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(This will work in exactly the same way as the Daily Logic Puzzle posted by r15. I just decided to copy him on a whim and make it with word puzzles instead, of which I have much experience in making.)

Every day this summer (From today up until September 7), I will be posting a puzzle having to do with words. Some of them may overlap with the theme of the logic puzzles that r15 makes, but I will do my best to try and not go into his domain too much, and to make sure that all my puzzles have strictly to do with words.

Each puzzle is worth a certain number of points, and some puzzles may have partial credit awarded for partial or partially correct answers. Unlike r15, I do not offer hints, nor do I offer solutions by PM to anybody, but I will post an answer along with a (perhaps) complete solution the day of the next puzzle.

Some of the puzzles will be effortlessly easy, and others will be diabolically difficult. The number of points a certain question is worth is directly correlated with the level of difficulty of a problem. So the easy ones may only be worth 1 or 2 points, but the really hard ones might be worth 7 or 8 or even more.

Send solutions by private message, preferably titled with "Daily Word Puzzle" followed by the puzzle number. Do not post solutions on this thread, otherwise you will lose all your points plus 10, and will only be able to gain half the number of points for correctly answered questions (rounding [i]down[/i], I'd suppose) until your score becomes zero or positive again.

If you are a fan of logic puzzles, you might like a few of the ones I offer here, but if word puzzles are your thing, then you'll (hopefully) also love this thread.

So, without further ado, here is the first puzzle:

[size=200]Puzzle #1[/size] (June 16, 2009)
LOD: Medium

Five of these words have something in common. Find out what it is, and determine whether [b]Wookiee[/b] would fit into the group.

[b]donkey cat goose elephant giraffe zebra plesiosaur goblin human grasshopper lion seal[/b]

This question is worth [b]4[/b] points.

*edit* Apparently, I brought out a lot of confusion by posting only 10 objects. I've changed this number to 12.

[quote="PowerOfPi"]Do we have to PM [i]solutions[/i] or can we just PM you our final [i]answers[/i]? I like the idea. Thanks![/quote]

No, just answers. I'll post solutions along with the answers, you guys just need to have the answers.

Eh, yesterday's puzzle turned out pretty badly. Everybody was confused as to whether there were 10 words or not. Anyway, here's the next puzzle:

[size=200]Puzzle #2[/size] (June 17, 2009)

The following sentence contains anagrammed words. The blanks are where the words go, and you need to fill in the sentence with the anagrammed words. (An anagram is rearranging the letters in a word, such as "rats" and "star".) Each underscore is one letter.

Martha was _ _ _ _ _ _ _ for her work as a nurse in the childbirth section of the hospital, but she was in _ _ _ _ _ _ _ because she had to change so many _ _ _ _ _ _ _.

This question is worth [b]5[/b] points.

[hide="Solution to Puzzle #1"]donkey, elephant, zebra, plesiosaur, and human all have an odd number of consonants. Wookiee only has two, so it does not fit.

[hide="Scorers"]None, because the question was badly phrased. Sorry! I promise this time there won't be any errors.[/hide]
[/hide]

From this puzzle on, I will post solutions to puzzles two puzzles back instead of one. This gives everybody at least one full day to submit answers.

[size=200]Puzzle #3[/size] (June 18, 2009)

This riddle gives clues as to the name of a famous mathematician.

My first is in Denmark, my second's in Spain;
My third is in Amsterdam, Burma, and Maine.
My fourth is in China, my fifth's in Japan;
My sixth's not in Kansas, but it is in Sudan.
My seventh's in Jordan but not Victoria,
Nor's it in Thailand or Greece or Pretoria.
My eighth is in Africa, Asia, and Laos,
And my ninth is in Norway but not in your house.

This puzzle is worth [b]3[/b] points.

[size=200]Puzzle #4[/size] (June 19, 2009)

The Daily Whodunit prints out a newsletter every day of the week except Sunday, but mysteriously this Friday, their punctuation marks have all disappeared! (Except for periods, evidently.) Help the hapless printer as he tries to manually add in the punctuation and use a photocopier instead.

I said you said said but you said that I said that which that other that didn't go along with very well.

This puzzle is worth [b]5[/b] points.

[hide="Solution to Puzzle #2"]Martha was [b]praised[/b] for her work as a nurse in the childbirth section of the hospital, but she was in [b]despair[/b] because she had to change so many [b]diapers[/b].

[hide="Scoreboard"]$ \begin{tabular}{|rl|r|} \hline & Name & Points\\
\hline 1. & cf249 & 5 \\
\hline 1. & K Sengupta & 5 \\
\hline 1. & r15s11z55y89w21 & 5 \\
\hline 1. & rd4593 & 5 \\
\hline 1. & vahalla & 5 \\
\hline 1. & yaofan & 5 \\
\hline \end{tabular}$
[/hide][/hide]

[size=200]Puzzle #5[/size] (June 20, 2009)

In a certain rule pattern, THREE comes before FOUR, but SEVEN comes before SIX. If NINE also comes before both EIGHT and SEVEN, and TEN comes before TWO, figure out whether FIVE comes before or after ELEVEN, and what the rule is.

This puzzle is worth [b]2[/b] points.

[hide="Solution to Puzzle #3"]The mathematician is [b]Ramanujan[/b].

[hide="Scoreboard"]$ \begin{tabular}{|rl|r|} \hline & Name & Points \\
\hline 1. & cf249 & 8 \\
\hline 1. & K Sengupta & 8 \\
\hline 1. & r15s11z55y89w21 & 8 \\
\hline 1. & rd4593 & 8 \\
\hline 1. & vahalla & 8 \\
\hline 2. & yaofan & 5 \\
\hline 3. & cyberspace & 3 \\
\hline \end{tabular}$
[/hide][/hide]

[size=200]Puzzle #6[/size] (June 21, 2009)

Complete the following sequences:

(2) 1. an, eb, ar, pr, ay, un, ul, ug, ep, ct, ov, ?
(2) 2. en, en, en, en, en, en, en, ty, ?
(1) 3. nd, nd, es, dn, ur, id, ?

This puzzle is worth [b]5[/b] points. Each sequence is worth the number of points posted to its left.

[hide="Solution to Puzzle #4"]I said, "You said 'said'", but you said that I said "that", which that other "that" didn't go along with very well.

Most of the answers got the first part wrong, and the one that did get it right got other stuff wrong.

[hide="Scoreboard"]$ \begin{tabular}{|rl|r|} \hline & Name & Points \\
\hline 1. & cf249 & 12 \\
\hline 1. & r15s11z55y89w21 & 12 \\
\hline 2. & K Sengupta & 11 \\
\hline 3. & rd4593 & 8 \\
\hline 3. & vahalla & 8 \\
\hline 4. & yaofan & 5 \\
\hline 5. & cyberspace & 3 \\
\hline \end{tabular}$
[/hide][/hide]

[quote="rd5493"]Have you forgotten about this thread?[/quote]

As a matter of fact, I have. I'm sorry, I've been too lazy (and perhaps busy) to keep this up.

[size=200]Puzzle #7[/size] (June 24, 2009)

Follow the instructions, in order, to get a certain phrase:

EACH TWO MILLION ACTRESS BETWEEN YOU TRACE RABBIT REACT GRAHAM ATRIUM LOST DORMOUSE THE CRATE SATURDAY AMONGST SIX AMERICA OF INSIDE GAME SHREW ABOUT UNIVERSAL THIRTEEN CARET

1. Remove all words representing species of rodents.
2. Remove all anagrams of the word CATER.
3. Remove all words starting with a vowel. (Y is not considered a vowel for this rule.)
4. Remove all numbers.
5. Remove all proper nouns.
6. Remove all prepositions.

This puzzle is worth [b]2[/b] points. A bonus of 2 applies to whoever's moral sanity isn't shattered after reading the final message.

[hide="Solution to Puzzle #5"]

FIVE comes before ELEVEN.

The order is determined by alphabetical order when the words are reversed.

[hide="Scoreboard"]$ \begin{tabular}{|rl|r|}\hline & Name & Points \\
\hline 1. & K Sengupta & 13 \\
\hline 2. & cf249 & 12 \\
\hline 2. & r15s11z55y89w21 & 12 \\
\hline 3. & rd4593 & 10 \\
\hline 3. & vahalla & 10 \\
\hline 4. & yaofan & 5 \\
\hline 5. & cyberspace & 5 \\
\hline\end{tabular}$[/hide][/hide]

*edit* Okay, due to confusion about what a rodent is, rule 1 has been changed into "Remove all animal names."

[size=200]Puzzle #8[/size] (June 25, 2009)

Find the names of seven countries hidden in the following paragraph:

The Grammar Corporation was at it again. They acted like they were still in diapers! They were trying to tell me how to spell stuff. For example, they told me to spell "yen" with a capital Y! I mean, it's bad enough that they managed to anger many a client, but this was newfangled of them. "Give us a call back," they said right before hanging up. "We'll see if we can adapt your profile to a pay-per-usage model." Well, that certainly gave me swollen glands. I'm never using their service again.

This puzzle is worth [b]4[/b] points.

[hide="Solution to Puzzle #6"]

1. ec. The rule is the second and third letters of each month in the year.
2. ne. The rule is the last two letters of the integers from 13 onwards.
3. tu. The rule is the third and fourth letters of each day of the week, starting from Sunday.

[hide="Scoreboard"]$ \begin{tabular}{|rl|r|}\hline & Name & Points \\
\hline 1. & K Sengupta & 18 \\
\hline 2. & r15s11z55y89w21 & 16 \\
\hline 3. & rd4593 & 14 \\
\hline 4. & cf249 & 12 \\
\hline 5. & vahalla & 10 \\
\hline 5. & cyberspace & 10 \\
\hline 6. & yaofan & 5 \\
\hline\end{tabular}$[/hide][/hide]

[size=200]Puzzle #9[/size] (June 26, 2009)

Let's play MAD GAB™, the game where you have to say it right!

Pronounce the following phrases. They are phonetically close to some certain well-known phrases. (For example: Mass Achoo Sits = Massachusetts.)

(2) 1. Are Dove Prop Plum Saul Bing
(4) 2. Nought Mike Uppity
(3) 3. Rid Them An Bull Use
(2) 4. A Nope An Buck
(4) 5. Yule Awe Stuck Aim...Toot I'm Snow.

This puzzle is worth [b]15[/b] points. Each phrase is worth the number of points to its left.

[hide="Solution to Puzzle #7"]

[hide="Scoreboard"]$ \begin{tabular}{|rl|r|}\hline & Name & Points \\
\hline 1. & K Sengupta & 20 \\
\hline 2. & r15s11z55y89w21 & 18 \\
\hline 3. & rd4593 & 16 \\
\hline 4. & cf249 & 12 \\
\hline 4. & vahalla & 12 \\
\hline 4. & cyberspace & 12 \\
\hline 5. & yaofan & 7 \\
\hline 6. & PowerOfPi & 1 \\
\hline\end{tabular}$
The bonus of 2 was a joke, sorry to anyone who took it seriously.[/hide][/hide]

[quote="louiethechewy123"]Do you need to solve all of them to get points? Or can you solve 1 or 2 and still get points?[/quote]

Yes, you can still solve 1 or 2. The number of points you'll get is stated to the left of each phrase.

[size=200]Puzzle #10[/size] (June 27, 2009)

The following well-known phrases have been "disemvoweled" (had their vowels removed) and "spaced out" (had their [b]space[/b]s taken [b]out[/b]). Put the vowels and spaces back and complete the poor phrases. (Note, the letter Y has been taken out where it is used as a vowel.)

(1) 1. ntrntnlMthlmpd
(1) 2. Ntbthhrnmchnnchnchn!
(1) 3. Sprclfrglstcxpldcs

This puzzle is worth [b]3[/b] points. Each phrase is worth the number of points to its left.

Also, sorry for not posting solution 7. But I can't edit the post now, so the solution is here.

[hide="Solution to Puzzles #7 and #8"]Puzzle #7: YOU LOST THE GAME

(No, that is the solution. I apologize for, but cannot claim any responsibility for, any disappearance of moral sanity.)

Puzzle #8:  India, Italy, Germany, USA, Canada, Peru, England

(Country names appear in bold.)

The Grammar Corporation was at it again. They acted like they were still [b]in dia[/b]pers! They were trying to tell me how to spell stuff. For example, they told me to spell "yen" with a cap[b]ital Y[/b]! I mean, it's bad enough that they managed to an[b]ger many[/b] a client, but this was newfangled of them. "Give [b]us a[/b] call back," they said right before hanging up. "We'll see if we [b]can ada[/b]pt your profile to a pay-[b]per-u[/b]sage model." Well, that certainly gave me swoll[b]en gland[/b]s. I'm never using their service again.

[hide="Scoreboard"]$ \begin{tabular}{|rl|r|}\hline & Name & Points \\
\hline 1. & K Sengupta & 24 \\
\hline 2. & r15s11z55y89w21 & 21 \\
\hline 3. & rd4593 & 20 \\
\hline 4. & cf249 & 15 \\
\hline 5. & vahalla & 13 \\
\hline 6. & cyberspace & 12 \\
\hline 7. & yaofan & 10 \\
\hline 8. & PowerOfPi & 1 \\
\hline\end{tabular}$
Nice try, cf249, Ithaca isn't a country. The one you were missing was [b]USA[/b].[/hide][/hide]

[size=200]Puzzle #11[/size] (June 28, 2009)

There are groups of two words listed below. Each group of words has another word that can fit between the two to create two new words, a clue for each of which is given next to the word.

E.g. for:

(the altitude at which plants stop growing) TREE ---- BACKER (a football player)

: the word is LINE.

1. (waking up drunk) HANG ---- LAY (cover)
2. (merchant) SALES --- AGE (take care of)
3. (comprehend) UNDER ----- BY (waiting for directions)
4. (hair care product) SHAM --- REST (most impoverished)

This puzzle is worth [b]4[/b] points, 1 per word.

[hide="Solution to Puzzle #9"]

1. Art of Problem Solving
2. Not My Cup Of Tea
3. Rhythm and Blues
4. An Open Book
5. You lost the game... two times now!

I once again apologize for anyone whose moral sanity has been shattered.

[hide="Scoreboard"]$ \begin{tabular}{|rl|r|}\hline & Name & Points \\
\hline 1. & K Sengupta & 39 \\
\hline 2. & rd4593 & 31 \\
\hline 3. & cf249 & 30 \\
\hline 4. & r15s11z55y89w21 & 29 \\
\hline 5. & cyberspace & 23 \\
\hline 6. & yaofan & 16 \\
\hline 7. & vahalla & 13 \\
\hline 8. & louiethechewy123 & 4 \\
\hline 9. & PowerOfPi & 1 \\
\hline\end{tabular}$[/hide][/hide]

Wow, it's been six days. I'm terribly sorry, but I'll be gone for another week at math camp (July 5 - July 11).

[size=200]Puzzle #12[/size] (July 3, 2009)

Here's one from Mensa. Don't worry, it's certainly a word puzzle.

Why are certain numbers above the line?

$ \frac {1\ 2\ \ \ \ \ \ \ \ \ 6\ \ \ \ \ \ \ \ 10}{\ \ 3\ 4\ 5\ \ \ \ 7\ 8\ 9}$

This puzzle is worth [b]2[/b] points.

[hide="Solution to Puzzle #10"]

1. International Math Olympiad
2. Not by the hair on my chinny chin chin!
3. Supercalifragilisticexpialidocious

[hide="Scoreboard"]$ \begin{tabular}{|rl|r|}\hline & Name & Points \\
\hline 1. & K Sengupta & 42 \\
\hline 2. & cf249 & 33 \\
\hline 3. & rd4593 & 31 \\
\hline 4. & r15s11z55y89w21 & 29 \\
\hline 5. & cyberspace & 23 \\
\hline 6. & yaofan & 17 \\
\hline 7. & vahalla & 14 \\
\hline 8. & louiethechewy123 & 4 \\
\hline 9. & PowerOfPi & 1 \\
\hline\end{tabular}$[/hide][/hide]

Because of the long break, I feel like I have to work overtime. I'll be posting a total of six puzzles today, to make up for the six-day hiatus.

I will be posting solutions to all the puzzles I post today on the 4th. (Oh, I just realized that that's Independence Day. XD)

[size=200]Puzzle #13[/size] (July 3, 2009)

The following math theorem has been expressed as words, in a very confusing manner indeed. Name the theorem.

[i]The sum of a mystery value, multiplied by itself and then a constant, and the same mystery value multiplied by another constant, is equal to the negative value of a third constant. The mystery value, then, is equal to the half-power of the difference between the second constant added to itself the number of times equal to the second constant, and the product of the first constant and quadruple the third constant, subtracted by the second constant, halved, and divided by the constant that was multiplied to the mystery value squared.[/i]

This puzzle is worth [b]3[/b] points.

[size=200]Puzzle #14[/size] (July 3, 2009)

You know how there are sometimes puzzles where you remove one letter from a certain word to get another?
Well, in this puzzle, all I'll give you is the letter that was removed!

To the left of the letter, will be a clue pertaining to what the original word is. Removing the letter, and rearranging the remaining letters will result in another word, a clue for which is given. Your job is to find both words.

(e.g. [temporary shelter] - C = [picture with roads] gives CAMP and MAP.)

(1) 1. [working] - Y = [long sandwich]
(1) 2. [fortune teller] - R = [Canadian sugary tree]
(2) 3. [Japanese massage] - S = [break]
(3) 4. [Russian comrade] - V = [warhorse car]

This puzzle is worth [b]7[/b] points. Each pair of words is worth the number of points to its left. If you only know one of the words in a pair, you won't get any points for that pair!

[size=200]Puzzle #15[/size] (July 3, 2009)

The following sentence contains two pairs of anagrammed words. The blanks are where the words go, and you need to fill in the sentence with the anagrammed words. (An anagram is rearranging the letters in a word, such as "rats" and "star".) Each underscore stands for one letter.

Goldfish was once the _ _ _ _ _ _ _ _ cheesy snack, but a _ _ _ _ _ _ _ version, which _ _ _ _ _ _ _ much better in today's market, came out under the name Catfish, when some anti-_ _ _ _ _ _ _ decided to _ _ _ _ _ _ _ _ the original recipe into something extraordinary.

This puzzle is worth [b]7[/b] points.

[size=200]Puzzle #16[/size] (July 3, 2009)

The following is a word train. A word train is a puzzle where one word is connected to another in a certain way.

For example:

1. The cardinality of the real numbers is 2 to the ____ null.
2. (1) is the first letter in the ______ alphabet.
3. (2) is the language of the Old Testament of the _____.

In this case, the words are ALEPH, HEBREW, and BIBLE, respectively. (n) stands for the answer to line n.

Complete the following word train:

1. If a Chinese person told you that you were in Dongjing, you'd be in the capital of _____.
2. (1) is home to ________, an old hanafuda card company that currently makes video games.
3. (2)'s most famous character, _____, is more well-known in the United States than Mickey Mouse.
4. (3) is a _______.
5. "Joe the (4)" was a well-known caricature used by ______ in his bid for presidency.

This puzzle is worth [b]5[/b] points, 1 per word. You must do the train in order!

[size=200]Puzzle #17[/size] (July 5, 2009)

It's time for... Rhyming Analogies!

Following this will be pairs of analogies in form of A is to B as C is to D. The D item in each line of the pair will be blanked out, but the two words will rhyme. Your job is to find out the rhyming pair.

For example:
SNOW is to WHITE as BANANA is to ______.
RAGE is to CALM as ANGRY is to ______.

In this case the rhyming words are YELLOW and MELLOW.

1. CARD is to DECK as PAPER is to _____.
DISK is to BYTE as TREE is to ____.

2. MUSIC is to NOTE as MATH is to ______.
DIM is to SHARP as CLUMSY is to ______.

3. DESK is to TABLE as CHAIR is to _____.
SCREW is to NAIL as HAMMER is to ______.

4. STREAM is to RIVER as LAKE is to _____.
5. FACE is to CREAM as SKIN is to ______.

This puzzle is worth [b]8[/b] points, 2 per pair.

[hide="Solutions to Puzzles #11, 12, 13, 14, and 15"]Puzzle #11:
1. OVER
2. MAN
3. STAND
4. POO

Puzzle #12:
The numbers above the line are three-letter words.

Puzzle #13:
The paragraph states: $ ax^2 \plus{} bx \plus{} c \equal{} 0$. Therefore, $ x \equal{} \frac { \minus{} b \pm \sqrt {b^2 \minus{} 4ac}}{2a}$. The theorem, therefore, is the [b]quadratic formula[/b].

Puzzle #14:
1. BUSY - Y = SUB
2. PALMER - R = MAPLE
3. SHIATSU - S = HIATUS
4. TOVARICH - V = CHARIOT

Puzzle #15:
Goldfish was once the [b]ultimate[/b] cheesy snack, but a [b]saltier[/b] version, which [b]retails[/b] much better in today's market, came out under the name Catfush, when some anti-[b]realist[/b] decided to [b]mutilate[/b] the recipe into something extraordinary.

I'm not keeping the scoreboard anymore, it's too tedious and there's no real point.[/hide]

[quote="PowerOfPi"]It has been about 10 days... Is this dead?[/quote]

As a matter of fact, it is. I was too busy trying to fix Linux, and doing other things.

Note: The new puzzles are solely for your enjoyment - do NOT send me private messages with solutions to the puzzle, I am not keeping a scoreboard. Because there is no longer a scoreboard, I will also be posting the solutions to puzzles one puzzle back instead of two.

Because the Daily Logic Puzzle has been discontinued, I might be incorporating some logic into the puzzles.

[size=200]Puzzle #18[/size] (July 24, 2009)

You know that "word game" thread that's going on in the Fun Factory right now? There's a puzzle version of that too.

Change the first word into the second word by changing one letter at a time, making valid English words each time. (E.g. CAT -> DOG is solved by CAT - COT - COG - DOG.)

1. KING -> KITE (Goal: 3 steps)
2. POLE -> DISK (Goal: 8 steps)
3. OPEN -> BOOK (Goal: 12 steps)

An exception to the PM's: If you have a solution that takes fewer steps than the goals, please share them with me.

[hide="Solutions to Puzzles #16 and #17"]Puzzle #16:
1. Japan
2. Nintendo
3. Mario
4. Plumber
5. John McCain

Puzzle #17:
1. SHEAF, LEAF
2. SYMBOL, NIMBLE
3. I forgot, I think it's "SEAT, BEAT".
4. OCEAN, LOTION
[/hide]

[size=200]Puzzle #19[/size] (July 25, 2009)

Welcome to the game of Caboose - just like the Geography game except that letters overlap by 2 instead of 1!

Basically, I'll be giving a series of words, where the last two letters of the previous word make up the first two letters of the next word. Each listed thing is a clue to the next word.

E.g.:

The first two letters are PL.
1. Three points define this.
2. Not far
3. Phoenix
4. Jesus' birth

The solution is PLANE, NEAR, ARIZONA, and NATIVITY.

The first two letters are EX.

1. The AIM-what? (000 to 999)
2. Toronto
3. Jupiter's moon
4. State
5. Dihydrogen monoxide
6. Mistake
7. Situation

[hide="Solution to Puzzle #18"]1. king, kins, kits, kite
2. pole, pore, pork, park, dark, dank, dunk, dusk, disk
3. open, opes, opts, outs, ours, oars, cars, care, core, cork, cook, book[/hide]

[size=200]Puzzle #20[/size] (July 27, 2009)

Complete the following letter sequences:

1. ni, ni, ni, ni, ni, on, ?
2. ve, ix, en, ht, ne, ol, ?
4. ty, ty, ty, ty, ty, ed, ?

[hide="Solution to Puzzle #19"]1. EXAMINATION
2. ONTARIO
3. IO
4. IOWA
5. WATER
6. ERROR
7. ORDEAL[/hide]

[quote="Adharsh Raghavan"][hide=" Solution to problem#20"]

1. ni, ni, ni, ni, ni, on,[b][u]on[/u][/b]

   They are the first two letters of : 95,96,97,98,99,100,[u][b]101[/b][/u]

2. ve, ix, en, ht, ne, ol, ?

   I know this has something to do with last two letters of : 5,6,7,8,9,?
   I just can't figure out what the next term is..

3. ty, ty, ty, ty, ty, ed,[u][b]en[/b][/u]

    They are the last two letters of : 50,60,70,80,90,100,[b][u]110[/u][/b]

[/hide][/quote]

erm.. you're supposed to pm your solutions, not post them.

[quote="sdkudrgn88"]Note: The new puzzles are solely for your enjoyment - do NOT send me private messages with solutions to the puzzle, I am not keeping a scoreboard. Because there is no longer a scoreboard, I will also be posting the solutions to puzzles one puzzle back instead of two.[/quote]

[quote="AIME15"][quote="Adharsh Raghavan"][hide=" Solution to problem#20"]
2. ve, ix, en, ht, ne, ol, ?

   I know this has something to do with last two letters of : 5,6,7,8,9,?
   I just can't figure out what the next term is..
[/hide][/quote]
[hide]
THe numbers are 1 googol - 5, google - 4, ..., googol, googol + 1, so it should be ne.[/hide][/quote]
Why "googol"? I mean, what is special about "googol"?

googol is defined as $ 10^{100}$ if i'm not mistaken, also

[url]http://lmgtfy.com/?q=googol[/url]

[quote=PowerOfPi]Why "googol"? I mean, what is special about "googol"?[/quote]

The fact that it ends in "ol" makes it the answer.

Alright, another puzzle.

[size=200]Puzzle #21[/size] (August 6, 2009)

Another Mad Gab series.

1. Ten Key Oof Or Them Use Ick
2. Ink Awed Wheat Rust
3. Lore Taw Furry Inks

[hide="Solution to Puzzle #20"]1. on. The answer is 95, 96, 97, 98, 99, 100, 101.
2. ne. The answer is (googol - 5), (googol - 4), (googol - 3), (googol - 2), (googol - 1), googol, (googol + 1).
3. en. The answer is 50, 60, 70, 80, 90, 100, 110.
[/hide]

[hide="number 18"]
I'm not sure if I'm allowed to use proper nouns, but here goes.
Pole
Polk
Pock
Puck
Duck
Dusk
Disk

Done!
[/hide]

This has nothing to do with the post, but I like your signature.


algebra function domain articles Harvard college geometry

[hide][url=http://members.tripod.com/worldupdates/islamintheworld/id27.htm]members.tripod.com/worldupdates/islamintheworld/id27.htm[/url][/hide]
I 've just checked that link,i wonder what you might think about its content  :) .

no comment till now?? i didn't know that it was that uninteresting :D

The internet is a wonderful place, where all crackpot theories can be aired for the price of a domain name ...

I didn't claim that what the link contained  was  true but u can't deny it either i guess the article talks about some references ,u'd better check them urself and then u can aprrove or disapprove it.

You asked what I thought about its content.  I think its content was complete nonsense.  I feel no particular need to fact-check every single outrageous claim I come across, especially those which have all of the major markings of silliness, fraud or pseudoscience.  In this particular instance, I decided to go out of my way to justify my initial instinct.
As far as I can tell there are no links to any of the cited sources (for example, a scholarly paper or historical document) in the article.  Of those individuals whose names are mentioned, their relevant works are (as far as I can tell) completely discredited.  For instance, the article cites on "Dr. Barry Fell (Harvard)."  Dr. Fell was a [i]marine biology[/i] professor at Harvard university; one can read a short critique of him and his work in this area at [url=http://cwva.org/ogam_rebutal/lesser_how_sci_works.html]here[/url].  Seperately, it references "A renowned American historian and linguist Leo Weiner of Harvard University[.]"  A quick search on Google for "Leo Weiner" brings up sites related to a (presumably unrelated) Hungarian composer/teacher.  Adding "linguist" brings up a large number of sites which have essays identical (or nearly so) to the one you linked to.  Adding "historian" in place of "linguist" brings up a similar but slightly different set of essays.  A great number of them refer to him as "renowned" -- if he is so renowned, why can I find no independent reference to him, positive or negative?

Really, this is such a classic form and style of nonsense that I don't know why I bothered to do this rather than shrugging it off and leaving you with your odd imaginings.  Not every essay that gets posted on the web is worthy of serious consideration.  This one, in particular, is of the class that are [i]certainly not[/i] worthy of such attention.

As a separate, personal request, please don't use "chatspeak."  I hate reading about how "u r goin 2 do sumthing" and related things.

[quote="JBL"] I don't know why I bothered to do this rather than shrugging it off and leaving you with your odd imaginings.[/quote]
I remind you that i didn't claim that the content was true and by posting the link here i aimed to discuss that content and not to defend it ,so that has nothing to do with my "odd imaginings" :) and finally let me tell you that you over-react here.

Regarding Dr.Fell the article has just quoted what he said in his book so nothing to blame,and regarding Leo Wiener here's an article that talks about his son ,but also about him :):
http://en.wikipedia.org/wiki/Norbert_Wiener
amazone web page regarding Dr.Fell book with some critics and a positive comment :):
http://www.amazon.com/gp/product/0812908473/104-2779166-4305500?v=glance&n=283155

[quote="MoHaMeD"]Regarding Dr.Fell the article has just quoted what he said in his book so nothing to blame[/quote]

Yes, and you just gave us a link to the website.  I challenge the idea that passing false information as if it were true, either for purposes of self-delusion or deluding others, is blameless.

This website really is an example of the worst sort of group-identity-aggrandizing tripe that one can find.

well i don't know why you do judge it as a false information as you haven't even read the book to judge the ideas yourself regardless the critics in the website you mentioned,moreover the ideas of the Dr.Fell aren't the basis of the whole article and the part related to the islamic state is Spain is true according to my knowledge "as i belong to a  moroccan family originally from Andalus=Spain" .Regarding Mr.Wiener i think the idea of the Hungarian teacher isn't true you have the link to wikipedia page and also two stamps (one of israel 1998 and the other of Moldova 2000) .
btw i appreciate our discussions even if i feel that you still over react though :).
it's 1:20 am i am going to sleep ,thank you for your comments :)

My God!  This is the kind of (for lack of a better word) crap that is killing us.  Which crackpot published this?

I don't know why u're so keen in denying its content meawhile hypothesis regarding the discovery of the continent by phenicians ,ancient egyptians and chinese are being discussed by so many historians  :huh:

the article seems unsure of what it wants to say, it focuses on Muslims coming prior to Columbus but if that's not right then at least there were muslims with columbus (undoubtably true-there were lots of undesirables (ie those who would have been subjected to the inquisition) with columbus who was possibly jewish himself).  if the article is so confident there were muslims prior to columbus why does it feel a need to argue about the muslims with him.

Huh, well I think there could be a chance that people in Africa managed to find the Americas before Columbus, but honestly, WHO CARES?


[hide]
Menelaus gives $ \frac{5}{5}\times\frac{16}{6}\times\frac{10\minus{}x}{x}\equal{}1$, where x is the length of AE. We have $ \frac{10\minus{}x}{x}\equal{}\frac{3}{8}\iff 80\minus{}8x\equal{}3x\iff x\equal{}\boxed{\frac{80}{11}}$[/hide]

I do not think this belongs in pre-olympiad, it is a simple application of a relatively well-known theorem.

[quote="luiseduardo"]The figure of the triangle ABC is equilateral.  AM = MB = 5 and CD = 6. The value of AE is:

[img]http://img98.imageshack.us/img98/2066/calsl.jpg[/img]

a) 76/11
b) 77/11
c) 78/11
d) 79/11
e) 80/11[/quote]
[hide]Using Menelaus's Theorem, and since M, E, and D are collinear (at least they look like it..)
we have that $ \frac{BD}{CD}\cdot\frac{CE}{EA}\cdot\frac{AM}{BM}\equal{}1$
From the information given this turns into:
$ \frac{8}{3}\cdot\frac{CE}{EA}\cdot\frac{5}{5}\equal{}1$. Set $ CE\equal{}x$, this means $ EA\equal{}10\minus{}x$.
$ \frac{x}{10\minus{}x}\equal{}\frac{3}{8}$.
$ x\equal{}\frac{30}{11}$.
$ EA\equal{}\frac{80}{11}$,  $ E$.[/hide]


From an old test: Show that $x^{2}\leq 4e^{x-2}$ for all $x\geq 0$. All the problems tend to have some quick trick to make the problem possible in the short time frame, so I'm assuming this has something to do with IVT or MVT or something.


MATHCOUNTS
probability
geometry
AMC
AIME
summer program
MathPath
\/closed

The AoPS [i][url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=10]Introduction to Number Theory[/url][/i] textbook is now available.

This books is ideal for students who have mastered basic algebra, such as solving linear equations. Middle school students preparing for MATHCOUNTS, high school students preparing for the AMC, and other students seeking to master the fundamentals of number theory will find this book an instrumental part of their mathematics libraries. Text: 336 pages. Solutions: 144 pages.

[quote="b-flat"]Yay!

Are all of the introductory books now complete?

These textbooks are great![/quote]

No, there will be at least one more introductory level book.

What are the other indroductory books going to be on?

Intro to Algebra?

There will be an Introduction to Algebra.  We're not sure how we might handle Pre-Algebra or Problem Solving yet.

[quote="MCrawford"][quote="mathjegal"]Would there be Intermediate books?[/quote]

Yes, intermediate books are in the works.[/quote]

In what topics?

The first intermediate book will almost surely be Counting & Probability, and the second is likely to be Algebra.  We also plan to write intermediate level Number Theory and Geometry books, and possibly more.

[quote="ShoeFactory"]Are the intermediate books coming before the start of 2007? (approximately)[/quote]

That's hard to say at this point.  There is an expected value of about .8 of them coming out by then, with a standard deviation of about .3.

I  would very much  like  to  get  the introductory  algebra book!!  im  looking forward to the intermediate  books  also!!

  please have  them   before  imo 2007!!!

    :)

[quote="MCrawford"]Er, the Introduction to Algebra isn't going to help you at the IMO.[/quote]
Maybe he put more emphasis on the intermediate books?  :huh:
 :omighty: :omighty: :omighty:
I Bow Down to you, O Mighty Genius AoPS ones!
*Majestic Music Plays*
May you always be there to grant us these 1337 books!

[quote="MCrawford"]Just to explain levels:

Introductory: Good for MATHCOUNTS through AMC 12 problems, and even a few AIME problems.

Intermediate: Good for hardest AMC 12, all AIME problems, and around a third of Olympiad problems.

Olympiad: All Olympiad level problems.[/quote]

Oh, okay, thank you. I was just about to ask what the book levels were when I saw your post.  :lol:

[quote="knexpert"]Does the Intro to Number Theory textbook cover other topics besides those mentioned in Volume 2?[/quote]

Yes, but the Volume 2 material is mostly at a higher level.  The Introduction to Number Theory book is a far more comprehensive treatment of number theory than the original AoPS books (which covered mostly other topics).  You can compare the tables of contents of the two books in our Bookstore pages.

[quote="MCrawford"][quote="PenguinIntegral"]Instead of asking you what will be covered in the intermediate books, may I ask you what you have decided to keep out of the intermediate books, and strictly in Olympiad territory?[/quote]

Yes, you may ask that question.

No, we can't give you a good answer yet, except that our classes are a rough guide, as are the forums.[/quote]Hmm, so would the Intermediate/Pre-Olympiad forums be approximately the difficulty of the intermediate books?

That would approximate the difficulty of the harder end of the books.

My dad ordered the textbook...I can't wait to receive it. It's probably going to be extremely beneficial for my number theory needs.

I am planning to buy it at mathpath, since I know that they are selling books at discounts there. If that particular book is not available, I will order it in early August and hopefully finish it before the school year starts again.

Would these books be helpful for mathcounts.

Yes, our Introductory books are designed with MATHCOUNTS and AMC students in mind.  In fact, there are scores of MATHCOUNTS problems throughout them.

I LOVE THE BOOK!!!!!

It's well written
It's easy to follow
You can catch on very fast

Thank you for making it, Mr. Crawford!

Hi,
I wanted to ask if there is a book available about calculus, that means an introduction into differential and integral calculus ?

There are many such books, but none of them have been written by anyone at AoPS.  You might look on the [url=http://www.artofproblemsolving.com/Forum/index.php?f=319]Books and Journals[/url] forum for past comments on various calculus texts.

[quote="JBL"]There are many such books, but none of them have been written by anyone at AoPS.  [/quote]

Yet ;)

If I have read and understood the AoPS Volume II Number Theory material, would it be better to wait for the Intermediate book?

[quote="MCrawford"][quote="buzzer11"]so can you give me like a sequence in which we should go though these books (age, grade, cooresponding courses, etc.), including the two origional AoPS books (Volume 1 and 2)?[/quote]

Really, the levels of the books overlap a bit.  However, our Introductory books are generally at the level of Volume 1, but go much deeper into the material and cover perhaps a bit of Volume 2 material.

For the Introductory books, they are probably easiest to hardest in this order:
Counting & Probability
Number Theory
Geometry

When the Introduction to Algebra comes out, it will probably be the easiest of the four, though also the longest.[/quote]

so should I fit any of these Introdutory books in between Volume 1 and Volume 2?

I would say that you should work through Volume 1 along with the Introductory subject books.


Given $ a, b, c$ is in $ [1, 2]$. Prove that: $ 3(ab \plus{} bc \plus{} ca) \geq\ 2(a \plus{} b \plus{} c) \plus{} a^2b \plus{} b^2c \plus{} c^2a$
When does equality occurs ?
 :)

your inequality is equivalent as:

$ \sum_{cyc}(ab \minus{} b)(2 \minus{} a)\geq0$

which is obviously true !! :lol:

Yes, you are right. Indeed, it's simpler than my proof  :oops:


Find $c$ if $a$, $b$, and $c$ are positive integers which satisfy $c=(a + bi)^3 - 107i$, where $i^2 = -1$.

[hide="solution"]Expanding and simplifying, we see that $c=a^3+3ia^2b-3ab^2-ib^3-107i=a^3-3ab^2+i(3a^2b-b^3-107)$. Because $c$ is a positive integer, the imaginary part of our exression must equal $0$. Now we have $3a^2b-b^3-107=0$, or $b(3a^2-b^2)=107$. The only $a$ and $b$ that satisfy this equation are $a=6$ and $b=1$. Plugging these values back into our original expression, $c=6^3-3(6)(1)+i(0)=\boxed{198}$.[/hide]

[hide=Solution]We begin by expanding $(a+bi)^3$:
$$(a+bi)^3=(a^2+2abi-b^2)(a+bi)=a^3+3a^2bi-3ab^2-b^3i$$
Since $c$ is a positive integer, meaning it is not imaginary or complex, we have
$$3a^2bi-b^3i=107i$$
$$3a^2b-b^3=107$$
Factoring the b out of the LHS gives
$$b(3a^2-b^2)=107$$
Since $107$ is prime, we have that $b$ and $(3a^2-b^2)$ are $1$ and $107$ in any order. It is easy to see that $b$ is $1$, which means that $a=6$. We then plug this into the non-imaginary part of our expression to find the value of $c$:
$$a^3-3ab^2=6^3-3(6)(1)^2=216-18=\boxed{198}$$[/hide]

[quote=shyong][quote="me@home"]I got stuck solving it this way, what went wrong:

By factoring $(a+bi)^3$ we have:
$c = (a + bi)(a^2 - b^2 - abi) - 107i$
We want all the terms with an $i$ to cancel so that we are left with a real number (a positive integer no less!).
Therefore we FOIL to find the coefficient of the $i$'s. It is:
$(a^2 - b^2)bi - a^2bi$ or $bi(-b^2)$, which means that $b^3=107$ which is obviously false. What is wrong??[/quote]

we are not dealing with factorisation .... and u apply the wrong thing also since

$a^3\pm b^3=(a\pm b)(a^2\mp ab+b^2)$[/quote]
Well I am 15 years too slow but thanks for the correction shyong! It looks like I was trying to factor $a^3+(bi)^3$ instead of $(a+bi)^3$ and that is where I went wrong. The correct factorization would be
$$
(a+bi)^3=(a+bi)(a+bi)^2=(a+bi)(a^2-b^2+2abi),
$$
which differs from what I wrote in the coefficient of $abi$.

[hide=Solution]
FOIL the last thing I wrote to find that the coefficient of $i$ in $(a+bi)^3$ is $3a^2b-b^3$. So $b(3a^2-b^2)=107$. Since $b$ is a positive so is $3a^2-b^2$, and both are integers. By primality of $107$ either $b=1$ or $b=107$. In the first case we find the solution $(a,b)=(6,1)$ and in the second case there is no solution, since $3a^2-107^2$ is congruent to $2\mod{3}$ and therefore cannot equal $1$. Thus $c=a(a^2-3b^2)=6\cdot 33=\boxed{198}$.
[/hide]

By expanding and evaluating, we get $3ba^2-b^3-107=0$, and, $b(3a^2-b^2)=107.$ It is easy to see that $a=6$, and $b=1$, doing computation, $c=\boxed{198}.$


Determine the number of integers $n\geq2$ for which the congruence
$x^{25}\equiv~x (mod n)$is true for all integers $x$.

I found integers $n$ satisfying $x^{24}\equiv~1(mod n)$ where $gcd(x,n)=1$.
A friend told me that there surely exists more solutions than those I found.

I cannot seem to find a good strategy for attacking this problem.
Any suggestions?


What is the largest possible number of distinct positive integers such that no four of them has a sum divisible by 4?

[hide]well you use mods here...

So you can have 3 [b]0mod4[/b] max or otherwise a 4th one would cause a sum divisible by 4.

You can also only have 3 [b]1mod4[/b], so I believe the answer is [u]6[/u][/hide]


probability geometry calculus AMC AIME USA(J)MO USAMO \/closed

The Art of Problem Solving Fall/Winter Schedule is available [url=http://www.artofproblemsolving.com/Classes/AoPS_C_Enroll.php]here[/url].  The classes we offer this fall and winter are listed below.

[b]Introductory level: Grades 6-10[/b]

*[url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=41]Algebra 1[/url]
*[url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=2]Introduction to Number Theory[/url]
*[url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=1]Introduction to Counting & Probability[/url]
*[url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=10]Introduction to Geometry[/url]
*[url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=42]Algebra 2[/url]
[url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=37]MATHCOUNTS/AMC 8 Basics[/url]
[url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=5]Advanced MATHCOUNTS/AMC 8[/url]
[url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=20]AMC 10 Problem Series[/url]

[b]Intermediate level: Grades 8-12[/b]

*[url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=4]Algebra 3[/url]
*[url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=3]Intermediate Counting & Probability[/url]
*[url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=11]Precalculus[/url]
*[url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=40]Calculus[/url]
[url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=13]AMC 12 Problem Series[/url]
[url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=14]AIME Problem Series A[/url]
[url=http://www.artofproblemsolving.com/Classes/classdetails.php?course_id=30]AIME Problem Series B[/url]

[b]Specialized Olympiad Training[/b]

[url=http://www.artofproblemsolving.com/Classes/AoPS_C_WOOT.php]WOOT[/url]!

Classes marked with a (*) have an accompanying textbook.  If you plan to enroll in one of these courses, you should do so well before the start of the course, so that you receive the text before the class starts.  (Note: The Precalculus and Calculus books are not yet available.)

All of our courses offer full transcripts for each class, so if you're going to miss a class or two, you'll still have access to everything that happened in class.

If you have any questions about a specific class, please contact us at classes@artofproblemsolving.com.

[quote="mathswimmer"]If a student has mastered (as in done the challenge problems) the 4 introductory books and mastered algebra 3, should he/she be ready to qualify for the USAMO?[/quote]

Yes, that's not unreasonable.  You'd be in a better position if you knocked off the Interm Counting and the Precalc, too.

[quote="batteredbutnotdefeated"]What course covers all of Alg 2 and (possibly) a little of precalc? As in all the Alg 2 covered in a regular school course because I want to take an Alg 2 exam and skip it.

I checked the algebra 2 course and it doesn't seem at all like the school course alg 2...[/quote]

Our classes don't line up to school classes quite like that.  We name our classes this way to give people guidance as to which of our classes you are ready for.  If you have taken Algebra 1 in school, then you should be ready for our Algebra 2 class.  If you have taken Algebra 2 in school, you should be ready for our Algebra 3 class.

If the AIME classes were a little hard, what intermediate courses should I have taken before the two classes? Would I be ready for both of the AIME classes if I have mastered Algebra 3 and Intermediate Number Theory?

[quote]
Our classes don't line up to school classes quite like that. We name our classes this way to give people guidance as to which of our classes you are ready for. If you have taken Algebra 1 in school, then you should be ready for our Algebra 2 class. If you have taken Algebra 2 in school, you should be ready for our Algebra 3 class.[/quote]

No, I mean if I want to take a SCHOOL Alg 2 final and score very high/ace it, which class would be the most appropriate? Or would I have to go to like a high school and take a course if I have to do that?

[quote="myyellowducky82"]even if we're taking Algebra 3, do you think it's okay to take Algebra 2 as a review?[/quote]

Haha, yes.  I did something similar.

[quote="myyellowducky82"]@battered: I have a similar story. I'm going into 8th grade and taking 10th grade Geometry, so I have to go to the high school. :P
I think you might have to go to a high school to take a course. :)[/quote]

The problem is, the county I'm currently in won't let me go farther than advanced geometry... That's why I either want to take precalc or calc as a freshman... (the only way I can do this is if I take Alg 2 (and possibly) precalc before summer starts in 2010)

[quote="batteredbutnotdefeated"]

No, I mean if I want to take a SCHOOL Alg 2 final and score very high/ace it, which class would be the most appropriate? Or would I have to go to like a high school and take a course if I have to do that?[/quote]

That depends.  "Algebra 2" means a lot of different things.  For most schools, our Algebra 1+2 classes together will cover most of what the school covers in their Algebra 1+2 classes (and do it in 1 year instead of 2).  Some schools integrate Algebra 2 with Precalc and race through to Calculus, skipping all sorts of good stuff that we cover in a variety of our Intermediate classes and Algebra 3.

I seriously doubt that's possible... I suggest you read intro to counting and probability instead of taking the class.

I've finished all the Intro Books and Volume 1, which order do you recommend doing Intermediate Algebra, Intermediate Counting&Probability, and Volume 2.?

Algebra before Counting.  You can probably work your way through Volume 2 on and off as you go through the Intermediate books -- it will give you extra practice on some of the harder problems, and show you a few things we don't address in the Intermediate books.

[quote="remy1140"]Does the pre-calculus course cover all the trig materials that I will need in competitions like AIME and ARML?[/quote]

Yes.

[quote="rrusczyk"]Intro Counting and Intro Geometry (summer session) are on different days.[/quote]
I forgot to add that I'm taking Algebra 3, as well. hmm, any class suggestions?

If you take like 4 classes at once, then you might be caught up in a lot of work... Especially if the classes continue through the beginning of the next school year. :|

I'm currently doing AoPS volume 2, and finishing all the problems in each section except for about 2.

I want to be good enough to make USAMO when that time comes around (and hopefully do well!), so which class would you say is better?: AIME problem series, or WOOT?

Also, after I finish volume 2, (hopefully with the same level of ability in each chapter, understanding, etc.),

what would you give my chances on the AIME?

Thanks.

narik - how do you fare on practice AIMEs now?  If you're shooting in 6+ range consistently, and will keep working between now and September, WOOT would probably be a good fit.

yellowduck -- check out the Intro Counting post-test.  If you can't tackle nearly all those problems, then that's probably a good class for you.

[quote="alexgnow"]If I just took the Mathcounts/AMC 8 Basics class, would you recommend continuing on to the Mathcounts/AMC 8 Advanced class?

Also, when would you suggest taking the WOOT class?[/quote]

I wouldn't recommend the WOOT class right after the MATHCOUNTS class.

As for taking the advanced class, that depends on whether or not you found the Basics class hard or easy.  If you found it pretty challenging, I recommend waiting a while before taking the Advanced class.

In any case, the Intro level of classes would be a good next step after the Basics class.

[quote="Blazer123"]The basics classes are Mathcounts right?

Also, if the one feels comfortable about the intermediate series, would WOOT be for them?

I'm just curious. :D[/quote]
There is a MATHCOUNTS/AMC8 Basics class and an Advanced MATHCOUNTS/AMC8 class.

I think W00t might be a challenge, but I haven't taken any classes so don't rely on that.

When will the Precalculus book be available? Will it be around September or November this year?

Also, would the Intro to Geometry course help us prepare for AIME, or would Olympiad Geometry be a better preparation (although we don't have it this fall/winter)?


[quote="tranngoctrung"][quote="jin"]What does it mean?the sum from 1 to n?[/quote]
  $\sum\limits_{i=1}^{n} f(i)=f(1)+f(2)+f(3)+f(4)+......+f(n-1)+f(n)$[/quote]
   why   no one prove it


[quote="Adriana N."]Can somebody recomand me some new books,pls?[/quote]

 But you in Romany have - from - the best olympiad books !!!

 Any way, do you need olympiad books and on which area ? In the last year I have not seen any new olympiad book for

senior level.But I will search again later.

 Babis

See [url=http://www.gil.ro][color=red][b]here[/b][/color][/url] (vezi [url=http://www.gil.ro][color=red][b]aici[/b][/color][/url] , succes !).


Let $S$ be some set. Suppose that $g : S \to S$ is a function. [b]Discuss[/b] the criteria on determining whether there exist a function $f: S \to S$ such that $f(f(x)) = g(x)$ for all $x \in S$.

Feel free to restrict to special cases, such as permutations, etc.

Depending on the set S we have quite interesting cases to look at:

case1: S is composed of finite # of elements . This one i find rather simple. I haven't worked on it, but first of all clearly if
g(x) be one to one such function f(x) exists.

case2. S is continous.

Proposition: Let g(x) be an increasing continous function, and let S be any set of form: S=(-inf, a) or (a,+inf) or (a,b)
Then there exists a continous function f(x) such that f(f(x))=g(x)
 The proof is simple. I post it if nessasary.

The general case has no constraint on continuty of g(x) and I find that to be the most difficult...
-Ali

[/b]

[quote="ali"]
case1: S is composed of finite # of elements . This one i find rather simple. I haven't worked on it, but first of all clearly if
g(x) be one to one such function f(x) exists.
[/quote]

Actually, no, it doesn't (not necessarily). $g$ is a permutation, and in order for it to be a square, it needs to be an even permutation, so not all bijections from a finite set to itself are squares in the group of permutations of that set.

I didn't give it too much thought, but I think we can characterize the bijections $g:S\to S$ which are also squares (i.e. $g=f\circ f$) in terms of their cycle desomposition (which can be performed just like in the finite case). I believe the functions we're looking for are those for which all the cycles have finite length, and for all even $k$, we either have infinitely many cycles of length $k$, or an even number of such cycles.


function
real analysis
real analysis unsolved

What is the limit for the sequence
$ a_n \equal{} (1 \minus{} \frac {1}{2})(1 \minus{} \frac {1}{2^2})(1 \minus{} \frac {1}{2^3})\dots(1 \minus{} \frac {1}{2^n})$
It is trivial to prove that it converges, but how do we calculate the limit?
Can we find out the closed form? I don't think so.

Nothing particularly elementary, you are right. This is called Euler's function (at the point $ 1/2$).

Put the value 0.2887880950866024212788997 into the Inverse Symbolic Calculator ... it yields only the product formula we started with.  So a "closed form" was unknown to them, at least at some time.

hello, Mathematica says
$ \prod_{i\equal{}1}^{n}1\minus{}\frac{1}{2^i}\equal{}\text{QPochhammer}[1/2,1/2,n]$
Sonnhard.

That's essentially the definition of the $ q$-Pochhammer symbol, so that says nothing.


AMC
AIME
geometry
logarithms
number theory
least common multiple
probability

The problems are found here: http://math.furman.edu/tournament/tournament.html

Remark: I really don't understand how the Furman test makers thought people could finish this specific test in 2 hours. Most of the questions were $\#20-25$ on the AMC-12 level and a few were even easy AIME level.

The problems I wasn't able to get are: #4,16,24,30,31,34,38

1.Using the rational root theorem, we see that $x=2$ is a root. We use synthetic division to factor the equation as: $(x-2)(3x^{2}+x-10)$. The roots of the second factor are $\frac{5}{3}$ and $-2$. $\frac{5}{3}=.6$, the answer is $6$.

2.Multiplying the denominator by $1-2i$ gives us that the expression is $2+3i$, the coefficient is thus $3$.

3.With the first cut, we must remove a $8 \times 8$ square leaving us with an $8 \times 5$ rectangle. We now remove a $5 \times 5$ square leaving us with a $3 \times 5$ rectangle. We now remove a $3 \times 3$ square and are left with a $2 \times 3$ rectangle. We now remove a $2 \times 2$ square and are left with a $1 \times 2$ rectangle. From this, we remove two $1 \times 1$ squares. We have $6$ total squares.

5.We have that $2b^{2}+3b+7=159 \Longrightarrow b=8$. So therefore $3(8^{2})+4(8)+5=229$

6.We have that $\frac{8}{3}(1+p)=\frac{10}{2}$. The closest to $p$ is $87$.

7.The area of the squares is $4^{2}+13^{2}+15^{2}=410$. Now after using Heron’s Formula, we have that the area of the triangle is $24$. Thus, the total area is $434$.

8.Using the change of base formula, we have $\frac{\log{x}}{\log{3}}=p$ and $\frac{\log{x}}{\log{7}}=q$. Also, notice that we are looking for the value of $\frac{\log{x}}{\log{21}}$ and $\log{21}=\log{3}+\log{7}$. From this we get the value we are looking for as $\frac{1}{1/p+1/q}$.

9.We use the Principle of Inclusion Exclusion. There are $[\frac{1000}{12}]+[\frac{1000}{18}]=83+55=138$. We must now the subtract the number of numbers divisible by their LCM which is $36$ which is $[\frac{1000}{36}]=27$. The answer is $138-27=111$.

10.There are $6(7)/2=21$ squares of length $1$. There are $8$ squares of length $2$, and there are $3$ squares of length $3$. The answer is $21+8+3=32$.

11.There are $\frac{10!}{2!3!5!}=2520$ arrangements.

12.We have:
$a_{1}=2(1)+3(2)=8$
$b_{1}=1-2(2)=-3$
$a_{2}=2(8)+3(-3)=7$
$b_{2}=8-2(-3)=14$
$a_{3}=2(7)+3(14)=56 \equiv 0 \mod{4}$

13.The roots of $p(x)$ are $1$ and $-2$. These are the values of $r$ we can use on $q(x)$. So we have $a+b+c=-1$ and $4a-2b+c=8$. From these, we have that $a-b=3$ and $c=2$ and the answer is $5$.

14. There are three possible cases to get a score of $5$.
Case 1: We have $1+2+2$. There are three flips of the coin and we must have one head and two tails. Thus the probability is $\binom{3}{1}(\frac{1}{2})^{3}=3/8$
Case 2: We have $1+1+1+2$. This has probability of $\binom{4}{3}(\frac{1}{2})^{4}=1/4$.
Case 2: We have $1+1+1+1+1$ which has a probability of $\binom{5}{5}(\frac{1}{2})^{5}=1/32$.

The total probability is $3/8+1/4+1/32=\frac{21}{32}$.

15.The area of the intersection is the area of a square of sidelength $1$ that is INSIDE the circle. After drawing in the radii of the circle, we see that this area is two $30,60,90$ triangles (with one of the legs of length $1$ and the hypotenuse as the radius). The total area is $\frac{1}{9}\pi+2(\frac{1}{2 \cdot \sqrt{3}}$ and $a+b=4$.

17.We use trial and error for each of the $n$ gons by splitting them into triangles and using the law of cosines to find out the sidelength of each of the sides. $7$ is the smallest $n$ that works.

18.Note that $\sin{\frac{23}{3}\pi}=\sin{\frac{5}{3}\pi}$. $\arccos{\frac{5}{3}\pi}=\frac{5}{6}\pi$ and $11$ is our answer.

19.There are $\binom{10}{0}+\binom{10}{1}+...+\binom{10}{10}=2^{10}$ total subsets that can be made. Now, we can find out the total number of subjects that can be made that DO NOT contain any elements of $B$. There are $\binom{7}{0}+\binom{7}{1}+...+\binom{7}{7}=2^{7}$ possible subsets. The number of subsets with atleast one element of $B$ is therefore $1024-128=896$.

20.With triangle inequality, we have that $c+19>17 \Longrightarrow c>2$. Also, using the fact that the triangle is acute, $c^{2}<17^{2}+19^{2}\Longrightarrow c<26$. So we have $2<c<26$ which gives us $23$ integer values for $c$.

21.Let $y=8^{x}$. The equation becomes $2y+32/y=65 \Longrightarrow 2y^{2}-65y+32 \Longrightarrow y=1/2, 32 \Longrightarrow x=5/3,-1/3$ and the sum of the solutions of $x$ is $4/3$.

22.The perfect cubes are $1,8,27,64,125$. Thus, our sum is $7(1)+19(2)+37(3)+61(4)=7+38+111+244=400$. Our answer is thus $(e)$.

23.We can expand the expression as such $((x+2y)+(3z+5w))^{8}$. Since we are looking for the coefficient of $xy^{2}$ and $z^{3}w^{2}$, we basically need to look at $\binom{8}{4}(x+2y)^{3}(3z+5w)^{5}$ (after using the binomial theorem and looking at different possibilities). Using the binomial theorem, the coefficient we have is $\binom{8}{5}\binom{3}{2}(2y)^{2}x \binom{5}{2}(3z)^{3}(5w)^{2}$ and the coefficient is $56 \cdot 3 \cdot 4 \cdot 10 \cdot 27 \cdot 25 \mod{10000}=6000$. The answer is $6$.

25.Using the law of cosines, we have $17^{2}=25^{2}+26^{2}-2(25)(26)(\cos{\alpha}) \Longrightarrow \cos{\alpha}=\frac{253}{13 \cdot 25}$. Now, using Heron’s Formula, we see that the area of the triangle is $204$, so $\frac{1}{2}\cdot 25 \cdot 26 \cdot \sin{\alpha}=204 \Longrightarrow \sin{\alpha}=\frac{204}{15 \cdot 13}$. The result follows.

26.$0\leq \sin{\theta}\leq 1$ which is at an interval of length $2$. But because a parabola contains two roots for each value of $y$, the interval length is $1$.

27.We have $2\sin{2\theta}\cos{2\theta}=\cos{2\theta}\Longrightarrow \sin{2\theta}=1/2 \Longrightarrow \theta=\pi/24, 25\pi/24$ (keeping in mind of the signs of cosine and sin. From there, we can get our answer by adding $2\pi$ to each of these base angles.

28.The line that goes through the two opposite vertices of the hexagon (in which the centers of the two circles are collinear to that line) is of length $2$ (from equilateral triangles). Let the radius of each of the circles be $r$. Drawing $r$ to the tangents of the circle, we form a $30,60,90$ triangle and the line of length $2$ can be written as $2r+2r/\sqrt{3}+2r/\sqrt{3}$. Thus $r=2\sqrt{3}-3$ and the answer is $-1$.

29.We can simplify the inequality to $\frac{x^{2}+3x-7}{x+3}>3 \Longrightarrow x^{2}>16 \Longrightarrow |x|<4$. The length of this interval is $8$ and thus $40%
$ of the given interval.

32.Trial and error shows that $x_{4}=12$ works.

33.The area of the triangle (after using Heron’s Formula) is $84$. It is obvious that the minimum area of the rectangle is when the altitude and the base of the triangle are the sidelengths of the rectangle. Thus the minimum area is $84 \cdot 2=168$.

35.Notice that $CF=CE$ and are the altitudes of the equilateral triangles. Thus, $CF=CE=\sqrt{3}$. Now, $\triangle{AFE}$ is equilateral (using similar triangles), so $FE=1$. It follows that $[\triangle{CFE}=\sqrt{11}/4$ and our answer is $4$.

36.$\sqrt{2}+\sqrt{6-2\sqrt{8}}=\sqrt{2}+\sqrt{6-4\sqrt{2}}=\sqrt{2}+\sqrt{(2-\sqrt{2})^{2}}=2$

37.Using similar triangles, we see that $PC=1$ and thus both of the triangles are $30,60,90$. The area that we are looking for is $[\triangle{ACO}]-[\triangle{PBC}]-[POA]=\sqrt{3}/2-\sqrt{3}/8-1/6\pi=\frac{9\sqrt{3}-4\pi}{24}$. The answer is $5$.

39.Shoelace Theorem gives us the area as $5/8$, thus the answer is $13$.

40.Use law of cosines to find out the cosine of the angle $\alpha$. Using heron's formula, the area of the triangle is $84$. Now use sin area to find out the sin of $\alpha$. The result follows.

[hide="4"]
The $x$-coordinate of the vertex equals $-\frac b{2a}$ and since we are given that $V$ is $(3,3)$ in the question we have $-\frac b{2a}=3\Leftrightarrow b=-6a$.  Thus $p(x)=ax^{2}-6ax+c$.

Since $(3,3)$ lies on $p(x)$ we have $9a-18a+c=3\Leftrightarrow c=9a+3$ giving $p(x)=ax^{2}-6ax+9a+3$.  To find the intersection points of this with $y=x^{2}$ we set the two equal $x^{2}=ax^{2}-6ax+9a+3\Leftrightarrow (a-1)x^{2}-6ax+9a+3=0$.  We want this to have one solution, so using the disciminant we have \[36a^{2}-4(a-1)(9a+3)=0\Leftrightarrow a=-\frac{1}{2}.\] [/hide]

[hide="16"]
Basically, take the pyramid apart and put it on a plane.  In particular you want to focus on two faces containing $A$ and $B$.  When they are put on a plane, they are just two equilateral triangles that share a side.  The shortest path is just a straight line.  We can find the length of this line by drawing a triangle with sides 1/2, 1 and angle between them $120^\circ$.  Using LoC we get $AB=\frac{\sqrt{7}}2$.[/hide]

[hide="23"]
We have \[\underbrace{(x+2y+3z+5w)(x+2y+3z+5w)\cdots(x+2y+3z+5w)}_{8}\] When you multiply this out, you pick one term from each of the 8 groups and multiply all these terms together.  To get an $xy^{2}z^{3}w^{2}$ term you need one $x$, two $2y$'s, three $3z$'s, and two $5w$'s.  Each of the $xy^{2}z^{3}w^{2}$ terms will have a coefficient of $2^{2}\cdot3^{3}\cdot5^{2}=2700$ and there will be $\frac{8!}{2!3!2!}=1680$. of them.  Thus when you add them up you will have a coefficient of $1680\cdot2700$ which has a thousands digit of 6.[/hide]

[quote="joml88"]

[hide="23"]
We have \[\underbrace{(x+2y+3z+5w)(x+2y+3z+5w)\cdots(x+2y+3z+5w)}_{8}\] When you multiply this out, you pick one term from each of the 8 groups and multiply all these terms together.  To get an $xy^{2}z^{3}w^{2}$ term you need one $x$, two $2y$'s, three $3z$'s, and two $5w$'s.  Each of the $xy^{2}z^{3}w^{2}$ terms will have a coefficient of $2^{2}\cdot3^{3}\cdot5^{2}=2700$ and there will be $\frac{8!}{2!3!2!}=1680$. of them.  Thus when you add them up you will have a coefficient of $1680\cdot2700$ which has a thousands digit of 6.[/hide][/quote]

much better solution than mine...:^o

:'(

[hide="24"]
Let there be $b$ black balls and $r$ red balls.  There are $br$ ways to select a black ball and a red ball if you draw two balls (disregarding the order the are chosen in) and there are $\binom{10}{2}=45$ ways to choose 2 balls overall.  Thus the probability of choosing a black and red ball is $\frac{br}{45}$ which the problem told us equals $\frac 8{15}$.  Thus \[\frac{br}{45}=\frac 8{15}\Rightarrow br=24.\] But $b$ and $r$ must be integers since they are the number of balls.  So (b,r) can be (24,1), (12,2), (8,3), or (6,4).  There are 10 balls overall so (6,4) is the only one that works.

So then we evaluate what they ask: \begin{eqnarray*}P(\text{at least one black ball}) &=& P(\text{one black and one red bull)+P(\text{two blacks}) \\ &=& \frac 8{15}+\frac 6{10}\cdot \frac 5{9}\\ &=& \frac{13}{15}\end{eqnarray*} BUT WHAT IS THIS...FURMAN DOES NOT KNOW HOW TO REDUCE FRACTIONS. OH MY...WELL C IS CORRECT EVEN THOUGH THEY CHOSE TO NOT CANCEL THE THREE
3
3
33
3
33
3[/hide]

[hide="30"]
[quote]
...[b]What the radius[/b]....[/quote]

WHAT THE RADIUS FOOL?

Ok, so you're looking for the radius of that small sphere with same center as the cube which is tangent to all the other spheres.

Just look at one of the space diagonals of the cube.  It consists entirely in two of the big spheres and in the small sphere.  It's length is equal to twice the radius of the big spheres and the diameter of the small sphere.  The radius of the big spheres is just 1. We can also compute the space diagonal using Pythagoras to be $2\sqrt{3}$.

Thus $2+2r=2\sqrt{3}\Rightarrow r=\sqrt{3}-1$. B[/hide]


Find all possible values of $ a$ for which the flowing system has a solution in $ \mathbb{R}^3$:

$ \begin{cases} x \plus{} y \plus{} z \equal{} 2 \\
xy \plus{} yz \plus{} xz \equal{} 1 \\
xyz \equal{} a \end{cases}$

First, note that $ P(m) \equal{} (m \minus{} x)(m \minus{} y) \equal{} m^2 \plus{} (z \minus{} 2)m \plus{} (z \minus{} 1)^2$ must have two real roots, which are $ x,y$. So the discriminant must be nonnegative, or
\[ z(4 \minus{} 3z)\ge0\implies0\le z\le\frac34.\]
By symmetry, this means that $ 0\le r \equal{} x,y,z\le3/4$. Also note that
\[ r^3 \minus{} 2r^2 \plus{} r \equal{} r(1 \minus{} r)^2 \equal{} a,\]
where $ r,1 \minus{} r\ge0$. But in the range $ [0,3/4]$, $ r\ge0$, so because $ (1 \minus{} r)^2\ge0$, $ a \equal{} r(1 \minus{} r)^2\ge0$. Furthermore, by AM-GM
\[ 2 \equal{} 2r \plus{} (1 \minus{} r) \plus{} (1 \minus{} r)\ge3\sqrt [3]{2a}\implies0\le a\le\frac4{27}.\]


hi
I want to add vertical lines between some cells in a table.  I've tried to use \vline but I cannot put the line right between 2 cells even with \hfill.

To put a vertical line between columns, put a | in the column declaration.

For example, \begin{tabular}{c|c} will produce a table with 2 columns with a line between them.

ok but I don't want to have a line between two columns, I want a line between two specefic cells.

Something like this?

\[ \begin{array}{|c|c|c|} blah & \multicolumn{2}{c}{blah} \\ blah & blah & blah \end{array} \]

[code]\begin{array}{|c|c|c|}
blah & \multicolumn{2}{c}{blah} \\
blah & blah & blah
\end{array}[/code]

ok, what I exactly want to do is a block diagram.  What I though is to create a table and show only the lines surronding the cells representing my block.  So if you know a good way in LaTeX to do block diagram, tell because what I did (shame on me) is to use Word to do it and then put it in jpg and insert it in my document.  It works but I don't like it.

I'm not sure exactly what you mean by a "block diagram", but you might want to use xypic.

[url]http://www.tug.org/tex-archive/help/Catalogue/entries/xypic.html[/url]

hello...

ive got the same problem...

im looking for a command similar to \cline{a-b} but instead of spanning along columns i need it to span along ROWS

suppose i want to draw a ''star'' from squares

\begin{tabular}{c|c|c|c}
\cline{2-3}
&&&\\
\hline
\vline&&&\vline\\
\hline
&&&\\
\cline{2-3}
\end{tabular}

that looks really HORRIBLE!

below is an example of what i need

Thanks

try this:

\begin{tabular}{|c|c|c|} \cline{2-2}
\multicolumn{1}{c|}{ }  & & \multicolumn{1}{|c}{ } \\ \hline
& &  \\ \hline
\multicolumn{1}{c|}{ }  & & \multicolumn{1}{|c}{ } \\ \cline{2-2}
\end{tabular}


I was wondering what I ought to do to prepare for Computer Science A that I'm taking independent study in May.  What books, what practice, any suggestions?

I used this:
http://www.ict.org/apcs.html

And then got a review book:
Be Prepared for the AP Exam in Java (something like this)

I did AB and got a 5 so it worked out well.


Show that for any positive integer n, the sequence
2, $ 2^{2}$, $ 2^{2^{2}}$ is eventually constant in mod n.

[hide]
We have $ 2^{2^k} \equiv 2^{k} \pmod{n}$ if $ 2^k \equiv k \pmod{\phi(n)}$. It follows that the sequence becomes constant for $ \phi(n)$, then it will also do so for $ n$. But $ \phi(n) < n$, so we can perform strong induction. It is trivial for $ n \equal{} 1$, so we're done.
[/hide]

Edit: Then the Chinese Remainder Theorem can be applied  :oops:

What if n is even?
Or if $ \phi(n)$ is even, etc.

The Totient theorem still holds in the sense that sequences of the form $ (a^k)$ are still [i]eventually[/i] periodic, and one just needs to observe that the exponents grow sufficiently large for this to be the case.

I'm sorry, but I just don't understand the solution given, maybe I'm missing a step.

Could you (or someone else who understands it) repost it in detail if you have the time?

Thanks,
-diophantient

[quote="t0rajir0u"]Let me state it in more generality so it makes more sense.  The key is the following sequence of deductions:

- The sequence $ a^k$ is eventually periodic $ \bmod n$ with period $ \phi(n) < n$.
- The sequence $ a^{b^k}$ is eventually periodic $ \bmod n$ with period $ \phi(\phi(n)) < \phi(n)$.
- The sequence $ a^{b^{c^{k}}}$ is eventually periodic $ \bmod n$ with period $ \phi(\phi(\phi(n))) < \phi(\phi(n))$.

What happens after $ \phi(n) - 1$ iterations?[/quote]
that explains it, thank you very much.


the final acceptor for hydrogen ions in aerobic cellular respiration is: oxigen?
am i correct?
thx

Yes, that's correct. Oxygen receives the hydrogen ions and electrons to form water.


[quote="tdl"]Let $ a,b,c > 0$ so that $ abc < 1$. Prove that: $ a \plus{} b \plus{} c < \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}$[/quote]
If
\[ a\plus{} b \plus{} c \ge  \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}
\]
we have

\[ a\plus{} b \plus{} c \ge  \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\ge \frac{a\plus{}b\plus{}c}{\sqrt[3] {abc}}
\]
Thus $ abc \ge 1$ contradiction

$ \Rightarrow\frac{2a}{c}\plus{}\frac{c}{b}\geq 3a ,     \frac{2b}{a}\plus{}\frac{a}{c}\geq 3b,   \frac{2c}{b}\plus{}\frac{b}{a}\geq 3c \Rightarrow 3(\frac{a}{c}\plus{}\frac{b}{a}\plus{}\frac{c}{b})\geq 3(a\plus{}b\plus{}c)\Rightarrow ...$  :|

We have
$ \frac{a}{b}\plus{}\frac{a}{b}\plus{}\frac{b}{c} \ge 3\sqrt[3]{\frac{a^2}{bc}} > 3a$
Similarly, we have the same evaluations for 2 other terms.
We are done! :)

[quote=tdl]Let $ a,b,c>0$ so that $ abc<1$. Prove that: $ a\plus{}b\plus{}c<\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}$[/quote]

Let $a=\frac{kx}{y},b=\frac{ky}{z},c=\frac{kz}{x}$,where $k^3=abc<1$.then,the ineqaulity becomes:
\[\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-(a+b+c)=\sum{\frac{yz}{x^2}}-k\sum{\frac{x}{y}}\]
\[=\frac{1}{3}\sum{\frac{z(x+2y)(x-y)^2}{x^2y^2}}+\frac{(x^2z+xy^2+yz^2)(1-k)}{xyz}\ge{0}\]

[quote=dogsteven][quote=sqing]Let $ a,b,c>0$ and $ abc\leq1$. Prove that$$ a\plus{}b\plus{}c\leq\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}.$$[/quote]

Case $a+b+c\geqslant ab+bc+ca$, by Cauchy-Schwarz's, we have: $\sum \dfrac{a}{b}\geqslant \dfrac{(a+b+c)^2}{ab+bc+ca}\geqslant a+b+c$
Case $a+b+c\leqslant ab+bc+ca$, we have: $\sum \dfrac{a}{b}\geqslant \sum ab^2\geqslant \dfrac{(ab+bc+ca)^2}{a+b+c}\geqslant a+b+c$[/quote]
Very nice.


Hi. How do you square root by hand. Does anyone know a fast and efficient way of doing it? :D

I think the only way would be by memorization. Although you could just guess and check. If you were asked to find the $ \sqrt{2}$ you notice that it is in between $ \sqrt{1}, \sqrt{4}$ therefore it is in between $ 1, 2$. Then I guess you could try to find the tenths digit which is 4 and so on. By the way $ \sqrt{2} \approx 1.414$

If you are talking about perfect squares you can also try to memorize them. For example $ 1 \equal{} 1 \cdot 1$ and $ 4 \equal{} 2 \cdot 2$ and so on.

You could try the Newton-Raphson method for polynomials with the irrational roots you are seeking. It's great because it converges very fast after a while. Probably not the best method if you only want a few decimal places though.

Well, brute force kinda takes a long time, but this isn't much faster... try this anyway.

[url]http://www.homeschoolmath.net/teaching/square-root-algorithm.php[/url]

I had to read it 2 times to understand it but it makes sense.

I learned that last method in school myself over 40 years ago - it was once commonly taught. Historically, it goes back to Arab mathematicians of the middle ages and maybe to Indian mathematicians before them. It takes a little practice, but with that practice it is reasonably quick and effective, nearly as fast as long division.

I learned a nice method recently that doesn't require much guessing, but it's pretty hard to explain...

I just found a good link: http://www.jimloy.com/arith/sqrt.htm

one way would be to use linearization...but you need calculus for that... :P

@Woohoo: There's a nice way to approximate it. The derivation of this formula uses calculus, but it's very simple:

$ \sqrt{x\plus{}\Delta x}\approx \dfrac{\Delta x}{2\sqrt{x}}\plus{}\sqrt{x}$ where $ \Delta x$ is a really small amount. Plugging in x=100 and $ \Delta x\equal{}1$, we get

$ \sqrt{101}\approx \dfrac{1}{20}\plus{}10\equal{}10.05$

Checking, we get that $ \sqrt{101}\equal{}10.04987562.....$, which is really really close to $ 10.05$.

@ridgers: There's a forum for testing latex. It's called the "test" forum. It's on the bottom of the index page. Also, try putting dollar signs around your code.

Yes, local approximations are nice in calculus, but once $ \Delta x$ is large enough, the method completely breaks down.  It's only good for extremely local approximations.  I once learned a method that involved some sort of division-like technique to compute square roots by hand, but I can't recall it.

[quote="JRav"]Yes, local approximations are nice in calculus, but once $ \Delta x$ is large enough, the method completely breaks down.  It's only good for extremely local approximations.[/quote]

Can't you then factor out the radical?

$ \sqrt {20} \equal{} 2 \sqrt {5}$

Edit: Wouldn't the $ \sqrt {x}$ limit to the perfect squares that you only know?

I don't know if you understood what I said.  Local linearity in calculus is only good if the differential is very small.  For example, finding $ \sqrt{120.9}$ with calculus will give you a pretty close number.  Finding $ \sqrt{115}$ will not be very good.  And finding $ \sqrt{100}$ based of off 121 will give you a terrible result.  This fact is based on the error of the Taylor polynomial.

Kinda off for classroom math, but...what's the calculus method?

Basically, you define your problem in terms of a function.  In this case, $ f(x)\equal{}\sqrt{x}$.  Then you pick a starting point (something that you know that is close to what you are trying to find) and follow this formula: $ f(a)\approx f(x)\plus{}f'(x)dx$, where $ dx$ is the difference between what you're trying to find and your starting point and $ f'(x)$ is its derivative.

Please, no more calculus or other advanced methods.  Remember, this is the classroom math forum.  You guys can take these conversations elsewhere.

Thanks.

Sorry  :blush: . A mod can move this to Other Problem Solving Topics (yes that is a forum) if needed.


[quote="thanhnam2902"][color=darkblue]Let $ a,b,c\in [0;1]$. Find maximum of $ \boxed {A \equal{} \sum_{cyc}{\frac {a}{b \plus{} c \plus{} 1} \plus{} (1 \minus{} a)(1 \minus{} b)(1 \minus{} c)}}$[/color][/quote]

$ 0\leq a\leq b\leq c\leq 1$ ----> $ (1\minus{}a^{2})(1\minus{}b^{2})\leq 1$ ----> $ (1\minus{}a)(1\minus{}b)(1\plus{}a\plus{}b\plus{}ab)\leq 1$ ---> $ 1\geq (1\minus{}a)(1\minus{}b)(1\plus{}a\plus{}b\plus{}ab)\geq (1\minus{}a)(1\minus{}b)(1\plus{}a\plus{}b)$----> $ \frac{1}{1\plus{}a\plus{}b}\geq (1\minus{}a)(1\minus{}b)$ ----> $ \frac{1\minus{}c}{1\plus{}a\plus{}b}\geq (1\minus{}a)(1\minus{}b)(1\minus{}c)$

$ A \equal{} \sum_{cyc}{\frac {a}{b \plus{} c \plus{} 1} \plus{} (1 \minus{} a)(1 \minus{} b)(1 \minus{} c)}\equal{}\frac{a}{1\plus{}b\plus{}c}\plus{}\frac{b}{1\plus{}a\plus{}c}\plus{}\frac{c}{1\plus{}a\plus{}b}\plus{}(1 \minus{} a)(1 \minus{} b)(1 \minus{} c)\leq \frac{a}{1\plus{}a\plus{}b}\plus{}\frac{b}{1\plus{}a\plus{}b}\plus{}\frac{c}{1\plus{}a\plus{}b}\plus{}(1 \minus{} a)(1 \minus{} b)(1 \minus{} c)\equal{}\frac{a\plus{}b\plus{}c}{1\plus{}a\plus{}b}\plus{}(1 \minus{} a)(1 \minus{} b)(1 \minus{} c)\leq \frac{a\plus{}b\plus{}c}{1\plus{}a\plus{}b}\plus{}\frac{1\minus{}c}{1\plus{}a\plus{}b}\equal{}1$

$ a\equal{}b\equal{}c\equal{}1$

[quote="hien"]This problem can solve quickly by applying Cauchy' inequality.[/quote]

[color=darkblue]Hi, Hien. You are true! This is my solution apply AM-GM inequality.

Suppose $ a \equal{} max(a,b,c)$, apply AM-GM inequality: $ (b \plus{} c \plus{} 1)(1 \minus{} b)(1 \minus{} c)\le \displaystyle \left[\frac {(b \plus{} c \plus{} 1) \plus{} (1 \minus{} b) \plus{} (1 \minus{} c)}{3} \right]^3 \equal{} 1$.

$ \Longrightarrow$ $ LHS\le \frac {a}{b \plus{} c \plus{} 1} \plus{} \frac {b}{b \plus{} c \plus{} 1} \plus{} \frac {c}{b \plus{} c \plus{} 1} \plus{} \frac {1 \minus{} a}{b \plus{} c \plus{} 1} \equal{} \frac {(a \plus{} b \plus{} c) \plus{} (1 \minus{} a)}{b \plus{} c \plus{} 1} \equal{} 1$

Then OK![/color]


Today is the first day of the new year 1427. I wish to all muslim members and to everyone a very happy new year.

[quote="nirvanatear"]there are few muslim in china ...so...
yah...[/quote]
There are a lot of Muslims in China.

不仅回族, 还有维族, 克族等等等等. 少说也有1000万.

有那么多?
感觉很多都被忽视掉了.
而且有些似乎不遵守教规的.
不知情的都不知道他们是MUSLIM


Find all primes $ p$ satisfying $ (p^3)|(C^p_{2p}\minus{}2)$


$P_n(x)$ is defined on $[-1,1]$ and $P_0(x)=0,x \in [-1,1]$
also such that $P_{n+1}(x)=P_n(x)+\frac{x^2-P^2_n(x)}{2}$
Prove $P_n(x)$ converge to $|x|$ when $n \to +\infty$

[quote="alekk"]think it converges pointwise and increasingly, so uniformly ?[/quote]
but how can we prove uniformly?


Is it possible to pair off the positive integers 1,2,3,....,2007,2008 in such a manner that the sum of each pair of numbers is a different prime number :?:

yes

2009 is prime

so take pairs p(i)
p(i)={i,2009-i}

for every 1<= i <=2008

The problem says that they should sum to different prime numbers, so they cannot all sum to 2009.

sorry , missed that part
will try again later
 :|

Claim: If the positive integers up to $ 2n$ can be partitioned into pairs such that the sum of elements in each pair is a different prime, then $ n\le64$.

[hide]If the integers can be so partitioned, then there are at least $ n$ primes up to $ 4n$. We know that only $ 2n$ of the integers up to $ 4n$ are odd; of these, at most $ \frac23$ are not divisible by $ 3$ ($ \plus{} 1$ for rounding up), and of these, $ \frac45$ ($ \plus{} 1$) are not divisible by $ 5$. Repeat once more multiplying by $ \frac67$ ($ \plus{} 1$), and now add $ 3$ for $ 3$,$ 5$,$ 7$ themselves ($ 2$ is not achievable). We have at most $ \frac {96n \plus{} 582}{105}$ primes up to $ 4n$. Using this, it is easy to verify that there are less than $ n$ primes up to $ 4n$ when $ n\ge65$.[/hide]
The original question corresponds to $ n \equal{} 1004$, so it is impossible.

In fact, it is really only possible when $ \pi(4n) > n$. Exhaustive search up to $ n \equal{} 64$ implies that in fact $ n\le29$. I can find examples for $ n\le6$, but that is all.
(1,2)
(1,2), (3,4)
(1,2), (3,4), (5,6)
(1,4), (2,5), (3,8), (6,7)
(1,4), (2,5), (3,8), (6,7), (9,10)
(1,4), (2,5), (3,8), (6,7), (9,10), (11,12)


[b]Now that the program has begun, these initial times have changed and this schedule is no longer completely accurate.[/b]

Please note the change in Sunday times.

Students may participate in 2 Office Hours per week and are strongly encouraged to attend at least 1 as the problems each week are selected to provide every student with a challenge at their own level.

Week 1:
Monday, January 10 from 8:00 PM to 9:00 PM
Wednesday, January 12 from 1:00 PM to 2:00 PM
Friday, January 14 from 7:30 PM to 8:30 PM

Week 2:
Sunday, January 16 from 3:00 PM to 4:00 PM
Tuesday, January 18 from 9:00 PM to 10:00 PM
Friday, January 21 from 1:00 PM to 2:00 PM

Week 3:
Monday, January 24 from 9:00 PM to 10:00 PM
Wednesday, January 26 from 1:00 PM to 2:00 PM
Thursday, January 27 from 7:30 PM to 8:30 PM
Friday, January 28 from 7:00 PM to 8:00 PM

Week 4:
Sunday, January 30 from 3:00 PM to 4:00 PM
Tuesday, February 1 from 12:00 PM to 1:00 PM
Wednesday, February 2 from 8:00 PM to 9:00 PM
Friday, February 4 from 1:00 PM to 2:00 PM

Week 5:
Monday, February 7 from 9:00 PM to 10:00 PM
Wednesday, February 9 from 1:00 PM to 2:00 PM
Wednesday, February 9 from 10:00 PM to 11:00 PM
Friday, February 11 from 7:30 PM to 8:30 PM

Week 6:
Monday, February 14 from 12:00 to 1:00 PM
Tuesday, February 15 from 9:00 PM to 10:00 PM
Wednesday, February 16 from 8:00 PM to 9:00 PM
Friday, February 18 from 1:00 PM to 2:00 PM

Week 7:
Monday, February 21 from 9:00 PM to 10:00 PM
Wednesday, February 23 from 1:00 PM to 2:00 PM
Thursday, February 24 from 8:00 PM to 9:00 PM
Friday, February 25 from 7:00 PM to 8:00 PM

Week 8:
Sunday, February 27 from 3:00 PM to 4:00 PM
Tuesday, March 1 from 8:00 PM to 9:00 PM
Wednesday, March 2 from 10:00 PM to 11:00 PM
Friday, March 4 from 1:00 PM to 2:00 PM

Week 9:
Monday, March 7 from 9:00 PM to 10:00 PM
Wednesday, March 9 from 1:00 PM to 2:00 PM
Thursday, March 10 from 10:00 PM to 11:00 PM
Friday, March 11 from 7:30 PM to 8:30 PM

Week 10:
Tuesday, March 15 from 8:00 PM to 9:00 PM
Thursday, March 17 from 8:30 PM to 10:30 PM
Friday, March 18 from 1:00 PM to 2:00 PM

Week 11:
Sunday, March 20 from 3:00 PM to 4:00 PM
Monday, March 21 from 12:00 PM to 1:00 PM
Wednesday, March 23 from 1:00 PM to 2:00 PM
Friday, March 25 from 7:30 PM to 8:30 PM

Week 12:
Monday, March 28 from 9:00 PM to 10:00 PM
Wednesday, March 30 from 10:00 PM to 11:00 PM
Thursday, March 31 from 7:30 PM to 8:30 PM
Friday, April 1 from 1:00 PM to 2:00 PM

Week 13:
Tuesday, April 5 from 8:00 PM to 9:00 PM
Wednesday, April 6 from 1:00 PM to 2:00 PM
Thursday, April 7 from 7:30 PM to 8:30 PM
Friday, April 8 from 7:00 PM to 8:00 PM

Week 14:
Sunday, April 10 from 3:00 PM to 4:00 PM
Tuesday, April 12 from 7:30 PM to 8:30 PM
Wednesday, April 13 from 10:00 PM to 11:00 PM
Friday, April 15 from 1:00 PM to 2:00 PM

Yes.  Again, these are tentative times.  We are considering moving a couple of them a little later, and would be happy to hear feedback on this.

Those ones that fall from 1-2 PM are perfect for me.  I have lunch at school from 1:14-2:04.  Now, we'll have to see if they'll let me use a computer....

[quote="jeffrey"]Well, about half of these fall during school time for me, and I can't make some of the other ones, so it would be difficult for me.[/quote]

Exactly one per week should be during school time and this is by design.  The other two are the two that most students are expected to attend.  This one hour is for foreign students, homeschoolers, and those who can get to a computer at school.  This allows us to concentrate our time better on fewer students during the evening hours and also open up the school to more students.

Like I said before, we could add even a few more timeslots depending on how large the program grows.  So far we have many more students enrolled this semester than last semester already.  We will see.  It would however be easiest for us if students let us know what times are best for them.

[quote="jeffrey"]Well, about half of these fall during school time for me, and I can't make some of the other ones, so it would be difficult for me.[/quote]

Wow-does your school really run until 5 or 6PM?  Note that the ones from 3:30 to 4:30 are on Sundays.

We have added a few more office hours.  Any student can attend, but these hours are added primarily to make it easier for students on the West Coast who can't get home until 6 PM or later.  Hopefully this schedule is now full enough to allow each student around 30 hours of Office Hour time.


Triangles $ABC$ and $ABD$ are isosceles with $AB = AC = BD$, and $\overline{BD}$ intersects $\overline{AC}$ at $E$. If $\overline{BD} \perp \overline{AC},$ then $\angle C + \angle D$ is

$\text{(A)} \ 115^\circ \qquad \text{(B)} \ 120^\circ \qquad \text{(C)} \ 130^\circ \qquad \text{(D)} \ 135^\circ \qquad \text{(E)} \ \text{not uniquely determined}$

[hide]
ABCD forms an isoceles trapezoid, ABE is 45-45-90
helpful to draw diagram..

$\angle BAE = \angle ABE = 45\\ \angle ABC = \angle BCE = \angle BAD = \angle ADE\\ \angle DAE + \angle BAE = \angle BAD, \angle CBE + \angle ABE = \angle ABC\\ \angle DAE + 45 = \angle BAD = \angle ADE, \angle CBE + 45 = \angle ABC = \angle BCE\\ \angle DAE + \angle ADE = 90 , \angle CBE + \angle BCE = 90\\ 2\angle BCE-45=90, 2\angle ADE-45=90\\ 2(\angle BCE + \angle ADE)=270\\ \angle BCE + \angle ADE = 135\\ \angle C+\angle D=135$

[/hide]

[hide]Let $\angle B$ and $\angle C$ each be $x^\circ$ and let $\angle A$ and $\angle D$ each be $y^\circ$. Adding up the angle measures, we get $2x+2y+90^\circ=360^\circ$. Simplifying gives $x+y=135^\circ$. Thus, $\angle C+\angle D=135^\circ$. $\boxed{D}$
[/hide]


vector
linear algebra
matrix
integration
calculus
derivative
invariant

let $u$ be a continuous symetric positive operator. prove that there exist (unique?) operator $v$ such that $u=v^2$

[quote="Kent Merryfield"]What is the context for this problem? Real scalars or complex scalars? Finite dimensional (matrices) or infinite dimensional (Hilbert spaces)? Of course, the theorem should be true in all of these contexts provided:

* The operator is linear, and

** By "symmetric" you mean "Hermitian" or "self-adjoint."[/quote]
1/no asumption about the dimension
2/real vector space
3/self adjoint

A postive operator (or matrix) may have many square roots. But the "right" square root would itself be a positive operator, and there's only one of those.

Here's the finite-dimensional version; I'll let someone else post the proof for general Hilbert spaces. Suppose $A$ is a real, positive, symmetric matrix. That is, $A^T=A$ and for all $x\ne 0,\,x^TAx>0.$ Then $A$ can be orthogonally diagonalized. That is, $\exists\,Q$ such that $Q^TQ=I$ and $Q^TAQ=D$ where $D$ is diagonal. The elements on the main diagonal of $D$ are the eigenvalues of $A$, and the eigenvalues of $A$ are positive. We can then uniquely define $D^{1/2}$ to be the diagonal matrix whose nonzero entries are the positive square roots of the corresponding entries of $D$. Then $A^{1/2}=QD^{1/2}Q^T$ is the positive square root we were looking for.

Hint for the infinite dimensional case: the theorem I quoted, namely that symmetric matrices can be orthogonally diagonalized and have real eigenvalues, could be called the "spectral theorem" for matrices.

thanx Kent  :)  . I prove the statement in infinite dimensional case(I forgot to say,as you mentioned it, that the square root has to be positive symetric) with series because I don't know anything of general theorem in infinite dimensional vector space. What is the spectral theorem for infinite dimensional case that you are talking about. I'm very interested...

Let X be a Hilbert space. Let T be a linear, continuous, self-adjoint ($<x,Ty>=<Tx,y>$), positive ($<x,Tx>\geq 0$) operator, $T:X\to X$. Then there exists an unique positive ,self-adjoint operator U such that $U^2=T$. Extra: U commutes with any linear, contiuous operator that commutes with T.

With modern analysis techniques not only can you show that there exists a solution but more importantly also "construct it". In particular, lets consider the realm of differential operators and consider the following illustration:

If $f \in {\cal S}({\Bbb R}^n)$ (i.e. Schwartz) we have,
\[
{\cal F}(\partial f) := \left( \frac{1}{2\pi}\right)^{-\frac{n}{2}} \int e^{-ix\xi }
(\partial f) d\xi
\]
\[
=
\left( \frac{1}{2\pi}\right)^{-\frac{n}{2}}
[ e^{-ix\xi} f ]_{-\infty}^{\infty} +
i\xi\left( \frac{1}{2\pi}\right)^{-\frac{n}{2}}
\int e^{-ix\xi }f \, d\xi.
\]
Therefore,
\[
{\cal F}(\partial f)= i\xi {\cal F}(f).
\]
So the Fourier transform turns differentiation into multiplication by $i\xi$.

Now, using this let consider the following more generate problem (at least assuming
we are dealing with differential operators:

We consider the differential operator, $P= \sum_{\alpha} a_{\alpha }
D^{\alpha}$, where $D=i \xi \partial $ and $\alpha$ is a multi-index.  Now if
\[
Pu=f, \, \, \mbox{then } \widehat{(Pu)} = {\cal F} f,
\]
where ${\cal F}$ denotes the Fourier transform.  That is
\[
\sum_{\alpha} a_{\alpha } {\xi}^{\alpha} {\cal F}u = {\cal F}f \,,
\]
so,
\[
{\cal F}u = \frac{ {\cal F}f}{\sum_{\alpha}a_{\alpha } \xi^{\alpha}},
\]
if $\sum_{\alpha} a_{\alpha} \xi^{\alpha}\neq 0$.

Therefore,
\[
u = {\cal F}^{-1}
    \left(\frac{ {\cal F}f}{\sum_{\alpha} a_{\alpha } \xi^{\alpha}}\right),
\]
if $\sum_{\alpha} a_{\alpha} \xi^{\alpha}\neq 0$, where ${\cal F}^{-1}$ denotes the inverse Fourier transform.

Now, ${\cal F}$ and ${\cal F}^{-1}$ are isomorphism's of the Schwartz class of distributions and hence $u$ can be interpreted in this sense.

The above formulae is important because of its philosophical significance rather than as a practical tool for finding solutions to differential operators as in practise the inverse Fourier transform is very hard to calculate. But the above formulae tells us that there always exist a solution to an (elliptic) differential operator which in general (at least) we cannot find explicitly. Therefore, we will try to do the next best things which is to find some of the solutions properties. I.e. modern analysis.

The subject of studying the Fourier Transfer is known as Microlocal Analysis. For the above problem also see Lemma 11.1, of:

http://www.mth.kcl.ac.uk/%7Eysafarov/Publications/pdoinput.pdf

which constructs the global invariant symbol of the square root of the Laplace operator plus first order operator on the space a half densities. The main aim of this paper is to establish a general calculus for a wide class of operators (including differential operators) which include besides other operators the square root.


Positive real numbers $ a,b,c$ satisfy inequality $ a\plus{}b\plus{}c\le \frac{3}{2}$ . Find the smallest possible value for:
$ S\equal{}abc\plus{}\frac{1}{abc}$

Use  ineq AM-GM:
  $ abc \plus{}\frac{1}{abc}   \ geq abc \plus{} \frac{1}{64abc}  \plus{} \frac{63}{64abc}  \geq \frac{1}{4} \plus{} \frac{63}{64\frac{(a\plus{}b\plus{}c)^3}{27}}  \geq \frac{1}{4} \plus{} \frac{63}{8}  \equal{} 8\plus{}\frac{1}{8}$


let  function continious   in R   such that : $ f(f(x))  \equal{}x$ forall$ x   \in R$
prove that : exist   $ x_0$  satisfying   $ f(x_0) \equal{}x_0$

Assume that tere is no such $ x_0$, that $ f(x_0)\equal{}x_0$. Then for all $ x\in R$ either $ f(x)>x$ or $ f(x)<x$(of course, if we have $ x_1$ and $ x_2$ such that $ f(x_1)\minus{}x_1>0$ and $ f(x_2)\minus{}x_2<0$, then by intermediate value theorem exist such $ x*$, that $ f(x*)\minus{}x*\equal{}0$).
But if $ f(x)>x$ for all $ x\in R$, then $ f(f(x))>f(x)>x$, and we have a contradiction. Similarly in case if $ f(x)<x$


In triangle $ABC$, tan $A$, tan $B$, tan $C$ are all positive integers. Find $A$, $B$, $C$

Consider three integers x,y,z such as x+y+z=xyz and 1<=x<=y<=z; then x=1, y=2, z=3 (the proof is obvious if we notice that, necessarily, xy<=3).
As tan(A)+tan(B)+tan(C) = tan(A)tan(B)tan(C), itfollows that {tan(A),tan(B),tan(C)} = {1,2,3}.
Kind regards.

This is a problem from 23rd Tournament of Towns, spring-2002, seniors, difficult variant, problem № 1. Author of the problem: A. Zaslavskiy.

By the way, there was no a condition that all tan are positive.

[b]The oficial solution.[/b] It can be find here (in Russian):
http://www.turgor.ru/23/reshTG23.ps (ps file).

Let $\alpha \geq \beta \geq \gamma$ be angles. Clearly, $0 < \gamma \leq \frac{\pi}{3}$. Then $0<\tan{\gamma} \leq \sqrt{3}$. Since $\tan{\gamma} \in \mathbb{Z}$, we immediately obtain $\tan{\gamma}=1$.

Then,
\[\tan{(\alpha+\beta)}=\frac{\tan{(\alpha)}+\tan{(\beta)}}{1-\tan{(\alpha)}\tan{(\beta)}}\] for $\alpha+\beta \neq \frac{\pi}{2}$,
\[\tan{(\pi)}=\tan{(\alpha+\beta+\gamma)}=\frac{\tan{(\alpha)}+\tan{(\beta+\gamma)}}{1-\tan{(\alpha)}\tan{(\beta+\gamma)}}=\frac{\tan{(\alpha)}+\tan{(\beta)}+\tan{(\gamma)}-\tan{(\alpha)}\tan{(\beta)}\tan{(\gamma)}}{1-\tan{(\alpha)}\tan{(\beta)}-\tan{(\gamma)}\tan{(\beta)}-\tan{(\alpha)}\tan{(\gamma)}}.\]
That is $\tan{(\alpha)}+\tan{(\beta)}+\tan{(\gamma)}=\tan{(\alpha)}\tan{(\beta)}\tan{(\gamma)}.$

Since $\tan{(\gamma)}=1$, we have
\[\tan{(\alpha)}\tan{(\beta)}-\tan{(\alpha)}-\tan{(\beta)}-1=0,\]
\[(\tan{(\alpha)}-1)(\tan{(\beta)}-1)=2=2 \cdot 1.\]

It is easy to understand that $\tan{(\alpha)}$ and $\tan{(\beta)}$ can't be negative at the same time. Then $\tan{(\beta)}=2$ and $\tan{(\alpha)}=3$.

The rest is simple verifying of existance the triangle with $\tan{(\alpha)}=3$, $\tan{(\beta)}=2$ and $\tan{(\gamma)}=1$.


[color=green]In the set of $ \binom{n}{2}$ line segments to connect $ n$ points in space, what is the maximum of equal line segments?[/color]

This equivalent to the unit distance problem proposed by Erdos.
http://en.wikipedia.org/wiki/Unit_distance_graph#Counting_unit_distances

Some related problems:
http://www.zblmath.fiz-karlsruhe.de/MATH/general/erdos/cit/56851011.htm
http://www.ics.uci.edu/~eppstein/junkyard/integer-distances.html

[color=green]Oh thank you very much! And for $ n\equal{}7$ it's is a beatiful problem with a nice solution. Please post the solution here.[/color]


[b]BMO 2007:[/b]

1: [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=827187#827187[/url]

2: [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=827178#827178[/url]

3: [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=827183#827183[/url]

4: [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=827192#827192[/url]

PS: thanks to stergiu who posted the problems :)

@Valentin: could you edit stergiu's post at problem 4, by writing the description "Balkan MO 2007" beneath the title, because he probably forgot :)

[quote="pohoatza"]

@Valentin: could you edit stergiu's post at problem 4, by writing the description "Balkan MO 2007" beneath the title, because he probably forgot :)[/quote]
I edited :D

hm...oops, i posted the problems too (i think too late)
here are the links (i`m sorry for the mess-up, but i`m paying this internet-cafe and it`s not really great :) )

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=827240#827240]problem 1[/url]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=827237#827237]problem 2[/url]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=827238#827238]problem 3[/url]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=827244#827244]problem 4[/url]

to mods : delete or merge them, sorry again. :)[/url]

no, problem maky, what did you do? and what did the rest of the romanian team do?

i will post here any results i know, when they will be ready (during tommorow, i think)

do you know which countries proposed the problems?

the fourth problem is proposed by romania.

this year the problems are more difficult!

nice exam!  :)

[quote="maky"]no, the fourth is not proposed by romania
from what i`ve heard,
1st - albania
2st - bulgaria
3rd - i don`t know
4th - turkey
tommorow will begin the coordination, i will try to keep the post updated[/quote]

sorry for my mistake  :oops:

romania results
$\begin{tabular}{clcccccc}rom 1 & adrian zahariuc (dzeta) & 10 & 10 & 9 & 10 & 39 & gold \\ rom 2 & lucian turea (me) & 10 & 10 & 10 & 10 & 40 & gold \\ rom 3 & livia ilie & 9 & 5 & 9 & 0 & 23 &silver \\ rom 4 & mihai dumitrescu & 10 & 2 & 10 & 8 & 30 & silver \\ rom 5 & andrei deneanu & 10 & 2 & 0 & 1 & 13 & h.m. \\ rom 6 & nicolae papara & 10 & 10 & 3 & 1 & 24 & silver \\ \end{tabular}$
i hope i got them right  :blush:

edit : obviously i had a mistake somewhere. i corrected it now

thanks  :blush:

congratulations for your perfect score, maky!

what about the other countries?

i hope everybody will post their results.
bulgaria was first with 184, romania second with 169, moldova third with 152.
also, iurie boreico from moldova (iura on mathlinks) has 40.

Our team's results:

Alexandr Zamorzaev aka Sasha 29 silver for sure
Andrei Frimu aka Freemind 25 silver for sure
Vladimir Vanovschi (Vova) 23 silver for sure
Stefan Sanduleanu 14 probably mention, little chances for bronze
Andrei Iliasenco 11 mention for sure

Transmission ended from internet stolen from the hotel Esperides

Iura

amazing maky ... that's big progress . congrats to dzeta and iura . congrats to Bulgaria to ... Nicolae Papara has shown big progress too

Results of Serbian team:

Marko Jevremovic 31 silver
Teodor fon Burg 30 silver
Mladen Radojevic 26 silver
Dushan Milijancevic 17 bronze
Marija Jelic 15 bronze
Boban Karapetrovic 13 h.m.

Serbia finished 4th.

What are the ranks of the other countries ?

Congratulations to the Romanian Team :) (Especially maky & dzeta.)

Will all results be available soon?

[quote="e.lopes"]ohhhhh, iurie boreico got a perfect score?! i'm surprisee!  :rotfl:  :rotfl:[/quote]

Just look at iura's results:

IMO 2005: 777777
IMO 2006: 777777
Moldavian NMO 2007: 77777777
BMO 2007: 10 10 10 10

 :rotfl:

$\mathbf{The\;Macedonian\;Results}$

$\begin{tabular}{clcccccc}MKD1 & Bodan Arsovski (me) & 10 & 5 & 4 & 1 & 20 & bronze \\ MKD2 & Viktor Simjanovski & 10 & 5 & 1 & 1 & 17 & bronze \\ MKD3 & Andrej Risteski & 10 & 5 & 1 & 0 & 16 & bronze \\ MKD4 & Matej Dobrevski & 2 & 1 & 0 & 0 & 3 &-\\ MKD5 & Aleksandar Ilioski & 10 & 4 & 1 & 0 & 15 & bronze \\ MKD6 & Djoko Kostadinovski & 2 & 0 & 0 & 0 & 2 &-\\ \end{tabular}$

Does anyone know the results of the Bulgarian team by people and problems. Or any other information. I heard about 3 golds, 2 silvers and a bronze here:
http://www.math.bas.bg/nt/

I wonder if Bulgaria has a perfect score :?:

Congratulations to all participants!!!

Results of the Bulgarian team:

(BUL1)Nikolai Beluhov - 10 8 10 10 - 38 (gold)
(BUL2)Tzvetelina Tzeneva - 10 10 10 2 - 32 (silver)
(BUL3)Alexandar Daskalov - 10 4 10 2 - 26 (silver)
(BUL4)Elina Robeva - 10 4 10 9 - 33 (gold)
(BUL5)Deyan Simeonov (that's me) - 10 10 4 10 - 34 (gold)
(BUL6)Borislav Valkov - 10 3 0 8 - 21 (bronze)

Total score: 184


Let $y= x^{x}$
 then $ln \ y = x \ ln \ x$

 differentiate this and put $y = x^{x}$
 to get

$\frac{dy}{dx}= x^{x}(ln \ ex)$

Write $x^{x}=e^{lnx\cdot x}$;
$(e^{lnx\cdot x})'=e^{lnx\cdot x}\cdot(lnx\cdot x)'=x^{x}(1+lnx)$


If  :theta:  is an angle instandard position such that (sin :theta: )(cos :theta: ) < 0 and sin :theta: =1/2, which of the following could be the value of  :theta: ?
A) 30 :^o:
B) 330 :^o:
C) 420 :^o:
D) 480 :^o:
E) 510 :^o:

The answer is E but could someone show me the setup and rules they follow to get to that answer?
Thanks :)

Well, since $(\sin \theta )(\cos \theta)<0$ and $\sin \theta =1/2$, $\cos \theta <0$.  This means that the angle is in the second quadrant since sine is postive and cosine is negative.  An angle of 510 degress is equivalent to an angle of 150 degrees, and this is the only angle in the second quadrant.

[hide="spoiler answer"]since $\sin \theta$ is positive and
$\left(\sin \theta \right) \left(\cos \theta \right) < 0$,
then $\cos \theta < 0$.

That means, by the rule $\cos^2 \theta + \sin^2 \theta = 1$,
$\cos \theta = \frac{-sqrt{3}}{2}$

the angle is originally 150, but because we cannot have 150 as an answer (its not shown), then we use 150 as a reference angle to find the angle in the given choice which is equivalent of 150.

since 510 - 150 = 360, 510 is our answer. E. [/hide]

[hide="hint"]
Use the induced fomula
[hide="answer"]
$E$
[/hide]
[/hide][/hide]


Can admins please make CS a forum with subforums "TI calculator Programming" and "Computer Science?"

There already is a CS forum.  Most TI programming threads are in the Other Problem Solving Topics forum.  Not every possible topic of discussion should have its own forum.


i want to start learning physics but i dont know how to start can someone plz help me because i dont want to w8 another year or two before i finally start to learn it in school

How much math do you know?  If you know calculus, then use the OCW on MIT's site.  The introductory classes have the lectures online, and they're really good, as long as you know basic calculus.