write the equation of a line and a plane that intersect at the point (1,4,-1)

if it helps the answer:they may vary,SAMPLE answer:x+2y+5z-4=0 and [x,y,z]=[-1,3,0]+t[2,1,-1]*****just don't understand how to get the answer

Because there is no such thing as "the answer" to this question. You are to [b]choose[/b] a line and a plane such that they both pass through (1,4,-1). The choice of normal vector of the plane and of the direction vector of the line is almost arbitrary - except they should not be orthogonal to each other, or you'll have a line lying in the plane.


superior algebra
superior algebra unsolved

Calculate

\[ \sum\limits_{i \equal{} 1}^n {\frac{1}
{{2i \minus{} 1}}}
\]

help me pls
thanks

This sum is $ H_{2n \minus{} 1} \minus{} \frac{1}{2} H_{n \minus{} 1}$, where $ H_n$ is a [url=http://mathworld.wolfram.com/HarmonicNumber.html]harmonic number[/url].  There is no simple closed form.


trigonometry
inequalities
probability
function
real analysis
real analysis solved

$p,q>0$ such that $p+q=1$
Prove that there exist $c>0$ such that for any $x\in [-\pi;\pi]$
$|pe^{iqx}+qe^{-ipx}|\leq e^{-cx^{2}}$

By direct calculation,

$|pe^{iqx}+qe^{-ipx}|^2=p^2+q^2+2pq\cos x$

$=1-2pq(1-\cos x) = 1-4pq\sin^2\left(\frac x2\right)$

And yes, one can find $c$ (necessarily depending on $p$ and $q$) such that that is $\le e^{-2cx^2}$ on $[-\pi,\pi].$

Use triangle inequality, $|pe^{iqx}+qe^{-ipx}|\leq |pe^{iqx}|+|qe^{-ipx}|=p+q=1$

A possible context for this problem:

Let $X$ be a random variable which takes on the value $q$ with probability $p$ and the value $-p$ with probability $q.$ (A single Bernoulli trial shifted so as to have mean zero.)

$E(X)=0$ and $E(X^2)=\text{Var}(X)=pq.$

Then the characteristic function associated with $X$ is:

$\chi_X(t)=E(e^{itX})=pe^{iqt}+qe^{-ipt}=1-\frac{pq}2t^2+O(t^3).$

One could, with effort, bring the inequality in this problem into a connection between this and the Central Limit Theorem.

I think the Kent's solution is true.
He will use the useful  result  :sin(x)/x>=2/pi for all x belong to (0,pi/2).And then the solution is trivial:D


Lucia flips a coin three times. What is the probability that she gets
Head-Tail-Head, in that order? Express your answer as a common
fraction.


Adam, Robert, Clifton, Stephen, and Brent are the five starters on a basketball team. Two are left handed and three right handed. Two are over six feet tall and three are under six feet. Adam and Clifton are of the same handedness, whereas Stephen and Brent use different hands. Robert and Brent are of the same height range, while Clifton and Stephen are in different height ranges.

The man who plays center is over six feet and is left handed. Who is he?

My answer: [hide]Stephen.[/hide]

[hide="I agree."]
Adam and Clifton must be RH, since between Stephen and Robert there will be one LH person. Thus, Robert is LH. Same logic for height; Robert and Brent are under 6', so Adam is over 6'. The only possible choice now is $\boxed{ \text{Stephen}}$.[/hide]

[hide="I also agree"]One of Stephen and Brent is left handed.

One of Clifton and Stephen is over 6 feet tall.

Stephen is over 6 feet tall and is left handed. Then we can catagorize each person:

LH: Stephen, Robert
RH: Adam, Clifton, Brent
over 6 feet: Stephen, Adam
under 6 feet: Clifton, Brent, Robert[/hide]


LaTeX articles videos AMC AIME USA(J)MO USAMO

Due to the locking of the last thread, I decided to re-make it with less loopholes that potentially cause spamming.

[u]Rules to always be followed:[/u]
1.  Each rule must follow every previous rule and the rule itself.
2.  Each rule must be a posting restriction for all future posts.
3.  You must post all previous rules in addition to the new rule.
4.  No rule may modify the rule number.
5.  No rule may make the rules difficult to read.
6.  No rule may forbid the posting of further rules.

[u]Scoring (Just to make it more fun)[/u]
1.  Every time a player posts a rule, they gain points equal to the rule number.  In other words, you are gaining points for how many rules you are following.
2.  If a player didn't follow a previous rule, they lose any points they had gained from it and a new set of rules begins.
3.  Scores may not be posted in $ \text{\LaTeX}$.

Rule 1: No rule following this will use the letter "a" in it.

$ \fbox{1 point total}$

[quote="miyomiyo"]
2. If a player didn't follow a previous rule, they lose any points they had gained from it and a new set of rules begins.
[/quote]
Does that mean they gain zero net points or that the points they [b]would[/b] have gotten are subtracted from their score?

[quote="miyomiyo"]
[u]Rules to always be followed:[/u]
1.  Each rule must follow every previous rule and the rule itself.
2.  Each rule must be a posting restriction for all future posts.
3.  You must post all previous rules in addition to the new rule.
4.  No rule may modify the rule number.
5.  No rule may make the rules difficult to read.[/quote]I'm totally not following rules 4 or 5.  They're there for no reason at all.

Rule 1: No rule following this will use the letter "a" in it.[color=darkblue]
Rule 2: Rules must be in different colors.[/color]

${ 0 + 2 = \fbox{2 points total}}$

Here's a historical note: I got the idea for The Mole task of this from [url=http://www.mafiascum.net/forum/viewtopic.php?t=7000]here[/url], only I simplified it a lot.

Rule 1: No rule following this will use the letter "a" in it.[color=darkblue]
Rule 2: Rules must be in different colors.[/color]
[b][color=darkred]Rule 3: All other rules following this one must bold.[/color][/b]

3 points! :D

Rule 1: No rule following this will use the letter "a" in it.[color=darkblue]
Rule 2: Rules must be in different colors.[/color]
[color=darkred][b]Rule 3: All other rules following this one must bold. [/b][/color]
[color=olive][b]Rule 4: There will be no more rules after this rule.[/color][/b]
ZhangPeijin: 4
Kimmystar94:3
MustItAlwaysBeMe: 2
Distracted523:1

Totally pwnt whoever posts next.

Rule 1: No rule following this will use the letter "a" in it.[color=darkblue]
Rule 2: Rules must be in different colors.[/color]
[color=darkred][b]Rule 3: All other rules following this one must bold. [/b][/color]
[color=olive][b]Rule 4: There will be no more rules after this rule.[/color][/b]
[b][color=indigo]Rule 5: You may put an exclamation mark at the end of your rule![/color][/b]
$ \mbox{ZhangPeijin: }\fbox{4}$
$ \mbox{Kimmystar94: }\fbox{3}$
$ \mbox{MustItAlwaysBeMe: 2\plus{}5\equal{}}\fbox{7}$
$ \mbox{Distracted523: }\fbox{1}$

Bam, it's not a rule 'cause it doesn't have to be done.

Rule 1: No rule following this will use the letter "a" in it.[color=darkblue]
Rule 2: Rules must be in different colors.[/color]
[color=darkred][b]Rule 3: All other rules following this one must bold. [/b][/color]
[color=olive][b]Rule 4: There will be no more rules after this rule.[/color][/b]
[b][color=indigo]Rule 5: You may put an exclamation mark at the end of your rule![/color][/b]
[b][color=orange]Rule 6: After this rule, you may not post any more new rules![/color][/b]
ZhangPeijin: 10
Kimmystar94:3
MustItAlwaysBeMe: 7
Distracted523:1

Get that one.

Okay, I'm not gonna try again.  And from now on, use $ \text{\LaTeX}$ when you are posting everyone's scores.  ...and you copied my awesome $ \text{\LaTeX}$ signature bug. COME ON!

Erm, unless you vow not to use my awesome signature technique again. :D

Dude, how did you get 4 lines?
Anyways, My Sig just screwed up your sig's formatting
:P
Meh, I changed it because you asked me, but it's still awesome.

I think we should add a rule to prevent what ZhangPeijin is doing just to keep the game running more smoothly

[quote]Well, rules 3, 4, 5, and 6 all break rule 1.[/quote]

True. We will begin again.
I propose a new global rule:
[b]No rule may be used again once it has been stated in the thread.[/b]

Rule 1:
All rules must be written in a way
such that they rhyme, okay?

randomdragoon: 1
MustItAlwaysBeMe: 2
Distracted523:1

(Next: One about iambic pentameter!)

uhhh...rule 3 breaks 2...wolf=animal

1. All rules following this one shall contain at least one word used three times. one

1. All rules following this one shall contain at least one word used three times. one
2.If I had had had to use correct grammar, i would feel vry bad, so two gramar errors and two spelling mistakes--  AT LEEST!

[hide]
1. reciprocity of zero: 109
2. pacman2812: 106
3. randomdragoon: 101
4. lingomaniac88: 83
5. Brut3Forc3: 66
6. distracted523: 45
7. lotsofmath: 44
8. 1=2: 41
8. hm29168: 41
9. 142857: 30
9. miyomiyo: 30
11. 33286: 28
12. hadasah: 25
13. cf249: 19
14. crazychinesemaniac: 17
15. andersonw: 14
15. erabel: 14
15. kimmystar94: 14
15. Yongyi781: 14
19. Klebian: 13
20. perfect628: 10
21. modeler: 9
21. Protestant: 9
21. abacadaea: 9
24. Flame: 8
24. Pi-Dimension: 8
26. Garyzx: 7
27. bowei: 4
27. ashpotter: 4
28. mz94: 3
29. mathboyzld: 2
29. asterday1: 2
32. MustItAlwaysBeMe: 1
33. InhibitorOfNostalgia: 1
33. BOGTRO: 1
33. Ihatepie: 1(¿?)[/hide]

awww... :(

1. All rules following this one shall contain at least one word used three times. one
2. If I had had had to use correct grammar, i would feel vry bad, so two gramar errors and two spelling mistakes-- AT LEEST!
3. pacman [b]is[/b] cool, so me think thtt if thy post number [b]is[/b] either odd or even, then all its letter [b]is[/b] in lowercaps

[hide=" do not click"]1. reciprocity of zero: 109
1. pacman2812: 109
3. randomdragoon: 101
4. lingomaniac88: 83
5. Brut3Forc3: 66
6. distracted523: 45
7. lotsofmath: 44
8. 1=2: 41
8. hm29168: 41
9. 142857: 30
9. miyomiyo: 30
11. 33286: 28
12. hadasah: 25
13. cf249: 19
14. crazychinesemaniac: 17
15. andersonw: 14
15. erabel: 14
15. kimmystar94: 14
15. Yongyi781: 14
19. Klebian: 13
20. perfect628: 10
21. modeler: 9
21. Protestant: 9
21. abacadaea: 9
24. Flame: 8
24. Pi-Dimension: 8
26. Garyzx: 7
27. bowei: 4
27. ashpotter: 4
28. mz94: 3
29. mathboyzld: 2
29. asterday1: 2
32. MustItAlwaysBeMe: 1
33. InhibitorOfNostalgia: 1
33. BOGTRO: 1
33. Ihatepie: 1(¿?) [/hide]

1. All rules following this one shall contain at least one word used three times. one
2. If I had had had to use correct grammar, i would feel vry bad, so two gramar errors and two spelling mistakes-- AT LEEST!
3. pacman is cool, so me think thtt if thy post number is either odd or even, then all its letter is in lowercaps
4. eef ze perzon'z rool ees zee some off eh milteple off 7 ahnd uh milteple off 2, than fastly yu moust tipe thee rool nombur owt ahnd eet moust bee decapitalized ahnd half uh "_" aphtre eet. Plous, mee beleives thees rool es harrd two tipe

2 grammar mistakes: fastly and "me believes"

[quote="pacman2812"]rule four broke rule one (?), start over

Rule one: you must type out every numbers at the rule (including scores!). If the numbers are on a nickname, you mustn't type out them.

[hide]one. pacman2812: one hundred ten
two. reciprocity of zero: one hundred nine
three. randomdragoon: one hundred one
four. lingomaniac88: eighty-three
five. Brut3Forc3: sixty-six
six. distracted523: forty-five
seven. lotsofmath: forty-four
eight. 1=2: forty-one
eight. hm29168: forty-one
ten. 142857: thirty
ten. miyomiyo: thirty
twelve. 33286: twenty-eight
thirteen. hadasah: twenty-five
fourteen. cf249: nineteen
fifteen. crazychinesemaniac: seventeen
sixteen. andersonw: fourteen
sixteen. erabel: fourteen
sixteen. kimmystar94: fourteen
sixteen. Yongyi781: fourteen
twenty. Klebian: thirteen
twenty-one. perfect628: ten
twenty-two. modeler: nine
twenty-two. Protestant: nine
twenty-two. abacadaea: nine
twenty-five. Flame: eight
twenty-five. Pi-Dimension: eight
twenty-seven. Garyzx: seven
twenty-eight. bowei: four
twenty-eight. ashpotter: four
thirty. mz94: three
thirty-one. mathboyzld: two
thirty-one. asterday1: two
thirty-three. MustItAlwaysBeMe: one
thirty-three. InhibitorOfNostalgia: one
thirty-three. BOGTRO: one
thirty-three. Ihatepie: one [/hide][/quote]

um no it didnt i used rule (rool) 3 times

oh, sorry, I hadn't seen. Anywise, your rule is very hard to understand. I will give you back the points, sorry :(

one. pacman2812: one hundred ten
two. reciprocity of zero: one hundred nine
three. randomdragoon: one hundred one
four. lingomaniac88: eighty-three
five. Brut3Forc3: sixty-six
six. distracted523: forty-five
seven. lotsofmath: forty-four
eight. 1=2: forty-one
eight. hm29168: forty-one
ten. 142857: thirty
ten. miyomiyo: thirty
twelve. 33286: twenty-eight
thirteen. hadasah: twenty-five
fourteen. cf249: nineteen
fifteen. crazychinesemaniac: seventeen
sixteen. andersonw: fourteen
sixteen. erabel: fourteen
sixteen. kimmystar94: fourteen
sixteen. Yongyi781: fourteen
twenty. Klebian: thirteen
twenty-one. perfect628: ten
twenty-two. modeler: nine
twenty-two. Protestant: nine
twenty-two. abacadaea: nine
twenty-five. Flame: eight
twenty-five. Pi-Dimension: eight
twenty-seven. Garyzx: seven
twenty-seven. mz94: seven
twenty-nine. bowei: four
twenty-nine. ashpotter: four
thirty-one. mathboyzld: two
thirty-one. asterday1: two
thirty-three. MustItAlwaysBeMe: one
thirty-three. InhibitorOfNostalgia: one
thirty-three. BOGTRO: one
thirty-three. Ihatepie: one

Rule one: you must type out every numbers at the rule (including scores!). if the numbers are on a nickname, you mustn't type out them.
Rule two: all rules must be completely in lowercase.
[hide]one. pacman2812: one hundred ten
two. reciprocity of zero: one hundred nine
three. randomdragoon: one hundred one
four. lingomaniac88: eighty-three
five. Brut3Forc3: sixty-six
six. distracted523: forty-five
seven. lotsofmath: forty-four
eight. 1=2: forty-one
eight. hm29168: forty-one
ten. 142857: thirty
ten. miyomiyo: thirty
twelve. 33286: twenty-eight
thirteen. hadasah: twenty-five
fourteen. cf249: nineteen
fifteen. crazychinesemaniac: seventeen
sixteen. andersonw: fourteen
sixteen. erabel: fourteen
sixteen. kimmystar94: fourteen
sixteen. Yongyi781: fourteen
twenty. Klebian: thirteen
twenty-one. ProtestanT: eleven
twenty-two. perfect628: ten
twenty-three. modeler: nine
twenty-three. abacadaea: nine
twenty-five. Flame: eight
twenty-five. Pi-Dimension: eight
twenty-seven. Garyzx: seven
twenty-eight. bowei: four
twenty-eight. ashpotter: four
thirty. mz94: three
thirty-one. mathboyzld: two
thirty-one. asterday1: two
thirty-three. MustItAlwaysBeMe: one
thirty-three. InhibitorOfNostalgia: one
thirty-three. BOGTRO: one
thirty-three. Ihatepie: one[/hide]

Bleh, you capitalized the R in Rule two... :P

Rulm 1: All M's must bm rmplacmd with thm lmttmr M.

[hide]one. pacman2812: one hundred ten
two. reciprocity of zero: one hundred nine
three. randomdragoon: one hundred one
four. lingomaniac88: eighty-three
five. Brut3Forc3: sixty-six
six. distracted523: forty-five
six. lotsofmath: forty-five
eight. 1=2: forty-one
eight. hm29168: forty-one
ten. 142857: thirty
ten. miyomiyo: thirty
twelve. 33286: twenty-eight
thirteen. hadasah: twenty-five
fourteen. cf249: nineteen
fifteen. crazychinesemaniac: seventeen
sixteen. andersonw: fourteen
sixteen. erabel: fourteen
sixteen. kimmystar94: fourteen
sixteen. Yongyi781: fourteen
twenty. Klebian: thirteen
twenty-one. perfect628: ten
twenty-two. ProtestanT: nine
twenty-three. modeler: nine
twenty-three. abacadaea: nine
twenty-five. Flame: eight
twenty-five. Pi-Dimension: eight
twenty-seven. Garyzx: seven
twenty-eight. bowei: four
twenty-eight. ashpotter: four
thirty. mz94: three
thirty-one. mathboyzld: two
thirty-one. asterday1: two
thirty-three. MustItAlwaysBeMe: one
thirty-three. InhibitorOfNostalgia: one
thirty-three. BOGTRO: one
thirty-three. Ihatepie: one[/hide]

Rulm 1: All M's must bm rmplacmd with thm lmttmr M.
Rulm 2: All rulms must havm at lmast 30 lmttmrs.

[hide]one. pacman2812: one hundred ten
two. reciprocity of zero: one hundred nine
three. randomdragoon: one hundred one
four. lingomaniac88: eighty-three
five. Brut3Forc3: sixty-six
six. distracted523: forty-five
six. lotsofmath: forty-five
eight. 1=2: forty-one
eight. hm29168: forty-one
ten. 142857: thirty
ten. miyomiyo: thirty
twelve. 33286: twenty-eight
thirteen. hadasah: twenty-five
fourteen. cf249: nineteen
fifteen. crazychinesemaniac: seventeen
sixteen. andersonw: fourteen
sixteen. erabel: fourteen
sixteen. kimmystar94: fourteen
sixteen. Yongyi781: fourteen
20. Klebian: 13
21. perfect628: 10
22. ProtestanT: 9
22. modeler: 9
22. abacadaea: 9
22. Garyzx: 9
26. Flame: 8
26. Pi-Dimension: 8
28. bowei: 4
28. ashpotter: 4
thirty. mz94: three
thirty-one. mathboyzld: two
thirty-one. asterday1: two
thirty-three. MustItAlwaysBeMe: one
thirty-three. InhibitorOfNostalgia: one
thirty-three. BOGTRO: one
thirty-three. Ihatepie: one[/hide]

Rulm 1: All M's must bm rmplacmd with thm lmttmr M.
Rulm 2: All rulms must havm at lmast 30 lmttmrs.
Rulm 3: All rulms with a numbmr that bmgins with a lmttmr from A-N aftmr this onm must contain thm namm of a US city in it

[hide]one. pacman2812: one hundred ten
two. reciprocity of zero: one hundred nine
three. randomdragoon: one hundred one
four. lingomaniac88: eighty-three
five. Brut3Forc3: sixty-six
six. distracted523: forty-five
seven. lotsofmath: forty-four
eight. 1=2: forty-one
eight. hm29168: forty-one
ten. 142857: thirty
ten. miyomiyo: thirty
twelve. 33286: twenty-eight
thirteen. hadasah: twenty-five
fourteen. cf249: nineteen
fifteen. crazychinesemaniac: seventeen
sixteen. andersonw: fourteen
sixteen. erabel: fourteen
sixteen. kimmystar94: fourteen
sixteen. Yongyi781: fourteen
twenty. Klebian: thirteen
twenty-one. perfect628: ten
twenty-two. modeler: nine
twenty-two. Protestant: nine
twenty-two. abacadaea: nine
twenty-five. Flame: eight
twenty-five. Pi-Dimension: eight
twenty-five. mz94: eight
twenty-eight. Garyzx: seven
twenty-nine. bowei: four
twenty-nine. ashpotter: four
thirty-one. mathboyzld: two
thirty-one. asterday1: two
thirty-three. MustItAlwaysBeMe: one
thirty-three. InhibitorOfNostalgia: one
thirty-three. BOGTRO: one
thirty-three. Ihatepie: one
[/hide]

Rulm 1: All M's must bm rmplacmd with thm lmttmr M.
Rulm 2: All rulms must havm at lmast 30 lmttmrs.
Rulm 3: All rulms with a numbmr that bmgins with a lmttmr from A-N aftmr this onm must contain thm namm of a US city in it
rulm 4: all rulms must namm a pokmmon in it, chicago

[hide]one. pacman2812: one hundred ten
two. reciprocity of zero: one hundred nine
three. randomdragoon: one hundred one
four. lingomaniac88: eighty-three
five. Brut3Forc3: sixty-six
six. distracted523: forty-five
seven. lotsofmath: forty-four
eight. 1=2: forty-one
eight. hm29168: forty-one
ten. 142857: thirty
ten. miyomiyo: thirty
twelve. 33286: twenty-eight
thirteen. hadasah: twenty-five
fourteen. cf249: nineteen
fifteen. crazychinesemaniac: seventeen
sixteen. andersonw: fourteen
sixteen. erabel: fourteen
sixteen. kimmystar94: fourteen
sixteen. Yongyi781: fourteen
twenty. Klebian: thirteen
twenty-one. perfect628: ten
twenty-two. modeler: nine
twenty-two. Protestant: nine
twenty-two. abacadaea: nine
twenty-five. Flame: eight
twenty-five. Pi-Dimension: eight
twenty-five. mz94: eight
twenty-eight. Garyzx: seven
twenty-nine. bowei: four
twenty-nine. ashpotter: four
twenty-nine. florinaapplezhu: four
thirty-two. mathboyzld: two
thirty-two. asterday1: two
thirty-four. MustItAlwaysBeMe: one
thirty-four. InhibitorOfNostalgia: one
thirty-four. BOGTRO: one
thirty-four. Ihatepie: one
[/hide]

Rulm 1: All M's must bm rmplacmd with thm lmttmr M.
Rulm 2: All rulms must havm at lmast 30 lmttmrs.
Rulm 3: All rulms with a numbmr that bmgins with a lmttmr from A-N aftmr this onm must contain thm namm of a US city in it
rulm 4: all rulms must namm a pokmmon in it, chicago
rulm 5: all rulms must not contain  thm twmnty-fourth lmttmr of thm alphabmt in it nmw york city mmvmm
[hide][hide][code]one. pacman2812: one hundred ten
two. reciprocity of zero: one hundred nine
three. randomdragoon: one hundred one
four. lingomaniac88: eighty-three
five. Brut3Forc3: sixty-six
six. distracted523: forty-five
seven. lotsofmath: forty-four
eight. 1=2: forty-one
eight. hm29168: forty-one
ten. 142857: thirty
ten. miyomiyo: thirty
twelve. 33286: twenty-eight
thirteen. hadasah: twenty-five
fourteen. cf249: nineteen
fifteen. crazychinesemaniac: seventeen
sixteen. andersonw: fourteen
sixteen. erabel: fourteen
sixteen. kimmystar94: fourteen
sixteen. Yongyi781: fourteen
twenty. Klebian: thirteen
twenty-one. perfect628: ten
twenty-two. modeler: nine
twenty-two. Protestant: nine
twenty-two. abacadaea: nine
twenty-five. Flame: eight
twenty-five. Pi-Dimension: eight
twenty-five. mz94: eight
twenty-eight. Garyzx: seven
twenty-eight-point-five. snoozingnewt: five
twenty-nine. bowei: four
twenty-nine. ashpotter: four
twenty-nine. florinaapplezhu: four
thirty-two. mathboyzld: two
thirty-two. asterday1: two
thirty-four. MustItAlwaysBeMe: one
thirty-four. InhibitorOfNostalgia: one
thirty-four. BOGTRO: one
thirty-four. Ihatepie: one[/code]

Rulm 1: All M's must bm rmplacmd with thm lmttmr M.
Rulm 2: All rulms must havm at lmast 30 lmttmrs.
Rulm 3: All rulms with a numbmr that bmgins with a lmttmr from A-N aftmr this onm must contain thm namm of a US city in it
rulm 4: all rulms must namm a pokmmon in it, chicago
rulm 5: all rulms must not contain  thm twmnty-fourth lmttmr of thm alphabmt in it nmw york city mmvmm
rulm 6: all rulms starting from this onm must contain at most 1337 charactmrs los angmlms pikachu
[hide][hide][code]one. pacman2812: one hundred ten
two. reciprocity of zero: one hundred nine
three. randomdragoon: one hundred one
four. lingomaniac88: eighty-three
five. Brut3Forc3: sixty-six
six. distracted523: forty-five
seven. lotsofmath: forty-four
eight. 1=2: forty-one
eight. hm29168: forty-one
ten. 142857: thirty
ten. miyomiyo: thirty
twelve. 33286: twenty-eight
thirteen. hadasah: twenty-five
fourteen. cf249: nineteen
fifteen. crazychinesemaniac: seventeen
sixteen. andersonw: fourteen
sixteen. erabel: fourteen
sixteen. kimmystar94: fourteen
sixteen. Yongyi781: fourteen
twenty. Klebian: thirteen
twenty-one. perfect628: ten
twenty-two. modeler: nine
twenty-two. Protestant: nine
twenty-two. abacadaea: nine
twenty-five. Flame: eight
twenty-five. Pi-Dimension: eight
twenty-five. mz94: eight
twenty-eight. Garyzx: seven
twenty-eight-point-two. ObjectZ: six
twenty-eight-point-five. snoozingnewt: five
twenty-nine. bowei: four
twenty-nine. ashpotter: four
twenty-nine. florinaapplezhu: four
thirty-two. mathboyzld: two
thirty-two. asterday1: two
thirty-four. MustItAlwaysBeMe: one
thirty-four. InhibitorOfNostalgia: one
thirty-four. BOGTRO: one
thirty-four. Ihatepie: one[/code][/quote]


$ a, b, c, d, x$ are real numbers.
$ x$ is the greatest number such that there exist numbers $ a, b, c, d$ such that:
$ a\plus{}b\plus{}c\plus{}d\plus{}x\equal{}8$ and $ a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}d^{2}\plus{}x^{2}\equal{}16$

Find $ 60x\equiv ? (mod 11)$ - What is the $ ?$  ?  Remember that $ ?$ is of course integer.

[hide]By Cauchy,

$ (a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}d^{2})(1^{2}\plus{}1^{2}\plus{}1^{2}\plus{}1^{2})\ge (a\plus{}b\plus{}c\plus{}d)^{2}$
$ 4(16\minus{}x^{2})\ge (8\minus{}x)^{2}$
$ 0\ge 5x^{2}\minus{}16x$
$ 0\ge x(5x\minus{}16)$.

Therefore, $ x\le 16/5$.  This is satisfied when $ a\equal{}b\equal{}c\equal{}d\equal{}6/5$.

Not sure what the mods have to do with this...[/hide]

The problem doesn't really make sense unless $ x$ is rational.  What are you trying to ask?


A man lives in the tenth floor of a building. Each day he takes the elevator to the first floor to go to his job. When he returns he goes to the seventh floor in the elevator, and from there he goes on the stairs the other three floors. Why does he do that?

You can ask for yes/no hints only! Good luck!

Does he take the elevator to the seventh floor and then [i]immediately[/i] take the stairs up the remaining three flights, or does he linger on the seventh floor?

Also, is the answer dependent in any way on his profession?

[quote="JBL"]Does he take the elevator to the seventh floor and then [i]immediately[/i] take the stairs up the remaining three flights, or does he linger on the seventh floor?[/quote]Yes and No. That means, he does not linger on the seventh floor

[quote="JBL"]Also, is the answer dependent in any way on his profession?[/quote]No, the answer is not dependent on his profession.

[quote="Bictor717"]I think I've heard this before, so I'll hide my question.
[hide]Does he do that on rainy days, when he has a nice long umbrella?[/hide] :D[/quote]No, those days he doesn't! I suppose you really have heard this one before, it is a direct question!

[quote="Bictor717"]I think I've heard this before, so I'll hide my question.
[hide]Does he do that on rainy days, when he has a nice long umbrella?[/hide] :D[/quote]

I've heard it, too. Very amusing clue! :P

[quote="xxreddevilzxx"]i dont want to give it all away, so i will put my partial answer
[hide]
he's too short... a midget, i suppose.
[/hide]

btw, i already knew this as well. howver, i could not remember it back until bictor reminded me :D[/quote]I don't know how partial is that, but it is correct, all what it needs is the explanation of why it is correct!

[quote="Shogia"][hide="solution"]Because he cant reach the button for the tenth floor without using the umbrella he takes with him on rainy days.[/hide][/quote]Correct!

that's clever! :wallbash_red:  :stretcher:


Express each sum or difference as a product of sines and cosines.

(1) sin(4x) + sin(2x)

(2) sin(x/2) - sin(3x/2)

[quote="Carcul"]Again? See my post [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=186099]here[/url].[/quote]

There is a significant difference in each of the problems that Sharkman has posted.

For (1), $ \sin(4x) \plus{} \sin(2x) \equal{} \sin(3x \plus{} x) \plus{} \sin(3x \minus{} x) \equal{} 2 \sin(x) \sin(3x)$, according to my post.

For (2), $ \sin\left(\frac{x}{2}\right) \minus{} \sin\left(\frac{3x}{2}\right) \equal{} \sin\left(\frac{x}{2}\right) \plus{} \sin\left(\minus{} \frac{3x}{2}\right) \equal{} \sin\left(\minus{} \frac{x}{2} \plus{} x\right) \plus{} \sin\left(\minus{} \frac{x}{2} \minus{} x\right)$

$ \equal{} 2 \sin\left(\minus{} \frac{x}{2}\right) \cos x \equal{} \minus{}2 \sin\left(\frac{x}{2}\right) \cos x$,

according to my post.


Just wondering, is there a reason the old F&G polls aren't being filtered out after the 7-day rule?

Also, are locked threads able to somehow escape the filtering?

Can't a moderator just filter out the old threads by hand? It does look as if the polls enjoy perpetual life, but they can still be deleted by a forum moderator, can't they?

Yes, i could manually delete them...

I was just wondering if there was a reason they're not being filtered out by auto-filtering...

And also, I realli dont want to mess with altering post count when possible...

Agreed Mith.  I have noticed it too but wasn't sure about it.


geometry
3D geometry
tetrahedron
octahedron

Two regular tetrahedrons $A$ and $B$ are made with the 8 vertices of a unit cube. (this way is unique)

What's the volume of $A\cup B$?

Label the regular tetrahedrons $T_1, T_2$ insdead of $A, B$, so that we can use $A, B$ to label points. Label A, B, C, D counter-clockwise the vertices of the bottom face of the cube and E, F, G, H counter-clockwise the vertices of the top face of the cube, so that ABFE (counter-clockwise) is the fromt face of the cube. Let $T_1 \equiv ACFH,\ T_2 \equiv BDEG$ be the 2 regular tetrahedrons. Let K, L, M, N, P, Q be the centers (i.e., diagonal intersections) of the cube faces ABFE (front), BCGF (right), CDHG (back), DAEH (left), ABCD (bottom), EFGH (top). Denote a = AB = ... = 1 the side length of the unit cube ABCDEFGH. Consider, for example, the cube with the body diagonal EO and side lengths $\frac a 2$. Let R, S, T be the midpoints of the cube edges AE, EF, HE, respectively. The vertices of the smaller cube are then RKONESQT. Since K, Q, N are the midponts of the edges EB, ED, EG of the tetrahedron $T_2 \equiv BDEG$, the plane KNQ is the midplane of this tetrahedron. Thus KNEQ is also a regular tetrahedron with 4 vertices common with the smaller cube RKONESQT. But the plane $KNQ \equiv AHF$ is the plane of one face of the regular tetrahedron $T_1 \equiv ACFH$, because the points K, Q, N are also the midpoints of its edges AF, AH, FH. Consequently, the common volume of the regular tetrahedrons  $T_1,\ T_2$ inside the smaller cube RKONESQT is the (non-regular) tetradedron KNQO. Similarly, by considering 7 cubes with the body diagonals AO, BO, CO, DO, FO, GO, HO and side lengths $\frac a 2$, we see that the common volume of the tetrahedrons $T_1, T_2$ is the regular octahedron KLMNPQ. The volume of this octahedron is

$V(T_1 \cap T_2) = V(KLMNPQ) = \frac{OP + OQ}{3} \cdot S(KLMN) =$

$= \frac{PQ}{3} \cdot KL^2 = \frac a 3 \left(\frac{a \sqrt 2}{2}\right)^2 = \frac{a^3}{6}$

The side length of the regular tetrahedron $T_1$ is

$d = AC = AB \sqrt 2 = a \sqrt 2$

its height

$h = \sqrt{d^2 - \left(\frac{d \sqrt 3}{3}\right)^2} = \sqrt{2a^2 - \frac{2a^2}{3}} = \frac{2a}{\sqrt 3}$

and its volume

$V(T_1) = \frac h 3 S(ACF) = \frac h 3 \cdot \frac{d^2 \sqrt 3}{4} = \frac{2a}{3 \sqrt 3} \cdot \frac{2a^2 \sqrt 3}{4} = \frac{a^3}{3}$

The volume of the tetradedron union is then

$V(T_1 \cup T_2) = V(T_1) + V(T_2) - V(T_1 \cap T_2) = 2V(T_1) - V(T_1 \cap T_2) =$

$= \frac{2a^3}{3} - \frac{a^2}{6} = \frac{a^2}{2}$

and since a = 1, $V(T_1 \cup T_2) = \frac 1 2$.


Ring Theory
superior algebra
superior algebra unsolved

Show that if $ R$ be a $ \mathbb{Z}$-graded ring with no nonzero homogenous prime ideal,then $ R_0$ is a field and $ R \equal{} R_0$ or $ R \equal{} R_0[x,x^{ \minus{} 1}]$.

This is an exercise in Charpter 2 of Eisenbud's Commutative Algebra with a View towards Algebraic Geometry.
No one post a proof,And I think I find it myself,please point out my mistake if there is,thank you.

Lemma:If $ I$ is a homogenous ideal of $ R$,then the maximal ideal contains $ I$ which is not $ R$ is a prime ideal.(Note that such a prime ideal may be not maximal.For example,the zero ideal in $ k[x,x^{ \minus{} 1}]$,it is the maximal homogenous ideal but it is not a maximal ideal of $ R$)

Proof of the Lemma:If $ P$ is such a ideal,to verify $ P$ is a prime ideal,it is only to prove if $ fg \in P$,then:$ f\in R$ or $ g \in R$.
Suppose not,so there exist $ f \notin P,g \notin P$,but $ fg \in P$,hence consider the homogenous ideal $ (f,P)$ and $ (g,P)$,by the maximum of $ P$,we must have $ (f,P) \equal{} (g,P) \equal{} 1$,hence $ (P,f)(P,g) \subset (P,fg) \equal{} P$,thus $ P \equal{} R$,contridiction.We prove the lemma.

For the original problem,from the lemma we know that $ R_0$ doesn't have a nontrivial ideal,because if $ I_0$ is a nontrivial ideal of $ R_0$,then $ \oplus_{i \equal{} \minus{} \infty}^{ \plus{} \infty} I_0R_i$ is a nontrivial homogenous ideal of $ R$.Then we can find a nonzero homogenous prime ideal by the lemma.Thus we proved that $ R_0$ is a field.

Following we will show $ R \equal{} R_0$ or $ R \equal{} R_0[x,x^{ \minus{} 1}]$.The first case obviously happens when $ R_i \equal{} 0,i \ne 0$.If not,then is some $ i \ne 0$ such that $ R_i \ne 0$,we may assume such $ i > 0$,first we can see from the lemma that every nonzero element in $ R_i$ is invertible because every nonzero homogenous ideal can generate an nonzero homogenous ideal.Then assume $ i$ is the minimal positive integer number such that $ R_i \ne 0$,then choose any nonzero element $ x \in R_i$,then $ x$ is invertible,for proving $ R \equal{} R_0[x,x^{ \minus{} 1}]$ we need to show $ R \equal{} \oplus_{k \equal{} \minus{} \infty}^{ \plus{} \infty} R_{ik}$ and $ R_{ik} \equal{} R_0x^k$,this is not hard and left for you.(By the definition of $ \mathbb{Z}$-graded ring.)


the answer was 6 I believe....  A friend told me you can prove it using combinatorics. I just brute forced it by using two sheets of paper to draw 4 lines.  :P

well, let's see:
[hide] 3 lines intersect at 3
4 at 6
5 at 10
6 at 15[/hide]

So let's see if we can find a pattern:)
Note: this was done by brute force, so numbers should be checked.

[quote="raccoon"]
Note: this was done by brute force, so numbers should be checked.[/quote]

Yep they should :) Try the 5 and 6 cases again.

Oops.  :blush:  Okay, this time I think it really is right!  I hope...

Now we're talking :)

I always have trouble with problems like this because after I draw a few lines, they begin meeting off the page and so on -- makes it more difficult :)

i just didn't draw my lines far enough.  I had enough room.

Yeah, that would be a problem as well, having the intersection points so close together that you can't count them properly.

Now, do you have any ideas for how to solve it for n lines?

[hide]
$\frac{n(n-1)}{2}$

1 line will have zero intersections
2 lines will have 1 intersection
3 lines will have 3 intersections
4 lines will have 6 intersections
...

Each time you add a line you add n intersections where n is the number of lines already there because that new line will cross all of the other lines in some arrangement at most once.
[/hide]

Here's an idea.  Start with $n$ parallel lines.  Now turn the 2nd one through a small angle, the 3rd one through a slightly larger angle, and so on (so that no three lines intersect at a point).  You notice that each turning adds $n$ intersections to the total, and we are doing this to $n-1$ of the lines

So the total is $n(n-1)$.  Are we done?

Well, no, not really.

We've actually counted each intersection twice - once when the first line turns into it, and once when we rotate the second line again.

So the total number is $\frac{n(n-1)}{2}$

Part "d" of the same problem (from some math contest)  (The above was part 'a"  followed by harder  'parts' 'b', and 'c' etc..)

If there are n points on a circle, and  all points are connected, what is the  number (maximum possible)  regions?

(When n=1, it is one. when n=2, it's 2, when n=3  it's 4  ( Inside of a triangle plus  3 crescents) )  etc..)

Heh, heh..

By the way, another way of doing the first problem:
[hide]
Every pair of lines intersects (at most) 1 time, right? How many pairs of lines are there?
[/hide]

[quote="hello"]Sorry for my bad math skills. But how does the answer follow from [quote]Every pair of lines intersects (at most) 1 time, right? How many pairs of lines are there? [/quote][/quote]
From n lines, you have nC2 = n(n-1)/2 pairs of lines... (choosing 2 in a group of n)
so n(n-1)/2 pairs that intersect at most 1 time.... so it follows that n(n-1)/2.

As n = 0, 1 , 2 , 3 , 4
f(n) = 0 , 0, 1 , 3 , 6

So the row of differences is 0,0,1

Following Newton's formula the generalized formula is:

0(n nCr 0) + 0(n nCr 1) + 1(n nCr 2) or

n(n-1)/2

[quote="Euler1001"]Part d is similar.

As n = 0 , 1 , 2 , 3 , 4
f(n) = 1 , 1, 2 , 4 ,

Then the row of differences is 1,0,1.
Using newtons formula:

1(n nCr 0) + 0(n nCr 1) + 1(n nCr 2):

(n 2 - n + 2) / 2[/quote]

I'm not sure what formula you are using there, but its not right. You haven't even used anything related to the problem - theres infinitely many sequences starting with 1, 2, 4... (or 1, 1, 2, 4 which you used).

You're right.  I was doing some bad math last night.
Had I extended the problem out past n = 3, it would have been obvious my wasn't going to work, but I didn't.

The formula I used can be seen here:

[url]http://mathworld.wolfram.com/FiniteDifference.html[/url]

[quote=jli][quote="hello"]What is the maximum number of points in which 4 lines cross?

Can this problem be generalized to n lines?[/quote]

[hide="solution"]the answer is 6

the equation is $\frac{n\cdot (n-1)}{2}$[/hide][/quote]

There is a formula, but proving it is hard.

[size=50][color=#f00]Sorry for bumping, the contest is coming up...[/color][/size]

12+ year bump is not necessary, just make a new thread

I'd like to know how you found this thread anyway

Also a line will intersect n lines in n spots at maximum. So each n line would intersect the other n-1 lines once, but since we double counted, divide by 2.


Triangle with vertices the mid-points of AB, BC, CA.

[quote="Cauchy_Schwartz"]Imi puteti spune va rog care este trinughiul medial asociat triunghiului ABC?[/quote]
     De fapt, se zice triunghi <median>, si este triunghiul format de liniile mijlocii ale triunghiului.Pe baza acestei probleme, demonstreza ca: fiind dat un triunghi ABC, demonstreza ca exista un triunghi inscris in ABC astfel incat sa avem relatia- S(triunghi inscris)=1/2^4012 S(ABC), unde S(XYZ) este aria triunghiului XYZ.


The measure of $ \angle\theta$ is $ 41^\circ$. Let $ \alpha$ be the supplement of $ \angle\theta$ and let $ \beta$ be the complement of $ \angle\theta$. What is the degree measure of $ \angle(\alpha\plus{}\beta)$?


A natural number $ n$ is called 3-partite if the set of its divisors can be partitioned into three subsets with the same sum of elements.
(a) Find at least one 3-partite number.
(b) Show that there exist infinitely many 3-parite numbers.

[quote="lambruscokid"]For part a) see http://www.mathlinks.ro/Forum/viewtopic.php?p=926647#926647

This shows us that $ 120$ is tripartite and also that it is minimal[/quote]
Part a> I had have solution.
Can anyone post solution problem part b?


You probably should hide your answers. .  .

special thanks to mathprincess09 and DarkKnight for showing their works! :)

[hide="Click to reveal hidden content...if you dare!"]1/2 is equivalent to 3/6, so 1/2+1/6=4/6=2/3. From here, 3*2/3 is obviously 2 and 2+2 is 4.[/hide]

thx 4 writin my whole SN
mostly no1 does tat online
they just say PRINCESS

[hide="To mathprincess09.  Off-topic for others."] [quote="mathprincess09"]Thanks for writing my whole screen name.
Mostly, nobody does that online.
They just say "princess".[/quote]
Since you bring it up, mathprincess09, most people on the Art of Problem Solving forums speak English rather than Chatspeak.  On these forums, you are quite likely to be referred to by your full login name.

"Chatspeak" is discouraged on the AoPS forums.  To quote administrator Valentin, from another forum:

[quote="Valentin Vornicu"][Chatspeak is] shortening words, eliminating vowels, making the text very hard to read for most people that don't use chatspeak. Something we deeply despise in Romania, and it is banned on all (most) Romania forums. :)[/quote]
I would strongly encourage you to avoid chatspeak on these forums.  You have posted many interesting comments, but I am disinclined to respond to them, because I know I will have to try to decipher a response in "chatspeak".  I suspect other people feel the same way.  Speaking English will elicit more responses to your thoughtful questions and responses.  Thank you.[/hide]

oh i am really sorry about that
you see i was chatting with
a lot of friends so i started
using chat language by mistake
i m really sorry :D

lol!(laugh out loud) atleast i m nicer than treething that i spelled it all out!  :D  :P


If Frederick A. Notacounter has n nickel(s) and d dime(s) worth a total of $5.50.  He has at least one of each coin.

(a) What is the least possible value of n+d?
(b) What is the greatest possible value of n+d?
(c) What is the total possible number of coins (n+d) that Fredercik A. Notacounter could have?

Can I answer this? I don't know how I can show my reasoning mathematically as I am still learning but you can highlight for my solutions.

[hide]

a) He has 56 total coins. 54 dimes totalling .10 he needs. So, 54 dimes + 2 nickels = 5.40 and he will need one more dime. So 1 dime brings the total to 109 and the 5.50.

c) The last question seems sort of vague to me, but I think that part b also serves to answer this question? If that is the case, then 109. I am prolly wrong on this though. EDIT: I realize what the last question asks for, give me a minute and I will solve it (or so I hope) Ok the total number of possibilities is 54 combinations. [/hide]

Does anyone want to tell me I'm right :D  or wrong  :( ?? Hehe, I keep checking to see if my solution is correct but noone is responding!

I think that is right.  For the last one, I checked by doing
109 - 54 + 1 = 56 (see the "quarters and dimes" problem in the "looking for answers" forum for a sort of strategy for checking this).

Mathfanatic, how did you know for sure that each and every one of those values that you tallied up worked?

what about me? :D  Is my answer right or wrong? I figured it was inclusive, so I added one but I am new to alot of this stuff so maybe I am wrong, I really want to know though!

Ripper1,

Yes, you were right on (a) and (b).

Mathfanatic,

I didn't notice before, but you made a slight miscalculation on the total number of solutions.  You have the right idea, just look back at the numbers you used as the boundaries.

Well, a and b were the easy ones.  It is c I'd like to hear is correct.  I used the same logic as Mathfanatic except I did 109-56+1 = 54.  So I really don't know what else someone can do. EDIT: Heh, did you read my edit on my solution, it gives the answer to c.

Your answer to (c) is correct, but I'll put the same question to you:

How do you know that all of the numbers from 56 to 109 work?

Well, i tried the first couple of values
from 108 to the 90's.  It worked so I assumed the pattern stayed true.

I think I know why it works.

Start with 54 dimes, and 2 nickels.  To keep the amount the same, remove a dime, and add two nickels.  Therefore the number of coins rises by one (2-1).  You can repeat that until you have 1 dime, and 108 nickels.  As each increase adds one more coin, you will get all numbers inbetween 56 and 109.

well of course, hehe i thought you meant you wanted to see a real algebraic proof or something.  That is how I figured out the 54..Oh well.

*smacks head on table*

I read the wrong numbers.  Yes, ripper1 is right, I just was thinking his answer to #1 was 54, and his answer to #3 was 56.  Just switch the numbers.

I feel stupid... grr....

hehe don't. I told Richard Rusczyk that log(base2)4 = log(base4)2, thinking for a second that log(base2)4 = 2 :^4:  and log(base4)2 = 4 :^2:. We all make silly mistakes.

cleanest method i think:

[hide]
say n nickels, d dimes.

5n + 10d = 550

n must be even or else the RHS is impossible.  say n = 2k.

10k + 10d = 550
k + d = 55, and any value of k and d satisfies the problem statement because we made the n=2k substitution.  now we can answer the questinos trivially

a) k=1, d=54.  ->> 1*2 + 54 = 56.
b) k=54, d=1 ->> 54*2 + 1 = 109.
c) k can vary from 1 to 54, so 54 possibilities.[/hide]


function
number theory
prime numbers
number theory unsolved

Consider the set $M_{n}$ of all numbers of the form $\{\frac{r_{1}}{p_{1}}+\frac{r_{2}}{p_{2}}+...+\frac{r_{n}}{p_{n}}\}\cdot p_{1}\cdot ...\cdot p_{n}$ , where $n>1,p_{1}=2,p_{2}=3,p_{3}=5,...$ is the sequence of prime numbers ,$1\leq r_{k}\leq p_{k},1\leq k\leq n$ and $\{x\}$ denotes fractional part of $x$. Prove that two smallest elements of $M_{n}$ are $1$ and $p_{n+1}$.

Almost obvious, $\{\frac{r_{1}}{p_{1}}+...+\frac{r_{n}}{p_{n}}\}=\{ \frac{x}{p_{1}...p_{n}}\}$ where $x$ goes through all numbers coprime to $p_{1}...p_{n}$. Hence $x=1$ and $x=p_{n+1}$ are the two smallest elements of $M_{n}$.

[quote="Megus"]$\{\frac{r_{1}}{p_{1}}+...+\frac{r_{n}}{p_{n}}\}=\{ \frac{x}{p_{1}...p_{n}}\}$ where $x$ goes through all numbers coprime to $p_{1}...p_{n}$.[/quote]
Why?  :maybe:


geometry
geometric transformation
rotation
conics
parabola
calculus
trigonometry

"sketch via rotation in standard form 9x^2+24xy+16y^2+80x-60y=0"

That's all the information that I had so I don't know exactly what axis it is supposed to be rotated around. I graphed the equation using the Wolfram Alpha web site and it's a parabola with directerix y = 4x/3 - 5/3 meaning that the parabola's axis is not parallel to the x, y or z axes. The surfaces I studied in calculus II and III were all derived from curves or conic sections that had axes parallel to either the x, y or z axes so I am at  a complete loss as to how to solve the problem. Does anyone have any ideas? Is it possible that it requires solving for y and then rotating that curve around the x-axis?

the $ xy$ term is responsible for the rotation. see the link http://www.mecca.org/~halfacre/MATH/rotation.htm

I think you are dealing with curves on the xy plane and not surfaces. (If its a surface, its parallel to the z axis). The curve $ f(x,y)\equal{}9x^2 \plus{}24xy \plus{} 16y^2 \plus{} 80 x \minus{}60y\equal{}0$ obviously passes through the origin. Suppose we rotate the xy system about the z axis by an angle $ \theta$. The equation of the curve in the rotated system is obtained by the transformation
$ x\mapsto x\cos\theta\minus{}y\sin\theta$ and $ y\mapsto x\sin\theta \plus{} y\cos\theta$
which transforms
$ f(x,y)\equal{}0$ to the equation $ f(x\cos\theta\minus{}y\sin\theta, \  x\sin\theta \plus{} y\cos\theta)\equal{}0$
Select $ \theta$ so that the term containing the product $ xy$ vanishes. Doing so, we obtain
$ \tan2\theta \equal{} \dfrac{24}{9\minus{}16}\equal{}\minus{}\dfrac{24}{7}$
Setting in the original equation and simplifying the equation (which requires a lot of patience) one obtains the curve as
$ 25(4x\plus{}y^2)\equal{}0$ i.e. $ y^2\equal{}\minus{}4x$

Thanks for the responses. The solution uses some (linear algebra type) transformations, right? Would this type of problem normally be found in a calculus II or III course? The questions are supposed to be limited to first year calculus problems and I don't recall these types of problems from first years calculus.


tuple
function
combinatorics
Subsets
IMO Shortlist

Let $n \geq 2, n \in \mathbb{N}$ and $A_0 = (a_{01},a_{02}, \ldots, a_{0n})$ be any $n-$tuple of natural numbers, such that $0 \leq a_{0i} \leq i-1,$ for $i = 1, \ldots, n.$
$n-$tuples $A_1= (a_{11},a_{12}, \ldots, a_{1n}), A_2 = (a_{21},a_{22}, \ldots, a_{2n}), \ldots$ are defined by: $a_{i+1,j} = Card \{a_{i,l}| 1 \leq l \leq j-1, a_{i,l} \geq a_{i,j}\},$ for $i \in \mathbb{N}$  and $j = 1, \ldots, n.$ Prove that there exists $k \in \mathbb{N},$  such that $A_{k+2} = A_{k}.$

Proof:
Induction with respect to $n$.
First, let us denote $A_i^{(m)} = (a_{i1}, a_{i2},\dots,a_{im}),\, m \le n $. Then $a_{i+1,j}$ depends only on $A_i^{(j)}$.
Suppose that the assertion is true for $n \le N$. Let $n=N+1$. Because $a_{i+1,j}$ depends only on $A_i^{(j)}, \, j=1,2,\dots,N $, then there exist $k$ such that

$A_{k+2}^{(N)}= A_{k}^{(N)}$.

Then $A_{k+3}^{(N)}= A_{k+1}^{(N)}$.

Denote $A^{(N)}=A_{k}^{(N)}, \, B^{(N)}=A_{k+1}^{(N)}$.
It holds:

$A^{(N)}=A_{k+2l}^{(N)}, \, B^{(N)}=A_{k+2l+1}^{(N)}, \, l=0,1,\dots$

Let $ f(A_s^{(N)}, a_{s,N+1})= Card \{a_{s,l}|a_{s,l} \ge a_{s,N+1},\, 1\le l \le N\}$. Notice that if $X^{(N)}$ is fixed, $f(X^{(N)}, x)$ is decreasing as a function of $x$.
Let $x=a_{k,N+1}$.

$ a_{k+2, N+1}(x)=\phi(A_k^{(N)}, x) = f(B_k^{(N)},  f(A_k^{(N)}, x)   ) $.

It is easy to see that $\phi$ is increasing as a function of $x$, because it is a superposition of two decreasing functions.
Let assume that $x \ge \phi(x)$. It follows $ x \ge \phi(x) \ge \phi^2(x) \ge \dots $
Because $\phi$ is bounded it will stabilize at the end. The case $x < \phi(x)$ is similar.

Thus, there exists $m$, so that $ \phi^m(x) = \phi^{m+1}(x) \Rightarrow a_{k+2m, N+1}=a_{k+2m+2, N+1}$ or for $ k_1= k+2m,\,\Rightarrow \, A_{k_1+2}= A_{k_1}$.


It says somewhere that your actual AMC score or some function of your USAMTS score, whichever is higher, counts as your AMC score. What function is used on your USAMTS score to determine what it counts as for the AMC?

Ah any USAMTS score above the USAMTS cutoff score for AIME is = 100 on AMC12.

Strange ... can you give a link as to where it officially says that?

There was a similar discussion about this but what beta said is true. You might think this is a "strange" since some people believe that USAMTS is much harder than AMC-12 (which is true on most problems but #24-25 are sometimes pretty tough) but you have to remember several things:

a) In USAMTS, you get five problems for a month. This is a fairly good amount of time to do five problems thoroughly if you can spend time as much as you need (sometimes you can't due to school but yeah). But in AMC-12, you only get 75 minutes to do the contest.
b) In USAMTS, there are "partial" credits given to the people who at least tried. In AMC-12, as you know, if you can't find the answer and guessed wrong, you don't get any points and it will hurt your score (don't worry.. You won't get negative scores like some people did in the past  :D ) so USAMTS is a lot better in that angle.
c) Note that you can use references in USAMTS (no other people though!). But in AMC-12, only some things are allowed (read the packet for this).

So, overall, USAMTS giving you AIME qualification as AMC 12 score of 10 is a fairly good deal. And just to make it easier, just do the AMC. It's [i]much[/i] easier.

I got a 24 on Round 1. If I get 24 or 25 on Rounds 2 and 3, will that be enough? How hard is it, really? Isn't the cutoff 68/75?

If you get a 24/25 on Round 1, I don't think you have to worry about passing the AMC (I got a 24 on Round 1, and I got a 132.5 on the AMC last year).

The one thing that makes me suspicious is this:

(Question 21 of USAMTS FAQ):
If I am invited to participate in the AIME by virtue of my performance in the USAMTS, how is my combined score calculated to qualify for the USAMO (the third step to qualifying for the USA Math Olympiad team)? Traditionally, a formula of my AMC score (the traditional first step) and my AIME score is used.
If you qualify via the USAMTS, your USAMO index is calculated based on a formula of your USAMTS score and AIME score. If you qualify via both the USAMTS and AMC, your USAMO index is [i]whichever is greater[/i] by the two ways of calculating it. (Thus it is to your advantage to qualify via both ways.)

If you qualify via both USAMTS and AMC, assuming as you said that a perfect score on USAMTS counts as 100 on the AMC, then how can your USAMTS score possibly be dominant?

The only possibility I see is this: you get less than 100 on the AMC, but you get counted as 100 for USAMTS, and then you qualify for the USAMO. In other words, you fail the AMC but you get a ridiculously good score on the AIME (it has to make up for your bad AMC score), thereby qualifying for USAMO. And that's just plain ridiculous.

I think they are encouraging people to take the AMC rather than USAMTS(in terms of qualification for AIME/USAMO).

Because it does count if you completely bomb the AMC (or if you miss both opportunities to take the AMC). That probably won't happen, but it's technically possible, therefore the organizers have to know what to do in case it does happen. If you say the USAMTS score does not count, then you're essentially saying you can't get to the USAMO if you qualify for the AIME by USAMTS alone. And that doesn't make sense.

[quote="Philip_Leszczynski"]I understand, but what really bugs me is why did they answer the question like that? Wouldn't it be easier to just say that the USAMTS score [i]does not count[/i] for the USAMO index?[/quote]

It would be easier, but not entirely acurate.  To qualify for the AIME through the AMC 12, you have to get a 100 OR score in the top 5 percent.  Recently, the top 5 percentile is easily over 100 but it may happen that this year there is a killer AMC and getting a 95 qualifies for the AIME.  If you just so happened to get a 95 on the AMC and a 100 on the USAMTS you would want that 100 to be counted.  Even though the possibility of this occurence is slim, it can happen.  Thus, the USAMTS website is being accurate and concise.


Is it possible to delete your aops account?

Not that i know of, but yes you can inactivate one.

http://www.mathlinks.ro/viewtopic.php?search_id=1978920067&t=272377
http://www.mathlinks.ro/viewtopic.php?search_id=1438112909&t=161324

You can't delete it. And mods can't delete other people's accounts. Next time, search for it :)


function
abstract algebra
group theory
real analysis
real analysis unsolved

Let $(\mathbb{R},+)$ be the additive group of reals. Find all continious characters from $(\mathbb{R},+)$ to the unit circle $\{z \in \mathbb{C}: |z|=1 \}$

They're of the form $t\mapsto e^{\alpha it}$ for some real $\alpha$.

It even suffices to know that a character is measurable in order to conclude this. The argument is very similar to the one employed for proving that the only measurable additive functions $f: \mathbb R\to\mathbb R$ are those of the form $x\mapsto ax$ for real $a$.

Three very closely related questions. In all cases, characters are continuous homomorphisms into the unit circle in $\mathbb{C},$ with multiplication as the group operation in the codomain.

1. Fix $P>0.$ Find all continuous characters on the group $(\mathbb{T}_P,+)$ where by $\mathbb{T}_P,$ I mean the group $\mathbb{R}/(P\mathbb{Z})$ with the operation of addition mod $P.$

2. Find all characters on $(\mathbb{Z},+).$ (We don't need to say continuous.)

3. Fix $N\in\mathbb{N}.$ Find all characters on $\mathbb{Z}/(N\mathbb{Z}).$

Second question: all the characters are of the form $f(n)=f(1)^n$.
3).all the characters are of the form $f(n)=\alpha^n, n\in \mathbb{Z_{N}}, \alpha$-$N$'th root of unity.
Is it ok?

[quote="eugene"]Second question: all the characters are of the form $f(n)=f(1)^n.$ [/quote]
With the provision that $|f(1)|=1.$

One way to think of this: $f(n)=e^{2\pi isn}$ for some real $s.$ However, $s$ and $s+1$ give the same character, so it is best to think of $s\in\mathbb{T}_1=\mathbb{R}/\mathbb{Z}.$

For the third question, you're right. If we let $\omega=e^{2\pi i/N},$ then we have $f(n)=\omega^{kn}$ for some $k.$ Since $k$ and $k+N$ give the same character, it is best to think of $k\in\mathbb{Z}/(N\mathbb{Z}).$

That still leaves the question of the continuous characters on $\mathbb{T}_P.$

The set of all continuous characters on a locally compact abelian group is itself a locally compact abelian group. (The group operation on the characters is a natural one - poinwise multiplication of the functions.)


Find all solutions to the equation $ a^3 \equal{} b^3 \plus{} b$.

I'm assuming you mean solve for integers.

[hide]
Let $ a\equal{}b\plus{}x$ for $ x$ an integer.  Then the equation becomes

$ b^3\plus{}3b^2x\plus{}3bx^2\plus{}x^3\equal{}b^3\plus{}b$

$ \Rightarrow 3b^2x\plus{}(3x^2\minus{}1)b\plus{}x^3\equal{}0$

This is a quadratic equation in $ b$.  Since $ b$ is integer, the discriminant must be a perfect square:

$ (3x^2\minus{}1)^2\minus{}4\cdot 3x\cdot x^3\equal{}n^2$

$ \Rightarrow 9x^4\minus{}6x^2\plus{}1\minus{}12x^4\equal{}n^2$

$ \Rightarrow 1\minus{}(6x^2\plus{}3x^4)\equal{}n^2$

Since $ 6x^2\plus{}3x^4$ is always greater than or equal to zero, the only possible values for $ n^2$.  Are $ 0$ and $ 1$.  But, $ 1\minus{}(6x^2\plus{}3x^4)\equal{}0$ has no integer solutions.  As for $ n^2\equal{}1$, this implies $ x\equal{}0\Rightarrow a\equal{}b$.  This only has solution $ a\equal{}b\equal{}0$.

yes?

[/hide]

[hide="Another solution"]Since $ b(b^2\plus{}1) \equal{} a^3$ and $ (b,b^2\plus{}1) \equal{} 1$ the numbers $ b$ and $ b^2\plus{}1$ must both be third powers. However, if $ b$ is a third power then so is $ b^2$; the only third powers with difference 1 are $ 0$ and $ 1$, so $ b\equal{}0\implies a\equal{}0$.[/hide]

Yes its supposed to be only integers.

I saw this somewhere and then was wondering how to solve it so the problem got rephrased a lot

[hide="Quite nice solution"]Obviously, $ a\equal{}b\equal{}0$ works. For $ b\neq0$, assume the contrary. $ \minus{}3b^2\plus{}2b\minus{}1<0$ and $ 3b^2\plus{}2b\plus{}1>0$ for all $ b$ because both have no real roots, so $ (b\minus{}1)^3<b^3\plus{}b\equal{}a^3<(b\plus{}1)^3$. Then $ b^3\plus{}b$, to be the cube of an integer, must be equal to $ b^3$. This violates that $ b\neq0$ so the only solution is $ (a,b)\equal{}(0,0)$.[/hide]


It seems that when I view a page in "Show preview" mode, the LaTeX does not work correctly (the math tags return a string like "UNIQ359679b21fde49b0-math-00000003-QINU").

Edit: Since this has been correctly been moved to the support subforum, just noting that this post was regarding the AoPS Wiki.

Also fixed the $ \$$ in the <nowiki> tags, and in <asy> tags in the Wiki.

Thanks.  :)

On a likely unrelated note, the double dollar sign ($ \$\$\ldots \$\$$) or the <cmath> tag no longer seem to be centering correctly. It might be a browser-related issue (IE6); but on recently edited pages (like [[2008 AIME I Problems/Problem 13]] doesn't center versus [[2008 AIME I Problems/Problem 11]] does) the <center> tag seems to be dropped from around the latex images.

I've dropped the center tag. It's not a needed markup. IE7, Firefox, Opera 9 and Safari 3 all support the CSS centering of images, so I'll go with that. Sorry IE6 users, go upgrade (it's free, even for those with pirated copies of Windows).


Let $k$ and $k'$ be concentric circles with center $O$ and radius $R$ and $R'$ where $R<R'$ holds. A line passing through $O$ intersects $k$ at $A$ and $k'$ at $B$ where $O$ is between $A$ and $B$. Another line passing through $O$ and distict from $AB$ intersects $k$ at $E$ and $k'$ at $F$ where $E$ is between $O$ and $F$.

Prove that the circumcircles of the triangles $OAE$ and $OBF$, the circle with diameter $EF$ and the circle with diameter $AB$ are concurrent.

yes yes my guess was true....

but u know,we wont prove the problem with inversion.we just use it to find the trusts in the problem.

$lemma$:( very nice and usefull)

let $(C1)$and $(C2)$ be two circles which intersects each other in points $A$ and $B$.

suppose a variable line goes through $B$ and intersects $(C1)$ and $(C2)$ in points$C$ and $D$,respectivly.

then: the triangles $AO1O2$ and $ADC$ are simmilar to each other.

its clear ,but we prove it very simply to do our best. ;)

$\angle AO1O2=\angle ACD$   and   $\angle AO2O1=\angle ADC$.

the lemma will also be true ,if $B$ is betwin$C$ and $D$.or if $D$ is betwin $B$ and $C$.

then: lets prove our problem by the lemma.

name the intersection of the circumcirclesof hthe triangles $OAE$ and $OFB$,$M$.

before that we use inversion.

an inversion with the center $O$ and the radius $DE.DF$ ,shows us that the angle betwin

the circumcircles of the triangles $OAE$ and $OFB$,is $90$.

because:the lines $A'E'$ and $B'F'$ which are the inversive shapes of the circles $OAE$ and $OFB$,

are prependicular to each other.and it shows us that the angle betwin the circumcircles of the triangles $OAE$ and

$OFB$,is $90$.

then acording to the lemma,$AB$ goes through one of the intersections of the two circles.(i mean $O$).

and we have:the triangles $MAB$ and $MO1O2$ are simmilar.then it shows that $\angle AMB=90$

by a simmilar way we can claime that $\angle EMF=90$.

then it shows that:a circle with the diameter $FE$ goes through $M$.

and a circle with the diameter $BA$ goes through $M$.

and the problem is proved... ;)

[quote="ashegh"]i think inversion will help...

i will think about it and will post the solution...

but im not sure with it by inversion... ;)[/quote]

Well, I have no idea how inversion (on a circle - as you probably meant) works (yes I know, I'm an uneducated geometrician), however I solved it by a mere angle-chasing (unspectacular but easy and stupit :D )

[hide]
Let $T$ be the point of intersection of the circumcircles of $OAE$ and $OBF$. Let $\angle AOE = \alpha$ and $\angle BOF = \beta$. We have $\alpha+\beta = 180$. $\triangle OAE$ and $\triangle OBF$ are isosceles, thus $\angle OEA = \angle OAE = 90-\frac{\alpha}{2}$ and $\angle OFB = \angle OBF = 90-\frac{\beta}{2}$.

$BFTO$ is cyclic, hence $\angle OTF = 180 - \angle OBF = 90+\frac{\beta}{2}$.
Also, $OATE$ is cyclic, hence $\angle OTE = \angle OAE = 90-\frac{\alpha}{2}$.
Thus, $\angle FTE = \angle OTF - \angle OTE = 90+\frac{\beta}{2} - 90 + \frac{\alpha}{2} = 90$, hence T lies on the circle with diameter $EF$.

Similary, we'll show that $\angle ATB = 90$:
$BOTF$ is cyclic, hence $\angle OTB = \angle OFB = 90-\frac{\beta}{2}$.
$AOET$ is cyclic, hence $\angle OTA = \angle OEA = 90-\frac{\alpha}{2}$.
Thus $\angle ATB = \angle OTA + \angle OTB = 90-\frac{\alpha}{2} + 90 - \frac{\beta}{2} = 90$, hence T lies on the circle with diameter $AB$.

$\Longrightarrow q.e.d.$
[/hide]

the best thing is to solve  easy problems easy...

the thing that u did... ;)


i think that \[a_{1}= \frac{x_{1}}{x_{2}}... a_{n}= \frac{x_{n}}{x_{1}}\] should help, tho i'm too lazy to think about it more...


Let $ a_n \equal{} 3^n \minus{} 2^n$ for every $ n\in\mathbb{N}$.
Prove that the sequence $ \left(a_1,a_2,a_3,...\right)$ does not contain three terms in geometric progression.

Assume that $a_k=\frac{a_m+a_n}{2}$ with $m<k<n$ (equivalent to $a_m<a_k<a_n$), thus $2a_k \geq a_n$. Now $k \leq n-1$ and already $2a_{n-1}=2(3^{n-1}-2^{n-1})<3^{n-1}+2(3^{n-1}-2^{n-1})=3^{n}-2^{n}=a_n$, so we get the desired contradiction.

But what the question asks for is geometric propgression, not arithmetic progression.

Then, the assumption should be : $a_k=\sqrt{a_m \cdot a_n}$ ;)

My first solution above is simply and totally wrong and shows that they are not in arithmetic progression.

Here is a brutal but generalising way to solve the problem:

We will use the Zsigmondy theorem:

[quote="Number Theory Formulary"]
[b]Zsigmondy's Theorem:[/b]
Let $ a > b \geq 1$ and be coprime integers. Then for any $ n \in \mathbb{N}$ there is a prime $ p$ dividing $ a^n \minus{} b^n$ but not dividing $ a^k \minus{} b^k$ for all $ k < n$ with two exceptions:
a) $ a \equal{} 2 , \ b \equal{} 1 , \ n \equal{} 6$
b) $ a \plus{} b$ a power of $ 2$ and $ n \equal{} 2$[/quote]

Well: by Zsigmondy's Theorem, every term $ a_m \equal{} 3^m \minus{} 2^m$ contains some prime factor not occuring in the previous terms $ a_l, l < m$.
Thus:
Assume $ a_k^2 \equal{} a_m a_n$. WLOG then $ m \geq k$, so let $ p$ be a prime dividing $ a_m$ but not $ a_l$ for $ l < m$. Then $ p|a_k$, thus $ k \equal{} m$, thus $ n \equal{} k$, thus no terms in true geometric progression.

This also works when we take any sequence $ a_n \equal{} c^n \minus{} d^n$ or $ \equal{} c^n \plus{} d^n$ with $ c > d\geq 1$.

Suppose that the assertion is wrong and $m \not=n$ ,$a_ma_n=a_k^2$. Suppose that $m>n$ and $n\geq 2, m\geq3 , k\geq 4$ we have
\[ 3^{m+n}(1-(\frac 23)^m)( 1-(\frac 23)^n)=3^{2k}(1-(\frac 23)^k)^2 \Longrightarrow 3^{2k-m-n}=\frac{(1-(\frac 23)^m)(1-(\frac 23)^n)}{(1-(\frac 23)^k)^2}  \].
But we have
\[ 3>\frac{1}{(1-(\frac 23)^4)^2}>3^{2k-m-n}>\frac{(1-(\frac 23)^2)(1-(\frac 23)^3)}{1}>\frac 13  \]
Then $3^1>3^{2k-m-n}>3^{-1} \Longrightarrow 2k-m-n=0 \Longrightarrow 2k=m+n$.
So that
\[ (3^m-2^m)(3^n-2^n)=(3^{\frac{m+n}{2}}- 2^{\frac{m+n}{2}})^2 \Longrightarrow 3^m\times 2^n+3^n\times2^m=2\times 6^{\frac{m+n}{2}} \Longrightarrow (3^{\frac m2}\times 2^{\frac n2}-3^{\frac n2}\times2^{\frac m2})^2=0 \Longrightarrow 3^{\frac{m-n}{2}}=2^{\frac{m-n}{2}} \Longrightarrow m=n  \] and it's contradiction.
Hence, $n=1$and $a_m=a_k^2$ ,($k\not=1$) then $k\geq 2$. If $k=2 \Longrightarrow a_m=\sqrt{5}$ which it's imposible ,so that $k\geq3$ and $m\geq 2$.
\[ 3^m-2^m=(3^k-2^k)^2 \Longrightarrow 3^{m-2k}=\frac{(1-(\frac 23)^k)^2}{1-(\frac 23)^m} \Longrightarrow 3=\frac{1}{1-\frac 23}>3^{m-2k}>(1-(\frac 23)^3)^2>\frac 13 \Longrightarrow m=2k  \] and it will reach to the contradiction. :D

my sol uses property of sequence: (Am,An)=A(m,n)


calculus integration MATHCOUNTS quadratics inequalities algorithm algebra

Find the largest possible value of $k$ for which $3^{11}$ is expressible as the sum of $k$ consecutive positive integers.

[hide]$kn+\frac{k(k-1)}{2}=3^{11}$, so $2kn+k(k-1)=2*3^{11}$.  If k is odd, $k|3^{11}$, i.e. $k=3^i$ for some nonnegative integral $i\leq 11$, so the greatest $k$ is $k=3^5=243$.  If k is even, $k=2j$ for some positive integer $j$, and $2jn+j(2j-1)=3^{11}$, so $j|3^{11}$, so the greatest $j$ is $j=3^5$, so the greatest $k$ is $k=2*3^5=486$.[/hide]

[hide]Consider the set \[ \mathcal{S}: = \{a, a+1, \ldots, a+k-1\}.  \] If the sum of the elements of $\mathcal{S}$ is equal to $3^{11}$, then \[ \frac{k(2a+k-1)}{2} = 3^{11} \Longleftrightarrow k(2a+k-1) = 2\cdot 3^{11}.  \] Since $2a-1>0$, we must have $2a+k-1>k$. So to maximize $k$, we must minimize $2a+k-1$ while keeping it greater than $k$; this means we want both factors as close as possible to $\sqrt{2\cdot 3^{11}}=\sqrt{54\cdot 81^2}\approx 7\cdot 81 = 567$. It is easy to see, then, that the largest possible value of $k$ is $2\cdot 3^5 = \boxed {486}$.[/hide]

[hide="Solution"]Let our $k$ consecutive numbers be $k+1,k+2,\ldots,n+k$. The sum of these numbers is $\frac k2[(k+1)+(n+k)]=3^{11}$. Thus, $k(2k+n+1)=2\cdot3^{11}$ and $k|2\cdot 3^{11}$. Additionally, $k^2<2\cdot 3^{11}$ because $k<2k+n+1$. We now seek the largest $k$ that satisfies these conditions ($k<\sqrt{2\cdot 3^{11}}$ and $k|2\cdot3^{11}$), so the answer is $k=2\cdot3^5=486$. [/hide]

[hide="any mathcounts student should be able to solve this"]
Let n be first term, k be the number of terms.

We have $n+n+1+n+2+...+n+(k-1)=3^{11}$.

$kn+\frac{(k-1)(k)}{2}=3^{11}$, or $2kn+k^2-k=2\cdot3^{11}$.

$k(2n+k-1)=2\cdot3^{11}$.

$2n=\frac{2\cdot3^{11}}{k}-k+1>0$.
Since we know that k>0, multiplying by -k reverses the sign:

$k^2-k-2\cdot3^{11}<0$. Solve it as an equality first.

The quadratic formula tells us $k=\frac{1\pm\sqrt{1+8\cdot3^{11}}}{2}$. The 1's are too small to consider, and k is positive, so we can have $2^{1/2}\cdot3^{11/2}\approx{600}$ (it's less than 729 and greater than 486).

Since the inequality was a "<", k is less than whatever $2^{\frac{1}{2}}\cdot3^{\frac{11}{2}}$ was and greater than that other negative root that we excluded.

The greatest integral factor of $2\cdot3^{11}$ that is less than 600 or so is 486. n is thus 244, which is positive and is okay.  [/hide]

$ n\plus{}(n\plus{}1)\plus{}(n\plus{}2)\plus{}...\plus{}(n\plus{}k\minus{}1)\equal{}\frac{2kn\plus{}k(k\minus{}1)}2\equal{}3^{11}$

$ k^2\plus{}(2n\minus{}1)k\minus{}2\cdot3^{11}\equal{}0$

$ k\equal{}\frac{\minus{}(2n\minus{}1)\pm\sqrt{(2n\minus{}1)^2\plus{}2^3\cdot3^{11}}}2$

$ (2n\minus{}1)^2\plus{}2^3\cdot 3^{11}\equal{}m^2$

$ m^2\minus{}(2n\minus{}1)^2\equal{}2^3\cdot 3^{11}\equal{}(m\plus{}2n\minus{}1)(m\minus{}2n\plus{}1)$

Note that the sum of $ m\plus{}2n\minus{}1$ and $ m\minus{}2n\plus{}1$ must be even, so we must have $ A\equal{}2^2\cdot3^x$ and $ B\equal{}2\cdot3^{11\minus{}x}$ be either factor.
Now is there a fast way to determine what $ x$ should be?

[quote="mathwizarddude"]$ n \plus{} (n \plus{} 1) \plus{} (n \plus{} 2) \plus{} ... \plus{} (n \plus{} k \minus{} 1) \equal{} \frac {2kn \plus{} k(k \minus{} 1)}2 \equal{} 3^{11}$

$ k^2 \plus{} (2n \minus{} 1)k \minus{} 2\cdot3^{11} \equal{} 0$

$ k \equal{} \frac { \minus{} (2n \minus{} 1)\pm\sqrt {(2n \minus{} 1)^2 \plus{} 2^3\cdot3^{11}}}2$

$ (2n \minus{} 1)^2 \plus{} 2^3\cdot 3^{11} \equal{} m^2$

$ m^2 \minus{} (2n \minus{} 1)^2 \equal{} 2^3\cdot 3^{11} \equal{} (m \plus{} 2n \minus{} 1)(m \minus{} 2n \plus{} 1)$

Note that the sum of $ m \plus{} 2n \minus{} 1$ and $ m \minus{} 2n \plus{} 1$ must be even, so we must have $ A \equal{} 2^2\cdot3^x$ and $ B \equal{} 2\cdot3^{11 \minus{} x}$ be either factor.
Now is there a fast way to determine what $ x$ should be?[/quote]

I did it this way too.  Is it a dead end, or can something be done after this?  Thanks!

I always remember that the sum of the numbers from a to b is $ \frac {(a \plus{} b)(a \minus{} b \plus{} 1)}{2}$, so we need $ (a \plus{} b)(a \minus{} b \plus{} 1) \equal{} 2 \cdot 3^{11}$.
Obviously $ a \plus{} b > a \minus{} b \plus{} 1$, so $ a \minus{} b \plus{} 1$ is maximized by letting $ a \minus{} b \plus{} 1 \equal{} 2 \cdot 3^5$, $ a \plus{} b \equal{} 3^6$.

[hide="testing"]$k(2a+k-1)=2\cdot{3^{11}}$

testing $k=3^6$ we find that it gives $2a+728=400$ and something, and since a is a positive integer this value of k is too high.

now we go down by 1 increment, i.e. $k=2\cdot{3^5}$, and this works.

Note that this is perfectly rigorous since we are told all the numbers are positive integers, so we can find a bound fo $k$ as above and then go down by 1 step at a time, the shortest interval between consecutive steps is:

test $3^i$ => find too high
test $2\cdot{3^{i-1}}$ => too high
test $3^{i-1}$

etc

so the shortest possible interval we can go by is $3^i-2\cdot{3^{i-1}}$

moreover this is a finite algorithm since the maximum value of k we can start with is 3^11 (this is the worst bound obviously, we can start educatedly somewhere at 3^6-3^7)[/hide]

One observation that may be useful.
[hide]
Let S be the sum of k consecutive integers where k is odd. Then k must divide S and the start of the sequence is $\frac{S}{k} - \frac{k-1}{2}$.

Since $S=3^{11}$, we know k if is odd, then $k=3^n$ for $n <=11$.

Now we have the requirement that consecutive integers be positive which mean that

$\frac{3^{11}}{3^n} - \frac{3^n}{2} > 0$ which implies $2*3^{11} > 3^{2*n}$

n = 5 is the largest value for the inequality is true which means $k=3^5$ is the largest sequence where k is odd.

Now need to think about where k is even...
[/hide]

feeling v sick and don't have packet on me :(

p1: Let our $k$ positive integers be $a, a+1, a+2, \ldots, a+(k-1)$. The sum of these is $ka + \frac{k(k-1)}{2} = 3^{11}$. Factoring gives $k\left(a+\frac{k-1}{2}\right) = 3^{11}$. If $k$ is odd, it's clear that $k$ has to be a power of $5$, and it must be less than $3^6$ since $\frac{3^6-1}{2} > 3^5$, so the maximum would be $3^5 = 243$. If $k$ is even, multiply our equation by $2$ to get $k(a+k-1) = 2 \cdot 3^{11}$ and let $k = 2p$, so $2p(a+2p-1)=2 \cdot 3^{11} \implies p(a+2p-1) = 3^{11}$. The maximum value of $p$ would now be $3^5$, so $k = 2 \cdot 3^5 = 486$. This is greater than $243$, so $\boxed{486}$ is the answer.


[hide]Do you accept mass points solutions?

Throwing on some weights (2,1,2,3,3,4,5 on A,B,C,D,E,F,K respectively) gives 3:2, I believe[/hide]

I wish I could, but thats all there is to mass points.  I could send you an explanation of how to use them that I wrote a while back, but that might be more than you wanted to know.   Might just wanna kick back and wait for a non-mass points solution :|  up to you

You can use the different form of Ceva's Theorem:

$\frac{KF}{BF}+\frac{KE}{AE}+\frac{KD}{CD}=1$

$\frac{1}{5}+\frac{2}{5}+\frac{KD}{CD}=1$

$\therefore \frac{KD}{CD}=\frac{2}{5} \Longleftrightarrow CK: KD=3: 2$


inequalities
algebra proposed
algebra
wiles orz

If $ x$, $ y$, $ z$, and $ n$ are natural numbers, and $ n\geq z$ then prove that the relation $ x^n \plus{} y^n \equal{} z^n$ does not hold.

By Last Fermat's, $ n<3$, so $ z\le 2$.

If $ n\equal{}1$, then $ x\plus{}y\equal{}1$, impossible

If $ n\equal{}2$, then $ x^2\plus{}y^2\equal{}1$, or $ x^2\plus{}y^2\equal{}4$, both impossible

No solutions

Are you kidding? Fermat Last theorem has been proved 9 years after the problem had been proposed. And I don't you would get a single point for this solution even nowadays...

Yes, it's true... But you are wrong: nowadays, I can use anything that's considered a theorem, Las Fermat is, you don't have to give me any point, but it's a full solution based on a theorem, not a conjecture (although in 1987 it was)

Ok, but I don't think if you were on a competition you would solve the problem like that.

Obviously $ z>x,y,1$

Then

$ 2(z\minus{}1)^n\ge x^n\plus{}y^n\equal{}z^n \Leftrightarrow 2\ge (1\plus{}\frac{1}{z\minus{}1})^n \ge 1\plus{} \frac{n}{z\minus{}1} > 2$ contradicton

Why does the first inequality hold?

$ 2(z\minus{}1)^n\ge x^n\plus{}y^n$?

$ z>x,y\Leftrightarrow z\minus{}1\ge x,y$
$ (z\minus{}1)^n\ge x^n$ adding $ (z\minus{}1)^n\ge y^n$ gives $ 2(z\minus{}1)^{n}\ge x^{n}\plus{}y^{n}$

[hide=Solution(not possible back then)]
We use Fermat's Last Theorem, which implies $n<3\implies z<3$.

Case 1: $n=1$.
Then $x+y=1$, a contradiction.

Case 2: $n=2$.
Then $x^2+y^2=4$, also a contradiction.  [/hide]


1)A student is allowed to select at most $n$ books from a collection of $(2n+1)$ books.If the total number of ways in which he can select a book is $63$,find the value of $n$.

2)In a certain test there are $n$ questions.In this test $2^{n-k}$ students gave wrong answers to at least $k$ questions,where $k=1,2,3\cdots,n$.If the total number of wrong answers given is $2047$,what is $n$?

[quote="shadysaysurspammed"]Anyone else?[/quote]

any help? I've made sth, but it is not enough...

[quote="tim1234133"][hide] 7C1+7C2+7C3=63, n=3[/hide][/quote]
[quote="shadysaysurspammed"][quote="surge"][hide="1"]
$C(2x+1,1)=63$
$\frac{(2n+1)!}{(2n)!\cdot(1!)}=2n+1$

$2n+1=63$
$n=31$

Note the wording [b]a book[/b], not $n$ books.
[/hide][/quote]

That's what i thought first but the answer is $3$....I dunno how...[/quote]
tim had it.  The number of ways you can choose [i]at most[/i] $n$ from a set of $(2n+1)$ is $\sum_{k=1}^n\binom{2n+1}{k}=63$.  In this case, $\binom{7}{1}+\binom{7}{2}+\binom{7}{3}=63$ so $n=3$


LaTeX
number theory proposed
number theory

Find all positive integers x,y,z such that

$5^x+3^y=1+7^z$.

Bomb

[color=red]Edited by moderator. Bomb, please learn LaTeX :)[/color]

I would appreciate whole soln, thanks.

Bomb


function
floor function
AMC
AIME
arithmetic sequence
arithmetic series

Let $ a<b<c$ be an arithmetic sequence of integers. Let $ d(n)$ be the number of positive divisors of $ n$. Given that $ a=-1$, $ d(b)=27$, and $ d(c)=15$, find the smallest possible value for $ \left\lfloor \frac{c}{10}\right\rfloor$.

[hide="hint"] $ b$ and $ c$ have an odd number of divisors.[/hide]

[hide]since a is odd, c must also be odd. (because c = a + 2x)
d(c) = 15, so c is of the form $ j^{4}*k^{2}$ or $ j^{14}$. Since (I hope) this is an AIME problem, c has to be less than 10,000, but $ 3^{1}4 > 10,000$. So $ c = j^{4}*k^{2}$.

d(b) = 27, so $ b = x^{2}*y^{2}*z^{2}$ or $ x^{8}*y^{2}$. The only way this can happen (for b to be less than 5,000) is if $ b = 2^{8}* 3^{2}= 2304$. c would then be $ 4609 = 11*419 \neq j^{4}* k^{2}$. Thus, $ b = x^{2}*y^{2}*z^{2}$

Since c must be odd, we can try some values.
$ 3^{4}* 13^{2}> 10000$
$ 5^{4}* 7^{2}> 10,000$ so we know that j = 4.

So we are left to do some fairly simple calculations:
$ 3^{4}* 5^{2}= 2025$
$ 3^{4}* 7^{2}= 3969$
$ 3^{4}* 11^{2}= 9801$
c $ \in$ {2025, 3969, 9801}

If we wanted to see possible values of b (which again, must be less than 5000, so $ xyz \leq 70$)
$ (3*5*7)^{2}> 5,000$ so x is 2.
We have 4 values left to check (remember, for a value of b, there has to be a c = 2b+1)
$ (2*3*5)^{2}= 900$ doesn't match up with a value of c
$ (2*3*7)^{2}= 1764$ also doesn't match up with a value of c
$ (2*3*11)^{2}= 4356$ also doesn't match up
$ (2*5*7)^{2}= 4900$ works! So c = 9801, so the answer is 980.[/hide]
I don't think this is the most elegant way to do this, and I am basing my guess that this would be an AIME problem, but it only takes 2-3 minutes to do.

$ b=y^{2}$; $ c=z^{2}$. Then we have an arithmetic series;

$ 2y^{2}=z^{2}-1\iff z^{2}-2y^{2}=1$. This is a simple pell equation. At this point, I guess you hope that one of the solutions has the correct number of divisors.


[b]Problem.[/b] Let $ a,b,c$ be non-negative, then

\[ \frac{a}{\sqrt{a\plus{}b}}\plus{}\frac{b}{\sqrt{b\plus{}c}}\plus{}\frac{c}{\sqrt{c\plus{}a}}\le \frac{5}{4}\sqrt{a\plus{}b\plus{}c}.\]

[i]Proof. [/i]Without loss of generality, assume that $ a\equal{}max(a,b,c)$ and $ a\plus{}b\plus{}c\equal{}1$, then

\[ \frac{b}{\sqrt{b\plus{}c}}\plus{}\frac{c}{\sqrt{c\plus{}a}}\le \sqrt{b\plus{}c}\equal{}\sqrt{1\minus{}a}.\]

Thus

\[ \frac{a}{\sqrt{a\plus{}b}}\plus{}\frac{b}{\sqrt{b\plus{}c}}\plus{}\frac{c}{\sqrt{c\plus{}a}}\le \frac{a}{\sqrt{a\plus{}b}}\plus{}\sqrt{1\minus{}a}\le a\plus{}\sqrt{1\minus{}a} \le \frac{5}{4}.\]

We are done!

haha. there is nothing wrong! very beautiful proof!!
I still think that given the time it took you to find such a solution this inequality remains one of the most challenging ones out there. :lol:

Edit: very beautiful wrong proof then  :D  I only checked the second line as the core to the "proof".

[quote="hungkhtn"]\[ \frac {a}{\sqrt {a \plus{} b}} \plus{} \frac {b}{\sqrt {b \plus{} c}} \plus{} \frac {c}{\sqrt {c \plus{} a}}\le \frac {a}{\sqrt {a \plus{} b}} \plus{} \sqrt {1 \minus{} a}\le a \plus{} \sqrt {1 \minus{} a} \le \frac {5}{4}.
\]
[/quote]
I am sorry, but why $ \frac {a}{\sqrt {a \plus{} b}}\le a$?. Note that $ a\plus{}b\le1$.

Ha ha, this solution is actually wrong  :D


i really liked this question,i'll give you a couple of hints

at time =0 ,area =0 (take this assumption)

if you notice carefully the inradius , r = vt

using this you can solve it,you'll get option <D>

|emf| =  $ \frac {d \phi}{dt}$

$ \phi \equal{} BA(t)$

$ \frac {d \phi}{dt} \equal{} B\frac{dA(t)}{dt}$

I think its simple math to find $ \frac{dA(t)}{dt}$ ; use the fact that $ A \equal{} \frac{\sqrt{3}}{4}a^2$

[quote="Conjurer"]|emf| =  $ \frac {d \phi}{dt}$

$ \phi = BA(t)$

$ \frac {d \phi}{dt} = B\frac {dA(t)}{dt}$

I think its simple math to find $ \frac {dA(t)}{dt}$ ; use the fact that $ A = \frac {\sqrt {3}}{4}a^2$[/quote]

I did the same thing. I assumed the intersection of any two rods to be my reference. That way, the rod, say the one at the extreme right will have a velocity $ 2v$ upwards ie $ \perp$ to lower rod.

But note that the resistance in the circuit too is changing.

So what I did is:

${ I = \frac{\frac{-d\phi}{dt}}{\frac{dR}{dt}}}
\\
\\
\\
=\frac{\frac{B_0\sqrt{3}a}{2}}{\lambda}$

Since $ \frac{da}{dt} = kv$
(where $ a$ is length of triangle and $ k$ is the co-efficient for the component of the velocity along any rod)

The $ a$ term remains. So this was a flop.

However, vivek_n's idea is good:

[quote="vivek_n"]at time =0 ,area =0 (take this assumption)

if you notice carefully the inradius , $ r = vt$

using this you can solve it,you'll get option $ D$
[/quote]

And $ D$ is the right answer!

See derivative of area gives $ a \frac {da}{dt}$ , we can find $ \frac {da}{dt}$ in terms of $ v$ and the $ a$ term cancels by the resistance term which is $ 3 \lambda$ a

[quote="Immanuel Bonfils"]Rajiv

Why deriving the resistor wrt time? You should just divide the variable efm by the variable resistor to get a constat current[/quote]
Ok! I understood! That would make it go wrong dimensionally. So the answer acc to my method turns out to be $ D$!

Thanks fellas!

[quote="Rajiv"]Three very long rods are placed as shown in the figure. Each rod moves with a speed $ v (v \perp$ length of rod$ )$ in a uniform magnetic field $ B_0$ directed into the plane of the paper (here it is your screen :D ). The triangle formed between the rods is always an equilateral triangle. The induced current in the triangle is (if resistance per unit length is $ \lambda$):

$ A) \; \frac {vB_0}{3\lambda} \\
 \\
B) \; \frac {2vB_0}{3\lambda} \\
 \\
C) \; \frac {vB_0}{\lambda \sqrt {3}} \\
 \\
D) \; \frac {vB_0}{\lambda}$[/quote]

[hide="My thoughts"]Ohm's law as derived from the expression for the Lorentz force is:

\[ \vec{J} \equal{} \sigma(\vec{E} \plus{}\vec{v}\times\vec{B})\]

Where $ \vec{J}$ is the current density, $ \sigma$ is the electrical conductivity, $ \vec{E}$ is the E-field if any, and $ \vec{v}$ and $ \vec{B}$ are self-explanatory.

In this case, $ \vec{E} \equal{} 0$.

Therefore $ I \equal{} \vec{J}\cdot\vec{A} \equal{} \sigma A v B$.

Resistance $ R \equal{}\frac{\ell}{\sigma A} \implies \frac{1}{\sigma A} \equal{} \frac{R}{\ell}\stackrel{\triangle}\equal{} \lambda$.

\[ \therefore I \equal{} \frac{v B}{\lambda}\][/hide]


With 3 numbers $ 2$ and all operations ($ \plus{}$, $ \minus{}$, x, $ /$, $ ln$, $ log$ ,$ n^{th}$ root,...), build an expression that has a result of $ 2008$ (Notice that 2008 is not important, it can be any numbers  :D )

what do you mean by all operations? anything bigger than $ 2^{2^2}$? or is $ e^x$ allowed. that seems "cheap" haha because you get an $ e$ for free...

Use tetration, $ ^2 2 \equal{} 16$ to get the value of 16 with just 2 2's, or direct exponentiation, $ 2^{2^2}$, to use all three 2's. Then, repeated usage of factorial (or, Gamma), squareroot, and floor/ceiling should eventually land you up with a solution, I suppose.

So the real question now is -- using some string of floor/ceiling, factorial/Gamma, and squareroot, is it possible to turn the number 16 into any other number?

Oh, and I assume that by "nth root" you mean n an integer. Because $ 2008 \equal{} 2^{12} \minus{} 2^{5} \minus{} 2^3$, if that works.

Unfortunately, tetration, like exponentiation, includes the base in its "count". So $ ^22$ is in fact $ 2^2$, which is 4. :(

Also, I think he means *just* using three 2's, nothing else. Wow, hard...

Also, "all operations" is ambiguous. There's basically no limit to the operations you can define. For the purpose of argument, I define $ a \& b \equal{} 2008$ for all $ a, b$. So my answer is $ (2 \& 2) \& 2$. Obviously, this is not the answer, but just saying...

[hide="Answer"]
$ -log_{2}(log_{2}\sqrt{\sqrt{\sqrt{...\sqrt{2}}}})=n$ where $ n$ is the number of $ \sqrt$
[/hide]
Hope you get what I mean :D (I'm not very good at English)


Let S be a non-empty set with an associative operation that is left and right cancellative (xy=xz implies y=z, and yx = zx implies y = z).  Assume that for every a in S the set {a^n : n = 0,1,2...} is finite.  Must S be a group?

I haven't had much group theory at this point...

All you really need is the definition. A group is a set with an associative operation, closed under that operation, for which there is a (two-sided) identity element and in which each element has a (two-sided) inverse.

In this case you have the associative operation, and I would interpret the hypotheses as including that it is closed under that operation. If you believe that this is true, then you must show the existence of an identity element and show the existence of inverses. Since "inverse" wouldn't mean much if you didn't have an identity, your first task would be to produce the identity.

That's if it's true. Note that the problem is noncommittal on this point. You might want to start at the other end - can you produce a counterexample? If you don't seem to be able to produce a counterexample, and you start to get a feeling of why not, then you can switch over to trying to prove it.

Yes, it must be a group.  Because the set is finite, some term must be repeated infinitely often, meaning we must have $a^{m}= a^{n}$ for some $m, n \in\mathbb{N}$ with $n-m\ge2$.  So, for all $x\in S$, we have $a^{m}\cdot a^{n-m}\cdot x = a^{n}\cdot x$, cancelling $a^{m}=a^{n}$, gives $a^{n-m}\cdot x = x$, and by a similar argument $x\cdot a^{n-m}= x$, so $a^{n-m}$ is an identity.  Now all that's left is to show the existance of an inverse; $a\cdot a^{m-n-1}= a^{m-n}$, and $a^{m-n-1}\cdot a = a^{m-n}$, and thus $a^{m-n-1}$ is a inverse of $a$, so $S$ is a group.

That's incomplete. What if $x$ does not belong to the set of powers of $a?$ What is $x^{-1}?$

[quote="Jimmy"]Why would that matter?  Isn't it enough to show that any $a\in S$ has an inverse?  Sorry if this is a dumb question, but I pretty new to this.[/quote]

You need to say more then: the way you've written it, it appears that you've fixed $a$.  It is possible to show that there is some $n$ associated to each $a$, etc., etc., but then you need an argument that the resulting identities are all the same (which is easy because of what an identity is).

So, could I say something like:  For any $a\in S$, we have $a\cdot a^{m-n-1}= a^{m-n}$, and $a^{m-n-1}\cdot a = a^{m-n}$, and therefore every $a\in S$ has an inverse?  And I'm still confused about what Kent said about $x$, why would we have to consider $x\not\in S$?

You found [i]an[/i] identity, namely $a^{k}$. But that only works for some particular $a$ (as far as we know). What if $b \in S$, and $b^{j}, j \ne k$ is the identity you get from the same argument? You need to show that $a^{k}= b^{j}$.

Let $S$ be a non-empty set with an associative operation that is left and right cancellative ($xy=xz$ implies $y=z$, and $yx=zx$ implies $y=z$). Assume that for every $a$ in $S$ the set $\{a^n\mid n=0,1,2,\ldots\}$ is finite. Must $S$ be a group?

It seems simple enough to show the existence of inverses and an identity, but I'm not sure how to tell whether or not the set is closed under the composition. If two elements can be composed to produce an element that is not in the set, then it wouldn't be a group.


vector
analytic geometry
conics
ellipse
geometry proposed
geometry

Given are $\triangle ABC$ with centroid $G$ with any $M$ on plane draw $MA',MB',MC'$ parallel with $GA,GB,GC$($A',B',C' \in$ line $BA,CA,AB$), resp, called $G'$ is centroid of $\triangle A'B'C'$ prove that $G'$ is midpoint of $MG$  :lol:

Let's use the complex numbers. Fix $A=0$, $B=1$, $C=z$, $M=m$; let $A'=a$, $B'=b$, $C'=c$.
The median from the vertex $A$ passes through $(z+1)/2$ (the midpoint of $BC$), so the parallelism between $GA$ and $MA'$ is equivalent to request that \[\frac{m-a}{z+1}\in\mathbb{R}\] or \[\frac{m-a}{z+1}=\frac{\overline{m}-\overline{a}}{\overline{z}+1}\quad (1)\] The condition that $A'\in BC$ means that $a=1+t(z-1)$ with $t\in\mathbb{R}$ so the equation (1) becomes \[\frac{m-(1+t(z-1))}{z+1}=\frac{\overline{m}-(1+t(\overline{z}-1))}{\overline{z}+1}\] which can be solved for t, leading to \[a=1+t(z-1)=1-\frac{1}{2}\frac{(z-1)(-m\overline{z}-m+\overline{z}+\overline{m}z+\overline{m}-z)}{z-\overline{z}}\] In a similar way, we find the equations \[\frac{m-b}{z-2}=\frac{\overline{m}-\overline{b}}{\overline{z}-2}\]  \[b=kz\quad\textrm{with }k\in\mathbb{R}\] which give an equation in k and solving it we get \[b=kz=\frac12\frac{-m\overline{z}+2m+\overline{m}z-2\overline{m}}{z-\overline{z}}\] So, we get the following for $c$, which has to be real in order to be on $AB$ \[\frac{m-c}{2z-1}=\frac{\overline{m}-c}{2\overline{z}-1}\] whose solution is \[c=\frac12\frac{-2m\overline{z}+m+2\overline{m}z-\overline{m}}{z-\overline{z}}\] Then $G'=(a+b+c)/3=$(after some algebra)$=z/6+m/2+1/6$. But $G=1/3+z/3$, so the midpoint of $GM$ is $1/6+z/6+m/2=G'$.

yes that's a good solution by complex number, I also have solution by vector but it isn't nice !

[quote="gemath"]Given are $\triangle ABC$ with centroid $G$ and with a point $M$ on plane draw $MA',MB',MC'$ parallel with $GA,GB,GC (A',B',C' \in$ line $BC,CA,AB)$, resp, called $G'$ is centroid of $\triangle A'B'C'$ prove that $G'$ is midpoint of $MG$[/quote][color=darkblue][b]The first  proof (with vectors).[/b] Exist $\{a,b,c,m,n,p,x,y,z\}\subset R$, where $x+y+z=1$ so that : $M=xA+yB+zC$, $A'=aB+(1-a)C=M+\frac{m}{3}(B+C-2A)$ a.s.o. $\Longrightarrow$ $A'=aB+(1-a)C=\left(x-\frac{2m}{3}\right)A+\left(y+\frac{m}{3}\right)B+\left(z+\frac{m}{3}\right)C$ a.s.o. $\Longrightarrow$
$x=\frac{2m}{3}$, $y+\frac{m}{3}=a$, $z+\frac{m}{3}=1-a$ a.s.o. $\Longrightarrow$ $m=\frac{3x}{2}$, $n=\frac{3y}{2}$, $p=\frac{3z}{2}$ and $a=y+\frac{x}{2}$, $b=z+\frac{y}{2}$, $c=x+\frac{z}{2}$  $\Longrightarrow$  $3G'=A'+B'+C'=(1-b+c)A+(1-c+a)B+(1-a+b)C=$ $\sum \left(1-z-\frac{y}{2}+x+\frac{z}{2}\right)A=$ $\frac{1}{2}\sum (2-z-y+2x)A=\frac{1}{2}\sum(1+3x)A=\frac{1}{2}(3G+3M)\Longrightarrow 2G'=G+M\ .$

[b]The second proof (with barycentrical coordinates).[/b] It is well-known that $A(1,0,0)$, $B(0,1,0)$, $C(0,0,1)$  and $G(\frac{1}{3},\frac{1}{3},\frac{1}{3})$. Denote $M(x,y,z)$, the middlepoint $D(0,\frac{1}{2},\frac{1}{2})$ of the segment $[BC]$ and $A_{1}\in BC\cap AM$. Thus, $A_{1}\left(0,\frac{y}{y+z},\frac{z}{y+z}\right)$ and from $MA'\parallel AD$ results $xD+(y+z)A_{1}=A'$, i.e. $A'\left(0,\frac{x}{2}+y,\frac{x}{2}+z\right)$ a.s.o. Therefore, $3G'\equiv\left(\frac{y+z}{2}+2x,\ \frac{z+x}{2}+2y,\ \frac{x+y}{2}+2z\right)=$ $\left(\frac{1-x}{2}+2x,\ \frac{1-y}{2}+2y,\ \frac{1-z}{2}+2z\right)=$ $\left(\frac{1+3x}{2},\ \frac{1+3y}{2},\ \frac{1+3z}{2}\right)$ and $(G+M)\equiv \left(\frac{1}{3}+x,\ \frac{1}{3}+y,\ \frac{1}{3}+z\right)=\left(\frac{1+3x}{3},\ \frac{1+3y}{3},\ \frac{1+3z}{3}\right)$. In conclusion, $2G'=G+M\ .$

[b]The third proof (syntetical).[/b] ....................... Soon !

[b]Generalization.[/b] Let $ABC$ be a triangle and let $P(m,n,p)$, $M(x,y,z)$ be two points (with the barycentrical coordinates w.r.t. the given triangle) wich belong to the same plane.
Define the points : $A'\in BC\ ,\ MA'\parallel AP$ ; $B'\in CA\ ,\ MB'\parallel BP$ ; $C'\in AB\ ,\ MC'\parallel CP$. Prove that the centroid of the triangle $A'B'C'$ belongs to the line $MP$ if and only if $xyp(n^{2}-m^{2})+yzm(p^{2}-n^{2})+zxn(m^{2}-p^{2})=0$, i.e. $\sum_{\mathrm{cyc}}\frac{1}{x}\left(\frac{p}{n}-\frac{n}{p}\right)=0\ .$ [/color]

[quote="Virgil Nicula"]
[b]Generalization.[/b] Let $ABC$ be a triangle and let $P(m,n,p)$, $M(x,y,z)$ be two points (with the barycentrical coordinates w.r.t. the given triangle) wich belong to the same plane.
Define the points : $A'\in BC\ ,\ MA'\parallel AP$ ; $B'\in CA\ ,\ MB'\parallel BP$ ; $C'\in AB\ ,\ MC'\parallel CP$. Prove that the centroid of the triangle $A'B'C'$ belongs to the line $MP$ if and only if $xyp(n^{2}-m^{2})+yzm(p^{2}-n^{2})+zxn(m^{2}-p^{2})=0$, i.e. $\sum_{\mathrm{cyc}}\frac{1}{x}\left(\frac{p}{n}-\frac{n}{p}\right)=0\ .$ [/quote]
Thank you Virgil Nicula, that is a very good idea, another consequence(like simson line) with any $M\in$ Steiner's Ellipse, draw $MA'\parallel GA,A'\in$ line $BC$ similar $B',C'$ we wiil have $A',B',C'$ are collinear (in the line called $l$, more image of $l$ by $h(M,2)$(homothetic) will be through $G$
Thank you again Virgil Nicula!


For $a, b, c, d >0$ prove the following:

\[\left(\frac{abc+abd+bcd+acd}{4}\right)^{\frac{1}{3}} \leq \sqrt{\frac{a^2+b^2+c^2+d^2}{4}}\]

probability 101, what's [hide]power mean[/hide]? and what's [hide]AM-GM[/hide]?

I didn't do it that way but they are well known inequalities.

AM-GM for two numbers $\Longrightarrow \frac{a+b}{2} \geq \sqrt{ab}$

I get it! AM stands for arithmetic mean, and GM stands for geometric mean.  I declare that I have now made up a new thm!  I call it HM-AM-GM (The HM stands for harmonic mean).   :)

[quote="bubala"]I declare that I have now made up a new thm!  I call it HM-AM-GM (The HM stands for harmonic mean).   :)[/quote]

You're a few hundred years too late. Sorry mate.

QM (Quadratic Mean) $\ge$ AM $\ge$ GM $\ge$ HM

Power Mean (off the top of my head, correct if I messed something up please) is as follows

Given positive reals $a_1, a_2, \ldots a_n$, $p \ge q$, then

$\displaystyle (\frac{a_1^p + a_2^p + \ldots a_n^p}{n})^{\frac{1}{p}} \ge (\frac{a_1^q + a_2^q + \ldots a_n^q}{n})^{\frac{1}{q}}$

As you can see, making the exponent and root 2, 1, 0, -1 give the QM, AM, GM, and HM respectively. (where GM is specially defined)

Since nobody actually posted a solution...
[hide]
From the Power-Mean Inequality, we have
$\sqrt{\frac{a^2+b^2+c^2+d^2}{4}} \le \sqrt[3]{\frac{a^3+b^3+c^3+d^3}{4}}$

So we need to prove $\left(\frac{abc+abd+bcd+acd}{4}\right)^{\frac{1}{3}} \le \sqrt[3]{\frac{a^3+b^3+c^3+d^3}{4}}$

or

$abc+abd+bcd+acd \le a^3+b^3+c^3+d^3$.

From here, we can multiply both sides by $3$ and rearrange to get $0 \le \sum (a^3+b^3+c^3-3abc) = \sum (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ which is true by the almighty trivial inequality.[/hide]

[quote="fuzzylogic"][quote="paladin8"]Since nobody actually posted a solution...

So we need to prove $\left(\frac{abc+abd+bcd+acd}{4}\right)^{\frac{1}{3}} \le \sqrt[3]{\frac{a^3+b^3+c^3+d^3}{4}}$
[/quote]
This is weaker than the original inequality[/quote]

No... its stronger by the Power-Mean Inequality...

This was discussed before.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=12892

No--although \[\left(\frac{abc+abd+bcd+acd}{4}\right)^{\frac{1}{3}} \le \sqrt[3]{\frac{a^3+b^3+c^3+d^3}{4}}\] since the original QM is less, it's still incomparable to the thing to which we wanted to compare it.

Whoops. Got my Power-Mean numbers mixed up.

Since I'm embarassed that I couldn't solve the problem the first time around (see fuzzylogic's link) I will redeem myself by offering a quick muirhead solution (I think I've posted this somewhere else before)

It's enough to show that

$\left(\frac {\sum abc}{4}\right)^2\leq \left(\frac {\sum a^2}{4}\right)^3$

Taking out constants, this is just

$4\left(\sum abc\right)^2\leq \left(\sum a^2\right)^3$

Expanding the LHS, we get

$4\left(\sum_4 a^2b^2c^2+2\sum_6 a^2b^2cd\right)$

Expanding the RHS, we get

$\sum_4 a^6+3\sum_{12} a^4b^2+6\sum_4 a^2b^2c^2$

But all of the symsums in the RHS are $\geq$ the individual symsums in the LHS, so the result follows immediately from Muirheads.  This way requires no ingenuity, and even symsums are easy to multiply out (e.g, I did them in my head and counted terms at the end to make sure that they matched)

[quote="tetrahedr0n"]Sorry, its more like

PM $\geq$ AM $\geq$ GM $\geq$ HM[/quote]
I think you can add to this
$PM \geq AM \geq LM \geq GM \geq HM$
where $LM$ stands for the logarithmic mean.
for two numbers $a$ and $b$, $LM=\frac{b-a}{\ln(b)-\ln(a)}$

[hide]By rearrangement, we get that

abc + bcd + acd + abd  <= a :^3: + b :^3: + c :^3: + d :^3:

Im sure you can go from there

JB[/hide]


help! probability with cards is really confusing

Two cards are drawn at random from a 52 card deck.  Find the probability of each event.
a) both are hearts
b) both are jacks
c) neither is red
d) neither is a spade

So, for the first one,do i find the probability of the card being red, then that both of them are red?  like $ \frac{(4c1)(13c2)}{(52c2)}$

If you have any tips/rules on probability, that would be greatly appreciated.

Probability with cards(however annoying) isn't that bad once you do tons of them.

Two cards are drawn at random from a 52 card deck. Find the probability of each event.
a) both are hearts
b) both are jacks
c) neither is red
d) neither is a spade

a) There are 13 hearts in a deck. So for the first card, the probability of choosing a heart is $ \frac{13}{52}$. To choose a second one is $ \frac{12}{51}$. Now we just multiply:

$ \frac{13}{52} \cdot \frac{12}{51} \equal{} \frac{1}{17}$

So our answer is $ \boxed{\frac{1}{17}}$

b) There are 4 jacks in a deck. For the first card we have $ \frac{4}{52}$ and for the second we have $ \frac{3}{51}$. Simply multiply:

$ \frac{4}{52} \cdot \frac{3}{51} \equal{} \frac{1}{221}$

We have $ \boxed {\frac{1}{221}}$.

Now, why don't you try the next two? (HINT: It might be easier to count the probability of not having a red by finding the probability of actually having a red and subtracting this from 1. The same goes for not picking a spade.  :wink: )

Please ask my any questions you have!  :)

Because the words "with/without replacement" are not stated in the problem, we use the combination formula to solve, since the absence of those words suggest that the cards are drawn at the same time. Therefore, we would solve the problems like this:

1. The probability that both cards are hearts is the number of ways to choose two hearts from the deck over the number of ways to choose two cards from the deck. Using combinations, this is (13C2)/(52C2), which simplifies to 1/17.

2. Using similar logic, the probability that both are jacks is (4C2)/(52C2), which simplifies to 1/221.

3. This is the same as both being black. So, the probability is (26C2)/(52C2), which is 25/102.

4. This would be (39C2)/(52C2), which simplifies to 19/34.

As a tip, I would suggest using combinations as much as possible. It helps to keep things organized. Also, remember that probability is always the number of favorable outcomes over the number of possible outcomes.

Without replacement: for #4

First chance of not getting a spade: $ \frac{3}{4}$
2nd chance of not getting a spade: $ \frac{38}{51}$
$ \frac{3}{4}*\frac{38}{51}$
$ \frac{19}{34}$

With replacement:
Total non-spade cards: 39. Pick 2 from them.
$ \binom{39}{2}$
Total possibilities: $ \binom{52}{2}$
$ \frac{39 \times 38}{2}\times\frac{52 \times 51}{2}$
$ \frac{19}{34}$


topology
geometry
function
complex analysis
complex analysis unsolved

edoo it's for you again! :)

$ \Omega$ is an open subset of $ \mathbb{C}$, $ f$ is analytic in $ \Omega$ and $ \gamma$ is a closed curve in $ \Omega$ and $ \gamma \sim_{\Omega}0. \ \ |f(\zeta)| \leq 1, \ \forall \zeta\in\gamma^{\star}, \ z_{0}\in\Omega$ and $ \eta(\gamma; z_{0})\neq 0.$

(i) If $ A$ is a connected component of $ \hat{\mathbb{C}}\backslash\gamma^{\star}$ which contains $ z_{0}$, then prove that $ A\subseteq\Omega$ and $ \partial A\subseteq\gamma^{\star}$

(ii) Prove that $ |f(z_{0})|\leq 1$.

($ \gamma^{\star}$ is the path of the curve $ \gamma$)

This problem is almost obvious (if you draw a picture for example) but I would like a detailed solution.

Alexandros

Since I have been called out, I will weigh in with my thoughts  :lol:

I think since $ \gamma \sim_{\Omega}0$, the curve must not self-intersect.  (This gets into algebraic topology which is antipodal to my area of interest; but I recall the "figure-8" not having a trivial fundamental group).  So if this is true, the Jordan curve theorem applies meaning that the curve separates the plane into two distinct connected components.  Should $ z_{0}$ lie in the unbounded component, then $ \eta(\gamma;z_{0})=0$ which contradicts your assumption.  Therefore $ z_{0}$ is in the other component which is bounded by the curve $ \gamma$ which means it is in $ \Omega$.  Additionally the boundary of this component is just $ \gamma$ which finishes up part (i).

For part (ii), this just follows immediately from the Maximum Modulus theorem given the result of part (i) with equality holding iff the function is constant.

I hope that was helpful.


See [url=http://www.mathlinks.ro/viewtopic.php?p=1333078#1333078]here[/url] post #5


TOday i come across this seemingly easy question:::

Find the number of solutions to the equation

tanx=2cosx+1

where x is between 0 to 360

without actually solving them.

How do you propose to solve it?

[hide]You can convert the two functions with the parametric formulas[/hide]

You can graph $y=\tan{x}$ and $y=2\cos{x}+1$ and count the number of intersections.

A graph can do?? (Yes, i think it does)

If resolve into t parametric, it will cost a lot of time, but it do the trick

thnks


algebra
polynomial
number theory unsolved
number theory

Let $ F(r) \equal{} \displaystyle\sum_{k \equal{} 1} ^{p \minus{} 1}k^r$, where $ 1\le r\le 2p \minus{} 1$, find $ F(r)$ mod $ p^2$.

$ 1^r, 2^r, 3^r, ... (p \minus{} 1)^r$ are the roots of the polynomial

$ x^{\frac {p \minus{} 1}{\gcd(p \minus{} 1, r)} } \minus{} 1 \equiv 0 \bmod p$

and the sum of the roots here is clearly $ 0$ (except when $ r \equal{} p\minus{}1$, in which case the sum of the roots is $ \minus{}1$). :)

It is true, if $ p\minus{}1\not |r$.
If r>1 odd, then $ p^2|F_r$.

[quote="Rust"]It is true, if $ p \minus{} 1\not |r$.
If r>1 odd, then $ p^2|F_r$.[/quote]
Yes, that's the original problem.
How do u prove this?

For the case where $ r$ is odd, $ 1\leq r\leq 2p\minus{}3$, and $ p\neq 2$, we have
\[ S\equal{}\sum_{k\equal{}1}^{p\minus{}1}k^r\equal{}\sum_{k\equal{}1}^{p\minus{}1}(p\minus{}k)^r\]
\[ \Rightarrow 2S\equal{}p\sum_{k\equal{}1}^{p\minus{}1} \left( k^{r\minus{}1}\minus{}k^{r\minus{}2}(p\minus{}k)\plus{}\cdots \plus{}(p\minus{}k)^{r\minus{}1}\right)\]
Modulo $ p$, that sum is congruent to $ r\sum_{k\equal{}1}^{p\minus{}1} k^{r\minus{}1}$. If $ r\equal{}p$, it's clearly 0, otherwise $ p\minus{}1$ does not divide $ r\minus{}1$.

We already had this problem about ten times before  :wink:

Solution is :$ p \plus{} (p \minus{} 1)!$ (mod$ p^2$)