The tallest building in the world is the Sears Tower in the city of Chicago. If the observation tower is 1450 feet above the ground, how far can a person see standing in the observation tower using the fact Earth's radius is 3960 miles?

NOTE: 1 mile = 5280 feet.

BTW The Sears Tower isn't the tallest in the world, my friend.

[hide="hint"]Draw a tangent from the top of the building to the earth, and treat it as a two dimensional problem. Also use the fact that the tangent and the radius from the point of tangency make a right angle.[/hide]

Yeah, the tallest building in the world is Taipei 101, as of today.

I've been to both. I went to sears tower in august in chicago. to the skydeck.

and taipei 101 last summer. also. I'm going this summer too.

now... Let's see how to solve the problem.  :lol:

Connect the point of tangency to the center and there's your right triangle.

Thanks for the good information about the tallest building.

In terms of the math tip provided, I tried drawing as suggested but have no idea where the right triangle actually is supposed to be in relation to the Earth's radius.

Help......

[quote="now a ranger"]I've been to both. I went to sears tower in august in chicago. to the skydeck.

and taipei 101 last summer. also. I'm going this summer too.

now... Let's see how to solve the problem.  :lol:[/quote]

way to spam...again.

OK Try this. Draw a circle with radius 3960. Then draw, a line coming out of it with length 1450/5280, whatever that is. It doesn't have to be to scale. Then, from the top of the line (representing the tower), draw a tangent line to the circle. The tangent is the line of your vision. Connect the point of tangency to the center and there's your right triangle.
Do you see where to go from there?

[hide="giveaway hint"]the hypotenuse is the radius + the tower, the long leg is the radius, the short leg is the distance you can see.[/hide]

This reasoning here is super.  Thanks a lot.

I'm not sure about Taipei 101 being the tallest currently...

I know Burj Dubai was something like 82 stories last november, and it has been under construction ever since, so it may be the tallest building right now... (if it isn't, then it will be soon)

Sorry about the off-topicness, but I just had to post that  :D

I had to post that too because of eagerness and that my relatives and family are all from Taipei...

oh and on the question. How do you know the tower is straight up. Couldn't it be a diagonal or something? not a perfect solid rectanglular prism or tower. so it would be different from sight...

Well that would be pretty safe to assume, seeing that it would probably fall over.

not if it is centralized and has a basement underground to support it. Buildings and Technological Advancements have made our generation more complicated living life.

Maybe it should be stated that it is a certain shape.

Just like people don't know that Yoplait Yogurt is the shape of a frustum.

Good information has been provided here.


[url=http://www.google.com/search?hl=en&client=firefox-a&rls=org.mozilla%3Aen-US%3Aofficial&hs=wwq&q=favicon&btnG=Search]google is your friend[/url]

EDIT: 1000th post!! xDD

[quote="Anaxerzia"]What happened to the AoPS favicon?[/quote]

What do you mean what happened? It's there for me. XP and Mozilla 3.0

Hmm its not working in my Firefox 3. Every other favicon shows up. It works in I.E., but not Mozilla (GASP!)

I don't have a favicon either...

EDIT: Never mind, I do.

[quote="myyellowducky82"][quote="#H34N1"][img]http://myfiles.zapto.org/ahem.jpeg[/img][/quote]
pic didnt show[/quote]
never mind it did.

[quote="Anaxerzia"]Okay the Favicon shows up for me now.

That was strange...[/quote]
it happens when it loads and you move throught a bunch of pages real fast, then the browser won't have much time to respond.

Please, let this thread die. It is over a year old. (I think).


[quote="Vasc"]If $a,b,c$ a non-negative numbers such that
$a^2+b^2+c^2=a+b+c$,
then
$a^3+b^3+c^3 \leq 3$.[/quote]
In any case it is very interesting!

Prove that if $\{a,b,c,d\}\subset\mathbb R$ and $a+b+c+d=a^2+b^2+c^2+d^2$ then :
$a^3+b^3+c^3+d^3\leq2+\frac{7}{2\sqrt3}.$
Equel for $a=b=c=\frac{1+\sqrt3}{2\sqrt3}, d=\frac{1+\sqrt3}{2}.$  :)

Sorry, Mildford. I reposted the correct statement.

If $a,b,c$ a non-negative numbers such that
$a^2+b^2+c^2=a^3+b^3+c^3$,
then
$a^4+b^4+c^4 \leq 3$.

Problem:
\[ (a^2+b^2+c^2)^4(a^4+b^4+c^4) \le 3(a^3+b^3+c^3)^4. \]
 :lol:
By SOS, it's done. But I will try another solution. It's very nice.

[quote="hungkhtn"]Problem: \[(a^{2}+b^{2}+c^{2})^{4}(a^{4}+b^{4}+c^{4}) \le 3(a^{3}+b^{3}+c^{3})^{4}.\] :lol:
By SOS, it's done. But I will try another solution. It's very nice.[/quote]

hungkhtn , how do you SOS this expression ? Maybe a hint?

**Edit :nevermind , i think I got it , it is something like

$\sum (X^{3}(a+b)+X^{2}Yab+Y^{3}a^{2}b^{2})(a-b)^{2}$

where $X=a^{3}+b^{3}+c^{3}$ , $Y=a^{2}+b^{2}+c^{2}$ .And we need to show $S_{a},S_{b},S_{c}\geq 0$

I proved this inequality for $n \leq 116$, but it is valid for $n \leq 157$ (by computer).


[quote="Victory.US"]find all positive integer $ n$ such that $ [ \sqrt [n]{111}]$ divides $ 111$[/quote]

$ 111\equal{}3 \times 37$

$ n\equal{}1$ is a obvious solution.

$ \sqrt [2]{111}<37$ so $ [ \sqrt [n]{111}]\equal{}37$ has no solution.

$ 3 \le  \sqrt [n]{111}<4 \Leftrightarrow \begin{cases} 3^n \le 111 \\ 4^n \ge 111 \end{cases} \Leftrightarrow \begin{cases} n \le 4 \\ n \ge 4 \end{cases}$ ,  so $ n\equal{}4$ is solution for this case.

So $ n\equal{}1,4$ are the only solutions.

you have a mistake in your proof  :lol:
what do you think about $ n \ge 7$ is solution


Patrick and Ravi are playing the smiley game. The first person to post 4  :D s wins. Each person tries to be the first to post the  :D  for a day, continuing until one person wins. Ravi is at his computer more often and has a 60% chance of winning the  :D    for a certain day. What is the probability that Patrick wins the smiley game? Express your answer as a common fraction.

this looks like an alcumus problem
if i was a mod, i would probably uhh...
uhh...
plz solve your own alcumus problems

[quote="avec_une_h"]It's not like we're going to do the problems anyway... (Sorry gauss)[/quote]
y? :(  :huh:  :o

Wow I'm up at 11:41 PM posting on this forum. O_O
...When I should be doing my AP World History essay. x_x
I think life is meann, except for the math meet on thursday. :)

and uhh yeah I'll solve it over the weekend if I have time, haha.

It's meet week!  And I have a Scholastic Bowl meet today.

AP World History?  I'm just doing honors (not AP) world history.  Although it is kind of like AP, we don't get as much homework or do as many papers.

Dude seriously??? I have a Scholbowl meet today too! ahaha
Meet tomorrow though, I'm going for another set of 25's. :D

Wow I just realized, HER (I got it right, hooray for me) screen name is "with an h" in French.

Yeah, my username is something from French camp.

Well, both meets were not so great.  My team won the first match of our Schol Bowl meet, and we lost the second by a bit.  And if I hadn't made 2 multiplication mistakes...  As for NSML, I copied #3 wrong onto my scratch paper and therefore messed it up.  But I got the rest correct.  Not a 3rd perfect though.  I did get to be a sophomore contestant because of the orchestra kids, but I only got two of the problems (including #5 strangely enough).  I made complementary probability mistakes on 1 and 3, and I still don't know how to do 4.  I suppose you could use trig.

We owned scholastic bowl, both varsity and JV.  We killed Naperville Central, our rivals, by 150 points, and West CHicago by 80.
Haha I failed number 5...we were missing people due to orchestra too.
Oh, number 4 you don't need trig, all you need to do is you know that the median of a right triangle from the right vertex is congruent to the two segments it divides the hypotenuse into.  You need to figure out what x degrees has to be so that AP=PB.  You want AP<BP, so then you take the angle measure x/90 to find the probability.  It should be 72 degrees, so then it would be 72/90=4/5.  Not a very good explanation, better with drawings.

It's kinda weird how we were in the same room, for freshman and sophomore contestant...I was the guy with the blue sweatshirt, heh.

lol this is weird that it's getting off topic in this thread's real intended topic :o

=O How unfair! Our season hasn't started yet. Our practices haven't even started yet. Come to think of it, our TRYOUTS haven't even started yet!

[quote="avec_une_h"]We lost to York.  Do you play varsity or JV?  I was on JV during the first meet, when varsity lost to Glenbard West and Hinsdale South, especially because they couldn't do math well.  Then, I was moved up to varsity and we've only lost to York.  Our JV team is undefeated so far.

Oh, that's how you do #4.  I completely forgot the thing about the median.  I was trying to use law of sines.  It took me forever to get #4 on the freshman test.  I kept messing up the equations.  But I didn't mind the snake one too much.  I wish we could do number bases again.  I found them really easy, and the rest of my team hated them.  Except for maybe Stephen.  He picked up how to do them pretty quickly.  Though there was one on candy bar, which they automatically gave to me.[/quote]

JV.  I think your school would be much better if you had more people involved.

The snake one took time, I just calculated it totally wrong, and I realized with 2 minutes left, so I didn't have enough time to make another table...#4 seemed confusing, but I made a rough of the equation, solved it, and then went back and made a differnet equation, and both worked.
The bases were easy too, but the test was also really easy.

We have a fair number of people involved.  We had 15 people at the last meet and more at our home meet.  There's just not that many smart people at my school.

Oh, that's too bad about the snake one.  Yeah, this is the first meet where I actually felt the time pressure.

The number bases test was easy, but I didn't have trouble with the much harder ones we were doing in practice.

D: north's varsity team beat us (central) by like 20 points, which i should have gotten cause i buzzer-jumped the first math question.

too excited >.<


Here is a game where you can fight with many weapons! Anyways, eight people can play at once. Everybody starts with 50 life points. You have 30 weapons to use, offensive, defensive, and healing. Each category has ten different weapons with each a different level. For example:

Offensive Weapons:

Fireball-1
Gun-2
Sword-3
Sniper Rifle-4
Big Gorilla Rampage-5
Bomb-6
Big Mace-7
Missle-8
Legendary Hammer-9
Elephant Rampage-10

These are the Defensive Weapons:

Wooden Shield-1
Steel Shield-2
Big Shield-3
Armor-4
Flaming Armor-5
Magical Armor-6
Force Field-7
Strong Force Field-8
Brick Wall-9
Military Epic Defense-10

And finally, here are the Healing Potions:
Soup-1
Potion-2
Super Potion-3
Elixir-4
Magic Elixir-5
Remedy-6
Magic Potion-7
Magic Spell-8
Sacred Spell-9
Holy Water-10

So there are ten numbers in each group of weapons. The rule is that every turn, you must use a number from either offensive, defensive, or healing, and once you use a number, you cannot use it again. For example, let’s say I use an offensive weapon of level 8, the missile. I cannot use defensive weapon level 8, strong force field, and I cannot use healing potion 8, magic spell. After ten turns, you should have used up all of your numbers and no longer be able to make a move. At that point, I will open up your numbers again for use, and the cycle repeats. When one person has zero life points, they are out, and the last stander wins. After the third cycle, the cycle when I have regenerated your numbers two times, whoever has the most life points wins. If there is a tie, the people who tied go for another cycle. If there is another tie, we cycle among these new people who tied, and so on until there is one winner. Oh yeah, and just to prove that you guys have read this, after whatever you say that states you join, add one and only one extra period. So that it looks something like: “I join..” I hid this message in the middle on purpose. Anyways, for offensive weapons, the amount of hit points it damages you is equal to twice the numerical correspondence. For example, since Big Mace is 7, it does 7*2 or it conflicts 14 hit points. For defensive weapons, it subtracts the amount of damage by twice the corresponding number that someone MIGHT do to you. If no one attacks you, then it is considered a waste to use the defensive weapon. Also, if your defense number is higher than the offense number, you do not regain health, the amount the attack conflicts is just zero. If multiple people attack you in one turn, the amount of hit points you take is the sum of the hit points everybody damaged you with. For healing weapons, you heal the amount of hit points that the potion shows, so Holy Water will only heal 10 hit points, while defense weapon military epic defense will reduce attacks on you by 20 hit points. You cannot get over 50 hit points no matter what. Each turn will last two days, all players must pm me to tell me their move, or what weapon from the three categories you would like to use. If it is an offensive weapon, you must tell me who you would like to attack. If somebody doesn’t pm me in two days, then they are automatically disqualified. After every turn, I will post the life points of each person, and I will tell when a person is disqualified. Do not worry; I will keep track of the weapons you used and your hit points. Post any questions you have here. Signups start today over here:
1.
2.
3.
4.
5.
6.
7.
8.

Read the instructions, please, spaceguy. I will not let you join until you do so.

I'm in.. and this message is most certainly not too small!

Sign up list:

1. 1=2
2. cyberspace
3. spaceguy524
4.
5.
6.
7.
8.

Sign-Up List:

1. 1=2
2. cyberspace
3. spaceguy524
4. ernie
5.
6.
7.
8.

Cough, look at that colon.  I meant for it to be a period. :P

cgyao15, you must read the instructions before you join. You are not counted in yet. Don't wonder how i knew, i just know.

Sign ups:
1. 1=2
2. cyberspace
3. spaceguy524
4. ernie
5. Chompy
6.
7.
8.

Its a bit long for instructions but I guess I'll join in

kianex apparently they were so long you didn't read them :wink:

This should be interesting....

I'm in..

Ok Yoshi, you're in, thanks for reading the instructions.

Sign ups:
1. 1=2
2. cyberspace
3. spaceguy524
4. ernie
5. Chompy
6. Yoshi
7.
8.

I Join..

Happiness runs in a circular motion.

it doesnt seem like theres much strategy involved, but oh well...anything with an elephant rampage is fun for me :)

i join.

[quote="algebra1337"]...i read the instructions...

btw mr. happy u cant turn people away because u [b]think[/b] they didnt read the instructions. thats just stupid. unless u have some psychic power. then its okay.[/quote]

You know, maybe there's something you didn't do that's in the instructions.

Maybe that's how he knows.

Ok, yes, you can join now. Sorry for the inconvenience anyways. I won't make this rule next round. Anyways, sign ups are over and here are the players:

1. 1=2
2. Cyberspace
3. Spaceguy524
4. Ernie
5. Chompy
6. Yoshi
7. Gogators
8. Algebra1337

Start p.m.-ing me your moves.

Ok, so I have a code. It's gonna look something like this: O5 or D3 for each person. The letter in front, O, D, or H, stands for Offense attack, Defense, or Healing, respectively. The number following means the numeric value they used. If it was Elephant Rampage, I would say O10. Also, if you did an offensive attack, I will write the name of the target person right after it. The final number is the damage dealt/defended/healed. I will show you everyone's move and target, and the updated health of each person.

[hide="The Moves"]

(username), (Move), (Target), (Damage dealt/defended/healed)

1=2	              O10 Ernie	     20
Cyberspace	 D1	                  2
Spaceguy524	 O4 Yoshi	                  8
Ernie	              O8 1=2	                  16
Chompy	              D5	                  10
Yoshi	              O3 Gogators	     6
Gogators	              O10 Algebra 1337	     20
Algebra1337	 O7 Cyberspace    	     14[/hide]

[hide="Health"]

1=2	             34
Cyberspace	38
Spaceguy524	50
Ernie	             30
Chompy	             50
Yoshi	             42
Gogators	             44
Algebra 1337	30
[/hide]

Start the next round, pm me. Remember, don't use the same number that you have already used!

Ok, now round 2 is over, sorry for the delay, i was quite busy this week. Anyways, there are the same codes, and here are the results:

The Moves:

1=2	             H8
Cyberspace	O7 algebra1337
Spaceguy524	D7
Ernie	             O10 1=2
Chompy	             D6
Yoshi	             O2 Chompy
Gogators        	D5
Algebra1337	O6 Gogators

Updated Health:

1=2	             22
Cyberspace	38
Spaceguy524	50
Ernie	             30
Chompy               	50
Yoshi	             42
Gogators	             42
Algebra 1337	16

Remember, you cannot use the same number you have used twice! PM me if you want to know what numbers you have already used!

By the way, me and some other players will be gone on school fieldtrips this week, and we will return on Sunday, November 1st. Pm me, but do not expect a reply until i come back. The game will be delayed until that date.
[b]ROUND 3 BEGINS! START PMMING ME![/b]

[quote="Mr. Happy"]
By the way, me and some other players will be gone on school fieldtrips this week, and we will return on Sunday, November 1st. Pm me, but do not expect a reply until i come back. The game will be delayed until that date.
[b]ROUND 3 BEGINS! START PMMING ME![/b][/quote]
November 1st.

Please PM me if anyone drops out, i want to join..

Um, November 4th, it's about time we started...

Yeah, sorry for the delay, there's still one person to wait for... what should we do about him?


inequalities
trigonometry
algebra
polynomial
function
real analysis
calculus

Prove that, for all positive integers $ n$,

$ ((n)(n + 1)(\sqrt {\sin^3 1^\circ} + \sqrt {\sin^3 2^\circ} + ... + \sqrt {\sin^3 n^\circ)}/2)^{\frac {2}{3}} \ge n \sqrt [n]{n! \sin 1^\circ \sin 2^\circ \sin 3^\circ ... \sin n^\circ)}$

By the way, I haven't tried solving this in an easy way, so I'm not sure if it's triv.

[hide="Step 1"]

$ n(n + 1)/2 = 1 + 2 + ... + n$

[hide="Step 2"]

$ (1 + 2 + ... + n)^2 = (1^3 + 2^3 + ... + n^3)$

[hide="Step 3"]

By Holder's Inequality:

$ (1^3 + 2^3 + ... + n^3)^{1/3}((\sin 1^\circ)^{3/2} + (\sin 2^\circ)^{3/2} + ... + (\sin n^\circ)^{3/2})^{2/3} \ge (\sin 1^\circ + 2 \sin 2^\circ + ... + n \sin n^\circ)$

By AM-GM:

$ (\sin 1^\circ + 2 \sin 2^\circ + ... + n \sin n^\circ)/n \ge \sqrt [n]{n! \sin 1^\circ \sin 2^\circ ... \sin n^\circ}$

[/hide]

[/hide]

[/hide]

It isn't that trivial

You had to do some work and manipulation to get to that final step.

I can't even solve that... :(

BTW, you're doing that for MOP, right?

[quote="Altheman"]It is an okay inequality but I'm not sure why you have the sines in there...they become negative for large enough $ n$ and the problem explodes.[/quote]

Oh, yeah, like for the nth root, you can have a negative inside of an even power, i see, nice thinking.

I'd say it is pretty trivial now that I look at it...It is easy to see that if say $ x_i \equal{} \sin i$...then equality holds for like $ \frac {x_i}{i} \equal{} k$ where $ k$ is a constant. Then the rest of the solution is very obvious when you do that...

You want to clear out the 3/2 powers on the $ x_i$s and you are like well I know the equality case, so we use cauchy. Then its like wow that worked out nicely since the factor that I added cancels out.

Edit: your proof shows the motivation okay. Nice solution!

AND

Post more MOP problems please.

i dont know wat type of inequality it is but i would like to say that for greater powers of n sin theta cannot be solved by me!
so get a solution next time otherwise...................
 :ninja:

haha that's not a mop problem, i wrote it in my spare time.

but here's one:

[b]Red Test 7, Problem 4[/b]
Let P(x) be a polynomial. Find the number of polynomials, in terms of n, such that each coefficient is either 0, 1, 2, or 3, and P(2) = n.

I'll post my solution to it when I have time (tomorrow?), but I gotta go for curfew.

hey dude thanx for these questions but could u help me with complex numbers
 :roll:

We are looking for the solutions to $ a_0 \plus{} 2a_1 \plus{} 4a_2 \plus{} ... \plus{} 2^ka_k \plus{} ... \equal{} n$ where $ a_k\in\{0,1,2,3\}$

The generating function for this is

$ (1 \plus{} x \plus{} x^2 \plus{} x^3)(1 \plus{} (x^2) \plus{} (x^2)^2 \plus{} (x^2)^3)...(1 \plus{} (x^{2^k}) \plus{} (x^{2^k})^2 \plus{} (x^{2^k})^3)...$

$ \equal{} (1 \plus{} x)(1 \plus{} x^2)(1 \plus{} x^2)(1 \plus{} x^4)...(1 \plus{} x^{2^k})(1 \plus{} x^{2^{k \plus{} 1}})...$

$ \equal{} \frac {1}{(1 \plus{} x)(1 \minus{} x)^2}(1 \minus{} x^{2^{k \plus{} 2}})$

For sufficiently large $ k$, we can ignore the $ x^{2^k}$ term since it does not contribute to the $ x^n$ terms..

$ \equal{} (1 \minus{} x \plus{} x^2 \minus{} x^3 \plus{} ... \plus{} ( \minus{} 1)^kx^k \plus{} ...)(1 \plus{} 2x \plus{} 3x^2 \plus{} 4x^3 \plus{} ... \plus{} (k \plus{} 1)x^k \plus{} ...)$ by the extended binomial theorem.

We are looking for the convolution of these two polynomials. If the coefficients are $ x_k$ and $ y_k$ of the polynomial, then the $ x^n$ coefficient is:

$ \sum_{t \equal{} 0}^n y_t x_{n \minus{} t} \equal{} \sum_{t \equal{} 0}^n (t \plus{} 1)( \minus{} 1)^{n \minus{} t}$

If $ n$ is even, this sum is $ \frac {n \plus{} 2}{2}$, if $ n$ is odd, the sum is $ \frac {n \plus{} 1}{2}$.

I haven't checked if this is right but I'm pretty sure that my method works...

:idea: youre damn good at it
kudos to u!
 :lol:

I just saw a different way to handle the expansion:

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

$ \equal{}(1\plus{}x\plus{}x^2\plus{}...\plus{}x^k\plus{}...)(1\plus{}x^2\plus{}x^4\plus{}...\plus{}x^{2k}\plus{}...)$

From here, it is much easier to derive the $ x^n$ term.

Here's my way:

$ P(x) \equal{} c_0 \plus{} c_1x \plus{} c_2x^2 \plus{} c_3x^3 \plus{} ...$

with

$ c_i \in \{0, 1, 2, 3\}$

The restriction on the coefficients resembles the base four expression of a number. So, we try to play with that.

$ P(2) \equal{} c_0 \plus{} 2c_1 \plus{} 4c_2 \plus{} 8c_3 \plus{} ... \plus{} 2^nc_n \plus{} ...$

$ \equal{} (c_0 \plus{} 4c_2 \plus{} 16c_4 \plus{} ... \plus{} 4^jc_{2j} \plus{} ...) \plus{} (2c_1 \plus{} 8c_3 \plus{} 32c_5 \plus{} ... \plus{} 2 \times 4^kc_{2k\plus{}1})$

$ \equal{} (c_0 \plus{} 4c_2 \plus{} 16c_4 \plus{} ... \plus{} 4^jc_{2j} \plus{} ...) \plus{} 2 \times (c_1 \plus{} 4c_3 \plus{} 8c_5 \plus{} ... \plus{} 4^kc_{2k\plus{}1})$

It is well-known that every non-negative integer can be uniquely expressed in any positive integer base (except base 1, which sucks), so the two expressions enclosed in parentheses above each represent non-negative integers.

$ P(2) \equal{} A \plus{} 2B \equal{} n$

with A, B non-negative integers.

B can take values from $ 0$ to $ \lfloor \frac{n}{2} \rfloor$, which is $ \lfloor \frac{n\plus{}2}{2} \rfloor$ values. A value of B determines a value of A, so there are

$ \lfloor \frac{n\plus{}2}{2} \rfloor$ such polynomials.


inequalities
AMC
AMC 10
algorithm
geometry
3D geometry
trapezoid

1) A charity sells 140 benefit tickets for a total of $2001.  Some tickets sell for full price (a whole dollar amount), and the rest sell for half price.  How much money is raised by the full price tickets?

(A) 782
(B) 986
(C) 1158
(D) 1219
(E) 1449

2) A number x is 2 more than the product of its reciprocal and its additive inverse.  In which interval does the number lie?
(A) -4=<x=<-2
(B) -2<x=<0
(C) 0<x=<2
(D) 2<x=<4
(E) 4<x=<6

3) how many positive integers not exceeding 2001 are multiples of 3 or 4 but not 5?
(A) 768
(B) 801
(C) 934
(D) 1067
(E) 1167

i have spent some time on these problems yet cannot quite solve them

For number three, how many numbers less than or equal to 2001 are divisible by 3? How about 4? both 3 and 4? 3 and 5? 4 and 5? all of 3, 4 and 5?

How do these numbers relate?

If you still don't get it try an easier problem like how many numbers less than 100 are multiples of 3 or 5?

I would try to help with the others but I have to go now.

Hello, demji!

Quote:3) How many positive integers not exceeding 2001 are multiples of 3 or 4 but not 5?
(A) 768
(B) 801
(C) 934
(D) 1067
(E) 1167[hide]

Here's my baby-talk approach to this problem . . .

Note: [x] is the "greatest integer" function.

How many numbers are multiples of 3? (every third number)

There are: [2001/3] = 667 of them.

How many are multiples of 4? (every fourth number)

There are: [2001/4] = 500 of them.

But there is an "overlap"? ... How many are multiples of both 3 and 4?

There are: [2001/12] = 166 of them.

Hence, there are: 667 + 500 - 166 = 1001 numbers not exceeding 2001

which are multiples of 3 or multiples of 4.

But how many of these are also multiples of 5?

Some are multiples of 3 and 5 (multiples of 15).

There are: [2001/15] = 133 of them.

Some are multiples of 4 and 5 (multiples of 20).

There are: [2001/20] = 100 of them.

But some are multiples of all three (multiples of 60).

There are: [2001/60] = 33 in the "overlap".

Hence, there are: 133 + 100 - 33 = 200 which are also multiples of 5.

Therefore: the number of positive integers not exceeding 2001

and are multiples of 3 or 4 but not 5 is:

1001 - 200 = 801 . . . Answer choice [B][/hide]

[quote="demji"]1) A charity sells 140 benefit tickets for a total of 2001 dollars.  Some tickets sell for full price (a whole dollar amount), and the rest sell for half price.  How much money is raised by the full price tickets?

(A) 782
(B) 986
(C) 1158
(D) 1219
(E) 1449
[/quote]

Let c be the cost of a full-price ticket and x be the number of full-price tickets sold.  From the given information, we know that 140-x half-price tickets were sold, so that
2001 = (cost of full-price)*(# full-price) + (cost of half-price)*(# half price)
2001 = cx + (c/2)*(140-x)
4002 = 2cx + c(140-x)
4002 = c(2x+140-x)
4002 = c(140+x)

We also know that c, x, and 140-x must all be positive integers, so that c > 0 and 0 $\leq$ x $\leq$ 140.  Thus $140 \leq x + 140 \leq 280$.

Since c and 140+x multiply to 4002, and both are positive integers, each must be a positive integer factor of 4002.  The positive integer factors of 4002 are 1, 2, 3, 6, 23, 29, 46, 58, 69, 87, 138, 174, 667, 1334, 2001, and 4002.  Of these, the only possible value for x+140 is 174 as a result of the inequality found above.  Thus c is 2004/174 = 23, and the amount of money raised by the full-price tickets, cx, is 23*34 = 782.  (A).

[quote]2) A number x is 2 more than the product of its reciprocal and its additive inverse.  In which interval does the number lie?
(A) -4=<x=<-2
(B) -2<x=<0
(C) 0<x=<2
(D) 2<x=<4
(E) 4<x=<6[/quote]

x = 2 + (1/x)(-x) = 2 + (-1) = 1.

1 is in interval (C).

(The only reason the question asks for the interval rather than giving five definite answer choices is so that you can't just try each answer to figure out which one satisfies the given property.)

Hello, demji!

Evidently, there is a typo in this problem . . .

Quote:1) A charity sells 140 benefit tickets for a total of $2001.
 Some tickets sell for full price (a whole dollar amount),
and the rest sell for half price.
How much money is raised by the full price tickets?

(A) 782
(B) 986
(C) 1158
(D) 1219
(E) 1449[hide]

Let H = number of half-price tickets sold.

Let F = number of full-price tickets sold.

Then we have: [1] F + H = 140

Let a = price for a half-price ticket.

Then 2a = price for a full-price ticket.

The total receipts were: [2] 2aF + aH = 2001

Multiply [1] by -a  .... -aF - aH = -140a

Add [2]  ................. 2aF + aH = 2001

and we get: aF = 2001 - 140a ---> F = (2001/a) - 140

Since F is a positive integer, a must be 3.

Hence: F = 527 . . . and H = -387 . . . What?

It seems that  they sold 527 full-price tickets and refunded 387 half-price tickets![/hide]

Here is the answer:

[hide]

1) A

Ok first I set up the equation b+s=140 (big tickets + small tickets) and that z*b + (z/2)*s=2001.  For z is some integer dollar amount.  Then I thought about the number 2001 and knew that it factors to 3*23*29 and I knew that z had to be between 29 and 14.  (29*0+14.5*140=2030 / 14*140 + 7*0=1960)  So I tried substituting 23 for z and got b=34 and s=106.

34*23=782[/hide]

nice job jason it checks out.  The other one above that looked like 23 * 34 but that doesn't work if the 23 is full tickets and 34 is price of full tickets.  Thx i could not figure it out

[quote="Soroban"]Hello, demji!

Evidently, there is a typo in this problem . . .

...

Let H = number of half-price tickets sold.
Let F = number of full-price tickets sold.

Then we have: [1] [b]F + H = 140[/b]

Let a = price for a half-price ticket.
Then 2a = price for a full-price ticket.
...
Hence: [b]F = 527[/b] . . . and [b]H = -387[/b] . . . [i]What?[/i]

It seems that  they sold 527 full-price tickets and [b]refunded[/b] 387 half-price tickets![/quote]

Soroban, there is no typo.  The problem lies in your assumption that the half-price tickets are a whole dollar amount (i.e. that $a$ is an integer); only the full-price tickets need be a whole dollar amount.  (In fact, if you look at my solution, I found that the full-price tickets were $23, so the half-price tickets were [i]not[/i] a full dollar amount.)

And demji, 23 is the price of the full-price tickets and 34 is the number of full-price tickets.  See if that solves your problem.

Hello, mathfanatic!

[i]Ha![/i] . . . I get it now ... silly me!

thx guys you really are good at these i wish i had the seeing eye.  What would you reccomend me to study in order to get better at problem solving.

demji,

the only way to get better at "seeing" things is to do more problems!!

theory will only take u thus far....

doing problems and carefully studying the solutions will  help u "see" things in later similar questions...

well, that's how i learn!!

btw, borrow some books with problems in them, instead of theory. as you progress, u should be able to do harder problems

For question 1...

I eventually ended up with mathfanatic's equation of x(a+140)=4002

and I was baffled at that point as there were 2 variables and one algebra to solve ....

So mathfanatic... I ask you; The main intension of this problem was to see whether if you can see the factors of 4002 rite?

I tried putting 4002 as multiples of prime numbers but than I got stuck after dividing it by 2 and 3...

How long did it take you to solve Q1?

[quote]The main intension of this problem was to see whether if you can see the factors of 4002 rite? [/quote]

Essentially.  A common problem-solving strategy for solving an equation in integers with more than one variable is to put it in the form xy = k, where k is some constant integer.  Then there are only a fixed number of values that x and y can take on, since they have to multiply to k...so yeah, you need to factor k.

It might be helpful to know the factorization of the current year (for example, 2004 = 2^2*3*167), since these problems come up frequently.  In this case, the problem came from the 2001 AMC10, so experienced problem-solvers knew the factorization of 2001 ahead of time.

As for how long it took...well, I've seen this problem before.  The first time, it took a long time though.  Don't worry too much about how long it takes you to solve problems.  Persistence is the most important quality.

thanx about the tip mathfanatic...

Ya, I agree with you about the "year" frequently coming out on AMC types of exams.

xy = k form

Thank you for the quick reply and kind xplanation

is there also an easy way to tell if a number is prime?  What are all those dividing rules again?  Such as if the sum of the digits is divisible by 3 then the number is divisible by 3

[quote]how can you quickly find the factorization of a year if you don't know it off the top of your head?[/quote]

The way you factor any number - divide by primes. Divisibility criteria can help you here for speed.

[quote]What are all those dividing rules again? Such as if the sum of the digits is divisible by 3 then the number is divisible by 3[/quote]

There are numberous of these divisibility criteria. The most obvious are the first few - divisible by 2 if a number is even, divisible by 3 if sum of digits is divisible by 3, divisible by 4 if last two digits are divisible by 4, divisible by 5 if last digit is 0 or 5. You can often construct your own rules for other integers.

[quote]is there also an easy way to tell if a number is prime?[/quote]

It isn't prime if you can find a factor of it that is not 1 or itself. Of course, if the number is large, this becomes extremely difficult. A lot of algorithms depend on the fact that it is very difficult to tell if very large numbers are prime or not. So, no.

Here are some more problems i am stuck on.

4) A right circular cylinder with a diameter equal to its height is inscribed in a right circular come.  The cone has a diameter 10 and altitude 12, and the axes of the cylinder and cone coincide.  Find the radius of the cylinder.
A) 8/3
B) 30/11
C) 3
D) 25/8
E) 7/2

5) In trapezoid ABCD, AB and CD are perpendicular to AD, with AB+CD=BC, AB<CD, and AD=7.  What is AB*CD?
A) 12
B) 12.25
C) 12.5
D) 12.75
E) 13

Answer to Q5)

Well this is a very easy one... I think you just didnt draw out the diagrams and tried to do all of it in your head

Just draw out the diagrams big enuff so that u'll understand

and use pythagorous theorm to work it out.

Heres the equation you should get:

$7^2+(b-a)^2=(a+b)^2$

where
$AB=a$
$CD=b$

and the answer is B 12.25

i understand your method but i don't see how you got B for your answer.

i have narrowed the equation you gave to 49-ab=ab     so 49=2ab
what do i do now

Letting AB = x and CD = y, we get (y-x)^2+49=(y+x)^2. Then expanding, we get y^2+x^2+49-2xy=y^2+x^2+2xy. So 49=4xy. Then xy=49/4, or 12.25.

[quote="demji"]4) A right circular cylinder with a diameter equal to its height is inscribed in a right circular come.  The cone has a diameter 10 and altitude 12, and the axes of the cylinder and cone coincide.  Find the radius of the cylinder.
A) 8/3
B) 30/11
C) 3
D) 25/8
E) 7/2[/quote]

We can just look at the cross-section of the figure to get a square inscribed in a triangle.
Then we draw the line of symmetry to get 4 triangles and two rectangles.
We can see that all four triangles are similar.
Calling the radius of the cylinder x, we find expression for the legs
of each of the right triangles.
Then because they are similar triangles, we can use proportions to find x.

yes i know i figured it out a while ago.  I forgot what the answer is but i have it written down.  Thx

[quote="demji"]how can you quickly find the factorization of a year if you don't know it off the top of your head?[/quote]

The only way I can think of is to keep dividing the number by prime numbers untill you get 1.


[quote="Nerd_of_the_Ages"]So how'd you guys do for the 1/26 SAT? Say if you're taking it for the CTY.

I took it for the CTY, and crapped up. I got 560 CR, 590 writing, and (:blush:) 700 math. I JUST qualified to SET, although, frustratingly, when I entered my Invitation ID on http://www.cogito.com on the register page, it keeps on saying my invitation ID is invalid. So yeah, that's not good.[/quote]
Eh CTY? This is AoPS, everyone goes to AMP, no CTY people :P lol. Okay I had a 750 which was good enough for SET but I never bothered submitting so whatever.

[quote="Nerd_of_the_Ages"]Sorry, what's AMP? Actually, in Hopkins, basically anybody who takes the SAT at all does it for CTY.[/quote]
AwesomeMath. I didn't go, but I really want to this year, but I have to do summer school. If I don't make MOP (which is likely, considering my noobiness) some time later in my high school career I'll go there. I sorta took SAT for CTY also, but then I decided it didn't mean anything.

i evidently did ok
although i can't see it myself
690 reading comp
800 math
690 writing (with a 7 essay, i have no idea how i managed that)

[quote="bowei"]!!! AMP !!!
amp rocks
i went there but prolly wont be going this year
as im going to cosmos (i know, blah)
amp is really beastly
i hope you get into track 1/2/3, serial
(the other tracks just dont match by beastliness)
hm
havent heard much about cty[/quote]

Uh what's cosmos.

And Circlesquared, those are beastly scores. You outperformed me by 60 points in reading, 50 in writing, and 1 on essay.

[quote="Ubemaya"]Uh what's cosmos.[/quote]
California State Summer School for Math and Science (don't ask me how csssms becomes cosmos)
http://www.ucop.edu/cosmos/apply.html
My parents wanted me to go, but

Well seeing as how the required GPA is 3.5 or above, and "advanced" is Geometry in 8th grade, that doesn't look very advanced. The topics sound interesting, though.

Is COSMOS easier than AMP? Also, who's going to AMP this year?

I got a terrible score  :mad:  :(

Math: 790
Reading: 590
Writing: 580

[quote="daermon"]i took it again even tho i dont need it for cty (international politics at loyola woot woot!)
math: 790 (+140 lol...i epic failed last year)
writing: 640 (+10)
reading: 630 (-60)[/quote]

Identical scores to me when I was in 8th grade, except I scored 10 higher on math.

[quote="CircleSquared"][quote="daermon"]i took it again even tho i dont need it for cty (international politics at loyola woot woot!)
math: 790 (+140 lol...i epic failed last year)
writing: 640 (+10)
reading: 630 (-60)[/quote]

oh you had to go to loyola THIS year eh
great just great[/quote]
lol sorry

[quote="davidyko"][quote="junggi"][quote="hardc0rerock"]wow... I failed compared to all of you..
Math: 640
Writing:540 (don'tlaugh i fail at that stuff...)
Reading: 560

At least I got into CTY I'm prolly gonna take C programming wat about anyone else in CTY?[/quote]
i sense that this CTY thing is a plot to make asians that make it in feel a sense of accomplishment without actually accomplishing anything[/quote]

I totally agree.[/quote]

Yah the requirements to get in r pretty low like 460 for each 1 or somthin its like u can fail the SAT and still get in.

[quote="hardc0rerock"][/quote][quote="davidyko"][quote="junggi"][quote="hardc0rerock"]wow... I failed compared to all of you..
Math: 640
Writing:540 (don'tlaugh i fail at that stuff...)
Reading: 560

At least I got into CTY I'm prolly gonna take C programming wat about anyone else in CTY?[/quote]
i sense that this CTY thing is a plot to make asians that make it in feel a sense of accomplishment without actually accomplishing anything[/quote]

I totally agree.[/quote][quote="hardc0rerock"]

Yah the requirements to get in r pretty low like 460 for each 1 or somthin its like u can fail the SAT and still get in.[/quote]
hmmm my post was intended to discourage you from going to this cty thing

[quote="junggi"][/quote][quote="hardc0rerock"][/quote][quote="davidyko"][quote="junggi"][quote="hardc0rerock"]wow... I failed compared to all of you..
Math: 640
Writing:540 (don'tlaugh i fail at that stuff...)
Reading: 560

At least I got into CTY I'm prolly gonna take C programming wat about anyone else in CTY?[/quote]
i sense that this CTY thing is a plot to make asians that make it in feel a sense of accomplishment without actually accomplishing anything[/quote]

I totally agree.[/quote][quote="hardc0rerock"]

Yah the requirements to get in r pretty low like 460 for each 1 or somthin its like u can fail the SAT and still get in.[/quote][quote="junggi"]
hmmm my post was intended to discourage you from going to this cty thing[/quote]
Shut up.  CTY was cool.  They send you motivating certificates.  And glossy magazines.  Hmm...i dunno about the camps though.

SOME Paly folk speak highly of CTY, I dunno...

Er, lessee:

Math: 740 (epicfail, i know)
Critical Reading: 660
Writing: 690

A total of 2090...okay I guess

Oh dang. I REALLY EPIC FAILED!!!!!!!!!
Math 780  :o
CR 730  :|
Writing 600  :huh:

[quote="AIME15"]I've never taken it except for practice tests:

Math: 800!  I BEAT JAMES OMG IN MATH!

Reading: 560 *blush* i phial

Writing: 580 *blush* phial again[/quote]

Good try, but practice tests do not count...

Dude that's worse than what I got in 8th grade, which was pretty damn pathetic (I don't personally know anyone that didn't beat it).

Yay for people taking SAT in 7th/8th grade. I'm slightly surprised at how freely everyone is displaying their scores for the world to see. It doesn't seem to me like the sort of thing you post on the internet.

[quote="AIME15"][quote="Ubemaya"][quote="AIME15"]I've never taken it except for practice tests:

Math: 800!  I BEAT JAMES OMG IN MATH!

Reading: 560 *blush* i phial

Writing: 580 *blush* phial again[/quote]

Dude, what grade are you in?
Those are the exact same scores I got in 7th grade.
*high five*[/quote]

OMG...i'm in 7th.  IM GONNA BECOME A MINI UBEMAYA~  NOOOOOOOOO!  MUST...NOT....GET....FAT![/quote]

WAIT, don't lose hope yet. WHAT SCORE DID YOU GET ON THE ESSAY? MAYBE YOU WON'T BECOME A MINIME IF YOU DIDN'T GET A 5.

That was a practice test, didn't get my essay graded.

YES I WILL MOST CERTAINLY PWN 5 ON THE ESSAY!

[quote="Ubemaya"][/quote][quote="AIME15"][quote="Ubemaya"][quote="AIME15"]I've never taken it except for practice tests:

Math: 800!  I BEAT JAMES OMG IN MATH!

Reading: 560 *blush* i phial

Writing: 580 *blush* phial again[/quote]

Dude, what grade are you in?
Those are the exact same scores I got in 7th grade.
*high five*[/quote]

OMG...i'm in 7th.  IM GONNA BECOME A MINI UBEMAYA~  NOOOOOOOOO!  MUST...NOT....GET....FAT![/quote][quote="Ubemaya"]

WAIT, don't lose hope yet. WHAT SCORE DID YOU GET ON THE ESSAY? MAYBE YOU WON'T BECOME A MINIME IF YOU DIDN'T GET A 5.[/quote]
I got 7...lol (I wrote too slowly so I had half an essay).


Am si eu o intrebare...Aceasta inegalitate geometrica : p^2 >= 3S sqrt 3, unde p = semiperimetrul si S = aria...poarta denumirea de inegalitatea lui Finsler...

Multumesc anticipat :)

daka purta numele de inegalitatea lui Finsler :| ...ma rog de fapt asta e a lui Mitrinovic...a lui Finsler este 4R + r > p radical din 3...sry de spam :|

asa cum zici tu :)
da sunt curios la ce iti trebuie sa stii toate denumirile astea ? oricum toate inegalitatile astea din triunghi ies cu medii sau ceva... asa scot si io ceva cu $S^{14}$ si $p^{69}$ si iaute inegalitate in triunghi.

Pooi...nu stiu ce sa zic dar aceste inegalitati sunt celebre ;) de exemplu...o problema de la IMO(nuj ce an) :

Demonstratzi ca a ^ 2 + b ^ 2 + c ^ 2 > 4 S sqrt 3.
Se inmulteste cu 3...iar 3 ( a ^ 2 + b ^ 2 + c ^2) > (a+b+c) ^ 2 si inlocuiesti cu 4p ^ 2 si cazi peste Mitrinovic...si gata problema de IMO  :blush:

pai...sunt si nu sunt celebre, adica daca nu le demonstrezi intr`un examen, risti sa iei 1-2 puncte cel mult pe ea (daca nu chiar 0)
cat despre inegalitatea asta, iese si fara inegalitatea de care zici tu.
daca notezi $p-a=x$, $p-b=y$, $p-c=z$, si ridici la patrat, obtii
$(\sum(x+y)^{2})^{2}\geq 48xyz(x+y+z)$.
spun ca tripletu $(p,q,r)$ majoreaza $(u,v,w)$ daca $p\geq u$, $p+q\geq u+v$ si $p+q+r=u+v+w$.
in stanga o sa iasa niste $(4,0,0)$, niste $(3,1,0)$, niste $(2,2,0)$ si niste $(2,1,1)$. (spun ca apare tripletu $(p,q,r)$ atunci cand in suma am toti termenii de tip $x^{p}y^{q}z^{r}$, cu toate permutarile dupa $p,q,r$ - sume simetrice adica)
in dreapta avem doar $(2,1,1)$, iar coeficientii ies cum trebe, ca asa e facuta inegalitatea.
deci se rade imediat cu muirhead, si oricum iese si banal cu medii dupa desfacut paranteze.
insa daca scrii de inegalitatea lu nustiucine, si nu o demonstrezi, risti sa nu primesti puncte aproape deloc

pai n-am zis ca nu se demonstreaza ma:)...ca nu e grea demonstratia pt Mitrinovic :|

Ca, tot vorbiti de inegalitatea Hadwiger-Finsler data de cei doi in anul 1937, am s-o postez odata pentru totdeauna si cateva forme echivalente:

[i] In orice triunghi $ABC$ este adevarata urmatoarea inegalitate:

              \[ a^{2}+b^{2}+c^{2}\geq4S\sqrt{3}+(a-b)^{2}+(b-c)^{2}+(c-a)^{2}.  \]

[/i]

 Demonstratii sunt multe la aceasta inegalitate sunt multe. In Engel-Problem solving strategies sunt 3, in R.M.T, in anul 2004 parca, domnul Titu Zvonaru da o demonstratie cu ajutorul inegalitatii lui Schur, insa pentru ca meniul sa fie complet am sa dau o demonstratie zic eu originala, publicata de subsemnatul in revista Arhimede, nr.9-10/2003.

[b] Demonstratie.[/b]

    Daca trecem suma aia de patrate in stanga si notam $p=\frac{a+b+c}{2}$, inegalitatea care vrem s-o demonstram se transforma in:

                                 \[ a(p-a)+b(p-b)+c(p-c)\geq 2S\sqrt{3}.  \]

    Aplicand, insa inegalitatea mediilor, vom avea \[ a(p-a)+b(p-b)+c(p-c)\geq 3\sqrt [3] {abc(p-a)(p-b)(p-c)}\Leftrightarrow  \]
    Acum  nu ne ramane decat sa demonstram ( bineinteles, asta dupa ce am folosit formula lui Heron) ca
                                      \[ 3\sqrt [3] {abc(p-a)(p-b)(p-c)}\geq 2S\sqrt{3}\Leftrightarrow  \]

                                                  \[ 3\sqrt [3] {abc\cdot\frac{S^{2}}{p}}\geq 2S\sqrt{3}\Leftrightarrow  \]

                                                          \[ p\leq\frac{3\sqrt{3}}{2}\cdot R.  \]

     Aceasta din urma este4 chiar inegalitatea lui Mitrinovic, ca sa vezi chestie a naibii. :)

  [b] Consecinte.[/b]

  1. \[ a^{2}+b^{2}+c^{2}\geq 4S\sqrt{3}.  \]
  2. \[ 4R+r\geq p\sqrt{3}.  \]

   si mai sunt multeeeee alteleeeeeeee..........o sa public un articol pe tema asta. Am chiar probleme de la baraje rezolvate cu Hadwiger-Finsler.

   P.S. O chestie de genul asta am prezentat la Simpozionul Arhimede 2005, la sectiunea "Inegalitati".

O inegalitate geometrica interesanta(se remarca inca de la inceput Inegalitatea lui Mitrinovic)
  Demonstrati ca in orice triunghi $\ ABC$ avem
$\frac{a^3}{a^2+bc}+\frac{b^3}{b^2+ca}+\frac{c^3}{c^2+ab}\geq\ 3\sqrt{3}r$.
       $\ Mircea\ Becheanu$

[hide]pai pentru inceput
$3\sqrt{3}r\leq p$ (*) $\iff$
$3\sqrt{3}rp\leq p^{2}$ $\iff$
$3\sqrt{3}S\leq p^{2}$ $\iff$
$27p(p-a)(p-b)(p-c)\leq p^{4}$ $\iff$
$27(p-a)(p-b)(p-c)\leq p^{3}$ $\iff$
$3\sqrt[3]{(p-a)(p-b)(p-c)}\leq \bigg( (p-a)+(p-b)+(p-c)\bigg)$, care este inegalitatea mediilor.
apoi,
$\sum\frac{a^{3}}{a^{2}+bc}\geq p$ (**) $\iff$
$2\sum(a^{3}(b^{2}+ac)(c^{2}+ab)\geq\bigg(\sum a\bigg)\prod(a^{2}+bc)$ $\iff$
$2\sum a^{5}bc + 2\sum_{sym} a^{4}b^{3} + 2\sum a^{3}b^{2}c^{2}\geq$
$\geq 2\sum a^{3}b^{2}c^{2} +\sum_{sym}a^{4}b^{3}+\sum a^{5}bc + \sum a^{3}b^{3}c + \sum_{sym} a^{4}b^{2}c$ $\iff$
$\sum a^{5}bc + \sum_{sym}a^{4}b^{3}\geq\sum_{sym}a^{4}b^{2}c+\sum a^{3}b^{3}c$,
(deja cred ca v`ati obisnuit...) adevarata conform inegalitatii muirhead.
acum (*) si (**) ne dau
$LHS \geq p \geq RHS$ , deci am terminat[/hide]

Mamaaaaaaaaaaaaaaaaa, ce naiba va place la inegalitatea asta? :? ........ va omoara la lot cu ea din cate am vazut.
Ma rog, eu urasc [b]Muie[/b]rhead asta. :)  Parea mea este ca rezolvarea problemelor folosind inegalitate omoara frumusetea inegalitatilor in general. Ma rog, hai sa dam o solutie, mai eleganta ,zic eu, acestei inegalitati.

[b] Problema. [/b]

    Fie $a, b, c$ trei numere reale strict pozitive. Sa se arate ca

    \[ \frac{a^{3}}{a^{2}+bc}+\frac{b^{3}}{b^{2}+ca}+\frac{c^{3}}{c^{2}+ab}\geq\frac{a+b+c}{2}.  \]

[i] Solutie. [/i]

  Observam ca \[ \frac{a^{3}}{a^{2}+bc}=\frac{a(a^{2}+bc)-abc}{a^{2}+bc}.  \]
  Astfel, avem

   \[{ \frac{a^{3}}{a^{2}+bc}+\frac{b^{3}}{b^{2}+ca}+\frac{c^{3}}{c^{2}+ab}=a+b+c-abc(\frac{1}{a^{2}+bc}}+\frac{1}{b^{2}+ca}+\frac{1}{c^{2}+ab})\Leftrightarrow  \]
   \[ \frac{a^{3}}{a^{2}+bc}+\frac{b^{3}}{b^{2}+ca}+\frac{c^{3}}{c^{2}+ab}\geq a+b+c-abc(\frac{1}{2a\sqrt{bc}}+\frac{1}{2b\sqrt{ca}}+\frac{1}{2c\sqrt{ab}})\Leftrightarrow  \]
   \[ \frac{a^{3}}{a^{2}+bc}+\frac{b^{3}}{b^{2}+ca}+\frac{c^{3}}{c^{2}+ab}\geq a+b+c-(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}).  \]
   Acum, ramane doar sa aratam ca \[ \sqrt{ab}+\sqrt{bc}+\sqrt{ca}\leq a+b+c,  \]
   consecinta a inegalitatii mediilor. :)

P.S. Sa nu credeti ca Muerhead este o inegalitate tare. Nici pe departe nu poate fi vorba de asa ceva. ;)

[quote="Cezar Lupu"]
    Fie $a, b, c$ trei numere reale strict pozitive. Sa se arate ca

    $\frac{a^{3}}{a^{2}+bc}+\frac{b^{3}}{b^{2}+ca}+\frac{c^{3}}{c^{2}+ab}\geq\frac{a+b+c}{2}$.  [/quote]

Mai putem folosi inegalitatea ajutatoare: $\frac{a^{3}}{a^{2}+bc}\geq \frac{4a-b-c}{4}$ si analoagele.
Prin sumare, obtinem inegalitatea data.


The unknown real numbers $x_1,x_2,...x_n$ satisfy $x_1<x_2<...<x_n$, where $n\geq3$. The numbers $s,t$ and $d_1,d_2,...,d_{n-2}$ are given such that\[s=\sum^n_{i=1}x_i\]\[t=\sum^n_{i=1}x_i^2\]\[d_i=x_{i+2}-x_i,i=1,2,...,n-2\]
For which $n$ is this information always sufficient to determine $x_1,x_2,...,x_n$ uniquely ?

South African Mathematical Olympiad 2001, I believe. Apparently you have been going through the South African questions recently since I see a lot of SAMO questions posted here.


function
limit
real analysis
real analysis unsolved

$f$:$]0,+\infty[\rightarrow ]0,+\infty[$ twice derivable such that
$f''(x)+f'(x)\geq f^{2}(x)$ for all x>0

Prove that $\lim_{+\infty}f$ exist ? Find its value

[quote=nbasrl]Does anyone have a solution to this ?[/quote]

I have discovered a truly marvellous proof but this margin is too narrow to contain it


The Art and Craft of Problem Solving, 2nd Edition
By: Paul Zeitz

[b]1. The problem statement, all variables and given/known data[/b]

1.3.5 (AHSME 1996)
Given that $ {{x}^{2}+{y}^{2} = {14x}+{6y}+{6}}$, what is the largest possible value that $ {{3x}+{4y}}$ can have?

[b]2. Relevant equations[/b]

None.

[b]3. The attempt at a solution[/b]

Hypothesis: Given $ {{x}^{2}+{y}^{2} = {14x}+{6y}+{6}}$, what is $ max({{3x}+{4y}})$?

Conclusion: The solutions $ \left({x}_{1}, {y}_{1}\right)$ to $ max({{3x}+{4y}})$ must satisfy $ {{x}^{2}+{y}^{2} = {14x}+{6y}+{6}}$.

Let,

$ {{3x}+{4y}} = C$
$ max({{3x}+{4y}}) = max(C)$

Consider,
$ {{3x}+{4y}} = C$

Solving for $ y$,
${ {y} = {\frac{-3x}{4}} + \frac{C}{4}}$

Since,
$ max({{3x}+{4y}}) = max(C)$

Then in order to find $ max(C)$ let,
${ \frac{C}{4}} = b$

So, that finding $ max(b)$ would lead to finding $ max(C)$

However, from here I am not sure what to do,...

Thanks,

-APSStudent

[quote="mchoi815"]When I saw the equation with x^2 +y^2, I thought of circles.  Rearranging yields:(x-7)^2 + (y-3)^2 = 8^2.
See if that helps.
 :)[/quote]

look for the line tangent to the circle in the form 3x+4y=a

Hey,

[quote="unimpossible"]
look for the line tangent to the circle in the form 3x+4y=a
[/quote]

Thanks for the reply unimpossible. However, how did you come to the conclusion that examining the line tangent to the circle of the form $ 3x \plus{} 4y$, would yield a max for $ 3x \plus{} 4y$?

That is, when approaching this problem how does one deduce that the maximum value of $ 3x \plus{} 4y$ occurs when the line $ 3x \plus{} 4y \equal{} C$ is tangent to the circle?

Thanks,

-APSStudent

[quote="timwu"]By Cauchy's
[hide]
Let $ M \equal{} 3(x \minus{} 7) \plus{} 4(y \minus{} 3)$, by Cauchy's Inequality
$ M^2 \leq (3^2 \plus{} 4^2)((x \minus{} 7)^2 \plus{} (y \minus{} 3)^2) \equal{} 25\cdot 64,$
$ M\leq 40$
with equality iff $ \frac {3}{x \minus{} 7} \equal{} \frac {4}{y \minus{} 3}$.

Hence, $ C \equal{} 3x \plus{} 4y \equal{} M \plus{} 33 \equal{} 73$.

(You can check to make sure there exists such one, either by solving for x and y or graphing.)
[/hide][/quote]

Oh wow nice solution.
let me post a link
http://www.math.uncc.edu/~hbreiter/AHSME/1996so.pdf
go down to problem 25.

[quote="timwu"]By Cauchy's
[hide]
Let $ M \equal{} 3(x \minus{} 7) \plus{} 4(y \minus{} 3)$, by Cauchy's Inequality
$ M^2 \leq (3^2 \plus{} 4^2)((x \minus{} 7)^2 \plus{} (y \minus{} 3)^2) \equal{} 25\cdot 64,$
$ M\leq 40$
with equality iff $ \frac {3}{x \minus{} 7} \equal{} \frac {4}{y \minus{} 3}$.

Hence, $ C \equal{} 3x \plus{} 4y \equal{} M \plus{} 33 \equal{} 73$.

(You can check to make sure there exists such one, either by solving for x and y or graphing.)
[/hide][/quote]
This solution is nice!But I think I should say a little bit about how you can think of it or whatever the coefficients of x and y are, you still can do it!
If we used the hypothesis, we easily calculate x in terms of y. That sounds simple but extremely important. Why?Because when you replace x by y, and try to find maximum as well as minimum(if possible) 3x+4y in terms of y, we see that it's a single-variable function. And how easy it is with derivative or using the combination of discriminant and the condition when the 2nd-degree equation has solution, that is the discriminant>=0 :) .It's much elementary than using derivative

This was already discussed [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=418742#418742]here[/url].

If you have new insights or questions, you can post them there.

APSStudent, you don't need to use a rubric like that, really, unless it's a homework problem (in which case it should be delineated as such).


The rate at which crickets chirp is thought to be related to the temperature by the equation $ c \equal{} t \minus{} 40$. where $ c$ is the number of chirps per 15 seconds and $ t$ is the temperature in degrees Fahrenheit. Using this relationship, what is the number of degrees Fahrenheit in the temperature when a cricket chirps 500 times in 5 minutes?


1.Fill all fucntion equation f: N into N, satisfy: f(x^2 +f(y))=xf(x) +y
 ( Mong moị người thông cảm vì mình đang học tiếng anh nên thực hành luôn)

Cho x=0 ta co: f(f(y))=y.
Tu do suy ra f la 1 song anh.
Cho x=1, y=f(0) ta co: f(1+f(f(0)))=f(1)+f(0) suy ra f(0)=0.
Vi f la 1 song anh nen chon a sao cho f(a)=1 suy ra f(1)=a.
Cho x=a, y=0 suy ra f(a^2)=a. ma f la don anh nen a^2=1 hay a=1.
Vay f(1)=1.
Cho x=1 ta co: f(1+f(y))=1+y. Bang quy nap ta chung minh duoc f(n)=n.

sao k gox latex zay!!!ma phai functional equations chu


geometry
circumcircle
incenter
rectangle
trapezoid
symmetry
geometry proposed

Let $ P$ be an arbitrary point inside a triangle $ ABC$, with cevian triangle $ A^{\prime}B^{\prime}C^{\prime}$ (i.e. $ A^{\prime}$, $ B^{\prime}$, $ C^{\prime}$ are the intersections of the lines $ AP$, $ BP$, $ CP$ with the sides $ BC$, $ CA$, and $ AB$, respectively). Denote by $ \mathcal{T}_{A}^{b}$, $ \mathcal{T}_{A}^{c}$ the circles tangent simultanously to $ AA^{\prime}$, $ A^{\prime}B$, $ \Gamma$, and $ AA^{\prime}$, $ A^{\prime}C$, $ \Gamma$, respectively, where $ \Gamma$ is the circumcircle of triangle $ ABC$. Similarly, define the circles $ \mathcal{T}_{B}^{c}$, $ \mathcal{T}_{B}^{a}$, $ \mathcal{T}_{C}^{a}$, $ \mathcal{T}_{C}^{b}$. Prove that the six circles $ \mathcal{T}_{A}^{b}$, $ \mathcal{T}_{A}^{c}$, $ \mathcal{T}_{B}^{c}$, $ \mathcal{T}_{B}^{a}$, $ \mathcal{T}_{C}^{a}$, $ \mathcal{T}_{C}^{b}$ are congruent if and only if $ P$ is the Nagel point of triangle $ ABC$.

Better version:

Let $ A^{\prime}$ be an arbitrary point on the side $ BC$ of a triangle $ ABC$. Denote by $ \mathcal{T}_{A}^{b}$, $ \mathcal{T}_{A}^{c}$ the circles simultanously tangent to $ AA^{\prime}$, $ A^{\prime}B$, $ \Gamma$ and $ AA^{\prime}$, $ A^{\prime}C$, $ \Gamma$, respectively, where $ \Gamma$  is the circumcircle of $ ABC$. Prove that $ \mathcal{T}_{A}^{b}$, $ \mathcal{T}_{A}^{c}$ are congruent if and only if $ AA^{\prime}$ passes through the Nagel point of triangle $ ABC$.

Let $(U)$ be the Thebault circle tangent to $\overline{AA'},\overline{A'B},\Gamma$ and $(V)$ the Thebault circle tangent to $\overline{AA'},\overline{A'C},\Gamma.$ $(U),(V)$ touch $BC$ at $D,E.$  $UV$ passes through the incenter $I$ of $\triangle ABC$ and $ID \perp UA',$ $IE \perp VA'.$ If $(U) \cong (V),$ then it follows that $\overline{UIV}$ is parallel to $BC,$ i.e. $UDEV$ is a rectangle. Thus, if $X$ is the tangency point of the incircle $(I)$ with $BC,$ then $IUDX$ and $IVEX$ are rectangles $\Longrightarrow$ $ID=XU,$ $IE=XV$ $\Longrightarrow$ $\triangle IDE \cong \triangle XUV$ $\Longrightarrow$ $\angle UXV=\angle DIE=\angle UA'V=90^{\circ}$ $\Longrightarrow$ $UVA'X$ is cyclic. Further, $UV \parallel BC$ implies that $UVA'X$ is indeed an isosceles trapezoid with legs $UX=VA'$ $\Longrightarrow$ $DX=EA'.$ Since $BD=CE,$ due to obvious symmetry, then $BX=CA'$ $\Longrightarrow$ $A'$ is the tangency point of the A-excircle with $BC.$

Dear Mathlinkers,
for this difficult and nice problem, you can see

http://perso.orange.fr/jl.ayme    vol. 20 "Shape and Mouvement" p. 20

Sincerely
Jean-Louis


Prove that $ \forall x,y,z>0$ we have:
$ \sum \frac{xy}{x^2+xy+y^2} \leq 1$.

here is what I found:
We observe that $ \frac{xy}{x^2+xy+y^2} \leq \frac{1}{3}$ which is equivalent to $(x-y)^2 \geq 0$ you do the same with the others and you have the ineq.
equality if $x=y=z$.

Do you have other solutions?

cheers! :D :D

[quote="Lagrangia"]Prove that $ \forall x,y,z>0$ we have:
$ \sum\frac{xy}{x^{2}\plus{}xy\plus{}y^{2}}\leq 1$.

here is what I found:
We observe that $ \frac{xy}{x^{2}\plus{}xy\plus{}y^{2}}\leq\frac{1}{3}$ which is equivalent to $ (x\minus{}y)^{2}\geq 0$ you do the same with the others and you have the ineq.
equality if $ x \equal{} y \equal{} z$.

Do you have other solutions?

cheers! :D :D[/quote]
By AM-GM,we have:
$ x^{2}\plus{}y^{2}\geq 2xy$,so $ \frac{xy}{x^{2}\plus{}xy\plus{}y^{2}}\leq\frac{xy}{3xy}\equal{}\frac{1}{3}$
Similar we have $ \sum\frac{xy}{x^{2}\plus{}xy\plus{}y^{2}}\leq 1$.

What about the following inequality:
for non-negative numbers $ x,$ $ y$ and $ z,$ no two of which are zeros, prove that:
$ \frac{xy}{x^{2}\plus{}xz\plus{}z^{2}}\plus{}\frac{yz}{x^{2}\plus{}xy\plus{}y^{2}}\plus{}\frac{xz}{y^{2}\plus{}yz\plus{}z^{2}}\leq\frac{x^{2}\plus{}y^{2}\plus{}z^{2}}{xy\plus{}xz\plus{}yz}$ $ ?$ :wink:


Prove that If $a,b,c,d$ are reals namber and $a+b+c+d=2$ then :

$a^4+b^4+c^4+d^4-4abcd \ge 2(a-b)(c-d)$

All I see is that it's very strange.

just a idea.
because your inequality is equivalent to:
$2(a^4+b^4+c^4+d^4-4)abcd \ge (a-b)(c-d)(a+b+c+d)^2$
and the equality happens when $a=c$,$b=d$
so I guess
we may can prove it by:
$2(a^4+b^4+c^4+d^4-4)abcd - (a-b)(c-d)(a+b+c+d)^2=x(a-c)^2+y(b-d)^2$
where $x,y \ge 0$
just guess :lol:

Do you find two numbers, Zhaobin? Because I don't use this method.

For this, we've some nice problem if replace a,b,c,d to -a,b,-c,d or somthing like that.

sorry,I haven't find the two numbers...
I was preparing my final exam :(
and What about your solution?

I'm not sure about your solution, there's not exists $x,y$. Bbecause maybe we must explant to:

$x(a-c)^2+y(b-d)^2+z(a-c)(b-d) \ge 0$

A hint for it . See that
http://www.mathlinks.ro/Forum/viewtopic.php?t=27041


Angel wants to sell 50 identical pencils in groups of 2 or 3. In how many ways can the pencils be grouped?

The $ LCM(2,3)\equal{}6$, so there is a combination of 2's and 3's every 6 numbers. $ \lceil\frac{50}{6}\rceil\equal{}\boxed{9}$.


trigonometry
geometry
cyclic quadrilateral
geometry unsolved

[size=150][color=red]Circle O cuts triangle ABC AB to D, cuts AC to C.   M is on the edge BC .          AM  CD pay points in N, prove : M is the midpoint of the BC necessary and sufficient conditions are vertical ON BC

 :) THANKS[color=red][/color][/color][/size]

I dont completely understand your problem,first of all does $ O$ pass thruogh $ A$ or not,and then is $ M$ an arbitary point on $ BC$ or is it the intersection of $ O$ and $ BC$?and also what does "vertical ON BC" mean? :huh:

[color=green][b] M[/b] is the midpoint of the [b]BC[/b] necessary and sufficient conditions are [b]ON [/b]and [b]BC[/b] are vertical

[b] O[/b] does not pass through

[b]M[/b] is an arbitary point on [b]BC[/b][/color]

and does [b]ON and BC are vertical[/b] mean $ ON \bot BC$?

Let us to restate the problem as follows:

[b][size=100][color=DarkBlue]PROBLEM. -A triangle $ \bigtriangleup ABC$ is given and let $ (O)$ be, the circle tangents to its sidelines $ AC,$ $ AB,$ at points $ C,$ $ D,$ respectively and we denote as $ N,$ the intersection point of $ CD,$ $ AM,$ where $ M,$ is the midpoint of $ BC.$ Prove that $ ON\perp BC,$ where $ O,$ is the center of $ (O).$[/color][/size][/b]

It is easy to show that the orthodiagonal quadrilateral $ ACOD$ is cyclic, inscribed in a circle $ (O')$ with diameter the segment $ AO,$ because of $ OC\perp AC$ and $ OD\perp AD$ $ ($ a quadrilateral is orthodiagonal, when its diagonals are perpendicular each other $ ).$

We denote as $ E,$ the intersection point of $ (O)$ from $ BC$ and as $ F,$ the one of $ (O')$ from the line through $ A$ and parallel to $ BC.$

Based on the below [b][size=100]Lemma,[/size][/b] we conclude that the points $ F,$ $ D,$ $ E,$ are collinear.

[b][size=100][color=DarkBlue]LEMMA. - A cyclic orthodiagonal quadrilateral $ ABCD$ is given, inscribed in a circle $ (O)$ with diameter $ AC.$ We draw the circle $ (C),$ centered at $ C,$ with radius $ CB = CD$ and we denote as $ E,$ $ F,$ the intersection points of $ (O),$ $ (C)$ respectively, from two arbitrary lines through $ A,$ $ D$ and parallel each other. Prove that the points $ E,$ $ B,$ $ F,$ are collinear.[/color][/size][/b]

Because of now, the diameter $ AO$ of $ (O')$ $ \Longrightarrow$ $ OF\perp AF$ and so, we have that $ OF\perp BC$ $ ($ because of $ AF\parallel BC$ $ ).$

We denote the point $ N'\equiv CD\cap OF$ and it is enough to prove that $ N'\equiv N\equiv CD\cap AM.$

$ \bullet$ Apply the [b][size=100]Menelaus theorem,[/size][/b] in triangle $ \bigtriangleup CDE,$ taken as its transversal the segment line $ OF,$ we have $ \frac{N'C}{N'D}\cdot \frac{FD}{FE}\cdot \frac{PE}{PC}= 1$
$ \Longrightarrow$ $ \frac{N'C}{N'D}= \frac{FE}{FD}\cdot \frac{PC}{PE}$ $ ,(1)$ where $ P\equiv BC\cap OF.$

Similarly from the triangle $ \bigtriangleup BCD,$ taken as its transversal the segment line $ AM,$ we have $ \frac{NC}{ND}\cdot \frac{AD}{AB}\cdot \frac{MB}{MC}= 1$
$ \Longrightarrow$ $ \frac{NC}{ND}= \frac{AB}{AD}\cdot \frac{MC}{MB}$ $ ,(2)$

From $ AF\parallel BC$ $ \Longrightarrow$ $ \frac{FE}{FD}= \frac{AB}{AD}$ $ ,(3)$

From $ (1),$ $ (2),$ $ (3)$ and because of the points $ M,$ $ P,$ are the midpoints of the segments $ BC,$ $ EC$ respectivey,

we have that $ \frac{N'C}{N'D}= \frac{NC}{ND}$ $ ,(4)$

From $ (4)$ we conclude that $ N'\equiv N$ and the proof is completed.

Kostas Vittas.

PS. I will post here later the proof of the above [b][size=100]Lemma[/size][/b] I have in Mind.

[quote="vittasko"][b][size=100][color=DarkBlue]LEMMA. - A cyclic orthodiagonal quadrilateral $ ABCD$ is given, inscribed in a circle $ (O)$ with diameter $ AC.$ We draw the circle $ (C),$ centered at $ C,$ with radius $ CB = CD$ and we denote as $ E,$ $ F,$ the intersection points of $ (O),$ $ (C)$ respectively, from two arbitrary lines through $ A,$ $ D$ and parallel each other. Prove that the points $ E,$ $ B,$ $ F,$ are collinear.[/color][/size][/b][/quote]

[b][size=100]PROOF.[/size][/b] – Because of $ AE\parallel DF,$ we have that $ \angle AED = \angle EDF$ $ ,(1)$

From cyclic quadrilateral $ AEBD,$ $ \Longrightarrow$ $ \angle AED = \angle ABD = \angle ADB$ $ ,(2)$ $ ($ because of $ AB = AD$ $ ).$

From $ (1),$ $ (2)$ $ \Longrightarrow$ $ \angle EDF = \angle ADB$ $ \Longrightarrow$ $ \angle ADE = \angle BDF$ $ ,(3)$

But $ \angle ADE = \angle ABE$ $ ,(4)$ and $ \angle BDF = \angle FBF'$ $ ,(5)$ where $ F'\equiv DF\cap AB$

From $ (3),$ $ (4),$ $ (5)$ $ \Longrightarrow$ $ \angle ABE = \angle FBF'$ $ ,(6)$

From $ (6)$ we conclude that the points $ E,$ $ B,$ $ F,$ are collinear and the proof is completed.

Kostas vittas.

[quote][color=darkred][u][b]An equivalent enunciation.[/b][/u] Let $ ABC$ be an isosceles triangle for which denote the points $ \{\begin{array}{c}D\in (BC)\ ,\ DB=DC\\\ E\in AD\ ,\ EC\perp AC\end{array}$.
For a point $ M\in (AB)$ define the points $ \{\begin{array}{c}L\in (MC)\ ,\ LM=LC\\\ N\in BC\cap AL\end{array}$. Prove that $ EN\perp MC$.[/color][/quote]
[color=darkblue][b]Proof I (thin metric).[/b] Denote $ \{\begin{array}{c}AB=AC=b\ ,\ BC=2a\\\ AM=x\in (0,b)\end{array}$. Observe that $ \{\begin{array}{c}EB=EC\\\ MB=b-x\end{array}$ .

Apply the [b]Menelaus' theorem[/b] to the transversal $ \overline{ALN}$ in $ \triangle BMC$ :

$ \frac{AM}{AB}\cdot\frac{NB}{NC}\cdot\frac{LC}{LM}=1$ $ \implies$ $ \frac{x}{b}\cdot \frac{NB}{NC}=1$ $ \implies$ $ \frac{NB}{b}=\frac{NC}{x}=\frac{2a}{b+x}$. Therefore,

$ EN\perp MC$ $ \Longleftrightarrow$ $ EM^{2}-EC^{2}=NM^{2}-NC^{2}$ $ \Longleftrightarrow$ $ EB^{2}+BM^{2}-EC^{2}=NM^{2}-NC^{2}$ $ \Longleftrightarrow$

$ MB^{2}+NC^{2}=NM^{2}$ $ \Longleftrightarrow$ $ MB^{2}+NC^{2}=MB^{2}+BN^{2}-2\cdot MB\cdot BN\cdot\cos\widehat{MBN}$ $ \Longleftrightarrow$

$ NC^{2}=BN^{2}-2\cdot MB\cdot NB\cdot \frac{BD}{AB}$ $ \Longleftrightarrow$ $ (\frac{2ax}{b+x})^{2}=(\frac{2ab}{b+x})^{2}-2\cdot (b-x)\cdot \frac{2ab}{b+x}\cdot \frac{a}{b}$, what is truly.

[b]Proof II (weak trigonometric).[/b] Denote $ \{\begin{array}{c}m(\widehat{BAD})=\phi\\\ m(\widehat{CEN})=\alpha\\\ m(\widehat{ACM})=\beta\end{array}$. Observe that $ \{\begin{array}{c}m(\widehat{AMC})=180^{\circ}-(2\phi+\beta )\\\ m(\widehat{BEN})=180^{\circ}-(2\phi+\alpha )\end{array}$.
Apply the [b]Sinus' theorem[/b] in $ \triangle ACM$ : $ \frac{AC}{AM}=\frac{\sin\widehat{AMC}}{\sin\widehat{ACM}}$ $ \implies$ $ \frac{AB}{AM}=\frac{\sin (2\phi+\beta )}{\sin \beta }\ \ (1)$.

Apply the [b]Menelaus' theorem[/b] to the transversal $ \overline{ALN}$ in $ \triangle BMC$ :

$ \frac{AB}{AM}\cdot\frac{LM}{LC}\cdot\frac{NC}{NB}=1$ $ \implies$ $ \frac{AB}{AM}=\frac{NB}{NC}=\frac{EB}{EC}\cdot\frac{\sin \widehat{NEB}}{\sin\widehat{CEN}}$ $ \implies$ $ \frac{AB}{AM}=\frac{\sin (2\phi+\alpha )}{\sin \alpha }\ \ (2)$.

From the relations $ (1)$ and $ (2)$ obtain : $ \sin \alpha\sin (2\phi+\beta )=\sin \beta\sin (2\phi+\alpha )$ $ \Longleftrightarrow$

$ \cos (2\phi+\beta-\alpha )-\cos (2\phi+\alpha+\beta )=\cos (2\phi+\alpha-\beta )-\cos (2\phi+\alpha+\beta )$ $ \Longleftrightarrow$

$ \cos (2\phi+\alpha-\beta )=\cos (2\phi+\beta-\alpha )$ $ \Longleftrightarrow$ $ \alpha =\beta$ $ \Longleftrightarrow$ $ \widehat{CEN}\equiv\widehat{ACM}$ $ \Longleftrightarrow$ $ EN\perp MC$.[/color]

[color=darkred][u][b]An extension.[/b][/u] Let $ ABC$ be an isosceles triangle for which denote the points $ \{\begin{array}{c}D\in (BC)\ ,\ DB=DC\\\ E\in AD\ ,\ D\in AE\end{array}$. For a point $ M\in (AB)$ define the points
$ \{\begin{array}{c}L\in (MC)\ ,\ LM=LC\\\ N\in BC\cap AL\end{array}$ and denote $ \{\begin{array}{c}m(\widehat{BAD})=\phi\\\ m(\widehat{AEB})=\psi\\\ m(\widehat{CEN})=\alpha\\\ m(\widehat{ACM})=\beta\end{array}$. Then $ \boxed{\tan \phi\cdot\tan \psi=\tan (\phi+\beta )\cdot\tan (\psi-\alpha )}$. If $ \phi+\psi =\frac{\pi}{2}$ then obtain
the proposed problem : $ \tan (\phi+\beta )\cdot \tan (\frac{\pi}{2}-\phi-\alpha)=1$ $ \Longleftrightarrow$ $ \tan (\phi+\beta )=\tan (\phi+\alpha )$ $ \Longleftrightarrow$ $ \alpha =\beta$ $ \Longleftrightarrow$ $ EN\perp MC$.[/color]

[quote]A triangle $ \bigtriangleup ABC$ is given and let $ (O)$ be, the circle tangents to its sidelines $ AC,$ $ AB,$ at points $ C,$ $ D,$ respectively and we denote as $ N,$ the intersection point of $ CD,$ $ AM,$ where $ M,$ is the midpoint of $ BC.$ Prove that $ ON\perp BC,$ where $ O,$ is the center of $ (O).$
[/quote]
[b]Another solution.[/b] Denote by $ I,\,K$ the intersections of $ AM$ with $ BC,\,CA,$ respectively.
Applying the Menelaus' Theorem to triangle $ ABM$ for the transversal $ CDN,$ we have
\[ \frac{NA}{NM}\cdot\frac{CM}{CB}\cdot\frac{DA}{DB}=1\implies\frac{NM}{NA}=\frac{DB}{2\cdot DA}\quad (1) \]
On the other hand, by Thales' Theorem, we have
\[ \frac{IM}{IB}=\frac{MK}{AB}\implies\frac{IM}{IB-2\cdot IM}=\frac{\frac{1}{2}\cdot DB}{AB-DB}\implies\frac{IM}{IC}=\frac{DB}{2\cdot DA}\quad (2) \]
From $ (1)$ and $ (2)$ we have $ IN\parallel AC\implies IN\perp OC$
Hence $ O$ is the orthocentre of triangle $ CIN,$ so $ ON\perp BC$

Thank you dear [b][size=100]April[/size][/b] for a neat solution, shows to me a specialist of management of a simple relation by [b][size=100]Thales theorem.[/size][/b]

I see only some typos in the text. The line segment $ AO$ ( instead of $ AM$ ), intersects $ BC,$ $ CD$ ( instead of $ AC$ ) at points $ I,$ $ K,$ respectively.

Also we have $ \frac{NA}{NM}\cdot \frac{CM}{CB}\cdot \frac{DB}{DA}= 1$ instead of $ \frac{NA}{NM}\cdot \frac{CM}{CB}\cdot \frac{DA}{DB}= 1.$

Let me now to try an explanation of the relation $ (2),$ you mentioned.

We have $ \frac{IM}{IB}= \frac{MK}{AB}= \frac{1}{2}\cdot \frac{DB}{AB}$ $ \Longrightarrow$ $ \frac{2(IM)}{IB}= \frac{DB}{AB}$ $ ($ because of $ DB = 2(MK)$ $ ).$

$ \Longrightarrow$ $ \frac{2(IM)}{IB-2(IM)}= \frac{DB}{AB-DB}= \frac{DB}{DA}$

$ \Longrightarrow$ $ \frac{IM}{IB-2(IM)}= \frac{1}{2}\cdot \frac{DB}{DA}$ $ \Longrightarrow$ $ \frac{IM}{IC}= \frac{1}{2}\cdot \frac{DB}{DA}$ $ (2)$

$ ($ because of $ IB-2(IM) = IB-II' = I'B = IC$ where $ I',$ is the symmetric point of $ I,$ with respect to $ M$ $ ).$

Best regards, Kostas Vittas.

[b]A generalization. (stated from another solution of my friend)[/b] A triangle $ \bigtriangleup ABC$ is given and let $ (O)$ be, the circle tangents to its sidelines $ AC,$ $ AB,$ at points $ D,$ $ E,$ respectively and we denote as $ N,$ the intersection point of $ DE,$ $ AM,$ where $ M,$ is the midpoint of $ BC.$ Prove that $ ON\perp BC,$ where $ O,$ is the center of $ (O).$

[b]Proof.[/b] Denote by $ d$ the line through $ A$, parallel to $ BC$. Since the pencil $ (d,\,AM,\,AB,\,AC)$ is harmonic, it follows the division $ (ZNED)$ is harmonic, where $ Z=d\cap EF$. Therefore $ N$ lies on the polar of $ Z$ w.r.t circle $ (O)$, so $ AZ$ is the polar of $ N$ w.r.t circle $ (O)$. Hence $ ON\perp AZ$, i.e. $ ON\perp BC$.

I think the best solution dear [b][size=100]April[/size][/b], thank you very much.

Kostas Vittas.


geometry
circumcircle
3D geometry
tetrahedron
geometry unsolved

[quote="Julia Robet"]Let $ABC$ be an acute triangle with altitudes $AD,BE,CF$ and let $O$ be its circumcenter . Let $M,N,P$ be the midpoints respectively of the segments $BC,CA,AB$. Let $D_1,E_1,F_1$ be the points of intersection of the circumsphere pf the tetrahedron respectively with $AG_1,BG_2,CG_3,DG_4$ .
Prove that: $\sum \frac{AG_1}{AA_1}\le \frac{8}{3}$.[/quote]

WTF?

  darij

WTF??
I don't understand your ideals. :(


Let ABC is a triangle   Prove
$ \frac{2(R\plus{}r)}{r}>\equal{}\sum\frac{b\plus{}c}{\sqrt{a(\minus{}a\plus{}b\plus{}c)}}$

Why didn't you post  ? this inequality is easy  By CBS and   tql04-->OK


Show how you would convert toluene into bromobenzene in the minimum number of steps.

That's correct, but how about just two steps?

1.$ \text{H}_{2},\text{Mo},\Delta$ $ (\text{Delalkylation})$
2.$ \text{Fe/Br}_{2}$

Here's what I had in mind:

1. Oxidation with hot, alcaline potassium permanganate, followed by acidification.

2. Hunsdiecker reaction: treat the acid with HgO and bromine in CCl4.


LaTeX
limit
real analysis
real analysis unsolved

for real numbers z_1, z_2, define
z_n+2 = (z_n + z_n+1)/2
prove that the sequence z_n converges and find its limit.
I know I need to learn software so i can post this in proper notation, but I haven't had time yet.

So writing out the sequence you can tell that the magnitude of the difference between two terms is half the magnitude of the difference between the last two terms. I don't know how to prove that though. I'm thinking that if I can prove that then I can prove the sequence is cauchy and then find the limit.

Is this the right direction to take this problem and if so, any suggestions? If not what would you suggest?

thanks for your help!

That's an excellent idea of an approach.  Try writing down algebraically the statement "the difference between two terms is half the difference between the previous two terms," and see if you can derive it from the given equation.  (In fact, there's a lot more theory that one can appeal to to solve problems like this, and in particular, writing down the explicit formula for $ z_n$ in terms of $ z_1, z_2$ is not very hard; but your idea works very nicely here.)

You don't need to download anything to use LaTeX on the forum: see http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690

[b]HINT:[/b] $ \mbox{If: }z_1 < z_2\Rightarrow$
$ z_1 < z_3 < z_5 < \dots < z_{2n \minus{} 1} < z_{2n \plus{} 1} < \dots < z_{2n \plus{} 2} < z_{2n} < \dots < z_6 < z_4 < z_2\\
\mbox{but }z_{2n}\minus{}z_{2n\minus{}1}\equal{}\frac{z_2\minus{}z_1}{2^n}\to 0\ (\mbox{if }n\to \infty)\Rightarrow (\exists)\lim_{n\to\infty}{z_{2n}}\equal{}\lim_{n\to\infty}{z_{2n\plus{}1}}\equal{}z.$

hello, one can prove that
$ z_n\equal{}\frac{1}{2}\left(\frac{4}{3}z_0\minus{}\frac{4}{3}z_1\right)\left(\minus{}\frac{1}{2}\right)^n\plus{}\frac{1}{3}z_0\plus{}\frac{2}{3}z_1$
and the searched limit is
$ \frac{1}{3}z_0\plus{}\frac{2}{3}z_1$
Sonnhard.

thanks for your help everyone!
Dr. Graubner, how did you derive that equation? I messed around with the algebra for quite some time and couldn't find it. Is there some trick or method to explicitly derive z_n that I don't know?

thank you

[quote="JBL"]In fact, there's a lot more theory that one can appeal to to solve problems like this, and in particular, writing down the explicit formula for $ z_n$ in terms of $ z_1, z_2$ is not very hard[/quote]

The key words to look for are "linear recurrence relation" to learn about solving equations of this form.  This particular one is "guessable," in the sense that one could come up with that formula just by playing around and prove it using induction.


Find all solutions of equation $ a^{2}\plus{}b^{2}\equal{}5c^{2}$, where $ c$ is positive integer and $ a$,$ b$ are integers from interval $ (c,2c)$

It is only a hint ,not solution . Consider all rational points on curve $ x^2\plus{}y^2\equal{}5$ , this curve contains a rational point $ A(2,1)$ , thus we only need to find out the intersect point of  lines form  $ ax\plus{}by\equal{}c$ with this curve where $ a,b,c$ are rational number by Viete's .  That's all solution of this equation .

That's not necessary.  Note that since $ a^2 \equiv 0, 1, 4 \bmod 5$, the assumption that the answers are primitive means that $ a^2 \equiv 1 \bmod 5, b^2 \equiv 4 \bmod 5$ (without loss of generality), hence $ a \equiv 1, \minus{} 1 \bmod 5$ and $ b \equiv 2, \minus{} 2 \bmod 5$.

[b]Case:[/b]  The signs match.  Then $ \frac {a^2 \plus{} b^2}{5} \equal{} \left( \frac {2a \minus{} b}{5} \right)^2 \plus{} \left( \frac {a \plus{} 2b}{5} \right)^2$ where the expressions in parentheses are integers.

[b]Case:[/b]  The signs don't match.  Then $ \frac {a^2 \plus{} b^2}{5} \equal{} \left( \frac {2a \plus{} b}{5} \right)^2 \plus{} \left( \frac {a \minus{} 2b}{5} \right)^2$ where the expressions in parentheses are integers.

Essentially, this is a Gaussian integer argument.  Now the question reduces to one about Pythagorean triples, and then one has to account for the bound.

We continue as I have said . In my post , I only consider all line in the form $ y\equal{}kx\plus{}b$ ,where $ k$ is different from 0 (some trivial case $ x\equal{}a$ or $ y\equal{}a$ is very easy )
A line which pass through a rational point  on the curve $ x^2\plus{}y^2\equal{}5$ has $ k,b$ are rational iff it pass through an other rational points on this curve .  This is obvious by using viete's law .
By using this method , I have found all solution of this equation is :
$ (x,y)\equal{}(\frac{\minus{}2(k^2\plus{}k\minus{}1)}{k^2\plus{}1},\frac{k^2\minus{}4k\minus{}1}{k^2\plus{}1})$ and some other ''trivial'' solutions is $ (\minus{}1,\minus{}2) ,(\minus{}1,2),(1,2),(1,\minus{}2)$ . Now we only need to find $ k$ satisfy $ 1<x,y<2$  .


Analyze the following argument:

I. At least one of the following two sentences is true.
II. The preceding sentence is false.
III. 40% of the digits of pi are odd.

Statement I is either true or false. If statement I is false, then both statements II and III are false. But if II is false then I is true -- a contradiction. Therefore statement I must be true.

If statement I is true then so is either statement II or III. But II must be false since I is true, so III must me true.

Okay.  Makes sense.  Anything you want us to do with it?

Does it really make sense to you?  Statement III can be replaced by any statement without changing the validity of the argument.  This is just a variant of the [url=http://en.wikipedia.org/wiki/Liar_paradox]Liar paradox[/url].


Design a synthesis of 2-methyltetrahydrofuran starting with 5-bromopent-1-ene.

Hydration(HgSO4/H20) followed by treatment with base.

I prefer Oxy mercuration and demercuration to prevent any rearrangements.

what rearrangement. its a secondary carbon

Hydration with HgSO4/H2O is usually made with alkynes, not with alkenes.


YAY!!!!! ITS THAT TIME OF YEAR AGAIN!!! math fair papers are due in like a week, at least for my skool, can someone suggest a topic for me? and maybe some online resources along with it.  Currently, i'm thinking of fair division or Comparing infinities.  And for any of you that have to give math fair papers too, what subjects are you ddoing?

Hi angelXSlayer, Everyone,

You’re lucky angelXSlayer.  When I was in grade school, I was never given an opportunity to participate in anything like a math fair.

The Canadian Mathematical Society ( [url]http://www.cms.math.ca[/url]) has a page for students who are looking for topics for a math fair.  Here’s the link: ( [url]http://www.cms.math.ca/Education/mpsf[/url] ).  I hope you find the resource useful.

Yours truly,

M.E.B.


I totally don't understand this problem presented in my Calculus book.

(a) Using vector methods, show that the distance between two nonparallel lines $ l_{1}$ and $ l_{2}$ is given by

$ d \equal{}\frac{|(\vec{v_{2}}\minus{}\vec{v_{1}})\cdot (\vec{a_{1}}\times\vec{a_{2}})|}{\parallel \vec{a_{1}}\times\vec{a_{2}}\parallel }$

where $ \vec{v_{2}}$, $ \vec{v_{1}}$ are any two points on $ l_{1}$ and $ l_{2}$, respectively, and $ \vec{a_{1}}$ and $ \vec{a_{2}}$ are the directions of $ l_{1}$ and $ l_{2}$.  [HINT: Consider the plane through $ l_{2}$ that is parallel to $ l_{1}$.  Show that the vector $ \frac{(\vec{a_{1}}\times\vec{a_{2}})}{\parallel \vec{a_{1}}\times\vec{a_{2}}\parallel }$ is a unit normal for this plane; now project $ \vec{v_{2}}\minus{}\vec{v_{1}}$ onto this normal direction.]

(b) Find the distance between the line $ l_{1}$ determined by the points $ (\minus{}1,\minus{}1, 1)$ and $ (0, 0, 0)$ and the line $ l_{2}$ determined by the points $ (0,\minus{}2, 0)$ and $ (2, 0, 5)$.

[hide="My approach"]Let $ \vec{p}\equal{}\vec{p}(s, t) \equal{} (\vec{v_{1}}\plus{}s\vec{a_{1}})\minus{}(\vec{v_{2}}\plus{}t\vec{a_{2}})$. Then the minimum length of $ \vec{p}$ is the distance between two line $ l_{1}$ and $ l_{2}$.

Since $ \left|\vec{p}\right|$ is minimized iff $ \left|\vec{p}\right|^{2}\equal{}\vec{p}\cdot\vec{p}$ is, we only have to investigate the global minimum of $ \vec{p}\cdot\vec{p}$.

Let $ (s_{0}, t_{0})$ be the point minizing $ |\vec{p}|$ and $ \vec{q}\equal{}\vec{p}(s_{0}, t_{0})$.

As $ (s_{0}, t_{0})$ is a critcal point, $ \frac{\partial (\vec{p}\cdot\vec{p})}{\partial s}(s_{0}, t_{0}) \equal{} 2\vec{a_{1}}\cdot\vec{q}\equal{} 0$ and $ \frac{\partial (\vec{p}\cdot\vec{p})}{\partial t}(s_{0}, t_{0}) \equal{}\minus{}2\vec{a_{2}}\cdot\vec{q}\equal{} 0$.

So $ \vec{q}$ is orthogonal to both $ \vec{a_{1}}$ and $ \vec{a_{2}}$ and $ \vec{q}$ is parallel to $ \vec{a_{1}}\times\vec{a_{2}}$.

Therefore, $ \left|\vec{q}\right|\; \equal{}\;\frac{\left|\vec{q}\cdot (\vec{a_{1}}\times\vec{a_{2}})\right|}{\left|\vec{a_{1}}\times\vec{a_{2}}\right|}\; \equal{}\;\frac{\left| (\vec{v_{1}}\minus{}\vec{v_{2}})\cdot (\vec{a_{1}}\times\vec{a_{2}})\right|}{\left|\vec{a_{1}}\times\vec{a_{2}}\right|}$
[/hide]


Apple,Banana,Candy,Dandy,and Elephant are sitting in a round table.Apple refuses to sit next to Banana,Banana refuses to Apple/Candy,.....Elephant refuses to sit next to Dandy.How many arrangements are there,and what if Farmer,Goat,and Hobo come in or for a general n number of people?

woops.my bad.Apple also refuses to sit next to Elephant and Elephant refuses to sit next to Apple. :D

Wait... I don't understand the last part about a farmer, a hobo, and a goat... Can you explain it more clearly?

oops.sorry.well,i'm just trying to ask for a general n number of people such that A refuses to sit next to B and the nth guy,B refuses to sit next to C/D,C refuses to sit next to B/D,....,(n-1)th guy refuses to sit next to the (n-2) guy and n guy,n guy refuses to sit next to the A guy and the n-1 guy how many arrangements are there.thanks in advance! :)


is it just me, or is the time messed up?  :maybe:

Its according to time zones, so it may or may not be messed up.

[quote="buzzer11"]It is 7:00 pm right now, US eastern.

my time zone thing (?) is set to US eastern.[/quote]

You can change it in your profile.


geometry
circumcircle
incenter
trigonometry
Euler
inradius
geometry proposed

Let $ABC$ be a triangle, and $O$ its circumcenter. And let $AC > AB > BC$. Two points $D$ and $E$ lie on the line segments $AB$, $AC$, respectively, such that $BD=CE=BC$. We call $I$ the incenter of the triangle $ABC$. Prove that $OI$ equals to the radius of the circumcircle of the triangle $ADE$.

Using what I proved in [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=14655]another geometry thread[/url], we have $DE = 2\sin A \cdot OI$, or, in other words, $OI=\frac{DE}{2\sin A}=\frac{DE}{2\sin \measuredangle DAE}$. But after the Extended Sine Law in triangle DAE, the circumradius of triangle ADE also equals $\frac{DE}{2\sin \measuredangle DAE}$. Hence, the length OI equals to the circumradius of triangle ADE. Proof complete.

  Darij

Could you please be more clear about what you call X ?

  Darij

Great! But I have 2 more solutions. Can you try more?

Pestich, your comment was probably about the other thread, but this time it was just a trivial typo. A typo in your first message: "X ( intersection of mid-perpendicular from A onto DE with circumcircle of ADE, or intersection of AI with the circle)". I was wondering what you understood under "[b]mid-perpendicular from A onto DE[/b]". Either mid-perpendicular or perpendicular from A ! You see, a simple misunderstanding because of a typo. Now as you have corrected the typo, I understand your posts.

Indeed, the quadrilateral IOO'X is a parallelogram. This is a nice observation, which, of course, solves the problem (since it implies O'X = OI, but O'X is the circumradius of triangle ADE), and is actually stronger than the problem. The proof of the observation can be done using the same auxiliary construction I used in http://www.mathlinks.ro/Forum/viewtopic.php?t=14655 .

  Darij

Use Euler formula $OI=\sqrt{R_{ABC}(R_{ABC}-2r)}$ and $r=(s-a)\tan{\alpha\over 2}$.

[b]Problem.[/b] Let $ABC$ have $BC=\min\{BC, CA, AB\}$. $D, E$ be the points lie on the side $AB, AC$ respectively so that: $BD=CE=BC$. Denote: $R(ADE)$ be the circumradius of triangle $ADE$. Prove that: $R(ADE)=OI$, where $O, I$ is the circumcenter and incenter of triangle $ABC$ respectively.
[b] Proof.[/b] Denote $R,r$ is the circumradius, inradius of triangle $ABC.$ We have: $\sum\cos A=1+\frac{r}{R}$ and
\[\begin{aligned}\overrightarrow{DE}&=\overrightarrow{DB}+\overrightarrow{BC}+\overrightarrow{CE}\\ \Longrightarrow DE^{2}&=3a^{2}+2\overrightarrow{DB}.\overrightarrow{BC}+2\overrightarrow{BC}.\overrightarrow{CE}+2\overrightarrow{CE}.\overrightarrow{DB}\\ &=3a^{2}-2a^{2}\cos{B}-2a^{2}\cos{C}-2a^{2}\cos{A}\\ &=a^{2}(3-2\sum\cos{A})\\ &=a^{2}(1-\frac{2r}{R})\\ \Longrightarrow 4\sin^{2}A.R^{2}(ADE)&=4\sin^{2}A.(R^{2}-2rR)\\ \Longrightarrow R(ADE)&=OI. \end{aligned}\]

[quote="April[color=darkred]"][b]Problem.[/b] Let $ABC$ have $BC=\min\{BC,\, CA,\, AB\}$. $D,\, E$ be the points lie on the side $AB, AC$ respectively so that: $BD=CE=BC$. Denote: $R(ADE)$ be the circumradius of triangle $ADE$. Prove that: $R(ADE)=OI,$ where $O,\, I$ is the circumcenter and incenter of triangle $ABC,$ respectively.[/color][/quote]

[color=indigo][b]Another solution.[/b] Denote $M,\,N$ be the projections of $I,\,O$ on $AB,$ respectively.

Contruct triangle $OPQ$ with circumcircle $C(I,\,IO)$ and $OQ\parallel AB,\,OP\parallel AC$

We have: $OQ=2MN=2(BN-BM)=2\left(\frac{c}{2}-\frac{a+c-b}{2}\right)=b-a=EA$
Similaly, we also have: $OP=AD,$ thus $\triangle AED=\triangle OQP\Longrightarrow R(ADE)=OI$[/color]


Is the first term (a^2+b^2)(2ab/(a+c)-[b]b[/b]) instead?

Then again, the inequality is going to be false since you can have a artibitary close to c-b on the left, and the second term will be artibitary negative

I'm terribly sorry, but I have again posted a great stupidity. The actual inequality is:
   (a^2+b^2)(2ab/(a+c)-c)+(b^2+c^2)(2bc/(a+b)-a)+(c^2+a^2)(2ac/(b+c)-b)>=0.
   Sorry again

harazi,cud u post ur proof for this inequality
it's intriguing .i used brute force and the expression has some cyclic terms and some symmetric. :(
thanx

The proof is easy:

  We will prove that a^3b/(a+c)+b^3c/(a+b)+c^3a/(b+c) \geq (ab 2 +bc 2 +ca 2 )/2. This follows from Cauchy-Schwarz inequality and the fact that a 2 b+b 2 c+c 2 a \geq 3abc. Indeed:
  (a 2 b+b 2 c+c 2 a) 2 =( \sum  \sqrt a^3b/(a+c)* \sqrt ab(a+c)) 2  \leq  \sum a^3b/(a+c)*( \sum a 2 b+3abc) and 3abc \leq  \sum a 2 b. Similarly, we have  \sum ab^3/(a+c) \geq 1/2* \sum ab 2 . Now add these 2 inequalities.

sweet,just the kinda proof i like :D .the problem with inequalities is that u can keep beating around the bush and not reach the crux :(  :(

[quote=harazi]
Prove that for all $a,b,c>0$ we have:$$(a^2+b^2)\left(\frac{2ab}{c+a}-c\right)+(b^2+c^2)\left(\frac{2bc}{a+b}-a\right)+(c^2+a^2)\left(\frac{2ca}{b+c}-b\right)\geq0.$$[/quote]

We have
\[LHS=\frac{1}{3}\sum{\frac{(4a^2c+2ab^2+5abc+5ac^2+bc^2+c^3)(a-b)^2}{(a+c)(b+c)}}\ge{0}\]

[quote=sqing][quote=harazi]Prove that for all a,b,c>0 we have:
  (a^2+b^2)(2ab/(a+c)-c)+(b^2+c^2)(2bc/(a+b)-a)+(c^2+a^2)(2ac/(b+c)-b)>=0.[/quote]
Prove that for all $a,b,c>0$ we have:$$(a^2+b^2)\left(\frac{2ab}{c+a}-c\right)+(b^2+c^2)\left(\frac{2bc}{a+b}-a\right)+(c^2+a^2)\left(\frac{2ca}{b+c}-b\right)\geq0.$$[/quote]
$\sum_{cyc}a^3(b+c)\geq 2\sum_{cyc}a^2b^2$,$\sum_{cyc}\frac{a^2+b^2}{c+a}\geq\sum_{cyc}a$,
$\sum_{cyc}a^3(b+c)+2abc\sum_{cyc}\frac{a^2+b^2}{c+a}\geq 2\sum_{cyc}a^2b^2+2abc\sum_{cyc}a\iff$
$\sum_{cyc}a\sum_{cyc}(a^2+b^2)\left(\frac{2ab}{c+a}-c\right)\geq0 \iff$
$\sum_{cyc}(a^2+b^2)\left(\frac{2ab}{c+a}-c\right)\geq0 .$


function
real analysis
real analysis unsolved

Find all continuous functions f:R->R, such that:
f(f(f(x))) = 8x, for each x in R.

Olympiad section is more appropriate for this question ;)


logarithms
absolute value
calculus
calculus computations

Know that the taylor series for ln(1+x) at 0 is x-x^2/2+x^3/3-....

find the taylor series at 0 for x^2*ln(1+x^3)

I was thinking that you just factor out an x^2 and replace x with x^3, can you do this?

Thanks.

We know that $ \ln(1 \plus{} u) \equal{} u \minus{} u^2/2 \plus{} u^3/3 \minus{} \dots$ (in the interval $ \minus{} 1 < u\le 1$). Substitute $ u \equal{} x^3$ to get $ \ln(1 \plus{} x^3) \equal{} x^3 \minus{} x^6/2 \plus{} x^9/3 \minus{} \dots$ (in the interval $ \minus{} 1 < x\le 1$). Multiply both sides by $ x^2$ to get $ x^2\ln(1 \plus{} x^3) \equal{} x^5 \minus{} x^8/2 \plus{} x^{11}/3 \minus{} \dots$

Now try to do this in $ \sum$-notation.

$ f(x)\equal{}\sum_{k\equal{}1}^{\infty} \frac{x^k\cdot (\minus{}1)^{k\minus{}1}}{k}\Leftrightarrow f(x^3)\equal{}\sum_{k\equal{}1}^{\infty} \frac{x^{3k}\cdot (\minus{}1)^{k\minus{}1}}{k}$

$ \therefore x^2\ln{|1\plus{}x^3|}\equal{}\displaystyle x^2\left|\sum_{k\equal{}1}^{\infty} \frac{x^{3k}\cdot (\minus{}1)^{k\minus{}1}}{k}\right|$

I think that's right.

I don't know why you put $ |\cdot|$ there... the formula would be correct without the bars, but now it's wrong.

Ah wait what's the interval of convergence...$ 0<x\le 2$ by tables for $ \ln{x}$ so the correct answer is without absolute value. For some reason I though it converged for all x and hence the absolute value to match what would be necessary in the actual equation, as logs cannot take a negative parameter.

However, I don't think my answer is 'wrong', its just not entirely correct. $ x^2\ln{(1\plus{}x^3)}$ is positive on that interval and so this shouldn't make a difference. The absolute value is around the [i]entire[/i] sigma so...

Well anyways I was trying to force it into agreement for all x, which is foolish.  :P

The interval of convergence of the series in this example is $ \minus{} 1 < x\le 1$, and $ \ln(1 \plus{} x^3)$ can be negative there.

Sorry, i'm not that good, but I'm a little bit confused, because some texts I've read say that the taylor series of $ \log(1\plus{}x)$ is the same as the taylor series of $ \ln(1\plus{}x)$. Are they the same, or is this a typo error, or something? If they aren't the same then what is the taylor series of $ \log(1\plus{}x)$? Thanks.

benni8888, you're confusion is arising from ambiguity in notation.

Some sources denote the natural logarithm by "$ \log$".  Others denote the natural logarithm by "$ \ln$".  In your case, these notations are meant to represent the same thing.  Do not confuse this with denotating the  logarithm with base $ 10$ (sometimes referred to as the "common logarithm").

Regardless, all logarithms are proportional to each other via a constant (as shown by the change of base formula for logarithms).


[hide="solution"] $\sqrt{14+6{\sqrt{5}}}= a+b\sqrt{c}$
Squaring both sides,
$14+6\sqrt{5}=a^{2}+2ab\sqrt{c}+b^{2}c$
Comparing radicals:
$c=5$
Comparing non-radicals:
$2ab=6$, so $ab=3$
$a^{2}+5b^{2}=14$
It's pretty clear that $a=3$ and $b=1$, so you finally get:
$3+\sqrt{5}=5.2361$ [/hide]

[quote="7h3.D3m0n.117"][quote="xxxxx"]simplify:
square root of (14 + 6 times the square root of 5)

(how do u do square root in LaTex?)[/quote]

[code]$\sqrt$[/code]

so the problem would be

$\sqrt{14+6\sqrt{5}}$

and how far do you have to simplify it? to decimals or the simplist radical form?[/quote]

 I don't know...It doesn't say how far to simplify it....

Ok, the 7th grade manual is filled with these kind of basic problems (In Romania), and i doubt they are at Mathcounts level...
To clear this up:
$\sqrt{A\pm\sqrt{B}}=\sqrt{\frac{A+C}{2}}\pm\sqrt{\frac{A-C}{2}}\\C^{2}=A^{2}-B$

[hide]
$C=\sqrt{196-180}\\C=4$

So

$\sqrt{\frac{14+4}{2}}+\sqrt{\frac{14-4}{2}}=3+\sqrt{5}$
[/hide]


analytic geometry
graphing lines
slope
geometry unsolved
geometry

here,
 triangle AXB is congruent to the triangle AYD
 Because
 <ABX=<YAD     -{1}
 as <BAX=90
 therefore
  <AXB=90-<ABX   -{2}
 AS, <BAD=90
  BUT <BAD=<BAY+<YAD
 hence <BAY=90-<YAD   -{3}
 therefore <AXB=<BAY
 [ of course <OAX=<YAD , <BAY=<BAO & similarly all]
AS 2 angles are same third must be
 hence
 triangles AOX & AOB are similar

[quote="FOURRIER"]Consider a square $ ABCD$
with $ X$ the midpoint of $ [AD]$
and $ Y$ the midpont of $ [CD]$
and let $ O$ be the intersection of $ (AY)$ and $ (BX)$

Prove that
$ \triangle{AOX}\sim \triangle{AOB}$

Thanks  :)[/quote]
My proof:
We have $ tan(\hat{ABX}) \equal{} tan(\hat{DAY}) \equal{} \frac {1}{2}$
=>$ \angle ABX \equal{} \angle DAY$
=>$ \angle DAY \plus{} \angle AXB \equal{} \angle ABX \plus{} \angle AXB \equal{} 90^o$
So $ \angle AOX \equal{} \angle AOB \equal{} 90^o$
then $ \triangle{AOX}\sim \triangle{BOA}$

[hide]We can easily find $ AY \perp BX$.  (Considering the slopes of lines $ AY$ and $ BX$ when the square is oriented with sides parallel to coordinate axes is one way to see it.)  From here we just note that $ AO$ is the altitude to the hypotenuse of right triangle $ XAB$ and thus it must divide the right triangle into similar right triangles.[/hide]

[quote="DuongChinh_K41SPHN"]prove that \ange{DAY}=\ange{ABX}[/quote]
Did you solve this problem?If so,there is no reason to post a hint,just post a complete,if it is just an idea,then hide it. :wink:


function
trigonometry
analytic geometry
graphing lines
slope
geometry
geometric transformation

If f(x) = x+sinx then find the inverse of f(x)

WolframAlpha says: [color=red]no result found in terms of standard mathematical functions[/color]. Believe it!

hmm but then using graphs a reasonable answer that i think is correct is g(x) = x-sinx .. It very much seems like the inverse function

But I am not sure. Can I have a confirmation??

Yes: I confirm that you are wrong and $ x \minus{} \sin x$ is not the inverse function of $ x \plus{} \sin x$ (and furthermore that $ x \plus{} \sin x$ does not have a nice inverse function).  Just plot their composition to see: http://www.wolframalpha.com/input/?i=(x+%2B+Sin[x])+-+Sin[x+%2B+Sin[x]]  (Actually, the two functions don't look even remotely like inverses: $ x \plus{} \sin x$ behaves like $ 2x$ near the origin, while $ x \minus{} \sin x$ looks completely flat instead of having slope $ \frac {1}{2}$ as would be necessary; $ x \plus{} \sin x$ has lots of points of slope $ 0$, but $ x \minus{} \sin x$ has no points of infinite slope.)

Are you sure you have plotted the correct graphs in Wolfram.

I checked it again. It seems that x+sinx is just the reflection of x-sinx about the line y=x confirming that it is the inverse of each other

see this
http://www.wolframalpha.com/input/?i=x%2Bsinx+%3D+x-sinx

asish2000: you seem to be using the wrong definition of "reflection".

In a true reflection across the line $ x \equal{} y,$ the reflection that gives the relationship between a function and its inverse, a point moves at a $ 45^{\circ}$ angle to the horizontal. The operation is $ (x,y)\mapsto(y,x).$ The point $ (3,2)$ goes to the point $ (2,3).$

The operation you seem to be using is "move vertically across $ x \equal{} y,$ moving the same distance beyond the line on the other side as you started on the one side." A point moves vertically - and that's not true reflection. You seem to be sending $ (x,y)\mapsto (x,2x \minus{} y)$ or sending $ (3,2)$ to $ (3,4).$

It's not the same thing at all.

And, as everyone else has said, the inverse function to $ f(x) \equal{} x \plus{} \sin x$ is not an elementary function.

$ f$ is a strictly increasing function on $ \mathbb{R}$ and is one-to-one and maps $ \mathbb{R}$ onto $ \mathbb{R}.$ So it has an inverse defined on all of $ \mathbb{R}.$

$ f'(x)$ ranges from $ 0$ (the value it takes on at every odd multiple of $ \pi$ to $ 2$ (the value it takes on at every even multiple of $ \pi,$ including $ 0.$) Hence the derivative of $ f^{ \minus{} 1}$ is always at least $ \frac12$ and does not exist at odd multiples of $ \pi.$ That is, $ f^{ \minus{} 1}$ is continuous on $ \mathbb{R}$ but not differentiable on $ \mathbb{R}.$ The graph of $ f^{ \minus{} 1}$ has a vertical tangent at each odd multiple of $ \pi.$


Squares are used to build the following sequence of drawings. If the length of a side of each square is one unit, how many units are in the perimeter of the 8th drawing?
[asy]size(250);
defaultpen(linewidth(0.8));
void drawSquare(Label l = "", pair A){
 draw((A)--(A.x + 1, A.y)--(A.x + 1, A.y + 1)--(A.x, A.y +1)--cycle);
 label(l,(A.x - 0.5,A.y + 0.5));
}

drawSquare("1st", (0,0));
drawSquare("2nd", (2,0));
drawSquare((2,1)); drawSquare((3,0));

drawSquare("3rd",(5,0)); drawSquare((5,1)); drawSquare((5,2)); drawSquare((6,0)); drawSquare((7,0));[/asy]

Find a pattern.

Figure 1: $ 4$

Figure 2: $ 8$

Figure 3: $ 12$

Obviously, $ F_n \equal{} 4n, n \in \mathbb{Z}^\plus{}$ <--- hope I stated that correctly

Thus, the answer is $ \boxed{32}$.


geometry
3D geometry
sphere
real analysis
real analysis unsolved

Does there exist one to one mapping between sphere and $ X$, $ X\subset\mathbb{R}^2$?

Of course there does; they have the same cardinality.  Do you mean a continuous mapping?

Thank you,
I mean only one to one mapping, not necessarily continuous.

Please give an exmple of such one to one mapping(with brief description).


well, i want to find information about ramsey theorem and ramsey numbers;do u have a link or can u inform me about that

I suggest you look at [url=http://www.math.uni-hamburg.de/home/diestel/books/graph.theory/GraphTheoryII.pdf]this book.[/url]

You may already know this, but one place to start is
[url=http://mathworld.wolfram.com/RamseysTheorem.html]wolfram[/url]or [url=http://planetmath.org/encyclopedia/RamseysTheorem2.html]Planetmath ..[/url]

You can also search on AoPS say :  [url=http://www.artofproblemsolving.com/Forum/topic-14738.html]here ..[/url]

try this:

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

[quote="Gyan"]You can also search on AoPS say :  [url=http://www.artofproblemsolving.com/Forum/topic-14738.html]here ..[/url][/quote]I really was starting to wonder if someone would actually use the search on this site to find stuff at one point :)


I'm not sure whether it has been posted before or not.

in triangle $ ABC$ let $ A',B',C'$ be tangency points of 9-point circle  with excircles wrt $ A,B,C$ in order.Prove that $ AA',BB',CC'$ are concurrent.

Only a idea: inversion!

If we choose $ M_a$ the midpoint of BC as inversion point, the radius of inversion can be chosen such that the excircle tangent to BC and the incircle does not change (as a whole, not pointwise). And the 9-point circle will be a line tangent to both circles (in fact, this is a way to demonstrate the NPC-tangency-theorem).

It appears to be more promising than a bunch of various circles. Then either we can use some sort of Menelaus-Ceva, or to calculate the intersecting point (analytical geometry - very bad idea- or some ingenious trick...)