$ points. Therefore, she needs $ 80$ more points, or $ 80/6 \equal{} 40/3 \equal{} 13.\overline{3}$. Rounding up, she needs 14 more correct answers. $ \mathrm{ (C) \ }$

[/hide]


a persons base is approx 25^2 centimetres. In 30^2 centimeter 30 raindrop will be droped on that spot. Whats the probability of that person to dodge the raindrops in a second.

Question 2: In 5 minuites, how many cubic centimeter of rain will be droped on the person if he stays on the same spot, asumming 1 drop=3cm^3?

Question 3: if this person were to run another 50metres to avoid all the rain asumming he is running at 5meter per second., how much more raindrops must he recieve?Whats the probability of this guy for not receiving any raindrops?

I honestly have no clue what you are talking about. If you could either link the source or copy verbatim, it would be appreciated.


[hide]
Let $y=x-3$ and simplify the equation into

$5y(y^2+6)=0$ $\implies y=0$ $\implies x=3$
[/hide]

are you sure
I get
[hide="sol"]
$a=x-3$
$2a(a^3+4a^2+30)=0$
Then solve for $a$
[/hide]

Riddler, how could you get that?  Your equation is a quartic, the given equation is a cubic.

[hide="sol"]
$2a(a^2+12)+2a(a^2+3)+a^3=0$
Sorry miscalculations.
[/hide]


Consider the following statements related to two sets A and B:

1) The Cartesian product AxB contains the pair (1,2)
2) A is subset of B.
3) The pair (1.2) is not an element of BxA.

Check the box sets which are consistent with these statements.

a) A = {1,2}, B = {1,2,3}
b) A = {1,2}, B = {2,3}
c) A = {1,6}, B = {1,2,6}
d) A = {1,3}, B = {1,3,6}
e) A = {1,3}, B = {1,2}

ANSWER: C

Please, if anyone knows how to help me explain myself in a more simple and easy to understand as possible. I'm still very new in this field. Another thing also did not understand when he meant subset.

A subset of some set is a set which contains some of the elements of the bigger ones-it might contain none of them, it might contain some of them, it might contain all of them.  For example, if $ A\equal{}\{1,2,3\}$, then all of the possible subsets are $ B\equal{}\emptyset$ (empty set), $ B\equal{}\{1\}$, $ B\equal{}\{1,2\}$, $ B\equal{}\{1,3\}$, $ B\equal{}\{2\}$, $ B\equal{}\{2,3\}$, $ B\equal{}\{3\}$, and $ B\equal{}\{1,2,3\}$.

Now let's look at choice C below.  $ A\times B\equal{}\{(a,b)\mid a\in A, b\in B\}$.  Since $ 1\in A$ and $ 2\in B$, $ (1,2)\in A\times B$.  Since $ 1,6\in A$, and $ 1,6\in B$, $ A\subset B$ ($ A$ is a subset of $ B$).  However, $ 2\notin A$, so $ (1,2)\notin B\times A$.  That is why C is correct.


Let $ n$ be a positive integer and $ S\equal{}\{1,2,\dots,n\}$. Find all one-to-one functions $ f: S\to S$ such that $ f(i)\minus{}f(j)$ divides $ i\minus{}j$ for all $ i\neq j$.

[hide="Is this enough?"]Let $ j \equal{} i \minus{} 1$ for all $ i > 1$. We find that $ f(i) \minus{} f(i \minus{} 1) \mid 1$, but since all values of $ f$ are integral, we have that $ |f(i) \minus{} f(i \minus{} 1)| \equal{} 1$.  This implies that the function $ f$ is linear (do we need to prove why this is true?). Thus, $ f(i) \equal{} i$ or $ f(i) \equal{} n \plus{} 1 \minus{} i$.  [/hide]

metafor, just with $ |f(i)\minus{}f(i\minus{}1)| \equal{} 1$ your function could oscillate. If you extend it to $ f(i)\minus{}f(i\minus{}2) \mid 2$ it works, since either $ |f(i)\minus{}f(i\minus{}2)| \equal{} 1$ or $ |f(i)\minus{}f(i\minus{}2)| \equal{} 2$. However, the former is impossible since $ f(i)\minus{}f(i\minus{}1)\equal{}\pm1$ and $ f(i\minus{}1)\minus{}f(i\minus{}2)\equal{}\pm1$. From here, continue as your solution goes.

[quote="Criticalline1859"]metafor, just with $ |f(i) \minus{} f(i \minus{} 1)| \equal{} 1$ your function could oscillate.[/quote]  Not if it's going to be one-to-one ;)  (Of course, this should be stated in the proof.)


[b]1[/b].Let $ z_1,z_2,z_3 \in C$, $ |z_1|=|z_2|=|z_3|=r>0$, $ z_k \neq z_l$, $ 1\leqq<l\leq 3$.Prove that if two of the numbers: $ z_1+z_2z_3$ ,$ z_2+z_1z_3$, $ z_3+z_1z_2$ are real numbers, then all three numbers are real and $ z_1z_2z_3=1$

[b]2.[/b]Let $ z_1,z_2,z_3 \in C$ such as $ |z_1|=|z_2|=|z_3|$ and $ \frac{z_k}{z_l} \neq \pm 1,\pm i$,for any $ 1\leq k <l\leq 3$.Prove that if two of the numbers $ z_1^2+\frac{z_2z_3}{z_1}$, $ z_2^2+\frac{z_3z_1}{z_2}$, $ z_3^2+\frac{z_1z_2}{z_3}$ are real numbers then all three of them are real and $ z_1z_2z_3=1$

1) wlog let z1 + z2z3 and z2 + z1z3 be real.

Let the arguments of z1,z2,and z3 in polar form be t1, t2, and t3 respectively.

From the first number being real, sint1 + sin (t2 + t3) = 0. sin (t2 + t3) = -sint1 = sin-t1. Also note that r=1 (from ri = r^2i).

t2 + t3 = -t1 + 2api
t1 + t3 = -t2 + 2api (the multiple of 2pi is also a, u see if u do substitution)
t1 + t2 + 2t3 = -(t1 + t2) + 4api
2t3 = -2(t1+t2) + 4pi(a+b)
-t3 + 2api = t1 + t2

We wanted to show that z3 + z1z2 is real, or sin-t3 = sin (t1 + t2 + 2kpi) so that's done.

Using the 3 equations:
t2 + t3 = -t1 + 2api
t1 + t3 = -t2 + 2api
t1 + t2 = -t3 + 2api

(t1 + t2 + t3) = 2api

z1z2z3 = r^3 (cos 2api + isin2api) = r^3. From before, r = 1, so z1z2z3 = 1.

2) If my method for 1) was correct, then it follows quickly from the same type of thing.


AMC
AIME
ratio
geometry
parallelogram
function
analytic geometry

To get to Olympiad, you get through AIME, a key step for many people.

I have decided to start a marathon covering problems at these levels:

1) Hard AMC 12 level (such as 20-25)
2) Easy AIME (1-5)
3) More difficult AIME Problems (6-9 or 10)
4) Very hard AIME Problems (basically the last 5)
5) Beginning Olympiad

Please do not post problems that frequently appear on national or international olympiads, and do not post too easy problems(such as AMC 10 1-5)

If possible please classify which category a problem is when you post it (listed above)

[hide="First Problem: Easy AIME"]Determine $ 3x_4\plus{}2x_5$ if $ x_1, x_2, x_3, x_4$, and $ x_5$ satisfy the system of equations below.
$ 2x_1\plus{}x_2\plus{}x_3\plus{}x_4\plus{}x_5\equal{}6$
$ x_1\plus{}2x_2\plus{}x_3\plus{}x_4\plus{}x_5\equal{}12$
$ x_1\plus{}x_2\plus{}2x_3\plus{}x_4\plus{}x_5\equal{}24$
$ x_1\plus{}x_2\plus{}x_3\plus{}2x_4\plus{}x_5\equal{}48$
$ x_1\plus{}x_2\plus{}x_3\plus{}x_4\plus{}2x_5\equal{}96$[/hide]

How about just use stars to represent the difficulty of the problem? For example, yours would be $ (\star\star).$

[hide="Solution"]
Setting $ S=x_1+x_2+x_3+x_4+x_5,$ we have \begin{align*}x_1+S&=6\\x_2+S&=12\\x_3+S&=24\\x_4+S&=48\\s_5+S&=96..\end{align*} Summing these 5 equations gives us \begin{align*}(x_1+x_2+x_3+x_4+x_5)+5S&=186\\6S&=186\\S&=31..\end{align*} Finally, \begin{align*}3x_4+2x_5&=3(48-31)+2(96-31)\\&=3\cdot17+2\cdot65\\&=51+130\\&=\boxed{181}.\end{align*}[/hide]

[hide="Problem 2"]
$ (\star\star)$

Find $ x^2+y^2$ if $ x$ and $ y$ are positive integers such that \begin{align*}xy+x+y&=71\\x^2y+xy^2&=880..\end{align*}[/hide]

I swear I've seen these problems before.

[hide="Solution #2"]
Note $ (x\plus{}1)(y\plus{}1)\equal{}72\equal{}2^3\cdot3^2$ and $ xy(x\plus{}y)\equal{}880\equal{}11\cdot2^4\cdot5$. It's easy to see WLOG $ x\equal{}11, y\equal{}5$. Thus $ x^2\plus{}y^2\equal{}\boxed{146}$.
[/hide]

Someone put up another problem; I'd like my original problems to go into mock AMC 12s.

[quote="RoFlLoLcOpT"]How about just use stars to represent the difficulty of the problem? For example, yours would be $ (\star\star).$[/quote]

Good idea.:)

Another thing: Please do not look at past tests to see the answers.

[hide="$ \star\star$"]Find the number of pairs $ (a,b)$ so no digits of a or b are zero and $ a\plus{}b\equal{}100$[/hide]

Oh, I don't think you can put the stars in a hide tag... I just put it above the problem. :)

Also, I think you should include the problem number in the hide tag.

[hide="Alternate Solution for two"]Let $ xy\equal{}a$ and $ x\plus{}y\equal{}b$. So we have $ a\plus{}b\equal{}71$ and $ ab\equal{}880$. Solving, we get either $ (a,b)$ as either $ (55,16)$ or $ (16,55)$. Obviously, the latter is impossible with integers, so it is the former. We can then see that $ (x,y)$ is either $ (5,11)$ or $ (11,5)$ which both yield the same answer of $ 146$. [/hide]

@toad, I assume you mean positive integers?

[hide="Solution"]So we have to divide 100 into 2. There are $ \binom{101}{1} \equal{} 101$ ways to do this. Out of these 101 solutions, we can't use the ones where $ a \equal{} 0,10,20,30,40,50,60,70,80,90,100$ That is 11 solutions. $ 101 \minus{} 11 \equal{} \boxed{90}$[/hide]
$ (\star\star\star\star)$ why do my stars look weird?
[hide="Problem 4"]The decimal representation of $ \dfrac{m}{n}$ where $ m$ and $ n$ are relatively prime positive integers and $ m < n,$ contains the digits, 2, 5, and 1 consecutively, and in that order. Find the smallest possible value of $ n$ for which this is possible.[/hide]

Plase correct me if i am wrong

Since the decimal representation is 251 in that order m/n is equal to some non negative integer.251. Let call this integer x. Now if this x is positive then m>n whch it cannot be and it cannot be negative anyways. Therefore x equals 0 and m/n is equal to 0.251 which equals 251/1000. 251 and 1000 are relatively prime so n must equal 1000.

If I am right I dont understand why bvecause shouldnt AIME problems be from 0 to 999

Anyways heres a problem
ILL rate it three stars just because its like a number 10 but
...This should be easy(I can solve it)

In a parlor game, the magician asks one of the participants to think of a three digit number (abc) where a, b, and c represent digits in base 10 in the order indicated. The magician then asks this person to form the numbers (acb), (bca), (bac), (cab), and (cba), to add these five numbers, and to reveal their sum,N . If told the value of , the magician can identify the original number, (abc). Play the role of the magician and determine the (abc) if N=3194

[quote="AIME15USAMO"]
$ (\star\star\star\star)$ why do my stars look weird?
[hide="Problem 4"]The decimal representation of $ \dfrac{m}{n}$ where $ m$ and $ n$ are relatively prime positive integers and $ m < n,$ contains the digits, 2, 5, and 1 consecutively, and in that order. Find the smallest possible value of $ n$ for which this is possible.[/hide][/quote]

i too cannot find a value of n less than 1000 that fits it. maybe it's m > n. or i just am not thinking about it hard enough.

but then again if m>n then cant there be a million different decimal representations

Does it say it has to be exactly $ 0.251?$ Unless I am mistaken, it's possible for the fraction to be $ \frac{m}{n}\equal{}\frac{251}{999},$ although that is very unlikely to be the smallest value of $ n.$

[quote="Ihatepie"]@toad, I assume you mean positive integers?[/quote]
Yes, sorry for not clarifying :blush:

[hide="New Problem"]An integer is called snakelike if its decimal representation a_1a_2a_3\cdots a_k satisfies a_i<a_{i+1} if i is odd and a_i>a_{i+1} if i is even. How many snakelike integers between 1000 and 9999 have four distinct digits?[/hide]

This is a $ \star\star\star$ problem (Medium AIME)

Please do not try to look up the solution!

[quote="Caelestor"]I've seen something like this before, so I'm at an unfair advantage  :ninja: Unfortunately, I shouldn't solve it, since I couldn't solve it by myself (this is an AIME #14 and I looked at the solution).

To answer a few questions, no it just says it has to contains 251 somewhere. (Heck, it doesn't need to have a repeating sequence of 251). Becomes much harder now, no?[/quote] Doesn't it? (because all fractions are repeating decimals.)

@toad we still have two problems going

Edit: I got 909 as a denominator, but I really doubt that is the lowest.

[quote="Ihatepie"]Edit: I got 909 as a denominator, but I really doubt that is the lowest.[/quote]

I agree. I got 800 easily (201/800 = 0.25125).

Whoops, I just realized my $ \binom42*5^2$ should be changed to $ 2(6!/3!2!)$, since first of all, even with my original logic, $ 5^2$ is incorrect when it should have been $ 5\plus{}\binom52$; secondly, this doesn't work when we have something like RIGHT(RIGHT)RIGHT & RIGHT RIGHT(RIGHT), which are the same but are counted as different.
$ \frac{\binom42*4^{2}\plus{}2(6!/(3!2!))\minus{}4\binom42}{4^{6}}\equal{}3/2^6$
By the way for the bus problem, what if we are looking for the exact number instead of just the number of digit?

I think all those problems have been solved...

[hide="New Problem"] $ \star\star\star$
A boy builds a 12-block-high tower, and wants to paint it with red and blue paint. However, he wants the tower to meet his aesthetic requirements, so he decides that he can have no more than three red blocks in a row. He also does not want any blue blocks to touch. Assuming that all the blocks are either red or blue, how many ways can the boy paint the tower?
[/hide]

[hide="messive ugly casework bash"]Case 1 (6 blue): trivially there are $ 7$ ways
Case 2 (5 blue): $ \binom82 \minus{} 4 \equal{} 24$, since after fixing 4 red inside the 4 gaps recreated by the 5 blue, there are 3 left, and there are only 4 cases with 4 or more red together (i.e, inside one of the 4 gaps between the blue)
Case 3 (4 blue): $ \binom95 \minus{} 2\cdot5 \minus{} 3\binom62 \equal{} 71$; almost the same as the previous case using complementary counting; after fixing the first 7, there are 5 red left, with first case of having 4 on either end with another 5 possible positions for the last one, and with the second case of having a 3 in one of he 3 gaps, and $ \binom62$ ways for the other 2
Case 4 (3 blue): $ 2\binom63 \plus{} 2\binom73 \minus{} 2 \minus{} 4 \equal{} 104$
_*_*_*_ where * = blue. again use complementay counting - there are 12-5 = 7 red left, and in order to have 4 or more red together, we can either have 4 or more red on one of the two ends, or a 3 or more in one of the 2 gaps:

1) 4 or more on either end: $ 2\binom63$
2) 3 or more in the gap : $ 2\binom73$

we have to subtract the cse with both 1) and 2) satisfied and the case where there are exactly 3 each for the two gaps: 4+2=6.

So the total number of ways is $ 206$ if I did not make any dumb mistake... is there a better way?[/hide]

On a road, we decide to give a name to each car that goes by, where a name consists of any set of letters. Each letter may only be used at most once. No two different cars may be given the same name. Two names which are merely permutations of each other are not considered different. Finally, each new name must have at least one leter in common with each of the names already given to cars before it. Let N be the number of cars it takes for us to run out of names. Find a way to calculate the exact value of N.

For those that want another one, try problem #8 on this page
[url]http://web.mit.edu/hmmt/www/datafiles/problems/1998/ADVTOP98.pdf[/url]

I used recursion to solve my problem. I find it to be easier and less time consuming. Also, you don't risk  missing or overcounting a case.

[hide="Solution"]
Count the number of nth-block-high towers where the nth block is blue.

$ n \equal{} 1$: Obviously, it can be blue or not, so there is only $ one$ tower where it can be blue.

$ n \equal{} 2$: Now, because no two blue blocks can touch, if the second block is blue, then the first must be red, so there is $ 1$ tower where the second block is blue.

$ n \equal{} 3$: If the third block is blue, then the first block can be red or blue. $ 2$ possibilities

$ n \equal{} 4$: If the fourth block is blue, then the third block is red. Then, the second block is red or blue. If it is red, the first block can be blue or red. If it is blue, the first block must be red. $ 3$ possibilities

Notice that for a given blue block block $ n$, the previous blue block can be $ n \minus{} 2$, $ n \minus{} 3$, or $ n \minus{} 4$ (it can't be $ n \minus{} 1$, that block is red!). So the number of distinct towers that blue block can be in is the sum of the towers that all the potential previous blue blocks can be in.

From this, we get the recursion $ a_n$ = $ a_{n \minus{} 2} \plus{} a_{n \minus{} 3} \plus{} a_{n \minus{} 4}$, where $ a$ is the number of number of distinct towers that meet our requirements.
We have the first four terms, so we can easily compute it up to $ n \equal{} 12$.
$ a_9 \equal{} 19$
$ a_{10} \equal{} 28$
$ a_{11} \equal{} 41$
$ a_{12} \equal{} 60$

Note that the solution is not $ 60$. We must add up the those four terms of our recursion, because any of them could be the last blue block in our stack. We get $ 148$.
[/hide]

EDIT: I just noticed a similar problem. AIME 2001 I number 14 is based on the same concept.
http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=2001

[quote="mathwizarddude"][hide="corrected solution"]Man, I just realized two really dumb mistakes: first in case 2, $ \binom82$ should be $ \binom83$, which changed the total in case 2 to be $ 52$ - I have no idea how it was possible to make such stupid blunder. Also in case 4, I forgot to subtract it from the total $ \binom{10}3$, despite saying COMPLEMENT counting... also in that case, I forgot to include $ (4,3)$ and $ (3,4)$ which contributes another 2 (I only considered $ (3,3)$), so the total for case 4 is
$ \binom{10}3 \minus{} 2\binom63 \minus{} 2\binom73 \plus{} 4 \plus{} 2 \plus{} 2 \equal{} 120 \minus{} 40 \minus{} 70 \plus{} 8 \equal{} 18$

So the grand total is $ 7 \plus{} 52 \plus{} 71 \plus{} 18 \equal{} 78 \plus{} 70 \equal{} 148$... Darn[/hide]
Anyway very nice recursive solution! (I actually have thought of using recursion but didn't find a nice way-should have thought of nth stack bring blue)
Here's a new problem from old HMMT: Find the smallest natural number which can be written as a sum of 3 cubes in two different ways. [url]http://web.mit.edu/hmmt/www/datafiles/problems/1998/ALGEBR98.pdf[/url][/quote]

[hide]Hmm I get 251=5^3+5^3+1^3=6^3+3^3+2^3. Yay for brute forcing[/hide]

[hide="New prob"]Let ABC be a triangle with $ \angle A\equal{}60^o$ and $ \angle B\equal{}80^o$. Let P be the point within $ ABC$ such that $ \angle PAB\equal{}\angle ACB$ and $ \angle PBC\equal{}\angle PCB$. Find the measure of $ \angle BPC$.[/hide]

dnkywin please show your full "brute-force" solution, not just the answer...
[hide="trig bash"]Let $ \angle PCB \equal{} x$. Applying trig Ceva we get

$ \frac {\sin(40 \minus{} x)}{\sin(80 \minus{} x)} \equal{} \frac {\sin20}{\sin40}\implies \cos20\sin(40 \minus{} x) \equal{} \cos60\sin(80 \minus{} x)$

Using $ 2\sin a\cos b \equal{} 0.5(\sin(a \plus{} b) \plus{} \sin(a \minus{} b))$ we get

$ \sin(60 \minus{} x) \plus{} \sin(20 \minus{} x) \equal{} \sin(140 \minus{} x) \plus{} \sin(20 \minus{} x)\implies 60 \minus{} x \equal{} 180 \minus{} (140 \minus{} x)\implies x \equal{} 10$
So $ \angle CPB \equal{} 160$[/hide]
Is there a synthetic solution?
[hide="Next Problem"]
Pyramid EARLY has rectangular base EARL and apex Y, and all of its
edges are of integer length. The four edges from the apex have
lengths 1, 4, 7, 8 (in no particular order), and EY is perpendicular to
YR.
Find the area of rectangle EARL with fully justified answer.
[/hide]

By brute force, he just means he tried different combinations until they added up to each other. There really is nothing to explain.

[hide]
    Let the center of EARL be O. Since EY is perpendicular to YR, Y must lie on the sphere centered at O with radius OA.  OA = OE = OR = OL, so the sphere intersects all vertices of EARL.  E, and A lie on opposite ends of a diameter, so AY is perpendicular to YL.  By the pythagorean theorem the length of a diagonal of EARL is$ (EY)^2\plus{}(YR)^2$.  The length of the other diagonal is the same, so $ (EY)^2\plus{}(YR)^2\equal{}(AY)^2\plus{}(YL)^2$.  We are giver that the edge lengths are 1,4,7, and 8.  To satisfy the equation, either EY and YR are 1 and 8, and AY and YL are 4 and 7, or EY and YR are 4 and 7, and AY and YL are 1 and 8.  Either way, the diagonal of EARL is $ \sqrt{65}$.
     EA and RL have to be {1,8} or {6,4} to satisfy the pythagorean theorem.   Note the lateral edge of length of length 1 is next to the lateral edge of length 6.  The base edge connecting the lateral edges cannot be 1 or 8 by the triangle inequality.  Therefore the base edges must have length 6, 4 in some order.  Either way the area of EARL is [b]24[/b].
[/hide]

Own problem:
$ a$, $ b$, and $ c$ are positive integers less than $ 200$, such that $ ab\equal{}\frac{(c^4\minus{}(a\plus{}b))^2}{4}$.  Find the greatest possible value of $ a\plus{}b\plus{}c$.

[hide="Solution"]First, simplify the expression, taking positive roots since we are given that $ a$, $ b$ and $ c$ are positive:
\begin{align*}  ab &=\frac{(c^{4}-(a+b))^{2}}{4} \\
\sqrt{4ab}&=c^4 - (a+b) \\
2 \sqrt{a} \sqrt{b} +a+b&=c^4 \\
\left(\sqrt{a}+\sqrt{b} \right)^2&=c^4 \\
\sqrt{a}+\sqrt{b} &= c^2 \end{align*}
The only possible values of $ a$ and $ b$ are perfect squares less than $ 200$, which are 196, 169, 144, 121, 100, 81, 64, 49, 36, 25, 16, 9, 4, or 1. These have square roots of 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, and 1, respectively. The largest possible sum of these is $ 28$, and the largest perfect square less than $ 28$ is $ 25$. Thus, $ a$ and $ b$ are 196 and 121 in any order or 169 and 144 in any order. The pair with the larger sum is $ 196$ and $ 121$, which have a sum of $ 317$ compared to $ 314$. We also found that the maximum value of $ c^2$ is $ 25$, so $ c=5$. Thus, the sum is $ 196+121+5=\boxed{322}$.[/hide]

[hide="New Problem"]Let the set $ S=\{8,5,1,13,34,3,21,2\}$. Susan makes a list as follows: for each two-element subset of $ S$, she writes on her list the greater of the set's two elements. Find the sum of the numbers on the list.[/hide]

[hide="Solution"]First, we write the numbers in increasing order:  1  2  3  5  8  13  21  34 . Now we know that

1 appears  0 times
2     "       1   "
3     "       2   "
5     "       3   "
8     "       4   "
13   "       5   "
21   "       6   "
34    "      7   "

So the sum of the numbers is

$ 1(0) \plus{} 2(1) \plus{} 3(2) \plus{} 5(3) \plus{} 8(4) \plus{} 13(5) \plus{} 21(6) \plus{} 34(7) \equal{} \boxed{484}$[/hide]

[hide="New Problem"]Given that $ \displaystyle\sum_{k\equal{}1}^{35} \sin5k \equal{} \tan \frac{m}{n}$ where angles are measured in degrees, and $ m$ and $ n$ are relatively prime positive integers that satisfy $ \frac{m}{n} < 90$ find $ m\plus{}n$.[/hide]

[hide="Solution"]For each element, the number of times it appears is simply the number of elements smaller than it, and so the answer is simply $ 34*7 \plus{} 6*21 \plus{} 5*13 \plus{} 4*8 \plus{} 3*5 \plus{} 2*3 \plus{} 1*2 \equal{} 484$[/hide]
[hide="Solution"]$ \Im\sum_{k \equal{} 1}^{35}\exp(5ik) \equal{} \frac {\sin5}{1 \minus{} \cos5} \equal{} \tan(5/2)$ after some simplifications. Please try not to repost AIME problems as I believe all of them have already gotten well-written solutions in the resource section or in AoPS Wiki... so try to post problems from other contests that are not in the resource sections already (like ARML or Mandelbrot, etc)  :) [/hide]
Let $ \Delta ABC$ be a triangle and let $ O$ be its incircle such that it is tangent
to $ BC$ at $ D$ and $ CA$ at $ E$. The bisectors of angle $ A$ and angle $ B$ intersect $ DE$
at $ F$ and $ G$ respectively. $ BF \equal{} 1$ and $ FG \equal{} GA \equal{} 6$. What's the radius of incircle $ O$?

It is just a bunch of trig sums that cancel out when you multiply the whole thing by sin 5. Then you get 1-cos175/sin175 which equals tan 175/2 which means the answer is 177.

Could you please post a new question? thanks!

hmmm

New Problem:

[hide]The centroids of the faces of a regualr tetrahedrom are connected to form another tetrahedron. What is the ratio of the area of the new tetrahedron to the old one?[/hide]

Can't you see my new problem at post #48?  :o It's the one just above varunrocks' post!
[hide="solution to ur problem anyway"]Consider the distance from the vertex to the centroid. For the big one, it's $ 3/4$. For the small one, it's $ 1/4$. So the surface area ratio is just $ 1/9$ (To prove 1/4 and 3/4, it's just a 3-D analog of 2-D version of median)[/hide]


Does anyone have any graph theory problems?


[hide] The chance of one employee being absent is x/100. Because  40.96/100 is the chance for all 4 employees to be absent, that means that is x^4 / 100. Additionally, that means x^4 = 4096, which boils down to x = 8. I kinda messed up on my method here, but I reasoned out that it would be an 80% chance from what I've done above. I'm sure someone could post a clearer solution.  :blush:   [/hide]

Divide 40.96/4 and you get 10.24% and that is the answer of any one of the employees of being late.

[quote="Treething"]Slick Sam has four employees at his business - Charlize, Shaniqua, Destiny, and Candy. On any given day, there is a 40.96% chance that every single one of his employees will be absent. Given that each employee has an equal chance of being absent, what is the probability that Shaniqua is absent?[/quote]
[hide]If x is the chance that one is absent, then $x^{4}$ is the chance that they are all abesnt. The chance that one is absent is $80 \%$. [/hide]
It would be a little trickier if you said that the chance that someone shows up is 59.04%. The second I saw 40.96 alarm bells rang in my head.
EDIT: At myspacesucks, by your reasoning if there was a 40% chance that any one employee was absent then there would be a 160% chance that they are all absent ;).

My bad, wasn't thinking and I was hungry, I'll vouch that its 80%.

[hide]The fourth root of .4096 is .80. So, there is an 80 percent chance that one will be absent, so the probability (and the answer) is 80%[/hide]


If Bill can mow 4 lawns in 1 hour and judy can mow 3 lawns in 45 minutes, how long would it take them to mow one lawn togehter?

they each work at the rate of 1 lawn per 15 minutes, so together they do 1 lawn in 1/8 of an hour... however many minutes that is...

1/4 of an hour is 15 mins
and half of 15=7.5

in this problem it is simple because they work at the same rate individually, what happens when they work at different rates? lets do it!

If Bill can mow 4 lawns in 1 hour and Judy can mow 3 lawns in 1 hour, how long would it take them to mow one lawn together? You can give answer in either amount of hours or minutes and you can leave the answer as a fraction if it is a fraction  :D

Enjoy!

1/7th of an hour... just add their rates and then get reciprocal

yea, thats what i wanted to stress out so that people understand better than just simple guessing

They both do 1 lawn in 15 minutes, so both together would be half of 15 minutes, which is 7.5 minutes...


Given a $ 1 \times 25$ rectangle divided into $ 25$ "boxes" ($ 1 \times 1$), is it possible to write integers $ 1$ to $ 25$ so that the sum of any two adjacent "boxes" is a perfect square?

Wow! Did you compute that or look it up in the internet?

I would have named it a "Magic" 'Row'."  "Rectangle" implies there is something
two-dimensional about and required of it.

Otherwise, the solution given is surprising to me.

How many total solutions are there, counting all reflections/rotations?

The steps for getting that solution are:

The possible squares from a sum of two numbers from 1 to 25 are: 4,9,16,25,36,49
You must start with 18, the number with only one possible neighbour.
And notice that some numbers have only two neighbours: 16 (9,20), 17 (8,19), 19 (6,17), 20 (5,16), 21 (4,15), 22 (3,14), 23 (2,13)
Then some combinations are must be in this order or the reverse one (9,16,20,5), (8,17,19,6)

[quote="Arrange your tan"]I would have named it a "Magic" 'Row'."  "Rectangle" implies there is something
two-dimensional about and required of it.[/quote]

Well, you have to fill in the "boxes" of the rectangle, so a certain sum is achieved (similar to the [url=http://en.wikipedia.org/wiki/Magic_square]magic squares[/url] concept).

No, you don't have to fill in the "boxes" of the/a rectangle, because a row with
its entries STILL satisfies a certain sum being achieved, hence the
"Magic" Row.

It does, but in this specific problem, they ask you to write the numbers down in a rectangle. >.>

Here is another solution: (which I worked out using the steps mc said - with the neighbors of numbers)

18
7
9
16
20
5
11
25
24
12
4
21
15
10
6
19
17
8
1
3
22
14
2
23
13

This happens to be the first solution with two sections reversed....pretty cool,...i was working to get the first solution but got this instead....


There will be eight characters in this game:

Sitka
Denahi
Kenai
Tanana
Rutt
Tuke
Koda
Tug

Out of them only seven are open for you (needless to say which seven). Once signups are done, we shall create fan fiction of sorts. And if you are not funny or seem to have no idea of the actual story whatsoever, you maybe replaced.

If you can't watch the film somewhere, try and read about it in the web!

[img]http://adorocinema.cidadeinternet.com.br/filmes/irmao-urso/irmao-urso-poster03.jpg[/img]

^jaw falls onto the ground^

aww... y glass aw ut ell onto da gwownd....

I mean come on!!! do you have to make 4 topics and counting on one thing only???!!!

But ... Colossal Knickerbockers! This is just a game!! And I have made only like two topics!

@Ignite: Koda is not open for registration  :mad:

i quit and i declare brother bear the most stupid movie ever.

and bubka plz watch better movies from know on ;)

Yeah whats up with this obsession with Brother Bear?


Let $ u_1 \equal{} 1,\; u_2 \equal{} 3,\; u_r \equal{} 4u_{r \minus{} 1} \minus{} u_{r \minus{} 2}\; (r\geq 3).$
Find $ \sum_{r \equal{} 1}^{\infty}{arccot \left(2u_{r}^2\right)}.$

hello, i have found $ u(n)\equal{}\frac{1}{6}\left((3\minus{}\sqrt{3})(2\plus{}\sqrt{3})^n\plus{}(3\plus{}\sqrt{3})(2\minus{}\sqrt{3})^n\right)$ but i can't evaluate the sum  $ \sum_{r \equal{} 1}^{\infty}{arccot \left(2u_{r}^2\right)}$.
Sonnhard.

$ \text{arccot} \ x\plus{}\text{arccot} \ y \equal{} \text{arccot} \ \frac{xy\minus{}1}{x\plus{}y}$.

Therefore,
\[ \sum_{r\equal{}1}^n \text{arccot} \ (2u_r^2) \equal{} \text{arccot} \ \left(\frac{\frac{\left(2\minus{}\sqrt{3}\right)^n\plus{}\left(2\plus{}\sqrt{3}\right)^n}{2}}{\frac{\minus{}\left(2\minus{}\sqrt{3}\right)^n\plus{}\left(2\plus{}\sqrt{3}\right)^n}{2\sqrt{3}}}\right)\]

Thus,
\[ \sum_{r\equal{}1}^\infty \text{arccot} \ (2u_r^2) \equal{} \text{arccot} \ \sqrt{3} \equal{} \frac{\pi}{6}\]

Because, $ \lim_{n \to \infty} \frac{\frac{\left(2\minus{}\sqrt{3}\right)^n\plus{}\left(2\plus{}\sqrt{3}\right)^n}{2}}{\frac{\minus{}\left(2\minus{}\sqrt{3}\right)^n\plus{}\left(2\plus{}\sqrt{3}\right)^n}{2\sqrt{3}}} \equal{} \sqrt{3}$


function
geometry
calculus
inequalities
3D geometry

I did a practice 20004 State Algebra II contest and when grading it, the solutions to each of the following 3 problems required graphing on a calculator (in the official solutions, that is).  I thought that solutions like that were only for people who don't really know how to do the problem or find it faster to use a calculator.

Anyway, here are the problems (I have not found a good "real" solution to them):

31.  The price of a computer over a 25 month period is described by $f(x)=1050-\sqrt[3]{9x^{5}+5x+503}$.  If the model continues to be accurate beyond the 25 month period, approximately when will the price reach half of its initial price?

32.  A luxury car's value is represented by the equation $y_{1}=50(0.822)^{x}$.  A sports utility vehicle's (SUV) value is represented by the equation $y_{2}=30(0.884)^{x}$.  In both functions, $y$ is the value (in $\$1000$'s) of the automobile after $x$ years.  Assuming one of each model is purchased on the same day, how long will it be before the SUV is worth $\$1000$ more than the luxury car?

35.  The Container Company is designing an open top rectangular box, with a square base, that will hold $108\text{ cm}^{3}$.  Estimate the minimum surface area for the box.

I am 99% sure that the first two are impossible without graphing on a calculator. As for the last one, you can't do it without calculus or some crazy inequalities, but you know a little, right? (not that anyone else in that contest would)

$x^{2}y=108$, so $y=\frac{108}{x^{2}}$. Now we want to minimize $x^{2}+4xy=x^{2}+\frac{432}x$. Differentiating, we want $2x-\frac{432}{x^{2}}=0$, or $x^{3}=216$, hence the minimum occurs when $x=6$ and $x^{2}+4xy=108$.

The fact that x is twice y leads me to believe that there may be some reason to see that... I dunno how.

Given that $x^{2}y=108$, we wish to minimize $x^{2}+4xy$

Now, $x^{2}+4xy=x^{2}+2xy+2xy$

This is at least thrice the cube root of $4x^{4}y^{2}=4 \cdot 108^{2}$

Also, if you're going to use calculus, you might as well use Lagrange multipliers.  They tend to be the quickest ways of killing these kinds of things, especially when the givens grow more complicated.

1) I started to use Lagrange Multipliers, but quickly realized that the basic calc 1 minimization was actually just as quick if not quicker.

2) One or two people in the algebra II contest likely know very basic calculus, but I bet nobody knows about Lagrange Multipliers.

Nice application of AM-GM. I must admit I never even think about not using calculus in such situations.

Bleh, approximation is no better than using a calculator. That's applied math, at best :wink:.

[quote="Teki-Teki"]umm...the first one asks you to approximate a solution to $9x^{5}+5x+503 = 505^{3}$, essentially.  since we want $x$ say to the nearest integer we have $x \approx \sqrt[5]{\frac{505^{3}}{9}}$ which you might not need a calculator for.  i mean, it's like the fifth root of 14*10^6 which is right around 26-28 without even doing a little binomial expansion to check the first few terms.  this took me like 5 seconds.  i would use this on the test.  (at least up until the fifth root of 140, which is 2.6-2.8 i think).

as it turns out, checking real quick on a scientific calculator $2.7^{5}= 143.489...$ so $x \approx 27$.  that's how i would solve it.  if they asked you for like 27.2, 27.4, 27.6 then you might want to graph it.  but other than that, no.  you don't really need a calculator.[/quote]

Well, two answers were 27 months, 19 days and 27 months, 22 days  :roll:

@jb05:  I do know the very basics of calculus, or at least enough to understand your solution.  And I do [b]kind of[/b] know what LaGrange Multipliers are, from WOOT.

On the computer problem (31), the solution says that the initial value is $\$1050$, which occurs at $<0$ days  :rotfl: .  I said it was $\$1050-\sqrt[3]{503}$, which did not give any of the answers, but was closest to an incorrect choice.


Find a closed form for the series

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

I think that series converge if and only if $|x|<1$ :)

Whoops, there was a mistake in the equation.  :oops_sign:

Correct version:

$\sum_{n = 1}^{\infty}{\frac{x^{n}}{x^{2n}-y}}$


How many distinct (noncongruent) right triangles are there with all integer sides and one leg of length 42?

The factors of 42 are:
1,2,3,6,7,14,28

So then, if there is a pythagorean triple with one of these factors as a leg, that counts as 1 distinct right triangle that includes 42, cause the triples can be proportional.

The ones I thought of:
3,4,5 - 42, 4x14, 5x14
7, 24, 25 - 42, 24x6, 25x6
21, 220, 221 - 42, 220x2, 221x2

So, I thought of 3, which seems to small to be the right answer, but oh well.

If you find some triangles that work, how can you be sure you haven't skipped any?

That's exactly the problem for me. I'm never sure of my answers. That's probably why I stink in probability stuff

I don't know how to do it either. Can one of y'all solve it?

Billly

[hide="my answer"] I got 4. The one I found in addition to Chinaboy'S numbers is 40, 42, 58.

I am pretty sure the solution is 4. Since 42 is an even mumber =2*3*7 and these factors are different, it makes the problem easier to solve. There are 5 pairs of even factors for 42^2. But 42*42 needs to be excluded. So there is a total of 5-1= 4 such
pythagorean triples.[/hide]


Given a sequence of 1999 reals, show that there exist some consecutive terms of the sequence, the sum of which differs from an integer by at most 1/1000.

Just consider the fractional parts of the numbers x_1,x_1+x_2,...,x_1+...+x_1999 and apply the pigeon-hole principle.


In the figure, P is external point to a circle with centre O such that OP is equal to the diameter of the circle. PA and PB are tangents to the circle. Show that tri. PAB is equilateral triangle.

[hide]Let $r$ be the length of the radius. This makes $OP=2r$ and $AO=BO=r$. Because of tangents, $\measuredangle{PBO}=\measuredangle{PAO}=90^\circ$. This makes 2 $30-60-90$ right triangles, so $\measuredangle{APB}=60^\circ$. Also, $\overline{AB}\perp\overline{OP}$, so $\measuredangle{PAB}=\measuredangle{PBA}=60^\circ$. Thus, $\triangle{PAB}$ is an equilateral triangle.[/hide]


decimal representation
number theory
algorithm
combinatorics
IMO Shortlist

Each positive integer $a$ undergoes the following procedure in order to obtain the number $d = d\left(a\right)$:

(i) move the last digit of $a$ to the first position to obtain the numb er $b$;
(ii) square $b$ to obtain the number $c$;
(iii) move the first digit of $c$ to the end to obtain the number $d$.

(All the numbers in the problem are considered to be represented in base $10$.)  For example, for $a=2003$, we get $b=3200$, $c=10240000$, and $d = 02400001 = 2400001 = d(2003)$.)

Find all numbers $a$ for which $d\left( a\right) =a^2$.

[i]Proposed by Zoran Sunic, USA[/i]

i think this is from ISL
the proof isnt elegant at all. if i remember correctly, it just involves considering lots and lots of cases, relating to the last digit of a. We get contradictions for most cases   :D  but i think we get 3 infinite solutions at the end.

I remember one infinite solution and some (three?) finite ones, but I am not sure. I don't know the solution any more (I have completely messed up the problem during the exam, and later, when Gronau showed us the solution, I have been sleeping).

  Darij

In what follows, we assume $a$ has $n+1$ digits, meaning $10^n \le a < 10^{n+1}$. In particular, $10^{2n} \le d < 10^{2n+2}$.

First, we analyze the last digit of $a$, say $x$. Suppose \begin{align*} 	a &= \overline{\dots x} \\ 	b &= \overline{x \dots} \\ 	c &= b^2 = \overline{y \dots} \\ 	d &= \overline{\dots y} \end{align*} so that $y = x^2 \pmod{10}$ (here $\dots$ denotes some irrelevant digits).

[b]Claim[/b]:  	$x \neq 0$.

[i]Proof.[/i]  	Otherwise, $b < 10^n$, $c < 10^{2n}$, so $d < 10^{2n}$. $\blacksquare$

Next, we make a large table by noting that if we know the leading digit of $b$, we can find all possible leading digits of $c = b^2$, hence all possible final digits of $d$. \[ 	\begin{array}{r|ccccc|cccc} 		x & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline 		x^2 \pmod{10} & 1 & 4 & 9 & 6 & 5 & 6 & 9 & 4 & 1 \\ 		\text{possible first digits of } \overline{x\dots}^2 & 123 & 45678 & 91 & 		12 & 23 & 34 & 456 & 678 & 89  	\end{array} \] From this, we conclude that $x \in \{1,2,3\}$, and we must have $y = x^2$. In particular, $10^{2n} \le c < 10^{2n+1}$, hence $10^{2n} \le d < 10^{2n+1}$.

Now we proceed to do the calculation: set \begin{align*} 	a &= 10m + x \\ 	b &= 10^n x + m \\ 	c &= 10^{2n} x^2 + 2 \cdot 10^n \cdot mx + m^2 \\ 	d &= 10\left( 2 \cdot 10^n \cdot mx + m^2 \right) + x^2 \end{align*} Since $a^2 = d$, we have \begin{align*} 	100 m^2 + 20 mx + x^2 &= 10\left( 2 \cdot 10^n \cdot mx + m^2  \right) + x^2  \\ 	10m + 2x &= 2 \cdot 10^n \cdot x + m \\ 	m &= 2x \cdot \frac{10^n-1}{9} = 2x \cdot \underbrace{1\dots1}_n \end{align*} In other words, $a$ should be one of the following shapes: $a = 2\dots21$, $b = 4\dots42$, $c = 6\dots63$. A quick check now shows the answers are: \[ a = \underbrace{2\dots2}_{n \ge 0}1, 	\qquad a = 2, \qquad a = 3. \]

[hide=Solution] This was pretty cool problem, [quote=ISL 2003 N2]Each positive integer $a$ undergoes the following procedure in order to obtain the number $d = d\left(a\right)$:

(i) move the last digit of $a$ to the first position to obtain the numb er $b$;
(ii) square $b$ to obtain the number $c$;
(iii) move the first digit of $c$ to the end to obtain the number $d$.

(All the numbers in the problem are considered to be represented in base $10$.)  For example, for $a=2003$, we get $b=3200$, $c=10240000$, and $d = 02400001 = 2400001 = d(2003)$.)

Find all numbers $a$ for which $d\left( a\right) =a^2$.
[/quote]
[b][color=#000]Solution:[/color][/b] Suppose the last digit of $a$ is $4$, then $b$ starts with $4$ $\implies$ $c$ should start with $1$ or $2$ contradicting the fact that $d$ ends with $6$. Similar can be repeated for other numbers and in the end, the possible options for last digit are $1,2,3$ only. Suppose, $a$ has $n$ digits. Hence, \begin{align*} a =\sum_{k=1}^n a_k 10^{k-1}=a_n10^{n-1}+a_{n-1}10^{n-2}+\ldots +a_1 \end{align*}
\begin{align*}  b= 10^{n-1}a_1 + \frac{1}{10} \left(\sum_{k=2}^n a_k 10^{k-1}\right) \end{align*}
\begin{align*}  c= 10^{2n-2}a_1^2 + \frac{1}{100} \left( \sum_{k=2}^n a_k 10^{k-1} \right)^2+2a_1 10^{n-2} \sum_{k=2}^n a_k 10^{k-1}\end{align*}
Further, it is important to note that: $a_1^2=1,4,9$, Hence, there are no carry-overs for the $(2n-1)$th place and hence by double counting: \begin{align*} d= \left( \sum_{k=1}^n a_k 10^{k-1} \right)^2=a_1^2+\frac{1}{10} \left( \sum_{k=2}^n a_k 10^{k-1} \right)^2 +2a_110^{n-1} \sum_{k=2}^n a_k 10^{k-1} \end{align*}
\begin{align*}\implies \frac{9}{10} \sum_{k=2}^n a_k 10^{k-1} = 2a_1(10^{n-1}-1) \end{align*}
\begin{align*} \left( \sum_{k=1}^n a_k10^{k-1} \right) -a_1 = a-a_1 = 20a_1 \frac{10^{n-1}-1}{9} \end{align*}
Plugging, $a_1=1,2,3$ and after further verifications: we have, \begin{align*}\boxed{1,2,3, 222 \ldots 21 } \qquad \blacksquare \end{align*}
[/hide]

R.

Let $a$ be a number that works. Let $a=10a_0+a_1$ where $a_0$ is a nonnegative integer and $b$ is a nonnegative integer from $0$ to $9$ inclusive. Then $b$ will have first digit $a_1.$

$d$ must have last digit equivalent to $a_1^2$ mod $10.$ So its first digit is equivalent to $a_1^2$ mod $10,$ all assuming $a_0\ge 1.$ We'll deal with $a_0=0$ later.

Now, we can compare the possible values of $a_1.$ For $a_1=4,5,6,\dots,9$ the possible values of the first digit of $c$ does not compute based on $b$ does not compute with the possible values of the first digit of $c$ based on $d.$ Thus, $a_1=1,2,3.$

Suppose $a$ has $n$ digits. Then $10^{n-1}\le a\le 10^{n}-1.$ We have $b=10^n\cdot a_1+a_0.$ Thus, \[c=10^{2n}\cdot a_1^2+a_0^2+2\cdot 10^n\cdot a_0a_1\]
We have $d=100a_0^2+20a_0a_1+a_1^2=10(10a_0^2+2a_0a_1)+a_1^2.$ Thus, \[c=10^{2n}a_1^2+10a_0^2+2a_0a_1\]
Thus, we have $9a_0=2(10^n-1)a_1$ so $a_0=2(111111...111)a_1$ with $n$ ones. Thus, $a$ is in the form $22222...21,$ $4444....42$ or $6666...63.$ One easily checks that the working ones are $2222...221,2,3$.

Now for $a_0=0:$ This is easy to prove impossible by the number of digits.


Let's say you have a very large odd number. Let's also assume you know that it is composite, but the factors are not divisible by $3$. Although, you don't know the factors. You know that there is an integer $x$ and an integer $y$, such that $(x+y)(x-y)=q$, where $q$ is the number you're trying to factor. So now you know that $x^{2}-y^{2}=q$, but you're trying to find an easy algorithm to find $x$ and $y$ that would make that algorithm work for all composite odd numbers without having to do any guess-and-check. You know that square numbers are the sums of consecutive odd numbers, so for an integer $k$, $q=2ky+k^{2}$. My question is, am I on the right track to finding an easy algorithm? Or am I simply wasting my time? If I am wasting my time, does anyone else have an easy algorithm to factor large composite odd numbers?

Every odd number is the difference of two squares (set $x = y+1$), so that doesn't help.

Integer factorization is very hard, and nobody knows an "easy" way to do it.

http://en.wikipedia.org/wiki/Integer_factorization
http://en.wikipedia.org/wiki/Integer_factorization#Factoring_algorithms

[quote="bigboy82892"]does anyone else have an easy algorithm to factor large composite odd numbers?[/quote]

If anyone does, they aren't telling.  Among other things, such an algorithm would allow you to break many of the public-key encryption schemes in wide use.

Unrelatedly, this seems to have been my 6000th (official) post.  Whee!


function
trigonometry
calculus
calculus computations

Let $ a,\ b,\ c$ be constants. Find the range of the function $ a\cos ^ 2 x \plus{} b\sin x\cos x \plus{} c\sin ^ 2 x$.

Use the identities $ \cos^2 x \equal{} \frac {1 \plus{} \cos 2x}{2}, \sin x \cos x \equal{} \frac {\sin 2x}{2}, \sin^2 x \equal{} \frac {1 \minus{} \cos 2x}{2}$; then we need to find the range of

$ \frac {(a \minus{} c) \plus{} (a \plus{} c) \cos 2x \plus{} b \sin 2x}{2}$

and it's well-known that the range of $ A \cos \theta \plus{} B \sin \theta$ is $ [ \minus{} \sqrt {A^2 \plus{} B^2}, \sqrt {A^2 \plus{} B^2}]$, for example by interpreting it as a dot product.


[quote="zhaobin"]just for $S_2 > S_3$,are you sure,have you noticed mongolia ineq should have some other condition?$x_1 \geq x_2 \cdots x_n$[/quote]

oops i missed the condition! Thanks.


The decisions for Brown, Harvard, Princeton and Yale are finally in!  Who got in where?

[quote="kokosai"]Brown, Harvard, Princeton and Yale[/quote]
as well as the other four members of the Ivy League (Dartmouth, Columbia, Penn and Cornell).

Princeton: Rejected.

Oh well...at least that makes my decision somewhat easier now :D

Princeton: Waitlisted. Not going to bother staying on it, though.

I didn't make it into Harvard :(.  Oh well, maybe there's always a possibility for a transfer?

[quote="ComplexZeta"]You can be upset about not getting into Harvard/Princeton/wherever, or you can work hard at any college you attend and show Harvard/Princeton/wherever that they were wrong not to take you. The latter sounds like the better option to me. You don't need them.[/quote]

Why are they mutually exclusive?  I think that disappointment is a natural response to being waitlisted.  Being upset and remembering how that feels is a good motivator for the future.

or apply again for grad school.  More important to attend a good school for your master degree.

Although this is not common i personally know an alum of harvard that initially got a rejection letter and sent them back a letter expressing how harvard was the only school for him and how he fit and what not. 48 hours later he got in....mind you that it was graduate school however.

I wish it were that easy for undergraduate admissions :).


geometry
geometric transformation
reflection
symmetry
circumcircle

Let $ ABC$ be a triangle with orthocenter $ H$. Consider the reflection of $ H$ with respect to each side of the triangles (labeled $ H_1, H_2, H_3$ below).

[asy]size(250);
pair A=(4,6),B=(12,0),C=(0,0),H=(3,3),I=(0,4),J=(3,-2),K=(7,7);
draw(A--B--C--cycle,linewidth(1));
draw(H--I,dashed);
draw(H--J,dashed);
draw(H--K,dashed);
label("$A$",A,NE);
label("$B$",B,SE);
label("$C$",C,SW);
label("$H$",H,SE);
label("$H_2$",I,SW);
label("$H_1$",J,SE);
label("$H_3$",K,SE);[/asy]

Okay then. I'll consider it. :P
Anyway... the rest of the problem is:

Show that these three reflected points all lie on the circumscribed circle of $ ABC$.

[hide]The the altitudes be AD, BE, CF. ADB = AEB, so ABDE is cyclic. Similiarly, BCEF and CAFD are cyclic. Also, AEHF, BFHD, and CDHE are cyclic. angle AHC = angle EHA + angle EHC = angle EFA + angle EDC = (180-EFB) + (180-EDC) = angle B + angle C, and the result follows.[/hide]

[hide="A little quicker"]It is sufficient to prove that $ ABCH_{1}$ is cyclic.  Using only the symmetry from our reflection and complementary angles from right triangles, we have

$ \angle CBH_{1} \equal{} \angle CBH \equal{} 90 \minus{} \angle ACB \equal{} \angle CAH_{1}$

so we're done.[/hide]


If $ 4$ points are randomly chosen from an $ a$ x $ b$ lattice grid, what is the probability that they do not form a convex quadrilateral?

I'm not exactly sure how to go about the problem, but Pick's theorem would help:

$ A \equal{} I \plus{} P/2 \minus{}1$ where $ A$ is the area of a lattice polygon, $ I$ is the number of interior points, and $ P$ is the number of lattice points on the boundary of the polygon.

After choosing three points, we can use this formula to find how many points are in the interior of the triangle formed by those three points.  Then the number of points we can chose for the 4th point that would not produce a convex quadrilateral is $ P\plus{}I$ (if points can be chosen twice) or $ P\plus{}I\minus{}3$ if points can only be chosen once.


limit
integration
calculus
function
real analysis
real analysis unsolved

Let $f: [0,1]\rightarrow \mathbf{R}$ and $g: [0,1]\rightarrow \mathbf{R}_+^*$ be continuous functions. Find:
\[ \lim\limits _{n\rightarrow \infty} \frac{\int_0^1f(x^n)dx}{\int_0^1g(x^n)dx}. \]

do you know dominated convergence for integral ? if so, it is straightforward. (and you just have to impose $g(0) \neq 0$ I think)

yeah allright and the limit is $\frac{f(0)}{g(0)}$

[quote="xirti"]I heard about this theorem. But I do not know it. Can you state this theorem and give a proof for my problem?.. Please..[/quote]
http://en.wikipedia.org/wiki/Dominated_convergence_theorem

The dominated convergence theorem is convenient but not required. We can prove this with tools at the level of the Riemann integral.

[b]Lemma 1:[/b] If $f$ is continuous on $[0,1]$ and$f(0)=0,$ then $\lim_{n\to\infty}\int_0^1f(x^n)\,dx=0.$

Proof: Given $\epsilon>0,\,\exists\,\delta>0$ such that $|f(x)|<\frac{\epsilon}2$ for $0<x<\delta$. Also, $\exists\,M$ such that $|f(x)|\le M$ for all $x.$

$\int_0^1f(x^n)\,dx=\int_0^{\delta^{1/n}}f(x^n)\,dx+\int_{\delta^{1/n}}^1f(x^n)\,dx$

$\left|\int_0^{\delta^{1/n}}f(x^n)\,dx\right|\le \int_0^{\delta^{1/n}}\frac{\epsilon}2 \,dx\le\frac{\epsilon}2\cdot\delta^{1/n}<\frac{\epsilon}2$

$\left|\int_{\delta^{1/n}}^1f(x^n)\,dx\right|\le M\left(1-\delta^{1/n}\right).$

Since $\delta^{1/n}\to 1$ as $n\to \infty,$ we can choose an $N$ large enough so that for all $n>N,\, M\left(1-\delta^{1/n}\right)<\frac{\epsilon}2.$

Adding that up, we get that for large enough $n,$ our integral is less than $\epsilon$ in absolute value.

[b]Lemma 2:[/b] If $f$ is continuous on $[0,1],$ then $\lim_{n\to\infty}\int_0^1f(x^n)\,dx=f(0).$

Proof: let $f(x)=f(0)+(f(x)-f(0))$ and apply Lemma 1 to $(f(x)-f(0)).$

We should be able to finish the problem from here.

and if we suppose that $f(0)=g(0)=0$, $f(x) \geq 0$, $g(x) \geq 0$. What conditions do we have to impose for the limit exists ?

Thank you Kent for proof. And about the theorem.. Does it hold for Riemann integrable functions?

For the original problem.. Can we do a proof lie this:
 Choose $g=M$ the bound of $f$. And $f(x^n)\to f(0)$ for $x\neq 1$ and is dominated by $g$, hence the limit is $\int_0^1 f(0)dx$?

That's the Dominated Convergence Theorem; my proof was an attempt to use more elementary tools than that.

My proof would work if $f$ were Riemann integrable (hence automatically bounded) and continuous at zero.

I tried to aply this theorem which is new to me.. I know that your solution is stronger. Thanks.


$ BD$ and $ CE$ are angle bisectors of the $ \triangle ABC$. If $ \angle BDE \equal{} 24^\circ$ and $ \angle CED \equal{} 18^\circ$, what are the degree measures of angles $ A$, $ B$, and $ C$.

See [url]http://www.ajorza.org/wimo/images/mathfiles/files/imoshortlist/scansod/imo1992s.pdf[/url].

One of the problems is this one.

hmm I actually solved this one so:

[hint=Hint]Which angles are external bisectors of which other angles? Bunch of angle chasing so have fun :) [/hide]

Anyone? :D

I'll wait a couple days before posting my solution


inequalities
function
algebra
domain
real analysis
real analysis unsolved

If $f$ is defined on a bounded open interval ($a, b$) and uniformly continuous, show that the limit of $f$ exists at endpoints.

$f$ is uniformly continuous, so $\forall\epsilon > 0\exists u_{\epsilon}>0$ such that $\forall x \in ]a+u,b[\ ,\ |f(x)-f(x-u_{\epsilon})|<\epsilon$
Make $x\rightarrow b$ and let $S_{b},I_{b}$ the superior,inferior  limit points (finite or infinite) of $f$ in $b-$.
Manifestly $f(x-u_{\epsilon})$ converges to $f(b-u_{\epsilon})$ because $f$ is continuous in $b-u_{\epsilon}$
It's easy to see that : $-\epsilon \leq S_{b}-f(b-u_{\epsilon}) \leq \epsilon$ ; $-\epsilon \leq f(b-u_{\epsilon})-I_{b}\leq \epsilon$
Manifestly $S_{b},I_{b}$ have to be finite . I add the inequalities and I obtain :
$\forall \epsilon >0,-2\epsilon \leq S_{b}-I_{b}\leq 2\epsilon$ , so $S_{b}=I_{b}$, and the limit of $f$ exists in $b-$.
In the same way you show that  the limit of $f$ exists in $a+$.
 :cool:

@countdownmath: I deleted your identical post in Computations and Tutorials. Please, post each problem only once. :)

The way I like to phrase it:

Continuous functions map convergent sequences to convergent sequences. But by "convergent sequence" I mean a sequence whose limit lies in the domain of the function. Hence, if $f$ is continuous on $(a,b)$ and $x_{n}\to a,$ then I don't call that a convergent sequence and we're not entitled to make any conclusion about the sequence $f(x_{n}).$

But [i]uniformly[/i] continuous functions map Cauchy sequences to Cauchy sequences. That way, if $f$ is uniformly continuous on $(a,b)$ and $x_{n}\to a,$ that that's a Cauchy sequence, so $f(x_{n})$ is a Cauchy sequence, so it has a limit, which we'll take that to be limit of $f(x)$ as $x\to a.$ (We still have a little more work to do, in showing that this is well-defined. That's not too hard.)

With this framework, we now know how to extend any uniformly continuous function to a function continuous on the closure (or completion) of the original domain. This generalization extends far beyond $\mathbb{R}.$


limit
factorial
function
induction
linear algebra
matrix
real analysis

[quote="Yuriy Solovyov"]Let we have sequences $ \{p_n\},\; \{q_n\}$ such that:

[b](1)[/b] $ p_{ \minus{} 1} \equal{} 0, \; q_{ \minus{} 1} \equal{} 1,\; p_{0} \equal{} q_{0} \equal{} 1$
[b](2)[/b] $ p_{n} \equal{} 2p_{n \minus{} 1} \plus{} (2n \minus{} 1)^{2}p_{n \minus{} 2}\; \forall n\geq 1$
[b](3)[/b] $ q_{n} \equal{} 2q_{n \minus{} 1} \plus{} (2n \minus{} 1)^{2}q_{n \minus{} 2}\; \forall n\geq 1$

Find the limit: $ \lim_{n\to\infty}{\frac {p_n}{q_n}}$[/quote]

we have, from $ (2)$,
$ p_{n\plus{}2}\minus{}2p_{n\plus{}1}\minus{}(2n\plus{}3)^2p_n\equal{}0\qquad (\; \forall n\ge \minus{}1 \; )$

auxiliary equation   is $ p^2 \minus{} 2 p \minus{} (2n\plus{}3)^2 \equal{} 0$ which when solved gives
$ p\equal{}1\pm \sqrt {\; 1 \plus{} (2n\plus{}3)^2 \; }$

$ \Rightarrow \quad p_n\equal{}A\;\left(1\plus{}\sqrt {\; 1 \plus{} (2n\plus{}3)^2 \; }\right)^n \plus{} B\;\left(1\minus{}\sqrt {\; 1 \plus{} (2n\plus{}3)^2 \; }\right)^n$
where $ A,B$ are constants to be determined from initial conditions.

since $ p_{\minus{}1}\equal{}0,\quad and\quad p_0\equal{}1$, we get, after putting  $ n\equal{}\minus{}1 \quad and \quad 0$, respectively,

$ p_n\equal{}\left(\dfrac{\sqrt 2\plus{}1}{2\sqrt 2}\right)\left(1\plus{}\sqrt { 1 \plus{} (2n\plus{}3)^2 }\right)^n \plus{}\left(\dfrac{\sqrt 2\minus{}1}{2\sqrt 2}\right)\left(1\minus{}\sqrt {1 \plus{} (2n\plus{}3)^2 }\right)^n$

in a similar fashion, from $ (3)$, only with [b]different[/b] initial conditions, we get

$ q_n\equal{}\left(\dfrac{\sqrt 2\plus{}1}{2}\right)\left(1\plus{}\sqrt { 1 \plus{} (2n\plus{}3)^2 }\right)^n \minus{}\left(\dfrac{\sqrt 2\minus{}1}{2}\right)\left(1\minus{}\sqrt {1 \plus{} (2n\plus{}3)^2 }\right)^n$

hence, $ \lim_{n\to\infty}{\frac {p_n}{q_n}}\equal{}\dfrac{\dfrac{\sqrt 2\plus{}1}{2\sqrt 2}}{\dfrac{\sqrt 2\plus{}1}{2}}\equal{}\boxed{\dfrac{1}{\sqrt 2}}$

[quote="misan"]

we have, from $ (2)$,
$ p_{n \plus{} 2} \minus{} 2p_{n \plus{} 1} \minus{} (2n \plus{} 3)^2p_n \equal{} 0\qquad (\; \forall n\ge \minus{} 1 \; )$

auxiliary equation   is $ p^2 \minus{} 2 p \minus{} (2n \plus{} 3)^2 \equal{} 0$ [/quote]

That's wrong. :wink:

Yes kunny, misan you can write characteristic equation of the form $ a_n \equal{} c_{1}a_{n \minus{} 1} \plus{} c_{2}a_{n \minus{} 2} \plus{} ... \plus{} c_{d}a_{n \minus{} d} \plus{} c$ only if all $ c_{i}$ are constants. It is the same with the characteristic equation of the differential equation.

Example: let we have sequence $ a_{n\plus{}1} \equal{} (n\plus{}1)a_{n}$ and $ a_1 \equal{} 1$. Using your method we get: $ a_n \equal{} \frac{1}{2}\cdot (n\plus{}1)^n$, but real solution is $ a_n \equal{} n!$.

[quote="Yuriy Solovyov"]Yes kunny, misan you can write characteristic equation of the form $ a_n \equal{} c_{1}a_{n \minus{} 1} \plus{} c_{2}a_{n \minus{} 2} \plus{} ... \plus{} c_{d}a_{n \minus{} d} \plus{} c$ only if all $ c_{i}$ are constants. It is the same with the characteristic equation of the differential equation. [/quote]

you're right, for a general $ n$, the formula that i have derived will not work. but that was OK with me at the time since what i had intended to find was merely an expression that tells me how $ p_n$ and $ q_n$ behave as $ n$ grows very large.

so, is the answer that i provided wrong ?  :maybe:

[quote="Yuriy Solovyov"]
Example: let we have sequence $ a_{n \plus{} 1} \equal{} (n \plus{} 1)a_{n}$ and $ a_1 \equal{} 1$. Using your method we get: $ a_n \equal{} \frac {1}{2}\cdot (n \plus{} 1)^n$, but real solution is $ a_n \equal{} n!$.[/quote]

i do not disagree with you. the method i used will not give you the right solution for any $ n$. but i have a feeling  it will give you a feel for the behavior as $ n$ grows very large.

in your "factorial-based" recurrence relation, the right formula is, as you have pointed out, is $ n!$.
now using stirling, which states $ n! \approx \sqrt {2\pi n}\, \left(\frac {n}{e}\right)^{n}$, we get
$ n! \approx \left(\dfrac{n}{e}\right)^n$ when $ n$ is very large.

  using the formula i employed, we get $ a_n \equal{} \frac {1}{2}\, (n \plus{} 1)^n$, which for large $ n$, can be approximated as $ n^n$.... telling me that for large $ n$, the method might yield a favorable approximation.

regardless of that, if you say my answer to your original problem is wrong, then i will have to go and try a different method, maybe use something like a generating function.....  :huh:

misan, answer of the problen is not $ \frac {1}{\sqrt {2}}$ and $ n!\ne \textbf O (n^n)$ because $ \lim_{n\to\infty}{\frac {n!}{n^n}} \equal{} 0$ :) My solution is not based on the generating function, but I think it should work here. I will give a prompt: try to use induction to find general formula of the $ q_{n}$. It depends on the first member of sequence. Then try to represent $ p_{n}$ throw $ q_{n}$. :) I'll post a solution in a few days if nobody solve it.

By induction for $ n\geq 0$:
$ q_{n}\equal{}\prod_{k\equal{}0}^{n}{(2k\plus{}1)}$
$ p_{n}\equal{}p_{n\minus{}1}(2n\plus{}1)\plus{}(\minus{}1)^{n}\prod_{k\equal{}0}^{n\minus{}1}{(2k\plus{}1)}$
Then:
$ \frac{p_n}{q_n}\equal{}\frac{p_{n\minus{}1}(2n\plus{}1)}{q_n}\plus{}\frac{(\minus{}1)^{n}\prod_{k\equal{}0}^{n\minus{}1}{(2k\plus{}1)}}{q_n}\equal{}\frac{p_{n\minus{}1}}{q_{n\minus{}1}}\plus{}\frac{(\minus{}1)^n}{2n\plus{}1}$
$ \frac{p_n}{q_n}\equal{}1\minus{}\frac{1}{3}\plus{}...\plus{}\frac{(\minus{}1)^n}{2n\plus{}1}$
$ \lim_{n\to\infty}{\frac{p_n}{q_n}}\equal{}\arctan(1)\equal{}\frac{\pi}{4}$

I have just discovered that by elliminating $ (2n \minus{} 1)^2,\ p_n,\ q_n$ satisfy the reccurence : $ r_n \minus{} 2r_{n \minus{} 1} \minus{} kr_{n \minus{} 2} \equal{} 0\ (n\geq 1)$ for some constant $ k\in\mathbb{C}$.
Solving this in using Matrix, we have that:
For $ n\geq \minus{} 1$,

$ p_n \equal{} n \plus{} 1,\ q_n \equal{} 1$ for $ k \equal{} \minus{} 1$

$ p_n \equal{} 2c_n \plus{} kc_{n \minus{} 1},\ q_n \equal{} (2 \plus{} k)c_n \plus{} kc_{n \minus{} 1}$ for $ k\neq \minus{} 1$, where $ c_n \equal{} \frac {\alpha ^n \minus{} \beta ^n}{\alpha \minus{} \beta}$ such that $ \alpha ,\ \beta \in {\lambda ^2 \minus{} 2\lambda} \minus{} k \equal{} 0$.

That seems at odds with what Yuriy posted.


all you guys taking the AIME I. Hope you do (moderately) well and reach your goals (set a low floor). Yeah, so good luck with the test.

Good luck to everyone!
I hope that all my AoPS friends will qualify!
AIME II for me......

Oh boy...ya know how your stomach sinks when the chain is pulling you to the top of the first hill of a roller coaster?  Well...the tension is not quite that intense :P But it's close :D

Good luck!!!

good luck everyone... :)

remmember to have phun :P

Yeah, good luck to everyone! Set a low floor so I can have a 1/100 chance of qualifying  :D

Good luck to you all!  :)

I could use some luck myself, lol. ;)

Good luck. Have fun. Don't Die. Oh wait, this is the AIME, not Starcraft. Be Careful Anyway.

Everyone good luck, and have fun

Hopefully we'll all make USAMO  :P

$\textsc{Good luck on your mission!}$
$\textsc{Acing the American Idiotic Martians Examination will make you ready to conquer Mars.}$

Bad luck to you all so I can qualify!!!!  :D

[hide]just kidding[/hide]

[quote="chess64"]$\textsc{Acing the American Idiotic Martians Examination will make you ready to conquer Mars.}$[/quote]I haven't heard of that one before.  :rotfl:  :rotfl:

Yeah good luck to all you people taking the AIME 1s. Don't fail. No pressure.

[quote="Max300"]This is a message to anybody who scored a 150 on the AMC 12. (or 140 something for that matter)
Score just high enough on the AIME to make USAMO to lower the floor.
You will be my bestest friend forever if you do  :D[/quote]

you see, the problem is what is "just high enough"?

[quote="JesusFreak197"]Try a $5$. :D[/quote]

That's cool. A six is fine too Amir.

Good luck to all. ANd don't forget me. Yes....I need more luck then all of you.....


The number 12148 has a unique feature: The sum of the first four digits equals the units digit. How many EVEN five-digit numbers have this unique property?

[hide]
Let the number be $abcde$.  From the conditions in the problem, we have $a+b+c+d=e$.

Since the number is even, $e$ can be 2, 4, 5, or 8.

For 2, $a+b+c+d=2$.  We want the number of solutions in non-negative integers, so the number of solutions is $C(5,3)=10$.  But since $a>0$, we have to subtract the number of solutions in which $a=0$.  That is, we need to subtract tne number of solutions of $b+c+d=2$, which is $C(4,2)=6$.  So the number of solutions for $e=2$ is $4$.

Similarly, the number of solutions for $e=4$ is $C(7,3)-C(6,2)=20$.  The number of solutions of $e=6$ would be $C(9,3)-C(8,2)=56$ and the number of solutions of $e=8$ would be $C(11,3)-C(10,2)=120$.

Adding those together, you get $4+20+56+120=200$.
[/hide]


function
trigonometry
real analysis
real analysis unsolved

Let $ f(x) \equal{} \sum 10^{\minus{}n} \cos (10^{2n}x)$

a) Prove that $ f$ satisfies the Holder 1/2 continuous. That is, there exists a constant $ C$ such that
$ |f(x\plus{}h)\minus{}f(x)| \le C\cdot h^{1/2}$

b) Prove that $ f$ is nowhere differentiable

for b)  cf     http://en.wikipedia.org/wiki/Nowhere_differentiable
with $ a\equal{}1/10,b\equal{}100,x\equal{}\pi{y}$.


Gauss
geometry
trapezoid
geometry proposed

[color=darkred]Let $ABCD$ be a convex quadrilateral for which denote the intersections $I\in AC\cap BD$, $E\in AB\cap CD$, $F\in AD\cap BC$ and the middlepoints $M,N,P,R$ of the sides $[AB],[BC],[CD],[DA]$ respectively. Define the line $d_{1}\parallel MP$ for which $E\in d_{1}$ and the line $d_{2}\parallel NR$ for which $F\in d_{2}$. Prove that the lines $d_{1}$, $d_{2}$ and the diagonal $BD$ are concurrently if and only if $IA=IC$.[/color]

i'm very interested in seeing a projective solution , as you can see, $IA=IC \Longleftrightarrow AC \| EF$, so $BD$ is the gauss line of $ABCDEF$, and so on.

[b][size=109]1).[/size][/b] – We suppose that $IA = IC,$ where $I\equiv AC\cap BD$ and we will prove that the concurrency point of the segment lines

$MN,$ $PR,$ lies on $BD.$

We consider the vertex $D$ of $ABCD,$ as an arbitrary point in the plane of the triangle $\bigtriangleup ABC$ and so, by Ceva’s theorem,

we have that $\frac{EA}{EB}\cdot \frac{FB}{FC}\cdot \frac{IC}{IA}= 1$ $\Longrightarrow$ $\frac{EA}{EB}= \frac{FC}{FB}$ $\Longrightarrow$ $EF\parallel AC$ $,(1)$

It is easy now,  to show that the triangles $\bigtriangleup BMN,$ $\bigtriangleup DPR$ are perspective, because of the points $E\equiv BM\cap DP,$

$F\equiv BN\cap DR$ and $\infty\equiv MN\cap PR$ $($ we say that two parallel lines, intersect each other at infinity $),$ are collinear.

So, by Desarques’s theorem, we conclude that the segment lines $BD,$ $MN,$ $PR,$ are concurrent at one point, so be it $S.$

[b][size=109]2).[/size][/b] – We suppose that the point $S\equiv MN\cap PR,$ lies on $BD$ and we will prove that $IA = IC,$ where $I\equiv AC\cap BD.$

Because of $S\equiv BD\cap MN\cap PR,$  we have that the triangles $\bigtriangleup BMN,$ $\bigtriangleup DPR,$ are perspective and so, by Desarques’s

theorem, we conclude that the points $E\equiv BM\cap DP,$ $F\equiv BN\cap DR$ and $Q\equiv MN\cap PR,$ are collinear.

Because of now, $MN\parallel AC\parallel PR,$ we have that $Q\equiv \infty$ $($ the point at infinity $),$ and hence, we conclude that

$EF\parallel MN\parallel PR.$

From trapezium $AEFC,$ it is easy to show that $IA = IC$ and the proof is completed.

Kostas Vittas.


geometry
geometric transformation
dilation
ratio

a dilation is a mapping of one figure to a similar figure. Scale factor is how much bigger or smaller it got. So if a triangle was mapped to a triangle with side length three times as big (meaning its area is $3^{2}=9$ times as big) then the scale factor would be 3.

For your specific problem, just find the side lengths of both figures (make sure they're corresponding sides!) and find the ratio.

What do you mean by

"For your specific problem, just find the side lengths of both figures (make sure they're corresponding sides!) and find the ratio"?

find side the length of side AB and then find the length of side A'B'. Then find the ratio $\frac{A'B'}{AB}$. That's your scale factor. You could also use side BC and B'C', or CA and C'A'.

I guess my question should now be:

How do I find the sides as you suggested?

Look at points A and B

A(0,0)
B(0,4)

It shouldn't be difficult to see that it is of length 4.

A'(0,0)
B'(0,10)

This is clearly of side length 10.

If the coorinates did not neatly have the same x-value, then we could have just used the distance formula, for find the distance between points $(x_{1},y_{1})$ and $(x_{2},y_{2})$

$\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

So since the points are (0,0) and (0,4)

$\sqrt{(0-0)^{2}+(4-0)^{2}}=\sqrt{4^{2}}=4$

[quote="Interval"]Under a dilation, triangle A(0,0), B(0,4), C(6,0) becomes triangle A'(0,0), B'(0,10), C(15,0). What is the scale factor for this dilation?

What is a dilation?

What is a scale factor?[/quote]
[hide]The scale factor would be found by taking the distance of a dialated side divided by the distance of the corresponding original side.
In this problem it would be:
$\frac{\left(\overline{A'B'}\right)}{\left(\overline{AB}\right)}$ by the distance formula, you can find that $\overline{A'B'}=10$ and $\overline{AB}=4$ so the scale factor is $\frac{10}{4}=\frac{5}{2}=2.5$.[/hide]
tjhance has answered your other questions...
[hide="another form of the distance formula"]
$\sqrt{(\Delta x)^{2}+(\Delta y)^{2}}$ where $\Delta x$ is the change in x.[/hide]


It is the season again!! I hope that everyone is prepared for the rollercoaster that is math!!

the first one except that i wont crush my opponents... :maybe:

I was going to third one... but then I decided not to vote...  :(

and don't worry pinky, i'm sure you WILL crush your opponents :)

Chose the 1st, but not crush opponents (physically or mentally).


Find $3x^2 y^2$ if $x$ and $y$ are integers such that $y^2 + 3x^2 y^2 = 30x^2 + 517$.

Uh...
this was really lucky...

[hide]
$y^2 + 3x^2 y^2 = 30x^2 + 517 \rightarrow y^2(1+3x^2)=10(3x^2+1) + 507 \rightarrow$
$y^2= 10+\frac{507}{1+3x^2}$

Then...I tested $x=2\rightarrow y^2=10 + \frac{507}{13} \rightarrow y^2 =10+39 \rightarrow y^2=49$

*EDIT*
Oh...just factor $507$ and $3\cdot 13^2$. For $(x,y)$ to be integers, $1+3x^2$ has to equal $3,13,39,169,$ or $507$. Uh, testing each one out, we see that $1+3x^2=13$ is the only solution avaliable.
*END-EDIT*

So...

$3x^2y^2= (3)(49)(4) = \boxed{588}$
[/hide]

[hide]

it is same as $(3x^2+1)(y^2-10)=3\cdot 13^2$

from here , we can easily check that $3x^2+1=13$ and $y^2-10=39$ satisfy the condition .So $3x^2y^2=588$
[/hide]

Editted some silly mistake :P

[quote="chess64"]He just forgot to multiply by 3.[/quote]

ohhhh

[hide]Given equations involving $xy$, $x$, and $y$, it's often a good idea to factor by grouping. Notice that $3x^2y^2-30x^2-10=(y^2-10)(3x^2+1)=517-10=3\cdot13^2$. From here, $3x^2+1=13$ and $y^2-10=39$ and our answer is $3x^2y^2=588$.
[/hide]


induction
inequalities
number theory
number theory solved

This is a problem from my number theory course in college, but you may be interested to solve it.
Prove, for all natural number n:
  (p1)(p2)(p3)...(pk)<4^n, while p1, p2, ..., pk are all primes smaller than or equal to n. While it's obviously true for large n due to lim (n->infinity) (log n)/prime(n)=1, it's not so for small n...

Oh this is quite well - known ...
You need to prove this lemma when proving Bertrand's postulate.

Is it that well known? I don't know how to prove it; could you post a soln?

I found it in an old notebook of mine, but I didn't write any soln there, just this:"use induction and Bertrand", then sth abt 2nCn/4^n->1 as n->infinity.

OK Here comes to proof. (Fundamentals in Number Theory, W.J. Leveque)

We use induction on n. The inequality is obvious for n = 1 or n = 2.

Suppose it is true for 1, 2, ..., n - 1, where n \geq 3.

Then we can restrict attention to odd n, since otherwise

\pip\leqnp = \pip\leqn-1p < 4n-1 < 4n.

So we can put n = 2m + 1. From its definition, the binomial coefficient

C(2m + 1, m) = (2m + 1)!/(m!(m + 1)!)

is divisible by every prime p with m + 2 \leq p \leq 2m + 1. Hence

\pip\leqnp \leq C(2m + 1, m) \pip\leqm+1p < C(2m + 1, m) 4m+1.

But the numbers

(2m + 1, m) and (2m + 1, m + 1)

are equal and both occur in the expansion of (1 + 1)2m+1 so that

C(2m + 1, m) \leq 1/2 22m+1 = 4m

and so

\pip\leqnp < 4m 4m+1 = 42m+1.

The result follows by induction on n. QED.


Solve in $\mathbb{R}$:
\[ x=\left(2007+\sqrt{x}\right) \left(1-\sqrt{1-\sqrt{x}}\right)^2.  \]

Let $u=\sqrt x$
$u^2=(2007+u)(1-\sqrt{1-u})^2$

Let $v=\sqrt{1-u}$
Then $u=1-v^2$.

$(1-v^2)^2=(2007+1-v^2)(1-v)^2$
$\implies (1-v)^2=0$--(1) or $(1+v)^2=2008-v^2$--(2).
Then $v=1$. The roots of (2) are rejected as they do not provide acceptable roots.
Finally, $v=1 \implies u = 0 \implies x = 0$.

Answer: The only solution is $x=0$.


I've always wondered about the following while LaTeXing: should I put $ symbols around pure numbers?  (For instance, if I wrote Find the length of the longest median of a triangle with sides of length 4, 7, and 9, should I put $4$ instead of 4, $7$ instead of 7, etc.?)
I know TeXnicCenter doesn't complain when one fails to put $ symbols around letters (I suppose it treats it like a normal block of text), and there is no noticeable effect the way that there is when you put $ symbols around letters.  But is there some stylistic rule that I am violating when I don't put $s around my numbers?

I don't put $...$ around numbers either. I don't think it makes any difference.

Right, dollar signs around numbers are not needed for ordinary purposes.  (I put them in to be foolishly consistent.)  For typesetting his book Concrete Mathematics, Don Knuth needed to put dollar signs around his numbers to achieve a particular effect.  I forget the exact reason.

[quote="ComplexZeta"]He has a neat font that was designed to give a handwritten effect. I wonder if I can get that font somehow.[/quote]
I believe it's in one of the AMS packages.

is anyone figures it out... could you post it here? thanks.

I believe the math font they used is called AMS Euler.  The text font is Concrete Roman.

In the following link, Knuth describes how he typeset Concrete Mathematics, including why he needed to put math numerals in dollar signs:
[url]http://www.tug.org/TUGboat/Articles/tb10-1/tb26knut.pdf[/url].

[quote="mathkid"]i thought $ .. $ around letters italicized the letters[/quote]
Mmmmhmmm, we're talking about numbers.

$...$ puts whatever within into math mode. Letters in math mode are automatically italicised, because they are assumed to be variables.

oops.. i misunderstood what mathfanatic was saying. nevermind.

[quote="mathfanatic"]I know TeXnicCenter doesn't complain when one fails to put $ symbols around letters (I suppose it treats it like a normal block of text), and there is no noticeable effect the way that there is when you put $ symbols around letters.[/quote]

Sorry for reviving an old thread but could someone explain to me how you change the font?  I wanted to see what it is like but I can't figure out what to do.

Could someone please explain how to implement AMS Euler in LaTeX. It sounds really cool!


hw question that i'm not sure where to go after i just subtitute formulas for nCr

Prove n[size=150]C[/size](r-1)    +   nCr = (n+1)[size=150]C[/size]r

combinatorial proof[hide]The question is in how many ways one can choose $r$ eggs from $n+1$ eggs  :D .
We can choose one egg say Number 1. Than we have to choose $r-1$ from the remaining $n$ so this makes $\binom{n}{r-1}$. Or we can exclude egg N1, and than we have to choose $r$ eggs from the remaining $n$ or $\binom{n}{r}$.[/hide]


modular arithmetic
number theory proposed
number theory

[quote="moldovan"]Show that $ 2005^{2005}$ is a sum of two perfect squares, but not a sum of two perfect cubes.[/quote]

$ 2005\equal{}41^2\plus{}18^2 \Rightarrow 2005^{2005}\equal{}(41 \times 2005^{1002})^2\plus{}(18 \times 2005^{1002})^2$

$ x^3 \equiv \minus{}1,0,1 \pmod{7}$ for all integer $ x$
$ \Rightarrow x^3\plus{}y^3 \equiv \minus{}2,\minus{}1,\minus{},1,2 \pmod{7}$ for all integers $ x$ and $ y$

But $ 2005 \equiv 1 \pmod{6} \Rightarrow 2005^{2005} \equiv 2005^{1} \equiv 3 \pmod{7}$

So $ 2005^{2005}$ cannot be sum of two perfect cubes, done.