trigonometry
integration
analytic geometry
geometry

I have a plate. The plate has a curved edge while the rest is a straight line. The equation of the edge of the curve is given by $ y\equal{}\sin x, \; 0 \to \pi$ if a corner is placed at the origin with the straight part along $ X\minus{}Axis$.

Find co-ordinates of the C.M of the plate if the density of material is uniform throughout.

this isn't really a physics problem, mostly calc.
[hide]
the x-coord is obviously on the axis of symmetry.

they y-coord is just the weighted average of all the little points on there (just like discrete CM)
which is \[ \dfrac{\int_0^{\pi} { \frac{1}{2} f(x) \cdot f(x) dx }}{ \int_0^{\pi} {f(x) dx} }\]
where the first bit represents the y-coord of the CM of a single vertical strip
the second bit represents the weight of that strip
and the denom represents the total weight of teh whole thing

which makes the whole expr. = the weighted average of the y-coords of the CMs of the strips, which is what we wanted.
[/hide]

If I did the calc right (big IF there), then it's

[hide]
$ \frac{\pi}8$
[/hide]

[quote="ffao"]If I did the calc right (big IF there), then it's
$ \frac {\pi}8$
[/quote]
Hello
Rajiv is asking for coordinates of CM.We can easily say just by seeing $ abscissa \equal{} \pi/2$.
Thank u.

[quote="kabi"][quote="ffao"]If I did the calc right (big IF there), then it's
$ \frac {\pi}8$
[/quote]
Hello
Rajiv is asking for coordinates of CM.We can easily say just by seeing $ abscissa = \pi/2$.
Thank u.[/quote]
Obviously the answer is ${ \left(\frac{\pi}{2}, y_1 \right)}$. The real challenge lies in finding the ordinate, $ y_1$, and that is what I couldn't do. I'll try $ \mathfrak {bowei's}$ method. Maybe I'll get the answer. If anyone does get the answer, please post it here.

[quote="bowei"]
\[ \dfrac{\int_0^{\pi} {\frac {1}{2} f(x) \cdot f(x) dx }}{\int_0^{\pi} {f(x) dx} }
\]
[/quote]
I cant understand what you did here. Pl tell me a general formula.

Thanks in advance,
-Rajiv.

I posted $ \frac {\pi}8$, because I thought the x-coordinate was obvious... Let me elaborate:

$ \overline{y} \equal{} \frac {\int_0^1 y(\pi \minus{} 2\arcsin{y})\ dy}{\int_0^\pi \sin{x}\ dx} \equal{} \frac {\frac {\pi}4}2 \equal{} \frac {\pi}8$

We take the area weighted average of all thin strips $ y$, which means

$ \frac {\sum y A_y}{A_{total}}$

Area of a thin horizontal strip: $ (\pi \minus{} 2\arcsin{y})\ dy$
Total area: $ \int_0^\pi \sin{x}\ dx \equal{} \minus{} \cos{\pi} \plus{} \cos{0} \equal{} 2$.

The rest is self-explanatory (unless, of course, my answer is wrong).

Wouldn't it be "easier"  bowei's solution? (obviously, $ f(x) \equal{} \sin x$)

Most definitely. It's just that I did the problem before he posted...


Post any brilliant poems here.

An acrostic:

Many
Epicly
Worthless
Tricky
Octopi

They don't have to make sense btw!

[url=http://www.youtube.com/watch?v=RSsJ19sy3JI]Rick Astley
In 1987[/url]
[url=http://tinyurl.com/c2g48b]C2g48b = tinyurl[/url]
[url=http://www.thekickback.com/rickroll/rickroll.php]Killing the tab causes a barrage of popups
Really? <sarcasm>
On Youtube
Led to many parodies[/url]
[url=http://www.youtube.com/watch?v=Kr2jlCyCJBI]Like this:[/url]

Ok, I know, it's not one word only.

Oh this is totally beast.

Nature's first green is gold,
Her hardest hue to hold.
Her early leaf's a flower;
But only so an hour.
Then leaf subsides to leaf.
So Eden sank to grief,
So dawn goes down to day.
Nothing gold can stay.

Last line familiar to anyone?

@justinahmann Yeah, its by Robert Frost, also in the Outsiders...

ok. I am a poetry best. like not even kidding. I phail at math and pwn at poetry now.

[u]Gold Digger[/u]
I never thought that I’d pride
Myself in hiding inside, quiet eyed
Then again,
Who knew you, too,
 could do few things truly true
And yet,
Me, me, me among debris
Disagree thee could flee, free.
You,
Czar, star, cigar-bizarre,
you mar, bar, your very own scar
You could
Hurt dirt, you alert and desert,
Exert goodness, yet convert and revert,
You live
In a cage, gauge age, guessing where the end page lay
Stage sage rage, for a man to sway
In order to
Get another old, told, fool with gold
Sold your last, when the hand hold fell cold
You said
You’d bring youth, but I’m a sleuth with a sweet tooth for the truth
And Ruth, oh Ruth, you run nothing more than a money booth.

[u]Tickling trees: Overwhelming Alliteration[/u]
To tell this tale, the Tickling trees
This task takes time, talent,
Take this time to tear thee, to turn
   Trenches that tell terrible tales tempt thee.

Running, really, rooted, rubbishly
Ruddy, River-dwelling rover rearing,
Reaching, reaching,
Rocky, rigid, Redwoods

Each eager, eating, eating
Earthy, ecru, ecosystem entity
Elite, egotistical, enchantments
Elevated ember enemies

Emancipated emblems,
Emphatic, emerald, emissaries
Esteemed, erupting
Evading evacuee

Standing, serious
Serviceable, serene, scarce
Scandalous specters
Still.

GO LIMERICKS. THEY ARE BEAST.

Doc

Chatting with my aunts
His hand down my pants
Doc said that I
Ought to and try
To keep from nuclear plants.

Tough Test

I said I had enough
Of the test that I found tough
Paper aflame
Me to blame
My hands are now in cuffs.

I'll leave you guys to interpret these.
My teacher said these were good and shes sending them to some contest.

[quote]
His hands down my pants
[/quote]

[size=200]W. T. F.[/size]

M olesting everybody all the time
E verytime he thinks pervertedly, the...
W orld blows up....
T his is an excellent poem that shows that Mewto is
O bvioiusly not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not
not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not
not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not
not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not
not not not not not not not not not not not not not not not not not not not not not not not not not not not
not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not
not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not not a Noob.

Meh you think that is an odd or even number of nots? My money is on even.

Also, I don't molest anyone...

It's an odd amount... it's 373 "not"s... I used word count...
Make a limerick!

$ \frac {12 \plus{} 144 \plus{} 20 \plus{} 3\sqrt4}{7} \plus{} [(5)(11)] \equal{} 9^2 \plus{} 0$
[hide]Sum a dozen a gross and a score,
Then add three times the square root of four,
All divided by seven plus five times eleven
Equals nine squared plus naught more[/hide]
My teacher made that up!

Hehe, mewto, you owe joyofpi and me 10 bucks [b]EACH[/b].

//Um your teacher didn't make that up. He/she found it on the internet, seeing as my algebra teacher also used that.//

haha PWNED.

Maybe he published it on the internet =O

O.o Fine lets edit that to I FOUND A PROBLEM... How are you going to send it over? by mail? plane? ship? jet? but then you'd have to be a stalker to find our addresses... I think...

i've a creeper van out back, wanna, uh, take a look?

Perhaps smileys count as an image file when parsed, so are much larger than they look.

[quote="dnkywin"][code][b][/b][b][/b][/code][/quote]
Oh please. You can check that I'm not doing this.

[quote="dnkywin"]
WHITE SPACE

[/quote]

what does your signature say?

[b]N[/b]ever stop failing
[b]I[/b] always fail
[b]K[/b]ing of the failz
[b]E[/b]veryday, I fail

The smilies below have the answer...
[hide=" :ninja: "][hide=" :D "][url=http://www.thekickback.com/rickroll/rickroll.php]:huh:[/url][/hide][/hide]


[color=darkblue]Given are a fixed circle $w$ and two fixed points $B,C$ which belong to the circle $w$. Denote the intersection $A\in BB\cap CC$. For a mobile line $d$, $A\in d$ and $d\cap w\ne \emptyset$ denote the intersection $d\cap w=\{M,N\}$. Ascertain the geometrical locus of the middlepoint of the segment $[MN]$.[/color]

$P$ be the midpoint of $MN$
$O$ is the center of circle $w$

$MN\perp PO$ since $MP=PN$, $OM=ON$ and $OP=OP$, $\angle OPN=\angle OPM$

$A$ and $O$ are fixed, and it is basic knowledge that since $\angle APO=90^\circ$, then then $P$ lies on the circle, $\omega$, with $OA$ as a diameter.

Now for the converse, all points on that circle are in the locus.

Consider a point, $P$, on $\omega$ different from $A$, $O$. Then $\angle APO=90^\circ$. Extend the line to hit the circle at $B$, $C$. Then $P$ is indeed the midpoint of $BC$ using HL congruency for $\triangle BOP$ and $\triangle COP$.

If $P=O$, then it clearly $P$ is the midpoint of $BC$ as it is a diameter.

If $P=A$, then we cannot draw line $PA$, however, consider the tangent to $\omega$ at $P$. Since $PO$ is a diameter, and $BC$ is tangent at $P$, $BC\perp PO$. Then $P$ is the midpoint of $BC$ so it is in the locus also

Thus the locus, $\omega$, is the circle diameter $OA$.


geometry
geometric transformation
reflection

I don't know how long this topic will last but I might as well see.  :|

Rules:

1. Discussing puzzles with other people is okay.
2. PM me the answers to the puzzles that are given.
3. Spam is [b][u]not[/u][/b] allowed in this forum.
4. [b]NO[/b] hints willl ever be said by me.
5. The right answer to a puzzle would be given in this forum.
6. Read rules 1 through 5.

To start, I will give the easiest kickself I have ever known:

What was the first human artifact to break the sound barrier?

This might possibly maybe belong in the puzzles and brainteasers forum...

another easy but famous one:

There is a river you need to cross, and you need to bring a wolf, a sheep, and a block of dirt with grass on it with you. To go across it, you have a boat that can only carry 1 object, not including yourself. The wolf eats the sheep, and the sheep eats the grass. If you leave 2 of them that eat the other on the other side, you will fail. How do you get all objects to the other side?

Bring the sheep over, go back.

Bring one of the others over, take the sheep back.

Take the third one over, go back.

Take the sheep over.

New puzzle:

[size=150][b][i]The Elusive Lift[/i][/b][/size]

I was on the second floor of a New York department store recently. "Let's take the lift," said my wife. "It won't be long; it's on the thrid floor and coming down." We waited, but to our annoyance it shot straight past us to floor one, then up to the fourth floor, straight back down to the first again and then back up to the fifth. "This is useless," I said. "Let's walk." "No, hang on for a few more minutes,"said my wife. "It willl probably continue straight up to the ninth floor and then come straight back down to us." We waited, and sure enough the lift did exactly as my wife had predicted.  How did she know? :ewpu:   :huh:

Was there anything special on the ninth floor?

[quote="mz94"]and 2 is next!
its the sequence of pi![/quote]
MAN YOU'RE GOOD HOW'D YOU GET THAT

[quote="mz94"]whadya mean? its b/c of yr thing i found that out.
if its abt how i knew that 2 was next, memorization helps[/quote]

Lol what he meant was he had already posted the solution, and you wrote it as if he had come up wtih it partially and you completed it. Anyway, never mind, it's good to see solutions and say them out loud.

An Isaac Aasimov classic:
There was a man called Sloan who had a robot called Teddy which hardly moved at all. His friend Bob had a robot which moved very slowly if it faced a sugar pile. Sloan entered Teddy in a race against Bob's robot (I forgot its name). Everyone thought that Teddy would lose, and so it seemed for some part of the race when Teddy sat tight while its competitor had started moving, albeit slowly. But Sloan seemed sure of his robot. When the other was halfway through, Sloan took a hammer and and told Teddy: "If you don't move now I'll smash you with this" and in a flash Teddy was across the finishing line, and the race was over.
How was Sloan so sure of Teddy's success?

[hide]The robot was voice-activated.  When Sloan said, "If you don't move now I'll smash you with this," that activated the robot and it moved very fast across the finish line.  Since people usually don't say something like that, the robot would hardly move.[/hide]

Nice try lingo, but the solution is much simpler really.

Probably Teddy didn't want to be smashed into pieces by Sloan so he raced to the finish line.

New puzzle:

[size=150][b][i]"Swiftly I Come"[/i][/b][/size]

by Jonathan Swift(1667-1745)

By something form'd, I nothing am,
Yet Wry thing that you can name;
In no place have I ever been,
Yet everywhere I may be seen;
In all things false, yet always true,
I'm still the same - but never new.
Lifeless, Life's perfect form I wear,
Can show a Nose, Eye, Tongue, or Ear,
Yet neither Smell, See, Taste, nor Hear.
All Shapes and Features I can boast,
No Flesh, no Bones, no Blood - no Ghost:
All colours, without Paint, put on,
And change, like the Cameleon.
Swiftly I come, and enter there,
Where not a chink lets in the Air;
Like thought, I'm in a Moment gone,
Nor can I ever be alone:
All things on earth I imitate
Faster than nature can create;
Sometimes imperial Robes I wear,
Anon in Beggar's Rags appear;
A Giant now, and strait an Elf,
I'm Wry one, but ne'er myself;
Ne'er said I mourn, ne'er glad rejoice,
I move my Lips, but want a Voice,
I ne'er was born, nor ne'er can die,
Then, prythee, tell me what am I?

Glossary:

Cameleon- Chameleon
Colours- Colors
Ne'er- Never
Prythee- ???
Wry- Everyone or Everbody

[hide]is it a shadow?

its got human's form, but isnt really there...[/hide]

next riddle: not veery hard...

Fragile Name

What is so fragile that when you say its name you break it?

mz94, that was not the answer to Swiftly I Come.  The real answer was a reflection in a mirror.

prythee means 'I pray to you', I came across it in julius caesar (ok it was prithee but the meaning is the same)
Answer to the Sloan riddle:
Sloan knew the old proverb - 'Sloan's teddy wins the race."  :D

I have another elevator riddle:

I live on the 30th floor of an apartment building.  When I go to school every day, I take the elevator down to the lobby.  On the way back from school, I ride up to the 25th floor and walk the remaining 5 flights.  On days that it's raining, however, I take the elevator all the way up.  Why?

[hide]
Umm... You're too short to normally reach the 30th button, so you go to the 25th floor, but on rainy days you have boots, so you can reach?
[/hide]

Or he has the raincoat ... hang on, how would that help? Maybe if it was hard enough he could roll it up and use it. :maybe:

If he was too short, why doesn't he just take a stick and poke the buttons  :D

He was not allowed to have one on his person, unless it rained.

Yes, the too short thing and the umbrella was correct.

Adam found a suit and tie in the attic. Because of that, 157 people died. Explain, please.

adam didn't find the bomb that blew up the garden party below minutes later.

Ummm...he wore the suit and tie. When he went outside everyone on his block died because he looked so good? :)

um... The scenario that I'm thinking of can be a real life scenario...

Inside one of the suit's pocket's there was a list of people who were terrorists (157 to be exact) and there was proof with it, adam took the proof and the list to the FBI, who sentenced the terrorists to a death sentence? :D

ok, it has something to do with terrorists, but that scenario isn't right.

hint 4: lifestyle plays a part in this.

It was the president's suit, and he wore it to a public location, a terrorist saw him, and blew him and other people up because he thought Adam was the president.  :huh:

[quote="bubka"]It was the president's suit, and adam was a terrorist, he wore it to a White House meeting (with a mask on probably), everyone though he was the president, and then he blew everything up.[/quote]

[quote="1=2"]1. The suit and tie was completely safe, and Adam was a sane person who didn't have murder on his mind.[/quote]

Fine. As soon as he found the suit he was jumped upon by an impostor, who tied him up and took the suit. Then on we call this new man Adam, and my story follows.

Did somebody mistake Adam for someone else?

My guess would be that terrorists thought that Adam was someone that he wasn't, and bombed his workplace.

Or, the suit was infested with smallpox :P


$ Given$ $ 0\leq a,b,c\leq1$

$ prove$ $ that$ : $ a^2\plus{}b^2\plus{}c^2\leq a^2b\plus{}b^2c\plus{}c^2a\plus{}1$

[quote="Evariste-Galois"]$ Given$ $ 0\leq a,b,c\leq1$

$ prove$ $ that$ : $ a^2 \plus{} b^2 \plus{} c^2\leq a^2b \plus{} b^2c \plus{} c^2a \plus{} 1$[/quote]

$ \Leftrightarrow a^2(1\minus{}b)\plus{}b^2(1\minus{}c)\plus{}c^2(1\minus{}a) \le 1$

$ LHS$ - convex by $ a,b,c\equal{}>$ for $ max(LHS)$ $ a,b,c$ must be  $ 0$ or $ 1$

So $ max(LHS)\equal{}1$ holdes for ($ a\equal{}b\equal{}1,c\equal{}0$) or ($ a\equal{}1 , b\equal{}c\equal{}0$)


$a,b,c>0$, prove that
\[\frac{1}{3}(\displaystyle\sum_{cyc}a)(\displaystyle\sum_{cyc}\frac{b+c}{b^2+bc+c^2}) \ge \displaystyle\sum_{cyc}\frac{b^2+c^2}{b^2+bc+c^2}.\]

[quote]It is not true. See a=0, b=c=1. [/quote]

But then a isn't positive. :P

Toodles,

Vasc is right. Take $a$ small enough, and the inequality is not true anymore (off by $\dfrac 23$). Maybe the following inequality is the one ment:

[quote="yptsoi"]$a,b,c>0$, prove that
\[\frac{1}{2}(\displaystyle\sum_{cyc}a)(\displaystyle\sum_{cyc}\frac{b+c}{b^2+bc+c^2}) > \displaystyle\sum_{cyc}\frac{b^2+c^2}{b^2+bc+c^2}.\][/quote]

[quote="RealZeta"][quote]It is not true. See a=0, b=c=1. [/quote]
But then a isn't positive. :P[/quote]
It doesn't matter. You can take $a=0.01$, $b=c=1$.


An 11-member committee makes its decisions by simple majority vote: if 6 or more of the members vote in favor of an issue, the issue is passed. In how many ways can 6 or more members vote to pass an issue?

There are $ 2^{11}$ total ways the committee can vote, because each person can say either "Yes" or "No".  Note that the number of ways $ 6$ or more members vote to PASS an issue is the same as the number of ways $ 6$ or more members vote to NOT PASS an issue.  Thus, we divide by $ 2$, to get $ \frac{2^{11}}{2}\equal{}2^{10}\equal{}\boxed{1024}$.


[hide]Tom and Huck paint 1/11 of the wall each hour together, so by the time Jim joins them, they have painted 4/11. They finished the remaining 7/11 in 2 hours, or 7/22 each hour. Since Tom and Huck and only work 1/11 an hour, Jim must have worked 7/22 - 1/11 or 5/22 every hour to get the job done. The whole job wouldn've taken him 1/(5/22) or 22/5 hours. [/hide]

[quote="4everwise"]Tom and Huck paint a fence for four hours, after which Jim helps them and they finish two hours later. If Jim had not helped them, it would've taken them five more hours to paint the fence. How long would it have taken Jim to paint the fence alone?[/quote]

they arent painting 1/11 every hour.. its 1/9 every hour. Tom and Huck both do 1/18 every hour. Jim works 3 hours worth of what they would work in 2 hours i think....  (5-2) (assuming they all work individually..)   they would work together 1/9.. and so in two hours he does 3/9 of the total work, or 6/18.. so each hour he works at 3/18.. it would take him//

[hide]6 Hours[/hide]

im not quite sure how I got it, but I got

[hide]

5hrs
[/hide]

[hide]Tom and Huck both do 1/18 every hour, jim does 3 hrs in what Tom and Huck would do in 2hrs together.  So in 2 hrs Jim does 3/9 of the work.  Half of that is 3/18, so the answer is [b]6[/b][/hide]


Ok, there are a lot of contest going in this forum.

So, here's a deal.

If you're interested on making a contest or currently in progress, please post below with your contest's name and brief description.

Thanks.

This is a request page. If it's not request on asking to put your contest up on here, DO NOT POST.

JMC: http://yashar.biz/jmc/

The Jifted Mathematical Competition

 :P

Anyone else??

If this is only test in here that's active, I'll remove this thread and just make the JMC thread stickied.

I'm considering posting a contest, but how many people are interested?
EDIT:
I haven't actually fully designed the contest yet, but this is the general format:

There will be 5 questions in each of the 3 groups A, B, and C. The group A problems will be on the level of the harder (around National) Mathcounts questions, and could be solved by a relatively gifted middle-schooler. All the group A questions will have numerical answers. You do not have to show work for any of the group A questions.

The group B problems will be slightly harder than the group A questions, and contain both proofs and numerical answer questions. You will have to show your work for the group B questions. These questions will be challenging, but I've seen harder on this forum.

The group C problems will be the hardest, and probably the level of the harder AIME questions, but will include proofs. Again, you'll have to show your work for group C problems.

You can choose one of the 3 groups of questions to complete, and PM the answers/solutions to me.
Part A will be marked out of 5 (you get 1 mark for a right answer, 0 marks for a wrong),

Part B and C will be marked out of 25 (you may get partial marks)

Of course, more clarification will be available when I post the contest.
So if you want to participate/have any suggestions, just reply. I want to have a good amount of how many people are interested before I actually post the contest.

[quote="white_horse_king88"]I suggest starting a new thread explaining your competition, and then waiting for responses.[/quote]

Exactly.

Also, Farazy is an exceptional case since he post the link to his website for the test but if you're posting here for the contest, you MUST put a brief explanation on it and before you even post here, post on the thread to see the popularity from other people.

Ok, I edited the post. Thanks for the reminder.

Ok.

But I want you to post this on DIFFERENT thread. I'm going to leave all the posts for just two more days. And then I'll remove all the threads except the one posted by Farazy.

People, I'll repeat this one more time.

To post in this thread you MUST:

1. Have a thread on the forum concerning about your competition (NOT ON THIS ONE!!)
2. You must have waited bit longer to see the people's attention, meaning posting on the forum doesn't necessarily mean your competition will be posted up here with hyperlink.
3. Brief description must be compiled (or if you want, post the link to the thread that has all these informations) if and only if you have done #1-2.

If this is not satisfied, your post will be deleted within one day.


[hide]

$5! = 120$ different combinations

[/hide]

im not absolutely sure, but is the answer[hide]$\frac{5!}{3!}=20$[/hide]????????

jorian

[quote="humblemastermind"][hide] Uh, it's just $\frac{5!}{2!\cdot 2!}=30$. This is like countdown level. [/hide][/quote]
[hide="hmmm"]He didn't specify 5 digit numbers.[/hide]
You might be right anyways though, I'm not sure what he meant.

[quote="humblemastermind"][hide] Uh, it's just $\frac{5!}{2!\cdot 2!}=30$. This is like countdown level. [/hide][/quote]

How come?

[hide]First, do $\binom{5}{3}$, which is $10$. This is the number of different combinations of 1, 2 and 3 there are. One of these, namely 123, has been counted 4 times. The actual number of combinations should be $7$. One of these 7, again 123, is the only combination with 6 permutations. The other 6 each have 3 permutations. $x=1\cdot6+6\cdot3=\boxed{24}$[/hide]

Umm I got 18 but since all the other posters got different answers I'm probably wrong.

[size=75][color=darkred]Edited by nebula42.[/color][/size]

Maybe if you posted your solution the "smarter ones" will be able to find your mistakes or correct their own? :)

Oops... I thought there were two 1's. Ubemaya and bpms are correct.

Ah thought it was how many 5 digit numbers...

[quote="anirudh"]How many different three digit numbers can be formed using the digits 1,2,2,3,3?[/quote]
[hide]
if the first digit is 1, then there are $(2)(2)$ ways, or 4.
if the first digit is 2, then there are $(2)(2)$ ways if 3 isnt picked for 2nd digit, and $3$ ways if 3 is picked for 2nd digit.
if the first digit is 3, there are also $(2)(2)+3$ ways.
so in all there are $4+4+4+6$, which is 18
[/hide]

Yes. I could do BPMS's method, but I was thinking of any oly combinatoric method? :maybe:


Last year I could have gotten an 8 or 9, instead of a 6, on the AIME if it weren't for careless mistakes.  Recently I tried a couple of old AIME's, and careless mistakes are a concern, making the difference between a 7 and a 12.  One thing I have started doing is writing out multiplication, addition and subtraction by hand in the margins of my scratch-work.

Any other suggestions on how to reduce the mistakes as much as possible?  This is a pretty significant obstacle for me, because I seem fairly often to make mistakes in algebra and arithmetic.  Also, my school makes us take the AIME at 7:45 in the morning and I am not a morning person.

Also, if I don't like how I do on the first one am I allowed to take both AIME's?

If you haven't yet, read the article [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Mistakes.php]here[/url]. Additionally [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=258239]this[/url] topic has a few good pointers. You cannot take both AIME's. As for becoming a morning person, wake up earlier and go to sleep earlier than you normally do the entire week before. Your body will become acclimated to the time difference.

You should definetly study a lot of past AIMEs as practice makes perfect. Also, try using a lot of scrap paper but write neatly and number your work to each corresponding question. It is very important to check and recheck. As the AIME is a 3 hour competition, do not rush, but wisely partition your time.

Totally agree. If you are talking just about the AIME, you should be checking a lot. In general, lots of practice can help you reduce and eliminate these stupid mistakes. Definitely read the article in the "resources" tab on the left section of the website.

It's almost certainly too late to start practicing this as it's hard to master, but periodically in every computation check to see if things just make sense. Do you have the right units digit? Right sum of digits? Right orders of magnitude? For instance: if you multiply two $ k$-digit numbers, you should only get a $ 2k$ or $ 2k \minus{} 1$ digit number.

Doing olympiad problems helps build this kind of intuition, which is why you will generally find high scorers on USAMO getting high AIME scores and seeming relatively unconcerned about stupid mistakes.

EDIT: That was not me testing you, that was me being :huuh:. I will console myself by thinking that if I had actually seen a $ k$ digit number and a $ 2k\plus{}1$ digit number I would have caught that immediately.

As good practice for catching careless mistakes, see if you can find the minor error in the previous post.

(MellowMelon was probably just testing us.)

I think I have it...[hide=":)"]Should be $ 2k$ and $ 2k\minus{}1$, instead of $ 2k$ and $ 2k\plus{}1$.[/hide]

[hide]$ 2k\plus{}1$ should be $ 2k\minus{}1$.[/hide]


When Mr. Smith cashed a check for x dollars and y cents, he received instead y dollars and x cents and found that he had two cents more than twice the proper amount. For how much was the check written?

[hide]
$ 100y\plus{}x\equal{}2(100x\plus{}y)\plus{}2$
$ 100y\plus{}x\equal{}200x\plus{}2y\plus{}2$
$ 98y\equal{}199x\plus{}2$
Consider the equation mod 98.
$ 3x\plus{}2\equiv 0 \mod{98}$
$ x\equal{}32$
$ y\equal{}65$

[/hide]

[quote="CircleSquared"][hide]
$ 100y \plus{} x \equal{} 2(100x \plus{} y) \plus{} 2$
$ 100y \plus{} x \equal{} 200x \plus{} 2y \plus{} 2$
$ 98y \equal{} 199x \plus{} 2$
Consider the equation mod 98.
$ 3x \plus{} 2\equiv 0 \mod{98}$
$ x \equal{} 32$
$ y \equal{} 65$

[/hide][/quote]
[hide="concluded (finishing off the thought)"]
Since $ x \equal{} 32$ and $ y \equal{} 65$, $ \boxed{\$ 65.32}$ was written on the check.
[/hide]


Prove that if $p = 4n+3$and $q = 8n+7$are both prime  then $q\mid{ 2^{p}-1}$.

Quadratic reciprocity.  Do you even have to know p is prime?


lets say we have the following two masses: $m_1,m_2$ on the incline with angle $\theta$ and coefficient of friction of $\mu$.
what is the range of angles that $\theta$ can change, s.t. the masses wouldn't move (answer should be in terms of $m_1,m_2,\mu$ and other stuff)?

[hide="My solution:"]For mass #2 to not move, the tension in the string must be exactly $m_2g$. The sum of gravitational force and tension of the string applied to mass #1 is $m_1g\sin\theta-m_2g$. This must be countered by the friction force or in other words, following inequality must hold:

$|m_1g\sin\theta-m_2g|<\mu m_1g\cos\theta$

From this you get two inequalities:

$m_1g\sin\theta-m_2g<\mu m_1g\cos\theta$ and
$m_2g-m_1g\sin\theta<\mu m_1g\cos\theta$.

These can be solved by substituting $\cos\theta$ with $\sqrt{1-\sin^2\theta}$ and you'll get two very unhealthy looking quadratic inequalities which are up to you to solve. Do you know of any better solution?[/hide]

[hide]I will post my solution later[/hide]

I got the same as bus. Thereafter ...

[hide]$m_2g > m_1g(\sin\theta - \mu \cos\theta) ---------------(1)$
$m_2g < m_1g(\sin\theta + \mu \cos\theta) ---------------(2)$

$\mbox{For (1), } \sin\theta - \mu\cos\theta < \frac{m_2}{m_1}$

Dividing everything by $V$,

$\sin\theta\cdot\frac{1}{V} - \cos\theta\cdot\frac{\mu}{V} < \frac{m_2}{m_1}\cdot\frac{1}{V}$

$\mbox{where, } \frac{1}{V} = \cos\phi,\ \frac{\mu}{V} = \sin\phi,\ \to \ \tan\phi = \mu.$

$\sin\theta\cos\phi - \cos\theta\sin\phi < \frac{m_2}{m_1}\cdot\cos\phi$
$\sin(\theta - \phi) < \frac{m_2}{m_1}\cdot\cos\phi$
$\theta < \phi + arcsin\left(\frac{m_2}{m_1}\cdot\cos\phi\right)$
=======================

And the other one is,

$\theta > -\phi + arcsin\left(\frac{m_2}{m_1}\cdot\cos\phi\right)$
========================[/hide]


Let $ \minus{}\sqrt 2\le x_0\le \sqrt 2\plus{}2$
$ x_{n\plus{}1}\equal{}1\plus{}x_n\minus{}\frac{x_n^2}{2}$
Prove that exist $ lim_{n\to \infty} x_n$

Consider $ f(x)\equal{}1\plus{}x\minus{}\frac{x^2}{2}$. If $ x_0\equal{}1\plus{}\sqrt 2\pm 1$, then $ x_n\equal{}\minus{}\sqrt 2 \ \forall n>0$.
Else $ lim_{n\to \infty}x_n\equal{}\sqrt 2$.

Set $ f(x)=1+x- \frac{x^2}{2}$
$ ->x_n+1=f(x_n)$
we can see that $ 0 \le x_n \le \frac{1}{2}$
$ ->f'(x) \le 1$
$ -> lim_{n\to\infty}x_n= \sqrt2$
$ done!$ :wink:

[quote="Rust"]If $ x_0 \equal{} 1 \plus{} \sqrt 2 \pm 1$[/quote]
I dont understand your solution since that lines,Sorry :blush:

[quote="Math pro"]
we can see that $ 0 \le x_n \le \frac {1}{2}$
[/quote]

Why?,I know I am very silly,but I can find it easyly(can you try $ x_0 \equal{} \minus{} \sqrt 2$ so $ x_n < 0$for all $ n$)
:

I know the technique which be used Lagrangre Theorem but I dont think it is useful here,It is a hard problem which appear in a TSTMO

[quote="Allnames"][quote="Rust"]If $ x_0 \equal{} 1 \plus{} \sqrt 2 \pm 1$[/quote]
I dont understand your solution since that lines,Sorry :blush:
[/quote]
Exactly if $ x_0\equal{}1\pm(1\plus{}\sqrt 2)$.

[quote="Allnames"]Let $ \minus{} \sqrt 2\le x_0\le \sqrt 2 \plus{} 2$
$ x_{n \plus{} 1} \equal{} 1 \plus{} x_n \minus{} \frac {x_n^2}{2}$
Prove that exist $ lim_{n\to \infty} x_n$[/quote]
Dear Allnames, we can solve as the following hint:
[hide="Hint"]$ u_{n \plus{} 1} \equal{} \frac {3 \minus{} (u_n \minus{} 1)^2}{2}$. Since $ \minus{} \sqrt 2\le x_0\le \sqrt 2 \plus{} 2$, we have that $ \minus{} \sqrt 2\le x_n\le \frac {3}{2}$  by induction.
After that, $ |u_{n \plus{} 1} \minus{} \sqrt {2}| \equal{} \frac {1}{2}|u_n \minus{} \sqrt {2}|.|u_n \plus{} \sqrt {2} \minus{} 2|$
$ \minus{} 2\le u_n \plus{} \sqrt {2} \minus{} 2\le 1$
$ \Rightarrow |u_{n \plus{} 1} \minus{} \sqrt {2}|\le |u_{n} \minus{} \sqrt {2}|$[/hide]

[quote="Mr Green"][quote="mr.danh"]$ |u_{n \plus{} 1} \minus{} \sqrt {2}|\le |u_{n} \minus{} \sqrt {2}|$[/quote]
I think problem is not end :D[/quote]
Thank mr danh for your hint
Mr Green,I think $ |u_{n \plus{} 1} \minus{} \sqrt {2}|\le |u_{n} \minus{} \sqrt {2}|$then $ |u_{ 1} \minus{} \sqrt {2}|\ge |u_{2} \minus{} \sqrt {2}|...|u_{n } \minus{} \sqrt {2}|\ge |u_{n\plus{}1} \minus{} \sqrt {2}|$so $ limu_n \equal{} \sqrt 2$
Thank mr danh

[quote="Allnames"][quote="Mr Green"][quote="mr.danh"]$ |u_{n \plus{} 1} \minus{} \sqrt {2}|\le |u_{n} \minus{} \sqrt {2}|$[/quote]
I think problem is not end :D[/quote]
Thank mr danh for your hint
Mr Green,I think $ |u_{n \plus{} 1} \minus{} \sqrt {2}|\le |u_{n} \minus{} \sqrt {2}|$then $ |u_{ 1} \minus{} \sqrt {2}|\ge |u_{2} \minus{} \sqrt {2}|...|u_{n } \minus{} \sqrt {2}|\ge |u_{n \plus{} 1} \minus{} \sqrt {2}|$so $ limu_n \equal{} \sqrt 2$
Thank mr danh[/quote]
Oh,no
If $ |u_{n \plus{} 1} \minus{} \sqrt {2}|\le \frac {1}{k}|u_{n} \minus{} \sqrt {2}|$ with $ k>1$ then $ |u_{n \plus{} 1} \minus{} \sqrt {2}|\le \frac {1}{k^{n}}|u_1 \minus{} \sqrt {2}|$
and $ Lim \frac {1}{k^{n}}|u_1 \minus{} \sqrt {2}| \equal{} 0$ then $ lim u_n \equal{} \sqrt {2}$
But $ |u_{n \plus{} 1} \minus{} \sqrt {2}|\le |u_{n} \minus{} \sqrt {2}|$ then $ |u_{ 1} \minus{} \sqrt {2}|\ge |u_{2} \minus{} \sqrt {2}|...|u_{n } \minus{} \sqrt {2}|\ge |u_{n \plus{} 1} \minus{} \sqrt {2}|$so $ limu_n \equal{} \sqrt 2$
It is wrong :D

sory,My solotion is wrong!
Mr green is true all name!
$ |u_{n \plus{} 1} \minus{} \sqrt {2}|\le |u_{n} \minus{} \sqrt {2}|$
you should remember that It is $ \le$ instead <
I think we should question mr danh!Can you give us some more hints,mr danh! :maybe:

[quote="Math pro"]sory,My solotion is wrong!
Mr green is true all name!
$ |u_{n \plus{} 1} \minus{} \sqrt {2}|\le |u_{n} \minus{} \sqrt {2}|$
you should remember that It is $ \le$ instead <
I think we should question Rust!Can you give us some more hints,Rust! :maybe:[/quote]
No,my friend,I think $ \le$or $ <$ is not important
With the hint of Rust,I think he only check with the case $ x_0 \equal{} 1 \plus{} \sqrt 2\pm 1$then he has $ x_n$=constant.But it is not enough,Mr danh,what is your opinion

I said, that if $ x_0\equal{}1\pm(1\plus{}\sqrt 2$, then $ x_i\equal{}\minus{}\sqrt 2, \ \forall i>0$.
If $ \minus{}\sqrt 2<x_0<2\plus{}\sqrt 2$,  then $ \minus{}\sqrt 2<x_n\le \frac 32 \ \forall n>0$ and $ \minus{}2\sqrt 2<u_n\equal{}x_n\minus{}\sqrt 2\le \frac 32 \minus{}\sqrt 2$.
Therefore $ U_{n\plus{}1}\equal{}u_n(1\minus{}\sqrt 2 \minus{}\frac{u_n}{2}$ satisfyed:
$ \frac{1\minus{}2\sqrt 2}{4}\le 1\minus{}\sqrt 2\minus{}\frac{u_n}{2}<1$ and $ |u_{n\plus{}1}|<|u_n|, n>0$ if $ u_n\not \equal{}0$.


Simplify $ \frac{(1\minus{}x)(x^5 \plus{} x^4 \plus{} x^3 \plus{} x^2 \plus{} x \plus{} 1)}{(1\minus{}x\plus{}x^2)(1\plus{}x\plus{}x^2)}$
[hide="I did"]
Multiplying out, I got
$ \frac{1\minus{}x^6}{x^4 \plus{} x^2 \plus{} 1}$.[/hide]

[hide="Hint"]Use grouping and then the sum of cubes formula.[/hide]

oh my god, I am so sorry, I lost track of the problems  :blush:


...i dont know how hard this is so im sticking to the getting started forum...

im going into my junior yr and for the past two yrs i've taken the AMC 12 and gotten around 110 each time, following a scoring guide that i devised for myself of the optimal # of questions to answer, when to guess, and when to erase an answer im unsure of... the aim of my guide was to pass(getting over 100), but both yrs i fell short of the USAMO by about 10-20 points. The AIME scoring is pretty straightforward, so now im trying to figure out how to maximize my score instead of just aiming to pass, any suggestions?

Now you have crossed the score optimization threshold into the threshold of doing more problems to boost your problem solving skill level. I think this forum is a great place to start. Do more old tests, and read the AOPS novels - then tests yourself again and watch as your score increases even without thinking too much about score optimization. The more you do problems, the more you will know when you have the correct or incorrect answer.

I'd imagine you'd like a straightforward answer like 'do as many problems as you can, and guess on the easiest 3/8 of the remaining ones' but unfortunately there is no answer like that.  (The best I could give is "do as many problems as you can, skip problems if you can't answer them quickly, check your answers, and guess if you can eliminate one or more answer choices" but you hear that from every standardized-test prep book and college counselor anyway.)

Instead, I'll offer this: read books like The Art of Problem Solving and do past problems from the AMC and other contests.  In other words, the best way to maximize your score is to learn and practice, rather than to determine an exact strategy.  (By the way, a helpful article for preparing for the AMC may be found [url=http://goodlymaths.0catch.com/contests/amc/amc_how_to_prepare.htm]here[/url].)

You should only guess if you have solved 24 problems and you don't know how to do the 25th.

Now that that's taken care of, I'll try to answer the question in a more helpful manner. If you think you can solve a problem and you start working on it, then solve it. If you spend five minutes on a problem and then give up, then you have achieved nothing but to waste five minutes. Since you can't afford to waste much time on the AMC12, you have to be very economical about these things. Take a look at a problem and decide if you can solve it quickly (where you can choose how ``quickly'' is defined). If you can, then solve it. If not, then skip it.


How many lines of symmetry can be drawn in a cube?  :lol:

[hide="???"]
none, since a cube wouldnt have a 'line' dividing it into two equal symmetric parts.. you would instead have what im guessing would be a 'plane' of symmetry?

I'm not sure really
[/hide]

how about how many planes of symmetry?  :lol:

[hide="Hint(s)"]
There are [hide="_"]$4$[/hide] lines of symmetry in a square.

There are [hide="_"]$6$[/hide] square faces of a cube.

There are [hide="_"]$4$[/hide] diagonals to a cube.
[/hide]

[hide]3+6=9

3 comes from the planes that are parallel to two ofthe faces

6 are the planes connecting opposite diagonals of the squares.[/hide]

did i miss any?

Yes, you missed some.  I don't have a 3D drawing program, but I'll try to describe it.

The front four corners from top left clockwise are $A_{1}, A_{2}, A_{3}, \text{and }A_{4}$ and the ones on the back are $B_{1}-B_{4}$ ($B_{i}$ is behind $A_{i}$).  Connect $A_{1}$ and $B_{3}$.  There's a plane of symmetry.

[quote="miyomiyo"][quote="jli"]how is that a plane of symetry?  you described a line :maybe:[/quote]
There is a plane passing through there that is a plane of symmetry.

[quote="mathnerd314"]Why is it called "a trick question"?[/quote]
Because the original question asked for a line of symmetry in a cube which is a trick question.[/quote]

i dont see it...

I'm bad at these 3D visualiation things so you may be right...


[quote="Peter"]Show that there exists a bijective function $ f: \mathbb{N}_{0}\to \mathbb{N}_{0}$ such that for all $ m,n\in \mathbb{N}_{0}$:
\[ f(3mn+m+n)=4f(m)f(n)+f(m)+f(n). \][/quote]
So $4f(3mn+m+n)+1=(4f(m)+1)(4f(n)+1)$
Let $\mathbb{N}_{a,b}=\{n: n\equiv a\mod b\text{ and }n\in\mathbb{N}\}$
Let $\mathbb{P}_{a,b}=\{n: n\text{ is a prime number congruent to } a\mod b\}$
Let the function $g: \mathbb{N}_{1, 3}\rightarrow \mathbb{N}_{1, 4}$ such that $g(3n+1)=4f(n)+1, \forall n\in \mathbb{N}_{0}$
Then $g((3m+1)(3n+1))=g(3m+1)g(3n+1)$

Define the function $h:\mathbb{N}\rightarrow\mathbb{N}$
Let $h$ be a bijection from $\mathbb{P}_{1, 3}$ to $\mathbb{P}_{1, 4}$
Let $h$ be a bijection from $\mathbb{P}_{-1, 3}$ to $\mathbb{P}_{-1, 4}$
Let $h(xy)=h(x)h(y), \forall x, y\in\mathbb N$
Then $h$ is a bijection from $\mathbb{N}_{1, 3}$ to $\mathbb{N}_{1, 4}$ such that $h(xy)=h(x)h(y), \forall x, y\in\mathbb N_{1, 3}$

By letting $g=h$, $f$ is a bijective function.

Hence there exists infinitely many bijective functions $f$.

Let a function $g:3\mathbb{N}_0+1\to 4\mathbb{N}_0+1$ be such that $$g(x)=4f(\frac{x-1}{3})+1 ~\forall x\in 3\mathbb{N}_0+1.$$ Surely $g$ is multiplicative.

Now for the example of such a $g.$ Let $S_1,S_2,S_3,S_4$ be the sets of primes $3n+1,3n+2,4n+1$ and $4n+3.$ Note that these sets are infinite. Also let $\phi$ be the bijection from $S_1\cup S_2$ onto $S_3\cup S_4$ that maps $S_1$ onto $S_3$ and $S_2$ onto $S_4$ bijectively. Construction: $$g(1)=1 \text{ and } g(x)=\prod \phi(p_i), \text{where } x=\prod p_i.$$ Clearly $g$ is bijective and multiplicative. Moreover, it is not hard to see that is well defined as well.

So such a $g$ exists implying the claim.


All my old posts still have my new signature on them, but no new posts do.  I am willing to and would like to change my signature that appears on my old posts to something less annoying, but I have lost rights over my signature in both post options and profile>signature options.  Could someone unlock my signature and please tell me why it was locked?

Thanks.

Most likely it was locked because it was too obnoxious and annoying (not to mention large).

I think you might be given a second chance, so don't violate the rules again (the admins are usually nice  :) ).

I've reactivated your signature. Make sure you listen to 7h3.D3m0n.117 above. Also put in a real email address (you can hide it from everyone else), that always helps.

Thanks.  I'll make sure I don't do that again. :)

On a tangentially related topic, Valentin, user (^_^)'s signature is really annoying -- could something be done about it?


function
geometry
geometric transformation
reflection
algebra unsolved
algebra

Find all continuous functions $f$, define on $R$,satisfying the condition :
$f(f(f(x)))+f(x)=2x$ for all $x\in R$.
Good Luck.

i don't know the solution but i just know the answer.

Just substitute $x = f^{-1}(x)$
then you will get $f(f(x)) + x = 2x$
which gives $f(f(x)) = x \to  f^{-1}(x) = f(x)$
since the inverse of a function is the reflection of the initial function about $f(x) = x$, so the only possibility where the inverse = initial function is on the graph $f(x) = x$ itself. so we got the answer

[quote="violetcraze"] so the only possibility where the inverse = initial function is on the graph $f(x) = x$ itself. so we got the answer[/quote]
This is not true. Check $f(x)=-x$ or $f(x)=\frac{1}{x}$ for $x\ne 0$ and $f(0)=0$, etc.
However, in our problem, the solution is $f(x)=x$, indeed. First, observe that $f$ is one to one (injective). If for some $x,y$ we have $f(x)=f(y)$ then $f(f(f(x)))=f(f(f(y)))$ and hence $2x=2y$, that is, $x=y$.
Next, a continuous one to one function is strictly monotonic and because $f(f(f(x)))$ and $f(x)$ have the same type of monotony, it follows that  $f$ is strictly increasing.
Finally, suppose, by way of contradiction, that $f(a) \ne a$ for some real $a$. If $f(a)>a$ then $f(f(a))>f(a)>a$ and $f(f(f(a)))>f(a)>a.$ We obtain that $f(f(f(a)))+f(a)>2a$, a contradiction. A similar contradiction is obtained if $f(a)<a.$


probability
advanced fields
advanced fields theorems

Quote: "Let $ G \equal{} <N, (A_i), (u_i)>$ be such a strategic game. We denote by $ \Delta (A_i )$ the set of probability distributions over $ A_i$ and refer to a member of $ \Delta (A_i )$ as a mixed strategy of player i; we assume that the players' mixed strategies are independent randomizations. "

So notationally there is a difference between pure and mixed strategies; however, how does this translate to a physical difference between a pure and a mixed strategy? Where does the Delta notation come from? (I'm reading a really abstract book on game theory)

===

EDIT:

From [url]http://www.virtualperfection.com/gametheory/1.StrategicFormGames.1.0.pdf[/url] ,
When A is a finite set, $ \Delta (A_i )$  is the set of all probability distributions over A. So how then, would we list the set of all probability distributions over A? Is the set infinite? (because you can assign any probability from [0,1] to the elements of A, and the range of the set has an infinite number of elements)

If any player has $ n \geq 2$ pure strategies, his set of mixed strategies is uncountable -- a single player's strategy space is the simplex whose vertices are his pure strategies.  If I have pure strategies $ S_1, S_2, S_3$ then my mixed strategies have the form $ pS_1 \plus{} qS_2 \plus{} rS_3$ for nonnegative reals $ p,q,r$ summing to one.  Does this answer your question?

Okay, yes, it mostly answers the second half. I already got the first half taken care of elsewhere.

Before you get into abstraction you should try some simple examples of games and mixed strategies.

For example. Suppose you have the two-person zero-sum game where each player picks a number (with replacement) from the set $ \{1,2\}$. If their numbers are equal, player 1 receives a payoff equal to the sum of the two numbers. If their numbers are not the same, player 1 receives a negative payoff equal to the sum of the two numbers.

a.) Suppose player 2 plays a mixed strategy of $ [.6, .4]$ (meaning they choose 1 60% of the time and 2 40% of the time). What is the optimal response strategy for player 1? Is it a pure strategy or a mixed strategy?

b.) What is the optimal strategy for player 1 if the strategy for player 2 not known, but they are assumed to be rational? Is it a pure strategy or a mixed strategy?


Find an equation for each ellipse below.  Then, graph the equation.

(1) Foci at (1, 2) and (-3, 2); vertex at (-4, 2)

(2) Center at (1, 2); vertex at (1, 4) and passing through the point (2, 2)

see the other topic. mod, this one (or the other) should be locked, they're essentially th exact same thing.

Yeah. sharkman, why did u post the same thing twice? once is good enough, kay?

dude..gauss..you don't have to get on his tail...it may have been a mistake...chillax dude.  :wink:


When a positive three-digit dividend with a units digit of 2 is divided by a positive one-digit divisor, the result is a whole-number quotient with a remainder of 1.  How many distinct values are possible for this three-digit dividend?

All three-digit dividends are of the form $ \overline{ab2} \equal{} 100a \plus{} 10b \plus{} 2$. In order for it to have a remainder of $ 1$ when divided by dividend $ d$, we use the division algorithm thing to see that $ 100a \plus{} 10b \plus{} 2 \equal{} kd \plus{} 1\implies100a \plus{} 10b \plus{} 1 \equal{} kd$ for some whole $ k$. To eliminate the possible $ d$, go one by one: $ d\ne1$, obviously, $ d\ne2,4,6,8$ because $ 100a \plus{} 10b \plus{} 1$ is odd, and $ d\ne5$ because $ 100a \plus{} 10b \plus{} 1\equiv1\not\equiv0\pmod5$. This leaves $ d \equal{} 3,7,9$ as possibilities. If $ 9|100a \plus{} 10b \plus{} 1$, then $ 3|100a \plus{} 10b \plus{} 1$, so we can ignore $ d \equal{} 9$ (they are included in $ d \equal{} 3$ because of this). For $ d \equal{} 3$, $ 100a \plus{} 10b \plus{} 1\equiv10a \plus{} b \plus{} 1\equiv0\pmod3\implies10a \plus{} b\equiv \minus{} 1\equiv2\pmod3$. Since $ 10a \plus{} b$ spans through all the two-digit integers, all $ 2 \plus{} 3k$ for $ 3\le k\le32$ work, giving $ 32 \minus{} 3 \plus{} 1 \equal{} 30$ values for this case. For $ d \equal{} 7$, $ 100a \plus{} 10b \plus{} 1\equiv0\pmod7\implies100a \plus{} 10b\equiv \minus{} 1\equiv20\pmod7$
$ \implies10a \plus{} b\equiv2\pmod7$. As seen in the previous case, all $ 2 \plus{} 7k$ for $ 2\le k\le13$ work, giving $ 13 \minus{} 2 \plus{} 1 \equal{} 12$ for this case. However, we are overcounting: some numbers satisfy $ 100a \plus{} 10b \plus{} 1\equiv0\pmod{21}\implies100a \plus{} 10b\equiv \minus{} 1\equiv20\pmod{21}$
$ \implies10a \plus{} b\equiv2\pmod{21}$. Again, all $ 2 \plus{} 21k$ satisfy this for $ 1\le k\le4$, and this is the amount we have overcounted. The answer is hence $ 30 \plus{} 12 \minus{} 4 \equal{} \boxed{38}$. This was actually somewhat nice for a MATHCOUNTS problem. :wink:  :P


A circle with radius $ 1$ is internally inscribed in a circle of radius $ 3$.  Find the area of the largest triangle whose vertices are the tangent point, a point on the smaller circle, and a point on the larger circle.

Mandelbrot $ 1991$

Do you have an answer key for this? I think I have the answer and method, but I'm not sure of the way to definitely prove it..


Suppse that $n$ be a odd integer. Show that \[ ((n-1)^n+1)^2 \mid n(n-1)^{(n-1)^n+1}+n.  \] And is it true for even numbers ?

It's easy to show that $(n-1)^n+1=n^2q$ where $q$ is an odd number and $gcd(n,q)=1$ so we must prove that \[ n^3q^2\mid (n-1)^{n^2q}+1.  \] And it's easy to show that $(n-1)^{n^2q}+1\equiv 0 (mod n^3).$ Then it's suffices to prove that \[ q^2\mid (n-1)^{n^2q}+1.  \] And \[ (n-1)^{n^2q}+1=[(n-1)^n+1][(n-1)^{n(nq-1)}-...-(n-1)^n+1]=n^2q[(n-1)^{n(nq-1)}-...-(n-1)^n+1].  \] We know that $(n-1)^n\equiv -1 (mod q)$ thus \[ (n-1)^{n(nq-1)}-...-(n-1)^n+1\equiv (-1)^{nq-1}-(-1)^{nq-2}+...-(-1)+1\equiv nq \equiv 0 (mod q).  \] Then for $n=2$ the assertion is true but for $n=4$ it's wrong. Because $4(3^{82}+1)$ and $(4-1)^4+1=82=2\times 41$  are not divisible dy $2^2\times 41^2$ , because $3^{41}\equiv 3(mod 41) \rightarrow 3^{82}\equiv 9(mod 41) \rightarrow 3^{82}+1\equiv 10(mod 41).$


Let $ a,b,c$ be positive real number
Prove that
$ \frac{a}{ab\plus{}b\plus{}1}\plus{}\frac{b}{bc\plus{}b\plus{}1}\plus{}\frac{c}{ca\plus{}c\plus{}1} \le 1$.

[quote="soruz"]Let $ a,b,c$ be positive real number
Prove that
$ \frac {a}{ab \plus{} b \plus{} 1} \plus{} \frac {b}{bc \plus{} b \plus{} 1} \plus{} \frac {c}{ca \plus{} c \plus{} 1} \le 1$.[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=3151

Let $ a,b,c$ be positive real number with $ a\plus{}b\plus{}c\equal{}1$

Prove that
$ \frac {ab}{c \plus{} 1} \plus{} \frac {bc}{a \plus{} 1} \plus{} \frac {ca}{b \plus{} 1} \le\frac {1}{4}$.

[quote="soruz"]Let $ a,b,c$ be positive real number

Prove that
$ \frac {ab}{c \plus{} 1} \plus{} \frac {bc}{a \plus{} 1} \plus{} \frac {ca}{b \plus{} 1} \le\frac {1}{4}$.[/quote]

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

[quote="soruz"]Let $ a,b,c$ be positive real number with $ a \plus{} b \plus{} c \equal{} 1$

Prove that
$ \frac {ab}{c \plus{} 1} \plus{} \frac {bc}{a \plus{} 1} \plus{} \frac {ca}{b \plus{} 1} \le\frac {1}{4}$.[/quote]
$ \frac {ab}{c \plus{} 1} \plus{} \frac {bc}{a \plus{} 1} \plus{} \frac {ca}{b \plus{} 1} \le\frac {1}{4}\Leftrightarrow\sum_{cyc}(2a^4 \plus{} a^3b \plus{} a^3c \minus{} 6a^2b^2 \plus{} 2a^2bc)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a \minus{} b)^2(a^2 \plus{} 3ab \plus{} b^2 \minus{} c^2)\geq0.$
Let $ a\geq b\geq c.$ Then $ S_c\geq0,$ $ S_b\geq0$ and $ S_a \plus{} S_b \equal{} 2c^2 \plus{} 3ac \plus{} 3bc\geq0.$
Who is author of this nice inequality, soruz? Thank you.

The author of this nice inequality is N.Vijitu and A. Zaharescu!
When  the equality occurs?

[quote="soruz"]
When  the equality occurs?[/quote]
$ a \equal{} b \equal{} c \equal{} \frac {1}{3}$ or when $ a \equal{} b$ and $ 2c^2 \plus{} 3ac \plus{} 3bc \equal{} 0,$ which can not be.

[quote="soruz"]Let $ a,b,c$ be positive real number with $ a \plus{} b \plus{} c \equal{} 1$

Prove that
$ \frac {ab}{c \plus{} 1} \plus{} \frac {bc}{a \plus{} 1} \plus{} \frac {ca}{b \plus{} 1} \le\frac {1}{4}$.[/quote]
$ \sum_{cycl}\frac{ab}{c\plus{}1}\le\frac{1}{4}\sum_{cycl}\left(\frac{ab}{a\plus{}c}\plus{}\frac{ab}{b\plus{}c}\right)\equal{}\frac{1}{4}\sum_{cycl}\left(\frac{ab}{a\plus{}c}\plus{}\frac{bc}{c\plus{}a}\right)\equal{}\frac{1}{4}$


Of course not.

Just do the Pre/Post Tests are see which level is appropriate for you

[quote="cubekid"]Is there an age limit for classes? I just recently discovered AoPS (like maybe a month ago), but I'm a senior in high school. I really wanted to take one of the number theory classes, but am concerned that I can't because I really won't be in high school in the summer.

Is this a problem?[/quote]

No age limit; we'd be happy to have you join us.

Yeah, I'm pretty sure I was like the only sophomore in the Introduction level Number Theory course lol.  Its not any different though.


1. In how many ways can two squares be selected from an 8-by-8 chessboard so that they are not in the same row or the same column.

2. In how many ways can we place r red balls and w white balls in n boxes so that each box contains at least one ball of each colour.

3. Find a formula for $ \binom{n}{0}^2\plus{}\binom{n}{1}^2\plus{}\binom{n}{2}^2\plus{}...\plus{}\binom{n}{n}^2$

Thanks :)

[hide="2 (this seems wrong)"]
Place one of each into each box to have $ r\minus{}n$ and $ w\minus{}n$ of each left over. This can be done in $ \binom{n}{r}\binom{n\minus{}r}{w}$ ways. The remaining can now be divided among the $ n$ boxes. Using balls and urns, we have $ r\minus{}n\plus{}w\minus{}n\equal{}r\plus{}w\minus{}2n$ to be divided with $ n\minus{}1$ dividers, so $ r\plus{}w\minus{}2n\plus{}n\minus{}1\equal{}r\plus{}w\minus{}n\minus{}1$, so we have $ \binom{r\plus{}w\minus{}n\minus{}1}{n\minus{}1}$

So: $ \boxed{\binom{n}{r}\binom{n\minus{}r}{2}\binom{r\plus{}w\minus{}n\minus{}1}{n\minus{}1}}$
[/hide]

[hide="3"]
Split a pile of $ 2n$ objects into two piles of $ n$ objects. Choose $ k$ from one pile, $ n \minus{} k$ from the other. Sum over all values of $ k$ to get $ \sum_{k \equal{} 0}^{n}\binom nk\binom n{n \minus{} k} \equal{} \sum_{k \equal{} 0}^{n}\binom nk^2$. This counts the total number of ways to choose $ n$ objects from $ 2n$ objects, which is also clearly $ \boxed{\binom{2n}n}$.
[/hide]

Oh, nobody did #1 yet?

[hide="1"]$ 64$ ways to pick the first square, and then $ 49$ ways to pick the 2nd square. Of course, order doesn't matter, so divide by $ 2!$ to get $ \frac{64\cdot49}{2!}\equal{}\boxed{1568}$[/hide]


dar mosallas e ABC  , F noghteye fermaye mosallas ast va zaviyeye A ,60 darage ast(<A=60)
agar M gharineye A nesbat be F bashad

sabet konid agar  <ACF + <ACM =120 :

2AF=MB+MC

rahnamai mikonam.......sabet konid ke M rooye dayereye mohitie ABC ast va bad say konid ba 1kare hooshmandane masalaro be batlamus tabdil konid.....!


inequalities
linear algebra
linear algebra unsolved

Let $A$, $B$, $C$ be real matrices. Prove the inequality $\mathrm{tr}(A(A^{T}-B^{T})+B(B^{T}-C^{T})+C(C^{T}-A^{T}))\ge0.$


The digits from 1 to 6 are arranged to form a six-digit multiple of 5. What is the probability that the number is greater than 500,000? Express your answer as a common fraction.

The last digit must be $ 5$.  Thus, we have $ 5$ ways to choose the first digit, but only one way will work ($ 6$).  Thus, the answer is $ \boxed{\frac15}$.

[quote=frogshot]Um I don't really understand what you mean?[/quote]

Is there something in particular you don't understand?

[quote=frogshot]Um I don't really understand what you mean?[/quote]

We know the last digit must be $5$ since the number is divisible by $5$... Then, you have the digits $1, 2, 3, 4,$ and $6$ left. Since 500,000 and our six-digit number have the same amount of digits, the number we make is greater than 500,000 only if the first digit is greater than $5$. In this case, only $6$ will work. Hope that made sense. :)

uhm, no

[hide=Un-Bogus Solution]The units digit must be five because the number is a multiple of five, and of the last arrangements, only 6 can be the first digit. Since there is 1 number you choose out of the 5 remaining, the answer is $\boxed{\frac{1}{5}}$.[/hide]

1/5 is wrong because the question never clarifies whether numbers can be repeated or not, so it can be safely assumed numbers can be repeated. for the first digit, 2/6 numbers can be used (5 and 6) and the last digit must be 5, leaving (2/6*1/6). there is a 1/18 chance that the number is greater than 500,000 and a multiple of 5

No. That is wrong because the question says "The digits from 1 to 6 are [b]arranged[/b]..." This implies that each number has to be used and can only be used once.

[quote=cool972][hide=Solution]
For the six-digit number to be a multiple of 5, it has to end with a 5. Therefore, $\boxed{\frac{1}{5}}$ is our answer.[/quote]

:rotfl:


abstract algebra
superior algebra
superior algebra unsolved

Let $ A$ be a ring. Let $ L_{0}, L$ are left ideals of A $ L \subset L_{0}$. Let $ H$ be a minimal left ideal of A. We have that $ L_{0}/L  \simeq H$ as a left A-modules.
Why there exist $ \phi : L_{0} \rightarrow A$ as A -modules $ \phi \neq 0$ such that $ \phi (L)\equal{}0$?

what do you mean?
$ \phi: L_0\to L_0/L\to H\to A$??


induction
combinatorics unsolved
combinatorics

$ \sum_{n \equal{} 0}^{\plus{}\infty} {n \choose \frac{1}{2}n} (pq)^{\frac{n}{2}} s^{n}$ where $ p\plus{}q \equal{} 1$ and $ s > 0$

I'm assuming you're using the convention $ {n\choose\frac {1}{2}n} \equal{} 0$ for odd $ n$?
[hide="solution sketch"]
Prove by induction that $ \frac {1}{\sqrt {1 \minus{} 4x}} \equal{} 1 \plus{} \sum_{k \equal{} 1}^\infty {2n\choose n}x^n$, then it follows that the sum we're looking for is $ \frac {1}{\sqrt {1 \minus{} 4pqs^2}} \minus{} 1 \plus{} {0 \choose 0}$[/hide]

Well I think I'd be better to just substitute $ n \equal{} 2k$, then we'll avoid that assumption.


inequalities
Euler
inequalities unsolved

[quote="Extremal"]Try to find max $C$
for wich we have
if ${a}\leq{a_{i}}\leq{b}$ and  ${c}\leq{b_{i}}\leq{d}$ and ${1/p}+{1/q}=1$ then
${C}({\sum^{n}_{0}{a^{p}_{i}}})^{1/p}({\sum^{n}_{0}{b^{q}_{i}}})^{1/q}\leq{\sum^{n}_{0}{a_{i}}{b_{i}}}\leq({\sum^{n}_{0}{a^{p}_{i}}})^{1/p}({\sum^{n}_{0}{b^{q}_{i}}})^{1/q}$
When we have equal?[/quote]

Few comment: right inequality is known It's calld Holders  inequality  , but it's know everyone.But you don't know left inequality, it's very nice and interesting. it's open many interesting things in real anaysis and not only here.It's can bu used to solve difficult problems posted in the extremal theories.Every problems wich solved by holders  inequality now we can use this  and give problems different face. I don't know is it known or not!! But I will post Sulution of this problem after Conferenc Eulers in Saint Petersburg.
The sulution is hard. Good luck.


It is known that the number of real solutions of the following system if finite. Prove that this system has an even number of solutions:

                                                        $(y^2+6)(x-1)=y(x^2+1)$

                                                        $(x^2+6)(y-1)=x(y^2+1)$

x=y give 2 solutions $x=y=2$ and $x=y=3$. If it had another solution $x=a\not =b=y$ then it had solution $x=b\not =a=y$. Therefore number of solutions is even.


can you solve the equation  Fx+Fy=2Fz  where Fn is Fibonacci sequence?

I'm not sure exactly what you mean... $\displaystyle F(n)$ is a Fibonacci term? So are you just looking for two distinct terms in the fibonacci sequence whose sum is twice some other term? I'm pretty sure there are many many (infinite?) solutions to this. For example:
\[\displaystyle F(2) = 1; F(3) = 3; F(4) = 5;\]
\[\displaystyle F(2) + F(4) = 2(F(3))\]
\[\displaystyle 1 + 5 = 2 * 3\]
\[\displaystyle x = 2, y = 4, z = 3\]

Or am I just missing something?  :blush:

go take a look at the proof of the fibonacci here, it will help...
take a look at my proof, then use the same idea.
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=34840&postorder=asc

[quote="Hikaru79"]\[\displaystyle F(2) = 1; F(3) = 3; F(4) = 5;\][/quote]
Just how much have you been sleeping this night? ;)

[quote="Hikaru79"]I'm not sure exactly what you mean... $\displaystyle F(n)$ is a Fibonacci term? So are you just looking for two distinct terms in the fibonacci sequence whose sum is twice some other term? I'm pretty sure there are many many (infinite?) solutions to this. For example:
\[\displaystyle F(2) = 1; F(3) = 3; F(4) = 5;\]

Or am I just missing something?  :blush:[/quote]
$F(2) = 1; F(3) = 2; F(4) = 3$
!!! :what?:  ;)


First things first Rakesh nice to see the question in fine $ \text{\LaTeX}$ :coolspeak:  :thumbup:
[hide="solution"]
let $ \vec{u}$ be the unit vector along the incident ray $ \vec{v}$ the vector along normal $ \vec{w}$ along the reflected ray
then it is clear that $ \vec{w} = \vec{u} - 2(\vec{u}.\vec{v})\vec{v}$ that's because the component of the reflected ray along the normal is just opposite in sign to the incident ray and that along the surface is same as the relected ray
here we have $ \vec{u} = \frac {\sqrt {3}\vec{i} - \vec{k}}{2}$ also $ \vec{v} = \vec{ - k}$
putting the values we get $ \vec{w} = \frac {\sqrt {3}\vec{i} - \vec{k}}{2} + \vec{k} = \frac {\sqrt {3}\vec{i} + \vec{k}}{2}$
so $ \boxed{\vec{w}=\frac {\sqrt {3}\vec{i} + \vec{k}}{2}}$
now we have for refraction $ \frac {\sin i}{\sin r} = \sqrt {\frac {3}{2}}$ now we have for incident ray makes angle $ i = \frac {\pi}{3}$ with the normal so we have $ \sin i = \frac {\sqrt {3}}{2}$ so we have $ \sin r = \frac {1}{\sqrt {2}}$ so $ r = \frac {\pi}{4}$ so we have the unit vector along refracted ray as
$ \boxed{\vec{s} = \frac { - \vec{i} + \vec{k}}{\sqrt {2}}}$ :)
[/hide]

:clap2:  :clap:  :10:

Man, u are cetainly one ingenious fellow pardesi. It took the pains out of me to find the refracted ray and for the reflected I saw that formula u gave to nitish so i was able to solve it.


Let x,y real numbers. Prove that

$3(x+y+1)^2 + 1 \geq 3xy$


geometry
analytic geometry
geometry proposed

A cool problem!

In fact, let S be the point of intersection of the tangents to (O) at A and at B. Then, the line CK always passes through S.

To prove this, we note that the line CS is the C-symmedian of triangle ABC. Hence, it remains to show that K lies on the C-symmedian of triangle ABC.

Well, let D and E be the midpoints of the segments BC and CA. Also, let the notation d(P; g) mean the distance from a point P to a line g. Since the line MK is parallel to CA, we have d(K; CA) = d(M; CA) = ME, and similarly d(K; AB) = ND. Now, triangles MCE and NCD are similar (for < MCE = < NCD and < MEC = < NDC = 90); hence, ME / ND = CE / CD. Thus we have

  d(K; CA) / d(K; AB) = CE / CD = (CA/2) / (BC/2) = CA / BC.

In other words, the distances from the point K to the sides CA and AB of triangle ABC are proportional to these sides themselves. Hence, K lies on the C-symmedian of triangle ABC, and the proof is complete.

By the way, the barycentric coordinates of the point K with respect to triangle ABC are

  $\left( a^{2}:b^{2}:2ab-c^{2}\right) $.

Equivalently, its trilinear coordinates are

  $\left( ca:bc:2ab-c^{2}\right) $.

  Darij

[quote="treegoner"]Let (O) be a fixed circle and A, B fixed points lying on it. C varies on (O). Denote M, N the intersections of the perpendicular bisectors of CA and CB and the bisector CD of ACB. Let K be the intersection of the line through M, N respectively parallel to CA, CB. Prove that CK always passes through a fixed point.[/quote]

I would like to give a different solution to the problem. I hope my english is good nough to understand my idea.

Let me give to the point K the name C'  ;) . Let us call similarly A' the intersection between the parallen from N to AC and the paralle from D to AB and finilly let us call B' the intersection between the parallen from M to BC and the paralle from D to AB. it easy to see that ABC and A'B'C are omothetic. As C move on (O) given that A, B and D (and therefore A' and B' since are the pole of AD and BD respectively) are fixed, so is the centre of omotetie (i.e. the intersecyion of AA' and BB')


algebra
polynomial
quadratics
quadratic formula

Find the exact values of all of the roots of f(x)=x^4 + 4x^3 - 17x^2 -42x + 90.
I found -5 and 3, but im not sure how to use synthetic division to find the other two roots.

[hide="Solution"]Since $-5$ and $3$ are roots, then $x+5$ and $x-3$ are factors.

Divide the equation by $x+5$ using synthetic division and then divide the resulting polynomial by $x-3$. The result should be

\[x^{2}+2x-6\]

Now use quadratic formula to get:

\[x = 1\pm\sqrt{7}\][/hide]

Theorem:
If c is a root of a polynomial $P(x)$ with real coefficients, then $(x-c)$ evenly divides that polynomial.
Converse:
If $(x-c)$ evenly divides a polynomial with real coefficients, then $c$ is a root of that polynomial.
You have stated that -5 and 3 are roots, therefore
$(x-(-5)) = (x+5)$ and $(x-3)$

The polynomial is then $(x+5)(x-3)(ax^{2}+bx+c)$, because it's a quartic.
Using synthetic division, or polynomial long division, we can see that the remaining factor is $(x^{2}+2x-6)$
This has roots at $-1\pm\sqrt{7}$, so those are the other two roots.

um, he said that he did not know how to use synthetic division, so the two solutions before mine are useless.

Try [url=http://www.purplemath.com/modules/synthdiv.htm]here[/url] for help on synthetic division.

then just use long division.  synthetic is just to make things faster...so i dont see how the first two solution arent valid.

[quote="DiscreetFourierTransform"]Theorem:
If c is a root of a polynomial $P(x)$ with real coefficients, then $(x-c)$ evenly divides that polynomial.[/quote]

The stipulation that the coefficients are real is quite unnecessary.  It's just that the quotient polynomial may not have real coefficients.


geometry
3D geometry
geometric sequence
arithmetic sequence

I understand Ramunajan's basic proof and the recurrence principle, but I'm having difficulty applying it to these problems. Any solutions?

For a we know that the answer is the constant cube root of n^2. So an important thing is that we need to find a number that is both a square and a cube which would be any square which is cubed: 1(not possible), 64, 729. so for 64 that would take out a 4 for 49. And thenn I am lost!!

this might be impertinent, but why is there a $ 3^3$ in c). Ramanujan's formula cannot be involved in these, because this is rather a geometric progression betw. the coefficients rather than the arithmetic progression Ramanujan invented the formula for(It would be helpful if someone could post the step-by-step method by which Ramanujan derived his formula for infintely nested radicals so we could all try to apply it to work on geometric progressions within radicals). I hope this isn't an old thread that I've revived.

oops, it was a typo! (I couldn't edit the post anymore, though...)


Εστω τριγωνο ΑΒΓ.Στις πλευρες ΑΒ,ΑΓ,ΒΓ θεωρω Δ,Ε,Ζ τετοια ωστε ΑΔ=ΓΕ=ΒΖ. Αν το ΔΕΖ ισοπλευρο τοτε και το ΑΒΓ ισοπλευρο.

http://www.mathematica.gr/forum/viewtopic.php?f=20&t=266

@@edit

Στο παραπάνω link η άσκηση που προτείνει ο Μανώλης λύνεται με τη βοήθεια τριγωνομετρικών αριθμών. Παρόλ'αυτα μου είπε πως ψάχνει μια καθαρά γεωμετρική απόδειξη :| .


$0<p<q<\infty$,

(1) $0< bn <2^{-n}$,
Is there a positive sequence $a_{n}$, such that
$\sum^{\infty}_{n=1}{a_{n}^{p}b_{n}}<\infty$ and $\sum^{\infty}_{n=1}{a_{n}^{q}b_{n}}$ doesn't converge?

(2) $1\leq  bn <\infty$,
Is there a positive sequence $a_{n}$, such that
$\sum^{\infty}_{n=1}{a_{n}^{p}b_{n}}$ doesn't converge and $\sum^{\infty}_{n=1}{a_{n}^{q}b_{n}}<\infty$ ?

Thanks for your help.

I assume that in both #1 and #2 that should be $b_{n}$ and not $bn.$ That is, in #1, $0<b_{n}<2^{-n}$ and in #2, $0<b_{n}<\infty.$

For #1, choose $a_{n}=b_{n}^{-1/q}.$ Then

$a_{n}^{p}b_{n}=b_{n}^{-p/q}b_{n}=b_{n}^{(1-p/q)}<2^{-n(1-p/q)}.$

Those are the terms of a convergent geometric series. But

$a_{n}^{q}b_{n}=1,$ so that series diverges.

I'll let someone else use similar ideas for an example on #2.


Simplify the following expression:

${\pi^{\pi^{(\pi^{-1})}}}$

[quote="anirudh"]it is pi. i saw it in h78greg's signature.[/quote]

Do you know why it is that way?

that's why i am waiting. i want someone to answer it, so i can see the reasoning. h78reg told me that it was pi(its in his signature)

The calculator evaluates it as approximately 5.196.   :?

Well without using your calculator, you just multiply the exponents, $\frac 1 \pi \cdot \pi = 1$ so $\pi^1 = \pi$

my signature is a true statement, though i cannot explain it. evidently, pi^(pi^-1)=1. if we use approximation and use 3 as pi, we can easily confirm the truth of my signature. however, that's just an approximation.

Actually, dakyru is correct about the approxamation. $\pi^\pi\ne1$. It equals 1.43 something. Because you work exponents from the top, your signature is false  :(

aidan pointed out an important point of the problem.
I think the reason people got different answers with calculators is because they interpreted the expression differently.

[quote="236factorial"]What was your key sequence? Mines was pi^pi^(pi^-1) which indeed comes out to 3.1415926.[/quote]
i pressed this:
pi^(pi^(pi^-1))

[quote]i pressed this:
pi^(pi^(pi^-1))[/quote]

With that you get about 5.196. Pressing this: (pi^pi)^(pi^(-1)), you get pi.

[quote]First, is  equal to  or ?[/quote]

pi is equal to the first expression that aidan mentioned. i just checked it on a calculator.

The calculator is probably programmed to take exponents correctly when no brackets are provided (which is from the top), so the answer should be approx 5.196.

weren't we supposed to do it without a calculator?  :?

i think so.  But since i couldn't get it, i used a calculator.


Define two polynomials $ p\left(x\right) \equal{} 4x^3 \minus{} 2x ^2 \minus{} 15x \plus{} 9$ and $ q\left(x\right) \equal{} 12x^3 \plus{} 6x^2 \minus{} 7x \plus{} 1$.

[b]a)[/b] Show that each of these two polynomials has exactly three distinct real roots;

[b]b)[/b] Let $ A$ be the largest root of $ p\left(x\right)$ and $ B$ the largest root of $ q\left(x\right)$. Show that $ A^2 \plus{} 3B^2 \equal{} 4$.

for the first part I have a very easy proof:

we can see that P(-2)<0, P(0)>0, P(1)<0, P(2)>0. Applying Rolle's theorem, P has three distinct real roots on the intervals (-2,0), (0, 1), and (1, 2). Similarly, Q has three distinct real roots on (-1, 1/6), (1/6, 1/4), and (1/4, 1).

for the 2nd part

 we have that 1<A<2, 1/4<B<1.

we can easyly see that  A is not equal to 3/4, and B=A/(4A-3).
It's easy to check that P(3/4) is nonzero. Thus, A=3/4 is impossible. Let c=A/(4A-3). To show B=c, we need to show that Q(c)=0 and c>1/4. Since 1<A<2, we have c=A/(4A-3)>2/5>1/4. (1)

with this in mind we continue:
we need to  show that  Q(c)=0 which is equivalent to 12c^3+6c^2-7c+1=0  which gives us:
12(A/(4A-3))^3 +6(A/(4A-3))^2 -7A/(4A-3) +1=0, which is equivalent
to (4A-3)^3 [12(A/(4A-3))^3 +6(A/(4A-3))^2 -7A/(4A-3) +1]=0
or 12 A^3 + 6A^2 (4A-3) - 7A(4A-3)^2 + (4A-3)^3 = 0
or -3(4A^3 -2A^2 -15A + 9)=0 or -3P(A)=0 (true because A is a real root of P(x))

Now, we show that A^2+3B^2=4.using (1) we have: A^2 + 3(A/(4A-3))^2 = 4 which gives A^2 (4A-3)^2 + 3A^2 = 4(4A-3)^2 <=> 16A^4-24A^3-52A^2+96A-36=0
Lets denote  H(x)=16x^4 -24x^3 -52x^2 +96x -36=0. We need to show that H(A)=0. We know that A is a real root of P(x). and we get that  H(x) = P(x)(4x-4). now we get that H(A)=(4A-4)P(A)=0.
all done!!! :D

cheers! :D hope it is okay!! :D

hmm I keep wondering from where did you come up with c=A/(4A-3) ?!

hey lagrangia
my proof is xactly the same.but i got it after fiddling with the problem for over a week.i aint that fast,u posted that solution within an hour :D .
there's not much u can do,unless u get sumthing of a brainwave. :D

4 Valentin Vornicu

about the idea of c=A/(4A-3) it just stroke me :? .. I don't know how I came up with that one... but it seems that it works... :D :D :D ...

cheers!!!

hi ,dieuhuynh
I use exactly the same method as yours,but I weren't patient enough to caculate it.Thank you to complete it.

And I wonder where is the another 5 problems of the Vietnam national


Let $ a,b,c$ be positive real numbers. Prove that
\[ \frac{a}{a\plus{}2b}\plus{}\frac{b}{b\plus{}2c}\plus{}\frac{c}{c\plus{}2a}\ge 1\]

hello, your inequality is equivalent to
$ \frac{2(\minus{}3abc\plus{}ba^2\plus{}ac^2\plus{}b^2c)}{(a\plus{}2b)(b\plus{}2c)(c\plus{}2a)}\geq0$ this is true because the of AM-GM inequality.
Sonnhard.

Yes this was a bit of an easy problem but I'm not sure a lot of people would have been wise to expand like that. I presume that the intended solution was along the lines of:

\[ [ a(a\plus{}2b)\plus{}b(b\plus{}2c)\plus{}c(c\plus{}2a) ]\cdot\left[\frac{a}{a\plus{}2b}\plus{}\frac{b}{b\plus{}2c}\plus{}\frac{c}{c\plus{}2a}\right]\ge (a\plus{}b\plus{}c)^2\]

by cauchy. Now note that $ a(a\plus{}2b)\plus{}b(b\plus{}2c)\plus{}c(c\plus{}2a)\equal{}(a\plus{}b\plus{}c)^2$. This is a strictly positive number so we can cancel it from both sides.

It is quite nice I think. As a comment, what if we replace $ 2$ with $ k\plus{}2$. What is the minimum of
\[ \frac{a}{a\plus{}(k\plus{}2)b}\plus{}\frac{b}{b\plus{}(k\plus{}2)c}\plus{}\frac{c}{c\plus{}(k\plus{}2)a}\]

where $ k\ge \minus{}2$ where $ a,b,c$ vary over positive reals?


Hi,

I had a test in which there was this question that had a octagon and a line extended from one of the sides (external angle). The question asked to find this external angle.

But there was no indication that the octagon was regular; it was not mentioned and no lines indicating equal sides were on the diagram.

Is this possible, can you find this particular exterior angle without it being regular???

(::note: in the test I assumed it was regular and just figured it out from there:: Was i right to do so?)

No; unless they indicated the interior angle for that particular exterior angle or some other indicator, you would not be able to find the exterior angle. Since the octagon was not regular, the exterior angles could be anything, as long as they summed to 360.


yeah - c'mon you embarass me with this problem :D

should I delete this topic?? just kidding! :D :D

c'mon it's not that easy! :D :D
cheers! :D

when does students in romania graduate?  12th grade?

yeah they do ... but we are now fetting off-topic ... :D it's such a simple problem - of course

[quote="Lagrangia"]Here is a problem which was proposed by Valentin Vornicu for the 10th grade last year for the national stage:

Let f be a polinomial form C[x] for which we have : f(X)=X^2003+a_2002*X^2002+...+a_1*X+a_0. We know that the polinomial has all it roots |xn|=1. If a_0 <>1 then the number (a_2002-a_1)/(1-a_0) is real.

cheers! :D :D[/quote]

This problem is really nice i am using Franois Viette
http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/Viete.html formulae relation between the roots and the coefficients of polynomial

Let A = (a_2002 - a_1)/(1-a_0)

A= ( \sum x_k  - \sum x_(i_1)...x_(i_(k-1)) )/(1-x_1...x_n)

|x_k| = 1 so x_k* = 1/x_k where x* is the conjugate of x

A* =   ( \sum x_k*  - \sum x_(i_1)*...x_(i_(k-1))* )/(1-x_1*...x_n*)

=  ( \sum 1/x_k  - \sum 1/x_(i_1)...1/x_(i_(k-1)) )/(1-1/x_1...1/x_n)

=(   \sum x_(i_1)...x_(i_(k-1)) - \sum x_k )/(x_1...x_n -1) =A

A is real

Is there some romanian tea ?

Actually a0=-(x1)(x2)...(x2003).. although the proof is similar
I assigned xk=e^(i ck), 0<=ck<2 \pi , i being the root of -1 and then the proof becomes straightforward by dividing  \sqrt (x1)(x2)...(x2003) on the denominator and nominator (spelling?) along with the fact e^iy+e(-iy) is real for real y

[quote="Charlie"]Actually a0=-(x1)(x2)...(x2003).. although the proof is similar
[/quote]

Thank you Charlie for the sign -

oubi


If $ x,y\in \mathbb{N}$ solve this equation :

$ (x^2\plus{}y^2\plus{}10)\left[\frac{x}{y}\plus{}20\right]\equal{}1995$

$ 1995\equal{}3 \times 5 \times 7 \times 19$

$ \left[\frac{x}{y}\plus{}20\right]\equal{}21$ or $ \ge 35$
If  $ \left[\frac{x}{y}\plus{}20\right] \ge 35$, $ x \ge 15y \ge 15$ $ \Rightarrow L.H.S.>1995$, impossible.

If  $ \left[\frac{x}{y}\plus{}20\right]\equal{}21$, then the question $ \Leftrightarrow$
$ \begin{cases} x^2\plus{}y^2\equal{}85 \\
2y>x\ge y \end{cases}$

So $ (x,y)\equal{}(7,6)$ is the only solution.


Consider a sequence of $N$ positive integers containing $n$ distinct integers. If $N \geq 2^n$, show that the sequence contains a nonempty consecutive block of integers whose product is a perfect square.

[i]Remark.[/i] By a [i]nonempty consecutive block[/i] in a sequence $\left(a_1,a_2,...,a_r\right)$, we mean a subsequence of the form $\left(a_k,a_{k+1},...,a_l\right)$ for some integers $k$ and $l$ satisfying $1\leq k\leq l\leq r$.

i dont understand this problem ... is it saying that if you have 2^n distinct integers then the product of some group of consecutive ones will be a perfect square? Why cant you jsut take 2^n distinct primes? What do you mean by a 'sequence'?

We are given n distinct positive integers...these n *distinct* integers compose a block of N integers... so, an example with n=2 and N=4 would be 3,5,3,5 because we have a block of 4 integers consisting of 2 distinct integers.

It's based on a very well-known pigeon-hole argument:

We will prove the claim for every sequence of $N\geq 2^n$ integers among which there are [i]at most[/i] (not necessarily exactly) $n$ distinct integers.

For simplicity we assume $N=2^n$ and let the sequence be $a_1,a_2,\ldots,a_N=a_{2^n}$. Since there are [i]at most[/i] $n$ distinct integers in this sequence, we can find $n$ (not necessarily distinct) integers such that each of $a_1,a_2,\ldots,a_N$ is among these $n$. Let $b_1$, $b_2$, ..., $b_n$ be these $n$ integers.

Now consider the blocks $B_1=(a_1),B_2=(a_1,a_2),B_3=(a_1,a_2,a_3),\ldots,B_{2^n}=(a_1,a_2,\ldots,a_{2^n})$. To each block we associate a string of length $n$ made up of $0$s and $1$s s.t. on the $k$'th position we have $0$ if $b_k$ appears in the block an even number of times and $1$ if it appears an odd number of times.

There are $2^n$ binary strings with $n$ digits. If all of our strings are different, then the string containing only $0$s must be somewhere in there, so one of the blocks we've chosen contains an even number of copies of each $a_k$. On the other hand, if two strings are equal, then assume the strings for $B_u,B_v$ are equal. In this case, the block $(a_{u+1},a_{u+2},\ldots,a_v)$ has an even number of copies of each $b_k$, and this concludes the problem.


[hide]
Use S + uL. Where u is parameter.Now this equation should represent a circle. So coeff of xy =0 and coefficients of x^2 and y^2 are equal and radius non zero. Does this fetch the answer?
[/hide]

[hide]
this does fetch the answer.
but doing this, the only thing that we can conclude is that when
$ u \neq  0$
then
$ \theta$ has to be
$ \pm \frac{\pi}{4}$

i had a doubt that is this the only condition, or should there be a condition for the radius of the circle too.
[/hide]


inequalities
Ross Mathematics Program
number theory theorems
number theory

Know that n is a natural number
Prove that there's at least a prime number between n and 2n.

Proof:
Lemma 1. when n \geq 5, the binomial coefficent nC(2n) satisfy:
2 2n /2n<nC(2n)<2 2n /4.
Proof: 2n*nC(2n)=2/1*3/1*4/2*5/2*..........*(2n-2)/(n-1)*(2n-1)/(n-1)*(2n)/n*(2n)/n>2 2n , that gives the first inequality.
the second inequality can be proved by mathematical induction.

Lemma 2. Product(10<p \leq b)p<2 2b .
Proof: Let ak be the minimum integer  \geq b/2 k.
It's easy to show thet a1 \geq a2 \geq ... \geq ak \geq .... (1)
and ak \leq 2a(k+1).                   (2)
let m be the largest integer such that am \geq 5.
By (2) we have am<10.
again by (2) and 2a1 \geq b we have the intervals
(am,2am], (a(m-1),2a(m-1)],......,(a1,2a1] covered entirely the interval (10,b].
Thus we have LHS \leq Product(a1<p \leq 2a1)p *..........*Product (am<p \leq 2am)p.
But for any a we have Product(a<p \leq 2a)p<aC(2a)<2 2n-2 (By lemma 1)
so LHS \leq 2 2(a1-1+a2-1+....+am-1) <2 2(b/2+b/4+.....+b/2 m ) <2 2b .

Suppose there is an n>50 such that there's no prime in the interval (n,2n).
for every prime p in (2n/3,n], we have p doesn't divide nC(2n).
Thus we have nC(2n) \leq Product(p \leq  \sqrt 2n)p rp *Product( \sqrt 2n<p \leq 2n/3)p \leq (2n)  \sqrt 2n  *2 4n/3 (Since  \sqrt 2n \geq 10, by Lemma 2)(rp=[log_p(2n)])
But by Lemma 1 we have nC(2n) \geq 2 2n /2n
so we have 2 2n <(2n)  \sqrt 2n  +1 *2 4n/3, and 2 2n/3 <(2n)  \sqrt 2n  +1 .

let 2n=x, we have 2 x/3 <x  \sqrt x  +1 .
So (x ln2)/3<( \sqrt x +1)lnx.
Thus ( \sqrt x  -1)ln2 /3<lnx, and
( \sqrt x -1)<3log_2 x=6 Log_2  \sqrt x.
let  \sqrt x=y we have y-1<6Log_2 y, thus
2 y  <2y 6 .    (3)
It's Obvious now that (3) cannot hold for suefficent large y(in fact, (3) doesn't hold when y \geq 32, that is, n \geq 512.)
for cases when n \leq 512, we can directly verify that the statement is true.

Q. E. D.

Well actually one can prove that there are two :P primes between n and 2n.

Anyway does anyone think that he can memorate / remember such a proof ? ( I mean I know the ideas, but it takes me hours to reproduce it, I've tried it once :D).

[quote="Valentin Vornicu"]Well actually one can prove that there are two :P primes between n and 2n. [/quote]

I don't think there are two between 2 and 4
neither between 3 and 6
neither between 1 and 2
neither between 5 and 10

so I don't know where you get those ideas from? :?

The stronger version of Bertrand's Postulate states that for n \geq 6, there exist at least 2 distinct primes in the interval (n, 2n). I remembered reading it in [i]Mathematical Gems III[/i], a book by Ross Honsberger. Valentin, could you write the proof out? I don't think there was a proof in the book and I have no idea on how to prove it.

In fact, if I remembered correctly from somewhere else, Bertrand's Postulate can also be further improved to: For large enough n, there exist a prime in the interval (n, 3n/2).

And I think that for any a>1 and suficiently large n there is a prime between n and an. I think it follows from prime number theorem.

There is an article of this theorem (Chebyshev's Theorem), with the proof of Paul Erds's first math paper.There are in Portuguese.

http://www.ime.usp.br/~brleite/arquivos/SemanaOlimp.pdf

I have found a very strong theorem that states that there is at least a prime number between x and  (9/8)*x, for any x=>48, but I can't proove it.  :D


how do you do the problem: when six coins are flipped what is the probability that at least five of them are heads. i know how to to find the probability when it asks for exactly amount of heads or tails, but im a little confused with "at least".  :ninja:

This means that you find the probability of 5 heads and 6 heads.
[hide]Then use the pascal's triangle to find 5 heads 1 tail and 6 heads 0 tails
thats 6+1 = [b]7[/b][/hide]
OR
[hide]If you don't know the pascal's triangle, then just do (6 c 5)+(6 c 6)
which is 6+1 = [b]7[/b][/hide]

I've heard that Pascal's Triangle can be used for probability, but how exactly does that work?

The pascal's triangle is
1
1,1
1,2,1
1,3,3,1
1,4,6,4,1
and so on
1 is the zeroth row. 1,1 the first row and so on.
1,1 stands for when you flip one coin there is 1 way to get a head and 1 way to get a tail
1,4,6,4,1 is the fourth row so 1 way to get  4 heads , 4 ways to get 3 heads 1 tail ,6 ways to get 2 heads 2 tails, and so on.

This can also be used for the expansion of (a+b)
(a+b)^2 is a^2 + 2ab + b^2
the coefficents are 1,2,1 as in the 2nd row
the exponents are 2,1,2 for 2 heads, 1 head 1 tail, and 2 tails

This can also be used for powers of two.  For the nth power of 2, add the numbers in the nth row.  2^4 = 1+4+6+4+1 = 16

Yes, the pascal's triangle is:

\[ \binom{0}{0}  \]\[ \binom{1}{0}\binom{1}{1}  \]\[ \binom{2}{0}\binom{2}{1}\binom{2}{2}  \]\[ \binom{3}{0}\binom{3}{1}\binom{3}{2}\binom{3}{3}  \]
and so forth.

But what's really interesting is [i]how[/i] it works.

Let's say we have $m$ objects and $n$ objects we need to pick.  Obviously, this are $\binom{m}{n}$ such sets.  Take one of the objects and call it $o$.  The number of the sets that don't have $o$ is $\binom{m-1}{n}$, since you take out $o$.  Now, you want to count the number of sets that [i]do[/i] contain $o$.  You already have $o$ in a set, and you want to choose $m-1$ more.  So there are $\binom{m-1}{n-1}$ sets that contain $o$.
Now, any set $\binom{m}{n}$ must either have $o$ or not have $o$.  So now you can say that $\binom{m}{n}=\binom{m-1}{n-1}+\binom{m-1}{n}$.

So that is why you can simply add the terms and why Pascal's triangle works.

:huh: Wow that was wierd.

I didn't know you could use pascal, but I guess it makes sense.
How would you do probability, add up all the numbers for the denominator, and the numbers you want for the numerator?

Also, if anyone knows a formula to find each number in the nth row, could they post it?

I think it means choose.

Go ahead and laugh at me, I don't know how to calculate that. :|

[quote="robinhe"][quote="Zenith Xan"]This may sound silly, but... I don't know what  $\binom{2}{0}$ means. I probably do in reality, but not in that form.[/quote]

That means how many ways to choose $0$ items from a group of $2$.

Of course, there is only one way to do that.  :)[/quote]

Ah. Is there just a specific quick formula?

for the probability, there are $2^6=64$ ways total (know your powers of two well)
7 working=$\frac{7}{64}$

Side note:
Ways to remember or get quickly your powers of 2.

$2^4=4^2=16$
$2^6$ = 64
$2^{10}$ = 1024

And try to fill in the gaps.

wow, i've only memorized to 4096. you must have a lot of time on your hands!  :)

Actually this was when I was real little and had no life

I thought most people started having no life when they were 17 or something.  Weird.

how many digits of $\pi$ do you have memorized?
I got 3.141592... I think that's right. I think.

$\pi$ is the greek alphabet for r. I think.
I'm guessing that stands for radius. I think.

this is what i have memorized:
$\pi$ is delicous

Depends on what kind.
I've memorized
3.141592653589793238462643383279
I've got a million digits if you want them.


if a line is drawn parallel to x axis from some point other than origin then its intercept on the x axis is given to be infinite . is it true . i think its is zero but answer given is infinite .

It is obviously true.The x-intercept is defined as the point on the x-axis where the line meets the x-axis.As the line parallel to the x-axis will meet the x-axis at infinity,the x-intercept will be infinity.


A stick of length A is broken at any arbitary point, into two parts of length a1 and a2. What is the probability that a1/a2 is less than k(where k is a constant)?

[hide]
Let the stick endpoints be $(0,0)$ and $(A,0)$. If $\frac{a_{1}}{a_{2}}< k$, then the break is between $(0, 0)$ and $\left(\frac{Ak}{k+1}, 0\right)$ (assuming $a_{1}$ is the piece on the left). So the probability of this is $\frac{k}{k+1}$.[/hide]


function
real analysis
real analysis unsolved

Let $f: [0,1]\to\mathbb{R}$ and $G(f)=\{(x,f(x))\,/\, x\in[0,1]\}$

Does $G(f)$ could have positive measure?
Another related and intersting question is,  if there is $f$ st $G(f)$ is not measurable?

What do you think :D?

We can restrict safely to bounded functions: if $G(f)$ has positive measure or is not measurable, $G(f)\cap [0,1]\times[-n,n]$ has positive measure or is not measurable for some $n$, and if we replace $f(x)$ by $n$ when it is large and $-n$ when it is large negative, the new graph has the same measure as $G(f)\cap [0,1]\times[-n,n]$.

Since we can fit infinitely many translates $G(f)+(0,c)$ into the finite box $[0,1]\times [-n-1,n+1]$, $G(f)$ cannot have positive measure. This still leaves the possibility of nonmeasurability.

You can adapt the construction [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=Myth&t=49300]here[/url] to get a function with non-measurable graph.


Find the direction angles of the vector given below.
Then write each vector in the form
v = ||v||[(cos A) i + (cos B) i + (cos Y) k ].

v = -6i + 12j + 4k

Basically just plot the vector in the $ xyz$-coordinate plane and use trig a few times.


geometry
circumcircle
ratio
perpendicular bisector
geometry unsolved

Dear Mathlinkers,

Let ABC be a triangle, 0 the circumcircle of ABC, P and Q two points on ABC, M the mediator of PQ, A', B', C' the points of intersection of M with BC, CA, AB, 1, 2, 3 the circumcircles of A'PQ, B'PQ, C'PQ and finally A", B", C" the second point of intersection of the mediator M with 1, 2, 3.
Prove that AA", BB", CC" are concurrent on 0.

Sincerely
Jean-Louis

Assuming that mediator $ m$ of $ PQ$ is the perpendicular bisector of $ PQ.$ It is sufficient to show that the lines $ AA'', BB''$ intersect on the circumcircle $ (O)$ of the $ \triangle ABC.$ Let $ m$ meet $ (O)$ at $ U, V.$ The circumcircles $ (O), (1), (2)$ of the $ \triangle ABC, \triangle A'PQ, \triangle B'PQ$ are coaxal $ \Longrightarrow$ powers of the points $ U, V$ to the circles $ (1), (2)$ are in the same ratio, $ \frac {\overline{UA'} \cdot \overline{UA''}}{\overline{UB'} \cdot \overline{UB''}} \equal{} \frac {\overline{VA'} \cdot \overline{VA''}}{\overline{VB'} \cdot \overline{VB''}}.$ It follows that the cross-ratios $ (U, V, A'', B'') \equal{} (U, V, B', A')$ are equal. Let $ AA''$ cut $ (O)$ again at $ X$ and let $ XB$ cut $ m$ at $ B_2.$ Then

$ \measuredangle UXV \equal{} \measuredangle UCV \mod \pi$

$ \measuredangle B_2XA'' \equal{} \measuredangle BXA \equal{} \measuredangle BCA \equal{} \measuredangle A'CB' \mod \pi$

$ \measuredangle VXA'' \equal{} \measuredangle VXA \equal{} \measuredangle VCA \equal{} \measuredangle VCB' \mod \pi$

$ \measuredangle A''XU \equal{} \measuredangle AXU \equal{} \measuredangle ACU \equal{} \measuredangle B'CU \mod \pi$

By projectivity of the circle $ (O)$ to the line $ m,$ the cross-ratios $ (U, V, A'', B_2) \equal{} (U, V, B', A')$ are also equal, which means that the points $ B_2 \equiv B''$ are identical, and $ AA'', BB''$ intersect at $ X \in (O).$


6 points are given in space such that the lengths of the 15 segments that join them in pairs are all different. Prove that one of these segments is the longest side of the triangles it is in and the shortest side of another of them

[quote="pbornsztein"]Posted before....

Color red each segment. Then, for each triangle, color blue its shortest side (which is not red anymore). Then, you have a bicoloration of the segments joining the six points. But, it is a well-known result that for any such coloration there must be a monochromatic triangle. Here, clearly, there is no red triangle. Thus, there is a blue triangle. For such a blue triangle, its longest side is blue and must be colored in blue because it is the shortest side of some other triangle.

Pierre.[/quote]
thank u very much .


The units-digit of the square of an integer is 9 and the tens-digit of this square is 0. Prove that the hundreds-digit is even.

If hundreds-digits is odd then number is congruent 5 (mod 8) so it can not be a perfect square which implies that hundreds-digit is an even number.

Let this number be $ x^2$ and let $ y$ be the number before that zero. Then

$ x^2\equal{} 100y\plus{}9$
$ (x\minus{}3)(x\plus{}3)\equal{}100y$

Left side is divisible by eight, so right side must also be divisible by $ 8$. Since $ 100$ is only divisible by $ 4$ , $ y$ must be divisible by $ 2$  .So its last digit must be evan.

$ \blacksquare$

Let the last three digits of number be : pqr
So, when we multiply ......pqr by itself that is we square it..
[b]Last Digit [/b]is : r^2 or (r^2-10n)
And r can be chosen from 0,1,2,3,4,5,6,7,8,9...
Also r^2  or r^2-10n = 9
So it can either be 3 or 7
So... r= 3 0r 7
[b]Second Last digit[/b] is : 2*(qr) + n  [ "n" is the carry from the last digit which can be 0 or 4 ]
so, 2*(qr) + n = 0
This is only possible if either of "q" or "r" is 0 and n =0..
"r" is not equal to zero.. so n=q=0.
and r =3
so, [b]Third Last digit [/b]is :  2* (3r) which is always even...

let  the original number be `k`.

then  `k`  is odd.   so  k^2  leaves  a remainder  of  1 when divided by 8.

so k^2-1  is divisible  by 8 .

a  number is divisible  by 8 only when the number  formed by the last three digits is divisible by 8.

since last 2 digits of k^2-1  are 0  and 8 , so hundreds digit should be even .


[quote="2424"]With $ a,b,c > 0$ such that $ abc \equal{} 1$.Prove that
     $ 3(a^2 \plus{} 1)(b^2 \plus{} 1)(c^2 \plus{} 1)\geq(a \plus{} 1)(b \plus{} 1)(c \plus{} 1)\sqrt {(a \plus{} b \plus{} c)(1/a \plus{} 1/b \plus{} 1/c)}$[/quote]
We have : $ \frac{a^2\plus{}1}{a\plus{}1} \ge \sqrt[3]{\frac{a^3\plus{}1}{2}}$,....
and $ \sqrt{(a\plus{}b\plus{}c)(1/a\plus{}1/b\plus{}1/c)} \le \frac{a\plus{}b\plus{}c\plus{}1/a\plus{}1/b\plus{}1/c}{2}$
So,we only need to prove:
$ 3\sqrt[3]{(a^3\plus{}1)(b^3\plus{}1)(c^3\plus{}1)} \ge a\plus{}b\plus{}c\plus{}1/a\plus{}1/b\plus{}1/c$
It 's true because:$ \sqrt[3]{a^3\plus{}1)(1\plus{}b^3)(1\plus{}c^3)} \ge a\plus{}1/a$

[quote="quykhtn-qa1"][quote="2424"]With $ a,b,c > 0$ such that $ abc \equal{} 1$.Prove that
     $ 3(a^2 \plus{} 1)(b^2 \plus{} 1)(c^2 \plus{} 1)\geq(a \plus{} 1)(b \plus{} 1)(c \plus{} 1)\sqrt {(a \plus{} b \plus{} c)(1/a \plus{} 1/b \plus{} 1/c)}$[/quote]
We have : $ \frac {a^2 \plus{} 1}{a \plus{} 1} \ge \sqrt [3]{\frac {a^3 \plus{} 1}{2}}$,....
and $ \sqrt {(a \plus{} b \plus{} c)(1/a \plus{} 1/b \plus{} 1/c)} \le \frac {a \plus{} b \plus{} c \plus{} 1/a \plus{} 1/b \plus{} 1/c}{2}$
So,we only need to prove:
$ 3\sqrt [3]{(a^3 \plus{} 1)(b^3 \plus{} 1)(c^3 \plus{} 1)} \ge a \plus{} b \plus{} c \plus{} 1/a \plus{} 1/b \plus{} 1/c$
It 's true because:$ \sqrt [3]{a^3 \plus{} 1)(1 \plus{} b^3)(1 \plus{} c^3)} \ge a \plus{} 1/a$[/quote]
yes,your solution is true!
And We also:
$ 3(a^3 \plus{} 1)(b^3 \plus{} 1)(c^3 \plus{} 1)\geq (a \plus{} b \plus{} c)(a \plus{} 1)(b \plus{} 1)(c \plus{} 1)$
$ 3(a^3 \plus{} 1)(b^3 \plus{} 1)(c^3 \plus{} 1)\geq(1/a \plus{} 1/b \plus{} 1/c)(a \plus{} 1)(b \plus{} 1)(c \plus{} 1)$ with $ a,b,c > 0$ such that $ abc \equal{} 1$ :)
It's easy! :P

[quote="manlio"][quote="quykhtn-qa1"]....So,we only need to prove:
$ 3\sqrt [3]{(a^3 \plus{} 1)(b^3 \plus{} 1)(c^3 \plus{} 1)} \ge a \plus{} b \plus{} c \plus{} 1/a \plus{} 1/b \plus{} 1/c$
It 's true because:$ \sqrt [3]{a^3 \plus{} 1)(1 \plus{} b^3)(1 \plus{} c^3)} \ge a \plus{} 1/a$[/quote]

Can you please conclude this step? :blush:

Thank you very much. :)[/quote]
It is Holder's inequality, manlio  :wink:
\[ \sqrt [3]{(a^3 \plus{} 1)(1 \plus{} b^3)(1 \plus{} c^3)} \ge (a\times 1\times 1 \plus{} b\times 1\times c) \equal{} a \plus{} 1/a\]

How, please, do you solve these two inequalities?

$ 3(a^3 \plus{} 1)(b^3 \plus{} 1)(c^3 \plus{} 1)\geq (a \plus{} b \plus{} c)(a \plus{} 1)(b \plus{} 1)(c \plus{} 1)$
$ 3(a^3 \plus{} 1)(b^3 \plus{} 1)(c^3 \plus{} 1)\geq(1/a \plus{} 1/b \plus{} 1/c)(a \plus{} 1)(b \plus{} 1)(c \plus{} 1)$ with $ a,b,c > 0$ such that $ abc \equal{} 1$ :)

Thank you very much. :)


analytic geometry
geometry
calculus
vector
percent

No offense meant but Rushil hasn't been active for quite some time so I think we need a new mod.
Maybe Pardesi will fit in all right :)
 :P

Good idea but the Indian forum is extremely inactive and I'm not sure if a mod is needed. I think this thread belongs elsewhere if you want to catch the attention of the seniors.

i would love to be mod but then we have to be active
probably we can post maths and physics problems here since we are all upto the same
since the physics is any way inactive and our kind of math problems are 'different' than normal olympiad problems
also we are all active  :D
if u both agree we can have a go at it.

We can post IIT problems in Math physics and Chem that might be of least interest to Olympiad people.
For example: Coordinate Geometry, Calculus (The Calculus forum seems way too high level except a few problems),Vectors and 3-D Geometry and Complex Numbers.
Also the physics Forum is highly inactive.
Many of my friends like Cyber-Shreyas etc. love physics and so you (Pardesi) too can be benifited.
As long as Carcul is active we can post in the Chemistry Forum itself and maybe put a link in the Indian Forum for the IIT problems.
Maybe we can give each other mock test of 10 problems once in a week (which we can attach to the Forum after compiling them) which will also be benificial to us.
I shall try to convince my friends to join AOPS and tell them how benificial it is (Chemrock was invited by me)
so that we can together form a productive community .

Hope you like my ideas  :P
 :D

Brilliant (pun intended  :D )
so just get people
Hope Carcul is active  :)
lets post daily 2-3 problems from physics and maths here
chemistry in chemistry forum and hope we can get someone to make this a sticky  :)

We can make a sticky only if we are a moderator (I think).
You (Pardesi) should contact gregx007 who's recently become the moderator of the Chemistry Forum and ask him how he became a moderator (PM him). Then you become the moderator and only then the Fun begins.

ok the first step for that is we must post the need of a moderator [url=http://www.mathlinks.ro/index.php?f=144]here[/url] specifically
Ok let the fun begin i will stsrt posting questions  :lol:

All this sounds nice theoretically but I can't really find that much activity in this forum :( and 3-4 of us cannot generate enough activity.

And even more sadly, many of the JEE toppers happen to be nerds of the zeroth order who consider (even) coming to AOPS a waste of time. But anyways, I appreciate your noble intentions......

Maybe we can request a separate sub-forum called JEE/Olympiad-Zone

Unfortunately or Forunately there are not much of Rushils left around
And I also heard Rushil rarely went to school  :maybe:  but we have lots more pressure in School
 :P

I PMed Mr.rrusczyk today .
See the conversation:
madness wrote:
Sir Can you make Pardesi the moderator of the Indian Forum.
The forum seems to be inactive as the old moderator seems to have not replied for quite some time.
Also can we start a subforum in India named IIT and Academics?
Please reply
Thank you

His reply:
Please contact Valentin Vornicu with these requests; he manages the forums. It may be a little while before he responds, as he is on vacation in Romania right now

thanks for that Madness
please PM that to Valentin .
Romania is his hometown and he will i think surely respond  :)
to some extent our school is ok we(atleast me) have been out of school for almost 2  months now

[quote="pardesi"]Hey shreyas Rushil will get angry if he reads ur topic he was AIR-7th in 2006 in the JEE went to ICho,IPho and IMO Tc's   and is a KVPY Scholar :lol: and is by far the indian with Highest number of posts and though he has been inactive for a almost two years now he has post rate of about 1.6
 :wow:

 and do u consider this as a waste of time .Not going anywhere else I get almost 99.99 percent of chemistry problems solved by Carcul.
Cheers[/quote]

You have totally misunderstood me. I DID NOT state that coming to AOPS is a waste of time. I just stated the common opinion that is circulating among nerds that going online is a waste of time. And the current IIT Pattern does not test your intelligence, it just tests how thorough you are... :((

And since Rushil is now in IIT (I guess he must be) he hasn't come online for a long time.

Rushil is one of a kind! Sadly we don't get him anymore. So let's hope to emulate. Maybe Pardesi will after he becomes mod;)