Let $a,$ $b$ and $c$  are positive numbers such that $a+b+c=1$ and $\alpha\geq0.$ Prove that:
$(ab+ac+bc)\left(\frac{a}{b^{2}+\alpha b}+\frac{b}{c^{2}+\alpha c}+\frac{c}{a^{2}+\alpha a}\right)\geq\frac{3}{1+3\alpha}.$

Very nice inequality Arqady .Here is my solution . Rename the $\alpha$ by $k$
Consider the function $f(x)=\frac{1}{b^{2}+kb}$ . It is convex so by the weighted Jensen we have
$af(b)+bf(c)+cf(a)\geq f(ab+bc+ca)$ or equivalently

$\sum\frac{a}{b^{2}+kb}\geq \frac{1}{(ab+bc+ca)^{2}+k(ab+bc+ca)}$

so it remains to prove that
$\frac{1}{(ab+bc+ca)^{2}+k(ab+bc+ca)}\geq \frac{3}{ab+bc+ca+3k(ab+bc+ca)}$

which is equivalent to $ab+bc+ca\leq \frac{1}{3}$ which is clearly true . :wink:

Arqady what is your solution ?

Very nice proof, silouan! :lol:
My proof:
by the Holder's inequality the following inequality holds:
$\sum_{cyc}ab\cdot\sum_{cyc}\frac{a}{b(b+k)}\cdot\sum_{cyc}a(b+k)\geq(a+b+c)^{3}.$
Id est, it remains to prove that $\frac{1}{ab+ac+bc+k}\geq\frac{3}{1+3k}.$
But $\frac{1}{ab+ac+bc+k}\geq\frac{3}{1+3k}\Leftrightarrow(a+b+c)^{2}\geq3(ab+ac+bc).$ :)

Also nice proof Arqady . Thank you for sharing this and the problem with us . :)


An  unlimited  supply  of  gasoline  is  available  at  one  edge  of  a  desert  800  miles  wide,  but  there  is  no  source  in  the  desert  itself.  A  truck  can  carry  enough  gasoline  to  go  500  miles  (this  is  called  the  "load"),  and  it  can  build  its  own  refueling  stations  at  any  spot  along  the  way.

What  is  the  minimum  amount  of  gasoline  the  truck  will  require  in  order  to  cross  the  desert?

you can check ur answer here:
[url]http://www.projecteureka.org/problem/question/97[/url]

[quote="uzurpatorul"]An  unlimited  supply  of  gasoline  is  available  at  one  edge  of  a  desert  800  miles  wide,  but  there  is  no  source  in  the  desert  itself.  A  truck  can  carry  enough  gasoline  to  go  500  miles  (this  is  called  the  "load"),  and  it  can  build  its  own  refueling  stations  at  any  spot  along  the  way.

What  is  the  minimum  amount  of  gasoline  the  truck  will  require  in  order  to  cross  the  desert? [/quote]

I can't answer the question, but here is a similar problem with a similar style:

An expedition has to travel 80 km on foot across a desert, moving at a constant pace of 10 km per day. Each traveller can only carry enough food for 5 days (water is no problem), but the travellers can give each other food. At the beginning of their 80-km trip, there is a city that has an unlimited supply of food.

What is the minimum number of travellers needed so that at least one person makes it to the end of the 80 km without anybody running out of food en route?

[quote="uzurpatorul"][hide]3[/hide][/quote]

Well, I know that, but I have another question.

What exactly is a refuelling station? Does it also have an unlimited supply of fuel? And in what units are the fuel counted in? Miles?

[quote="SnowHawk"]Well refueling station is a place where the truck can fill up its tank.... right?

And since fuel is a liquid, wouldn't it be measured in gallons?[/quote]

But the problem states nothing about the fuel efficiency of the truck.

Well yes, but then the problem is retarded. You could just go 500 miles, build a refueling station, and fill up for the next 300 miles.

Yes but the fuel stations don't have fuel, so the truck has to put it there itself.
I've heard of this question before, but it was with a donkey and carrots.

I'm not too sure what the first riddle means; it's a bit ambiguous.

About sdk's riddle, Can someone explain this? I've heard the around the world thing, but you get 3 because you can go from the opposite side of the world.

[quote="AIME15"]To uzurpatorul: Stop advertising your website.[/quote]

There's nothing wrong with advertising, as long as the post has something else besides the advertisement in it. And you should PM this instead.

Regarding my initial post:

1. A "load" means full tank
2. With the "load" a truck can go up to 500 miles
3. however u(the driver) can choose to go 100 miles - leave 300miles worth of fuel at ur newly constructed refuel station (3/5 of load) and go back to get more.
4. The MPG for the truck is the same during the whole trip

Let me know if the problem makes more sense now.

This is another variant of the same problem:
You have 3,000 bananas and a camel which can carry at most 1,000 bananas at a time. The camel eats a banana before moving a unit. You want to transport the bananas 1,000 units. What is the maximum number of uneaten bananas that you can move 1,000 units?

Of course the solution can be verified here: [hide]http://www.projecteureka.org/problem/question/105[/hide]

Hints are here: [hide]http://mathforum.org/dr.math/faq/faq.camel.html[/hide]

Here's my solution to the first problem, which apparently is wrong...  It might give someone else an idea for a solution, though.

[hide="solution"]Step 1: Get 700/3 and drive until mile 100/3, depositing 6 each mile continuously.

Step 2: Drive back with an empty tank, using the fuel spread on the ground.  From here, there is 5 layers left

Step 3: Get 500 and pick up gas continuously, until mile 100/3, then start depositing 4 each mile until you run out of gas at 400/3.  We use 4 for 100 miles, plus fuel, giving 500.

Step 4: Return to the start, picking up fuel along the way.  There are 3 layers of fuel from 0 to 400/3.

Step 5: Get 500 and pick up gas continuously, until mile 400/3, then start depositing 2 each mile until you run out of gas at 300.  We use 2 for 500/3 miles, plus fuel, giving 500.

Step 6: Return to the start, picking up fuel along the way.  There is 1 layer of fuel from 0 to 300.

Step 7: Get 500 and pick up gas continuously until mile 300.  From here, use the 500 to drive across the desert and win.

To get to 800, need 500.
To get to 300, need 500.
To get to 400/3, need 500.
To get to 100/3, need 700/3.
This totals 5200/3=1733.33[/hide]

More hints to the crossing the bridge problem (also want to clarify, the answer is in loads):
[hide]

 with 1 load the track(T) goes 500 miles
 with 2 loads , T can build a refueling station somewhere at distance x, T wants to have a full load there so it can go another 500 miles, then T needs 3 trips to go from one edge of the dessert to the newly made refueling station to have a full load hence x = 500/3 miles, consequently the track can go 500 + x = 666.66 miles (1st trip T ----> X (1/3 load is consumed during this trip, leaves 1/3 load to the refueling station), X ----->  T (second trip, T goes back for more fuell), T ------> X (3rd trip, consumes 1/3 load, 2/3 load remained + 1/3load (stored in the station) = 1 load
 with 3 loads, T can build a refueling station that holds 2 loads at distance y, T needs 5 trips from the edge of the dessert to y , 500/5 = y , hence the truck can go 100 miles + 666.66 = 766.66

[/hide]


function
logarithms
calculus
calculus computations

How do I find the inverse Laplace transforms of the following functions:

(a) $ F(s) \equal{} \frac {2e^{\minus{}3s}}{(s \minus{} 1)^2} \plus{} 3$

(b) $ F(s) \equal{} \ln \left( \frac{s^2 \minus{} 1}{s^2} \right)$

Any help would be appreciated!

For the first one:

Separate the terms. The $ 3$ is the the transform of a Dirac delta. As for the first term: $ \frac1{s^2}$ is the transform of $ t.$ Make that $ s\minus{}1$ instead of $ s$ and multiply by $ e^{t}.$ Multiply by $ e^{\minus{}3s}$ and shift the thing to the right by 3 units. Put that together:

$ f(t)\equal{}2H(t\minus{}3)\cdot(t\minus{}3)e^{t\minus{}3}\plus{}3\delta(t).$

(I'm using $ H(t)\equal{}\begin{cases}1&t>0\\0&t\le 0\end{cases}$ for the Heaviside unit step function.)

For the second one, my first instinct is to expand out the power series for the logarithm and then deal with that series term by term.

Thanks!

For the second one, there exists (in the inverse Laplace tables) that the inverse of $ \ln \frac {p\plus{}b}{p\plus{}a} \equal{} \frac{1}{x}(e^{\minus{}ax} \minus{} e^{\minus{}bx})$.

Could i use this fact to split (b) into $ F(s) \equal{} \ln \left( \frac{s\minus{}1}{s} \right) \plus{} \ln \left( \frac{s\plus{}1}{s} \right)$ and apply the inverse transform separately on both?

I would then get $ \frac{1}{x}(1 \minus{} e^x) \plus{} \frac{1}{x}(1 \minus{} e^{\minus{}x})$.


The house where Toby lives is equipped with an elevator.  It has four floors, including the basement and the attic. The elevator is very unusual: it has only two buttons, "A" (toward the attic) and "B" (toward the basement) and goes only one floor up or down at a time.  It takes 30 seconds between the time you press the button until the elevator reaches the corresponding floor above or below (at which time you can press a button again).  Furthermore, the elevator doesn't work if you press two consecutive identical strings of two or more buttons.  For example, suppose you start in the basement and proceed with the sequence AAABBAB.  Then the elevator will stall, for it you then press A, you have BA two times in a row, and if instead you press B, you have ABB two times in a row.  If he starts in the basement, can Toby ride the elevator for more than 8 minutes before he is stymied by a stalled elevator?  What if he starts on the first floor?

I would just like to paraphrase the question to make sure that I understand it correctly:

"Are there any binary sequences of length 16 are there which do not contain two consecutive identical sub-strings of length greater than 1, such that the number of 1's at before any place in the sequence is greater than the number of 0's?"  (where that last condition is slightly different starting on the first floor).  Right?

I think not quite, since you can't have >3 1's in a row.

Okay, right -- at any point in the sequence, 0 :le: #1's - #0's :le: 3.  You couldn't have more than 3 in a row, anyhow, since that would be a duplication of the string 11.  But yes, you can't have 111011.

I believe both are possible.  Sequences that I used were 1. AAABBBAAABBAABBA.  2. BAAABBBAAABABBAA.  Someone tell me if I'm wrong.

jobro:
AAABBBA[b]AABBAABB[/b]A
BAAABBBAA[b]ABAB[/b]BAA

very interesting question...

How about these? 1. AAABBBAAABBAABAB   and 2. AABBBAAABBAABABB.

1. AAABBBAAABBA[b]ABAB[/b]   and
2. AABBBAAABBA[b]ABAB[/b]B.

Ahhhhhhhhh!!!! I'm going CRAZY  (edit: CRAZIER  :twisted:  !)

Something tells me that maybe guess and check isn't the best method....

And I'll add a subquestion, to which I don't know the answer:
How long can Toby ride the elevator for?  How long if there were 5 floors?  3?

I'll give 10 cool points to anyone who finds a solution to this problem.

Do you have a solution? I haven't tried the problem since my last post.

I don't have any idea of where to even start with this problem.  My gut instinct is no, otherwise it'd be sort of a lame problem.  It also feels like a contradiction proof.  Other than that I got nothing.

Edit: Unless you're asking MdClark, in which case I'm in no position to answer.  :-)

I think I got a possible solution.

ABAAABBBAAABBAABA

[quote="Coder"]I think I got a possible solution.

A[b]BAAABB[/b][u]BAAABB[/u]AABA[/quote]

Nice try, though -- that took me a long time to catch :)

Has noone given any thought to the possibility that it can't be done.

There is a solution.  He can ride the elevator for a full 9 minutes before it stalls.

It doesn't matter if the building has any extra floors.  9 minutes is the maximum ride possible.

I think the 9-minute solution is unique.  And I think once you see it, you will probably say "duh".

Second try, and it even fits the 9 minute one:

ABAABBAAABBBAABBAB

Shoot, I tried something like that.  Clever.  10 cool points.


[quote="Rep123max"]President Bush said today that the U.S. will begin to investigate whether Iran played a part in 9/11. That means the U.S. and President Bush will make some stuff up, say it over and over for probably 2 months, then go to war right before the election and then Bush will win because no wartime president has ever lost.[/quote]

From what I read a link to Iran seems far more credible than a link to Iraq especially given reports that most of the hijackers travelled through Iran shortly before the plan was executed and passports were not stamped.  It is being reported that the Iranian government is allowing people who might be terrorists to cross the border unquestioned and without passport stamping.  It also seems far more likely that Iran has a nuclear weapons program.

All that said, I think it's far from clear what the best course of action would be...

[quote="Rep123max"]President Bush said today that the U.S. will begin to investigate whether Iran played a part in 9/11. That means the U.S. and President Bush will make some stuff up, say it over and over for probably 2 months, then go to war right before the election and then Bush will win because no wartime president has ever lost.[/quote]

Only because LBJ chose not to run.

Also, unless Iran did something between now and our attacking, I think the US invading Iran would be a huge political loser for Bush.  (But I could be way wrong about that - the swaying political passions of this country are tough to predict.  For example, if we invaded Mexico, I think this country would be against it up until the minute Chirac publicly condemned the invasion.)

[quote="soilent green"]Being from a military community I know there are not enough military to support another war.  I'm sure this is just talk at least for now although stranger things have happened.  I do not think that Iran has ever been an innocent country- remember 1979-1980.[/quote]

I think this is the biggest reason we won't see an invasion of Iran.  We would be overextending ourselves if we did.  We don't even seem have enough troops in Afghanistan or Iraq to do the intended job as cleanly as our miliary leaders say they would like.

[quote="Magnara"]I say war on Saudi Arabia.  Not that that would devestate our economy for a few years or anything....[/quote]

NO!! NOT ANOTHER WAR. Can't we have peace?  As long as we are on high alert, terrorists are not going to do anything.

[quote="beta"]As long as we are on high alert, terrorists are not going to do anything.[/quote]

??

Please. Not another war! It's like, the best way to peace in the Middle East is to take it all over. What!?

[quote="Rep123max"]Key word their is "shoud." Because we definetly didn't for Iraq.[/quote]

I'm curious, what would YOU have done if you were in Bush's position?

[quote="Rep123max"]You mean assuming 9/11 happened? First I would have sent as many troops as possible to Afganistan. Then, I would have helped out the 9/11 commission as much as possible. Then, if I got intelligence that said Iraq was a threat, I would have gotten the U.N. to have inspectors in until they were satisfied if there were/weren't (Bush got them out after a month or two). If they were a threat, I would have made sure that the whole world understood it and we actually had a coalition instead of just Americans and a few other countries. But I wouldn't have gone to war because there would have been no smoking gun like they said there was.[/quote]
Your argument seems awfully uneducated. Saddam only allowed inspectors when he realized that there was no way left for him to avoid war. He had a long history of kicking inspectors out of his nation and refusing to comply with them even when they were there. The UN is not equipped to force a country to do anything - even with peacekeeping troops, the nation may order them to leave and they must comply (the nation may lose some face on the global scene, but it can certainly be done.)

[quote="JBL"]With respect to an invasion of Iran, I think the most relevant comment that can be made is just a twist on an old put-down:

Us and whose army?[/quote]Maybe Morocco would offer some monkeys again.

Seriously, though, another war is unlikely to be manageable without a draft, which is highly unlikely to be implemented before the election.

I cannot help but think of this James Carville quote:
"Back in 2000, a Republican friend of mine warned me that if I voted for Al Gore and he won, the stock market would tank, we'd lose millions of jobs, and our military would be totally overstretched. You know what? I did vote for Al Gore, he did win, and I'll be damned if all those things didn't come true."

[quote="h_s_potter2002"]War should ONLY be used as a last resort.  If nothing else works, and if the problem NEEDS to be solved quickly, only then should we go to war.[/quote]

I don't see where it is possible to draw the line between further diplomatic action and military action under that philosophy. How do you decide when you NEED to go to war? Can't you always postpone action another month? What is a "last resort?"

[quote="Rep123max"]http://www.costofwar.com/index-pre-school.html

Imagine if we spent that money on fighting terrorism?[/quote]
Just imagine - we did!

How many civiliians died in Iraq [url]http://www.iraqbodycount.net/[/url]

i think that really iran might have more cause for a war, but we shouldn't wage one.  for one thing, theres like no more soldiers, which means a draft, which means my brother might join the army.  and, i don't want gwb to win, so that's another reason.

gwb would probably <i>not</i> win if a draft were instituted, since that would anger a lot of families who wouldn't have voted otherwise.

[quote name="AlienGirl05" url="/community/p29035139"]
[quote=tigeryong]curious what everyone's thoughts on this are, now that it's been 19 years. Since this all happened, I'm mainly talking about the aftermath.
As in, not obeying the Powell doctrine, and Trump pulling out of Afghanistan (not Iran, but directly related) a few days after the election results were announced. In my opinion, it was a jerk move to make Biden's office assumption hard.[/quote]

My current thoughts: You should run for president. /hj.
[/quote]

NO PLEASE.
I always take leadership positions, intentionally or not (2 4h clubs essentially forced me in as president), but president? No way. Too much responsibility, and besides. Name one person in congress that isn't far right or far left?
Strike that, name the percentage of americans that can comprehend what 'politically neutral' means.
That's POH-lih-tih-call-E nu-trawl

[quote name="AlienGirl05" url="/community/p29035210"]
Alright, never mind.

Personally, I believe that we should never begin wars. Other countries may have done horrible things to your country, but commencing a war for the sake of revenge is a bad route. During most wars, thousands of innocent civilians who have absolutely nothing to do with the atrocities that their country's government committed are being killed.
[/quote]

Exactly. I think that going after al-quaeda was ok, since they hit the twin towers, but putting Hussein in power then ruining iraq fighting him...

[quote=AlienGirl05]Go after terrorist organizations, not entire countries. It's like going after a person because their family members are awful people. There are quite a few insufferable doorknobs in my family, but I don't have anything to do with what they have done.[/quote]

This comes into discussion now with the beginning of a [purported, but I think it's legit] Israeli invasion of Gaza. Whether their intentions align with their actions or not, the events remain as follows:

[b]Hamas[/b] attacks Israel for no reason
Israel punishes the [I]Palestinians[/I] as a byproduct of destroying Hamas.

Now nobody should validate Hamas' actions. They're not the ones suffering. In fact, they perpetrate the suffering by hiding behind civilians and making them take all the harm for their own actions. We can draw many parallels with this. Al-Qaeda attacks the US for no reason, the US obliterates Al-Qaeda but also destroys Afghanistan in the process.

What's the point?

[quote=tigeryong][quote name="s12d34" url="/community/p29057157"]
Well it seems like finally, after two decades, the biggest dream of the neocons is finally coming true; that is, war against Iran. Once Israel begins the ground invasion of Gaza and Hezbollah intervenes in the north, the US will use that as a pretext to launch a war against Iran, which will probably start WW3 because Russia will not sit and let the US destroy Iran and then China will not sit and let the US destroy Russia.
[/quote]

WWIII has started. Nobody calls it a world war until we're well into it.[/quote]

Indeed. Just like how WW1 was considered just another regional European conflict until the UK intervened on the behalf of Belgium or how the Japanese invasion of Manchuria is considered to be under the same umbrella of events relating to WW2, even though it happened 8 years before WW2.

[quote=AlienGirl05]Go after terrorist organizations, not entire countries. It's like going after a person because their family members are awful people. There are quite a few insufferable doorknobs in my family, but I don't have anything to do with what they have done.[/quote]

Yeah.  Like, don't nuke a forest to kill a few wolves (I don't think that wolves should be needlessly killed either, but I remembered this from somewhere so I just used it).


Just out of curiosity, were the people that did Alcumus during the pre-release, randomly selected?

Or students of a few outstanding teachers we know through MATHCOUNTS.


Two regular polygons of the same number of sides have sides 48m and 55m in length respectively. What is the length of the side of the another rectangular polygon of the same number of sides, if its area is equal to the sum of the two?

oh, yeah, that's true....so can anyone find the answer?

[hide="I think there is a better explained solution and easier solution..."]The ratios of the areas of two similar polygons is the square of the ratios of their lengths. We have that the ratio of side lengths of the two polygons is $ \frac {55}{48}$, so the ratio of their areas is $ \frac {55^2}{48^2}$. Since the area of the third polygon is the sum of the areas of the two polygons, we have a ratio between the third polygon and the polygon with side length $ 48$:
\[ 1 \plus{} \frac {55^2}{48^2} \equal{} {\left(\frac {x}{48}\right)}^2 \implies \boxed{x \equal{} 73}
\]
[/hide]

[hide="Solution"]We don't really care what kind of polygon it is as long as know it's a regular polygon. The area will be expressible as Area = $ cs^2$, where $ c$ is a constant is $ s$ is the length of a side. For triangles, this constant c is $ \frac{\sqrt{3}}{4}$, for squares it is $ 1$, for pentagons it is $ \frac{5\tan{54^\circ}}{4}$, etc.

$ 48^2c \plus{} 55^2c \equal{} x^2c$
$ 48^2 \plus{} 55^2 \equal{} x^2$
$ x \equal{} \pm73$, but we're dealing with side lengths, which are positive, so we're just left with $ \boxed{73}$[/hide]


The traffic on a certain east-west highway moves at a constant speed of 60 miles per hour in both directions. An eastbound driver passes 20 west-bound vehicles in a five-minute interval. Assume vehicles in the westbound lane are equally spaced. Which of the following is closest to the number of westbound vehicles present in a 100-mile section of highway?

$\text{(A)} \ 100 \qquad \text{(B)} \ 120 \qquad \text{(C)} \ 200 \qquad \text{(D)} \ 240 \qquad \text{(E)} \ 400$

[hide="Answer"]If the opposite direction was not moving, he would have passed half as many cars - $10$. It was a five-minute interval, so he traveled $5$ miles. Therefore, in a $100$-mile interval, he would pass $10\cdot 20=200$ vehicles. $\boxed{C}$.[/hide]

[hide]If traffic is traveling at 60 mph in both directions, the reference frame can be equivalently set up such that he is traveling east-west at 120 mph and traffic west-east is standing still. If that is the case, he travels 10 miles in 5 minutes (120 mph * (5/60) hours = 10 miles) during which he sees 20 cars. There are 10 10 mile segments in a 100 mile journey, therefore (given that the cars are equally spaced), he will see 10*20 = 200 cars over the whole journey => B[/hide]

[hide="Click for solution"]
Since the vehicles moving in each direction are traveling at $ 60$ miles per hour, the eastbound traffic is actually traveling at $ 120$ miles per hour relative to the westbound traffic.  Thus, in five the eastbound driver will actually count the number of westbound vehicles in a ten-mile section of highway.  Since $ 20$ vehicles are in a ten-mile section, there will be $ 200$ vehicles in a $ 100$-mile section of highway, or $ \boxed{\textbf{(C)}}$.
[/hide]

eeshking - your solution is correct, however, 200 cars corresponds to $ \textbf{(C)}$, not $ \textbf{(B)}$.


limit
trigonometry
calculus
integration
function
calculus computations

What is the general method for evaluating statements such as $ \lim_{n \to \plus{} \infty} \frac{1}{n} \cdot\left( \sin \frac{\pi}{n} \plus{} \sin \frac{2\pi}{n} \plus{} \sin \frac{3\pi}{n} \plus{} ... \plus{} \sin \frac{n\pi}{n} \right)$ and $ \lim_{n \to \infty} \frac{1}{n} \cdot \left( \left( \frac{1}{n} \right)^{2} \plus{} \left( \frac{2}{n} \right)^{2} \plus{} \left( \frac{3}{n} \right)^{2} \plus{} ... \plus{} \left( \frac{n\minus{}1}{n} \right)^{2} \right)$.
They are obviously integrals whose intergrands are evident.  Is there a method to doing these?

If you accept that Riemann sums give definite integrals, write these as Riemann sums.  Are you asking for a proof that Riemann sums give definite integrals?

In general, if a function is Riemann integrable on the interval $ [0,1]$ then you have:

$ \lim_{n \rightarrow \plus{}\infty} \frac{1}{n} \sum^n_{i\equal{}1}f\left(\frac{i}{n}\right) \equal{} \int^1_0f(x)\,dx$

Thanks milin.  :)

I understand it now, it was just a way of interpreting Reinman sums, in a way.


number theory
greatest common divisor
number theory unsolved

Prove that $gcd(f_m, f_n) = f_{gcd(m, n)}$ where $f$ is fibonacci sequence.


How can I fill in a semicircle?  Specifically, I'm trying to recreate the picture attached with Asymptote.

OK let's see...

[asy]pair A=(-2,0), O=origin, C=(2,0);
path X=Arc(O,2,0,180), Y=Arc((-1,0),1,180,0), Z=Arc((1,0),1,180,0), N=X..Y..Z..cycle;
filldraw(N, black, black);
draw(reflect(A,C)*N);
draw(A--C, dashed);

label("A",A,W);
label("C",C,E);
label("O",O,SE);
dot((-1,0));
dot(O);
dot((1,0));[/asy]

There we go

[geogebra]9de88efdeadc3f327faf61cd9df1a0850b94b56d[/geogebra]

A geogebra picture displaying the same thing.


superior algebra
superior algebra solved

How can I show that any finite Boolean ring is isomorphic to $\mathbb{Z}_{2}\times \mathbb{Z}_{2}\times \mathbb{Z}_{2}\times ... \times \mathbb{Z}_{2}$?

[quote="grobber"]There are several proofs [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=boolean&t=45041]here[/url].[/quote]

harazi's proof is cute! :D

thanks for digging up the thread for me


geometry
3D geometry
tetrahedron
circumcircle
trigonometry
trig identities
Law of Sines

Prove that if the circumcircles of the faces of a tetrahedron $ABCD$ have equal radii, then $AB=CD$, $AC=BD$ and $AD=BC$.

(A picture might help follow)

Let <CAB=a, <ABC=b, <BCA=c, so that a+b+c=180. From Law of Sines, sin(BAC)=BC/(2R)=sin(BDC), so either <BDC=a or <BDC=180-a. Let's dispose of the second case.

Assume <BDC=180-a. We must also have <ADB=c or 180-c and <ADC=b or 180-b. We can't have both 180-b and 180-c, cause then at vertex D the three angles would add up to 360, i.e. vertex D would be flat. Furthermore, we cant have both b and c because then at vertex D, b+c=180-a, i.e. vertex D is flat again. So, WLOG, assume it is 180-b and c.

We must have (180-a)+c>(180-b), which (remembering that c=180-a-b) is equivalent to a<90. Likewise, b is acute. Also, we must have (180-a)+(180-b)+c<360, which is equivalent to c<90. Since <BDC=180-a and <ADC=180-b are obtuse, <DBC and <DAC are both acute, and since sin(DBC)=sin(DAC), the two angles are the same. Call their measure d. Also let <DCB=e and <DCA=f, which are both acute. Since (180-a)+d+e=180, e=a-d, and likewise, f=b-d. Let's look at triangle DAB.

If <DAB=e and <DBA=f, then c+(a-d)+(b-d)=180 implies d=0, which isn't possible. Also, if <DAB=180-e and <DBA=180-f, then triangle DAB has two obtuse angles, not possible. If <DAB=180-e and <DBA=f, then c+(180-e)+f=180 is equivalent to a=90, contradicting the fact that a is acute. So we have reached our contradiction, showing that <BDC=a and not 180-a.

In the same way, <ADC=b, <ADB=c, <BAD=e, and <ABD=f. We have 2(d+e+f)=(a+d+e)+(b+d+f)+(c+e+f)-(a+b+c)=180+180+180-180=360, implying d+e+f=180. Thus, f=180-d-e=a, and likewise c=d and b=e. So triangles BAD and DCB have the same angles and the same corresponding side BD, so they are congruent, so AB=CD. Likewise, AC=BD and AD=BC.

augh, i can't visualise these things  :(


Say I have a function f(x). Lets make it $ f(x)= 2x^{6}+9x^{5}+10x^{4}-13x-5$ just because its the first equation i see in my textbook  :|
Now, $ f^{\prime}=12x^{5}+45x^{4}+40x^{3}-13$ and $ f^{\prime}^{\prime}=60x^{4}+180x^{3}+120x^{2}$

Now, I know that if f(x) is concave up on I iff $ f''(x) \ge 0$ on I.
Now what?

Thank you in advance.

What is your question?  Are you looking for the set on which $ f$ is concave up?  If so then as you already calculated $ f'' \equal{} 60x^2(x \plus{} 1)(x \plus{} 2)$, which is non-negative iff $ (x \plus{} 1)(x \plus{} 2)\geq 0\Leftrightarrow x\in ( \minus{} \infty , \minus{} 2]\cup [ \minus{} 1,\infty)$.

Oh. I'm sorry...I forgot to write in the question... :oops:
I have to determine all the intervals on which the graph is concave up, concave down and I have to find all the inflection points.


[quote="silouan"]Could you post a solution dear Arqady ?[/quote]
The hint: still Cauchy! ;)

I don't catch it .....Could you help Arqady ??


analytic geometry
graphing lines
slope
function

....[hide]
OBJECTIVE: SLOPE FORMULA = $\frac{\Delta X}{\Delta Y}$

-consider y a function of x.

$\frac{y}{x}= \frac{f(x)}{x}$, but lets not stop yet.

true slope = $\frac{\Delta y}{\Delta x}= \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

because [i]y is a function of x[/i], a change in x will result in a change in y.
lets call this augmentation distance (which can be either negative or positive) $h$, so $x_{1}= x, x_{2}= x+h$.

Considering y a function of the value x, not just point x;
$y_{1}= f(x_{1}) = f(x)$
$y_{2}= f(x_{2}) = f(x+h)$

RESULTING IN:
$\frac{f(x+h)-f(x)}{x+h-x}$

$\frac{d(y)}{d(x)}= \frac{f(x+h)-f(x)}{h}$[/hide]

Please don't call that $\frac{dy}{dx}$. "d" has the specific connotation of something that is infinitesimally small, $\Delta$ is the right symbol.

lol sorry, i looked at wiki for the functions name or whatnot.

<-- precalc student.

but ah... bro?    :huh:

I think you'll need to tone it down on the HW prawbs.
It's hard to believe that anyone who frequents this forum can brainstorm such monotonous problems. That being said, [i]so many[/i].

But it seems to me that if you really are having trouble solving them, perhaps make one post that has ten questions. Not ten posts of one.

Thanks bro.


function
real analysis
real analysis unsolved

f is increasing and continuous on [0,1]. Let $ g$ be $ f^{\minus{}1}$
prove that
$ \sum_{i\equal{}1}^9  [f(\frac{i}{10})\plus{}g(\frac{i}{10})] < 9.9$

It is not true . For example: $ f(x) \equal{} 100x$
Must be $ f(x): [0,1]\to\ [0,1]$

oh yup!! im sorry :blush:  must be from [0,1]-->[0,1]


Which of the following describes the behavior of y = cubicroot(x + 2) at x = -2

(A) differentiable
(B) corner
(C) cusp
(D) vertical tangent
(E) discontinuity

well i graphed the function, and i'm not sure.. i know for sure it isn't E... because f(-2) = 0... it has a value..  uhmm.. as for the rest i'm not sure.. i don't even know what a corner and a cusp is.

I don't know either. It's not A, D or E for sure, so it rests on what the heck corners or cusps are.

ok i figured out that it isn't A, it isn't C, it is D, and it isn't E... i don't know about B... that's the only one i need help with

Oh... well I guess it's B then.  :blush:

And exactly how did you think you ruled out (D), vertical tangent?

We have $y=\sqrt[3]{x+2},$ which we can rewrite as $x=y^3-2.$

We're interested in the behavior of this curve at the point $(-2,0).$

Instead of finding $\frac{dy}{dx},$ how about $\frac{dx}{xy}$ ?

$\frac{dx}{dy}=3y^2.$ At the point $(-2,0),\,\frac{dx}{dy}=0.$

That's a vertical tangent.

I said I made a mistake, but then I made a mistake in my mistake-post. Jeez.


geometry
ratio
IMO
concurrence
Triangle
IMO 2005

Solution:
[hide]
I will assume that the verices lie in the orders
$BA_1A_2C$
$CB_1B_2A$
$AC_1C_2B$

Claim: $C_2B=A_2C$
Proof: Suppose that $C_2B>A_2C$
This implies $BA_1<CB_1 \implies$
$AB_2<CA_2<BC_2 \implies$
$AC_1>CA_2>BC_2 \implies$
$AC_1+C_1C_2+C_2B>BA_1+A_1A_2+A_2C$
a contradiction.

By the same reasoning,
$A_2C=B_2A=BC_2$
and
$BA_1=AC_1=CB_1$

It follows that $\triangle AC_1B_2 = \triangle CB_1A_2 = \triangle BA_1C_2$, and a rotation of $2\pi/3$ radians around the center of $\triangle ABC$ brings one triangle to another.

It follows that $\triangle C_1B_1A_1$ and $\triangle B_2A_2C_2$ are equilateral and centered at the center of $\triangle ABC$

Consider the perpendicular bisectors of the sides of $\triangle A_1B_1C_1$ Clearly $A_2$, $B_2$, $C_2$ lie on these perpendicular bisectors, and
$B_2$ lies on the perpendicular bisector of $B_1C_1$, etc.
Since $A_1,B_1,C_1$ also lie on these perpendicular bisectors, we have proved that $A_1B_2,B_1C_2,C_1A_2$ are the perpendicular bisectors of the sides of $\triangle A_1B_1C_1$, and thus are concurrent at the center of $\triangle ABC$[/hide]
EDIT: The argument I use to prove the first claim is wrong--it fails when the hexagon has acute angles. Many other solutions in this thread have similar flaws, see Darij's post later in the thread: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=44478&postorder=asc&start=26

Please tell me if this hunch is on the right track:
[hide]Show that the incircle of ABC is tangent to every vertex of the hexagon, then use Brianchon's Theorem?[/hide]

Using schulmannerism's claim, one can also easily finish Ceva's Theorem using ratio of areas (half of product of two sides and the sine of the angle in between.)  :)

Strange why most participants I have met used inequalities to solve this problem. My solution, at least, doesn't explicitely make use of them (of course, it uses the arrangement of the points, which can be considered as a kind of inequalities, but I guess there is no way to avoid using it...).

So here is, depending of how soon we will quit the internet cafe, my solution or just a rough outline of it:

Since the hexagon $A_1A_2B_1B_2C_1C_2$ has equal sidelengths, we have $C_2A_1=A_1A_2$; thus, the triangle $C_2A_1A_2$ is isosceles, so its base angle is $\measuredangle A_1C_2A_2=\frac{180^{\circ}-\measuredangle C_2A_1A_2}{2}=\frac{\measuredangle C_2A_1B}{2}$. Similarly, $\measuredangle C_1A_1C_2=\frac{\measuredangle A_1C_2B}{2}$. But by the sum of angles in triangle $C_2BA_1$, we have $\measuredangle C_2A_1B+\measuredangle A_1C_2B=180^{\circ}-\measuredangle C_2BA_1=180^{\circ}-\measuredangle ABC$. Finally, < ABC = 60°, since the triangle ABC is equilateral. Hence, if the lines $C_1A_1$ and $C_2A_2$ intersect at a point K, the exterior angle theorem in triangle $C_2KA_1$ yields

$\measuredangle A_2KA_2=\measuredangle A_1C_2K+\measuredangle KA_1C_2=\measuredangle A_1C_2A_2+\measuredangle C_1A_1C_2$
$=\frac{\measuredangle C_2A_1B}{2}+\frac{\measuredangle A_1C_2B}{2}=\frac{\measuredangle C_2A_1B+\measuredangle A_1C_2B}{2}$
$=\frac{180^{\circ}-\measuredangle ABC}{2}=\frac{180^{\circ}-60^{\circ}}{2}=60^{\circ}$.

So the lines $C_1A_1$ and $C_2A_2$ intersect at an angle of 60°. Similarly, the lines $A_1B_1$ and $A_2B_2$ intersect at an angle of 60°, and so do the lines $B_1C_1$ and $B_2C_2$. Hence, the triangle $A_2B_2C_2$ is the image of the triangle $A_1B_1C_1$ under a spiral similitude with rotation angle 60°. [On the exam, I avoided speaking of "angles between lines" without properly defining the orientation of these angles, and thus I rewrote the above argument in a non-transformational way.]

Now, let Z be the center of the spiral similitude mapping the triangle $A_1B_1C_1$ to the triangle $A_2B_2C_2$. Then, since the rotation angle of the similitude is 60°, the triangles $ZA_1A_2$, $ZB_1B_2$, $ZC_1C_2$ are all directly similar, and have the angles $\measuredangle A_1ZA_2=\measuredangle B_1ZB_2=\measuredangle C_1ZC_2=60^{\circ}$. Actually, since $A_1A_2=B_1B_2=C_1C_2$ (as the hexagon $A_1A_2B_1B_2C_1C_2$ has equal sidelengths), these triangles are even congruent. Thus, for instance, $ZA_1=ZB_1$. Together with $A_1A_2=B_1A_2$ (again since the hexagon $A_1A_2B_1B_2C_1C_2$ has equal sidelengths) and $ZA_2=ZA_2$, the triangles $ZA_1A_2$ and $ZB_1A_2$ are congruent, so that $\measuredangle B_1ZA_2=\measuredangle A_1ZA_2$. Since $\measuredangle A_1ZA_2=60^{\circ}$, we thus have $\measuredangle B_1ZA_2=60^{\circ}$. Hence,

$\measuredangle A_1ZA_2+\measuredangle B_1ZA_2+\measuredangle B_1ZB_2=60^{\circ}+60^{\circ}+60^{\circ}=180^{\circ}$.

And thus, the point Z lies on the line $A_1B_2$. Similarly, the point Z lies on the lines $B_1C_2$ and $C_1A_2$. And we are done.

Oh my goodness, I guess 75% of my above argumentation were redundant. But actually, the whole thing took me 25 minutes, and I found it too much (I expected a geometry problem 1 to be doable in 10 minutes, like the one last year, but this one was harder), so I did not try to simplify but rather wrote it up as quickly as possible to have the time for the other two, especially since problem 2 seemed to be absolutely hard :D . Anyway, a good first day, although not for me, since I could have done the third one...

  Darij

The whole point of the problem is to prove that $A_1 B_1 C_1$ and $ A_2 B_2 C_2$ are equilateral.
To prove that let angle $C_1 B_1 B_2 = b$, $C_2 C_1 A_1 = c$ and $A_1 B_1 A_2 = a$.
Then we easily get : $ A_1 C_1 B_1 = 60 +b-c$, $ C_1 A_1 B_1 = 60 + c-a$ and $ A_1 B_1 C_1 = 60 + a-b$.
We can also prove in an easy way that : $ |a-b| < 30 $... and $ a<60$ ... respectivly.
Then by the cosine rule to the triangle $C_1 C_2 B_1$ we get that : $(C_1 B_1)^2 = (2x cos(b))^2 $ ...
By the sine rule to the trianlge $A_1 B_1 C_1$ and some calulation, we get the following system of equaltions :
(1) $ sin(a+b-c)\cdot cos(60 +b-a)=\frac{1}{2} \cdot sin(c) $
(2) $ sin(b+c-a)\cdot cos(60 +c-b)=\frac{1}{2} \cdot sin(a) $
(3) $ sin(a+c-b)\cdot cos(60 +a-c)=\frac{1}{2} \cdot sin(b) $
Now, if $a=b$ we easily get from (1) that $a=b=c$.
Let , wlog, that $a>b$. then from (1) we get that $sin(c) < sin(a+b-c)$ or $ 2c<a+b$ so $c<b$.
If $a>c$ then similarly from (3) $ 2b <a+c$ . Contradiction.
If $b>c>a$ then from (2) $sin(b+c-a) > sin(c)>sin(a)$ so $\frac{1}{2} > cos(60+c-b)$ so $60+c-b > 60$ or $c>b$ . Contradiction.
So $a=c$ but then from (3) we get $a=b=c$. Contradiction.
At last, we proved that $a=b=c$.
But then $A_1 B_1 C_1$ is equilateral and angle $C_1 C_2 A = B_1 A_2 C = A_1 C_2 B$. Simirly for the triangle $A_2 B_2 C_2$ and finally it is obvious that triangles $A C_1 B_2 = B A_1 C_2 = C A_2 B_1$.
The rest follows easily.

it seemed to me that this problem is harder than the 2nd, but only because I was looking for a more complicated solution  :huh:   :)

denote |A(1)A(2)|=d, |AB|=a, |A(2)C|=x, |B(2)A|=y, |C(2)B|=z.
let's assume x>y, then a-d-y>a-d-x, and from triangles BA(1)C(2) and CB(1)A(2) follows that z>x, then a-d-x>a-d-z and similarly from the 2 triangles BA(1)C(2) and AC(1)B(2) it follows that y>z.
So we have x>y>z>x, a contradiction. Similarly for x<y. So x=y and by symmetry x=y=z.
Hence A(1)B(1)C(1) and A(2)B(2)C(2) are equilateral and the lines from the problems are their heights (medians, bisectors.. ) and all meet in one point.

Some say it is not an IMO problem? Well, I agree, since even I am able to solve it.

My solution to problem one:

[color=blue]Lemma(Poland 1965--1966)
If the diagnals AD, BE, CF half the area of the convex hexagon, then AD, BE and CF are concurrent.[/color]
Proof: Assume the do not meet at one point.
Let's say $O= AD \cap BE$, $P= CF \cap BE$, $Q= AD \cap CF$
Since AD, BE, CF half the area, so:
\[\triangle{AOB}=\triangle{DOE}, \triangle{EFP}=\triangle{BCP}, \triangle{CQD}=\triangle{AQF}\]
then:

$1=\frac{\triangle{AOB}}{\triangle{DOE}} \cdot \frac{\triangle{EFP}}{\triangle{BCP}} \cdot \frac{\triangle{CQD}}{\triangle{AQF}}= \frac{AO \cdot BO}{DO \cdot EO} \cdot \frac{EP \cdot FP}{BP \cdot CP} \cdot \frac{CQ \cdot DQ}{AQ \cdot FQ}$

$= \frac{AO}{AQ} \cdot \frac{BO}{BP} \cdot \frac{CQ}{CP} \cdot \frac{DQ}{DO} \cdot \frac{EP}{EO} \cdot \frac{FP}{FQ} $

$ \neq 1$

Hence they must be concurrent.

Now get back to this problem.

Since $C_1C_2 = B_1B_2= A_1A_2$, and $A=C=B=60^o$, so we shall have:
$R_{BA_1C_2}=R_{AC_1B_2}=R_{CA_2B_1}$
($R$ is the circumradius.)

Assume $A_2C > B_2A$. Since $BA_1+A_2C=CB_1+B_2A$, so $BA_1 < CB_1$.

Since $2R=\frac{BA_1}{\sin{BC_2A_1}}=\frac{CB_1}{\sin{B_1A_2C}}$

So $\angle{BC_2A_1} < \angle{CA_2B_1}$
So $ \angle{BA_1C_2} > \angle{CB_1A_2}$

Use the law of sines again, we get $BC_2 > CA_2$.
So $BC_2 > B_2A$. Hence $\angle{BA_1C_2} > \angle{AC_1B_2}$,
Hence $\angle{BC_2A_1} < \angle{AB_2C_1}$
Hence $BA_1 < AC_1$.
Because we already have $C_2B > CA_2$, so add them togather we have $AB-C_1C_2 > BC- A_1A_2$.

It is impossible.

By symmetry, we can know that $CA_2=BC_2=AB_2$, $A_1B=C_1A=B_1C$.

The rest is easy. We can prove that
$\triangle{AC_1B_2}=\triangle{CB_1A_2}=\triangle{BA_1C_2}$
$\triangle{A_1C_1C_2}=\triangle{B_1A_1A_2}=\triangle{C_1B_1B_2}$
$\cdots \cdots \cdots$
It is easy to prove that the diagnols half the area of the hexagon.
Use our [color=blue]lemma[/color] now...

Cheers!
How much point can this proof recieve?

I shall also add a pic for the lemma.
P.S. How do I add two pictures in one post?

This question is very easy.
Step 1: prove AC1=BA1=CB1.

          Shift AC1 to A'C'', C2B to C''B', AB2 to A'B'', B1C to B''C'. Then it is easy to show that  B'A'' and A''C' are also shifted BA1 and A2C.  Since triangle A'B'C' and A''B''C'' are both  equilateral triangles, A'C''=B'A''=C'B'' => AC1=BA1=CB1

           Step 2: A2C1 is the middle line of C2B2 and B1A1 and  so on. It  is easy.

           Step 3: triangle B2C2A2 is an equilateral triangle so the three lines meet.

I have beautiful solution:

[tex]I[/tex] is point inside convex hexagon [tex]A_1A_2B_1B_2C_1C_2[/tex] such that [tex]IA_1A_2[/tex] is regular triangle. It's easy to prove that [tex]IC_1B_2[/tex] is regular triangle. Let the lines $A_1B_2$ and $A_2C_1$ meet at $M$. So [tex]I, M, B_1[/tex] are collinear (because they lie on the perpendicular bisector of segment [tex]B_2A_2[/tex]); and [tex]I, M, C_2[/tex] are collinear (because they lie on the perpendicular bisector of segment [tex]A_1C_1[/tex]). So [tex]C_2, M, B_1[/tex] are conlinear or [tex]A_1B_2, B_1C_2, C_1A_2[/tex] are concurrent.

[color=red][Moderator edit: Corrected some language mistakes and defined point M.][/color]

I think only Darij 's proof and Anto 's proof are valuable .
  Check your proofs again ! Something is vague .

check out this one:

as the three vectors A1A2, B1B2, C1C2 are similar (in the same scale) to BC, AC, AB respectively,
we have A1A2+B1B2+C1C2=0, so A2B1+B2C1+C2A1 is also a zero vector,
so we can build an equilateral triangle from vectors A2B1, B2C1, C2A1, which has its sides parallel
to the sides of the triangle made by the prolongations of A2B1, B2C1, C2A1,
so this triangle is also equilateral (denote its vertices as A',B',C').

as we have angle B1B'C1=60=B1AC1, B1B'AC1 is cyclic, so angle B'B1A=B'C1A, so the angles
B1 and C1 of the hexagon are equal.

analogically we can prove that the angles of the hexagon A1=B1=C1=alpha, A2=B2=C2=beta,
so A1B2, B1C2, C1A2 are the axes of symmetry of the hexagon....

Steps of the proof:

1 - Construction of the hexagon.
2 - Show that [tex]A_1B_1C_1[/tex] and [tex]A_2B_2C_2[/tex] are equilateral triangles.
3 - Show that the point of concurrence is the circuncenter of the triangle [tex]ABC[/tex].

1 - It's easy to see that we can construct the hexagon described by the problem by the following manner:

Draw the equilateral triangle and its circuncircle. Choose other three points on the circuncircle and draw another equilateral triangle, different from [tex]ABC[/tex], named [tex]\overline{A}\overline{B}\overline{C}[/tex]. The points where the triangles concur are exactly the vertices of the hexagon.

2 - By marking some anlgles and with the known fact of congruence, it's very easy to see that the triangles [tex]A_1B_1C_1[/tex] and [tex]A_2B_2C_2[/tex] are equilateral.

3 - Call [tex]P[/tex] the circuncenter of the triangle [tex]ABC[/tex]. Draw [tex]AP,C_1P, B_1P[/tex] and [tex]\angle{AC_1B_2}=\alpha[/tex]. We know that [tex]\overline{C_1P}[/tex] bissects [tex]\angle{B_2C_1C_2}[/tex], and then [tex]\angle{B_2C_1P}=\frac{\pi-\alpha}{2}[/tex] and using the fact that [tex]C_1A\overline{A}B_1[/tex] are concyclic e conclude that [tex]\angle{AB_1P}=\frac{\pi-\alpha}{2}[/tex], since [tex]\angle{AC_1P}+\angle{AB_1P}=\pi[/tex] then, this quadrilateral is cyclic to, which proves that [tex]\overline{PC_1}=\overline{PB_1}[/tex]. By the same way, you prove the same for the other segment, which makes us conclude that [tex]P[/tex] is the circuncenter of [tex]A_1B_1C_1[/tex]. Analogously, you can prove the [tex]P[/tex] is circuncenter of [tex]A_2B_2C_2[/tex]. Since we know then that [tex]P[/tex] belong to the mediatrice of the equilateral triangle, we conclude that the segments concur in [tex]P[/tex]

If you find any mistake, I'll be glad to be noticed =]

thanks =]

a sketch of a very weird proof where i missed like 3 different places i could have been done earlier!

[asy]
import olympiad;
    size(10 cm);
    defaultpen(fontsize(10pt));
    pair A = dir(330);
    pair B = dir(90);
    pair C = dir(210);
    draw(A--B, black);
    draw(B--C, black);
    draw(A--C, black);
    dot("$A$", A, dir(330));
    dot("$B$", B, dir(90));
    dot("$C$", C, dir(210));
    pair A_1 = B + dir(240)*0.5;
    dot("$A_1$",A_1, dir(180));
    pair C_1 = A + dir(120)*0.5;
    dot("$C_1$",C_1, dir(330));
    pair B_1 = C + dir(0)*0.5;
    dot("$B_1$",B_1, dir(270));
    draw(A_1--B_1--C_1--A_1,red + dashed);
    pair A_2 = extension(midpoint(A_1--B_1),circumcenter(A_1,B_1,C),B,C);
    pair B_2 = extension(midpoint(B_1--C_1),circumcenter(B_1,C_1,A),A,C);
    pair C_2 = extension(midpoint(A_1--C_1),circumcenter(A_1,C_1,B),A,B);
    dot("$A_2$",A_2,dir(0));
    dot("$B_2$",B_2,dir(270));
    dot ("$C_2$", C_2,dir(0));
    draw(A_2--B_2--C_2--A_2,green + dashed);
    draw(A_1--A_2--B_1--B_2--C_1--C_2--A_1, blue + opacity(2));
    draw(B_1--C_2,cyan+dotted);
    draw(B_2--A_1,cyan+dotted);
    draw(C_1--A_2,cyan+dotted);
    pair O_1 = extension(B_1,C_1,C_2,B_2);
    pair O_2 = extension(A_1,C_1,C_2,A_2);
    pair O_3 = extension(A_2,B_2,B_1,A_1);
    dot("$O_1$",O_1,dir(0));
    dot("$O_2$",O_2,dir(0));
    dot("$O_3$",O_3,dir(0));
    pair Q = extension(A_1,B_2,C_1,A_2);
    dot("$Q$",Q,dir(0));
    draw(circumcircle(O_2,C_2,C_1),orange+dotted);
    draw(circumcircle(O_2,A_2,A_1),orange+dotted);
    draw(circumcircle(B,A_1,C_2),purple + dotted);
[/asy]

let $O_2$ be the intersection of $C_1A_1$ and $A_2C_2$ and let $Q$ be the miquel point of quad $BA_1C_2O_2$.

claim: $BA_1C_2O_2$ cyclic
proof: angle chase to get $C_1O_2C_2$ is $60^{\circ}$. (let $\angle A_1A_2C_2$ be $x$, and write $\angle C_1C_2A_2$ and $\angle A_1C_1C_2$ in terms of $x$)

now $Q$ lies on $A_2C_1$ by well known miquel point properties

now angle chase to get $A_2B_2C_2$ similar to $A_1B_1C_1$ (let $\angle A_1A_2C_2$ be $x$, $\angle A_2B_2B_1$ be $y$, and $\angle C_1B_2C_2$ be $z$ and just spam angles until you find)

now let $Q'$ be the center of the spiral sim sending the red triangle to the blue triangle, clearly it exists since any two directly similar figures have a spiral sim between them
note that if the spiral sim sends line $A_2C_2$ to $A_1C_1$, then it must lie on $(A_1O_2A_2)$ and $(C_2O_2C_1)$ by well known spiral sim lemma. but this coincides with $Q$, therefore $Q'$ lies on $A_2C_1$, and similarly, it lies on $B_2A_1$ and $B_1C_2$, so we are done.

remark: similar to 2020 USEMO 5; main motivation for my proof was that problem. also i have no clue how i missed the equilateral triangles but whatever

Let $O$ be the center of $\triangle ABC$. Consider the unit vectors $A_1A_2,A_2B_1,B_1B_2,B_2C_1,C_1C_2,C_2A_1$. Since $\vec{A_1A_2}+\vec{B_1B_2}+\vec{C_1C_2}$, we also have $\vec{A_2B_1}+\vec{B_2C_1}+\vec{C_2A_1}=0$ so these vectors are all at $120^\circ$ to each other. Then $\angle B_1A_2C=\angle A_1C_2B$, and a similar equality (and the properties of the hexagon) imply that $\triangle B_1A_2C\cong\triangle A_1C_2B$. Thus rotation about $O$ by $120^\circ$ yields the same diagram with different point labels. Then $O$ is the Miquel point of $C_2B_2C_1B_1$. This yields $\triangle OC_2C_1\cong\triangle OB_2B_1$ thus $OC_2=OB_2$ and $OC_1=OB_1$. Similar information yields that $O$ is both the circumcenter of $\triangle A_1B_1C_1$ and of $\triangle A_2B_2C_2$. Now it suffices to show that $C_2B_1$ is the perpendicular bisector of $B_2A_2$ and symmetric results. Indeed, $B_2B_1=A_2B_1$ by the givens and $\triangle B_2OC_2\cong\triangle C_2OA_2$ because $C_2O=B_2O=A_2O$ and $\angle B_2OC_2=120^\circ=\angle C_2OA_2$.

[quote=Valentin Vornicu]Six points are  chosen on the sides of an equilateral triangle $ABC$: $A_1$, $A_2$ on $BC$, $B_1$, $B_2$ on $CA$ and $C_1$, $C_2$ on $AB$, such that they are the vertices of a convex hexagon $A_1A_2B_1B_2C_1C_2$ with equal side lengths.

Prove that the lines $A_1B_2$, $B_1C_2$ and $C_1A_2$ are concurrent.

[i]Bogdan Enescu, Romania[/i][/quote]

I found this problem to be pretty interesting.
Let $A_1, B_1, C_1$ to be the point closer to $B$, $C$ and $A$ respectively.

Take $X$ to be a point inside the hexagon such that $\triangle XA_1 A_2$ is equilateral. We know that $XA_1 \parallel C_1 C_2$ and $XA_2 \parallel B_1 B_2$. Therefore by the condition of the problem, we get that $XA_1 C_2 C_1$ and $XB_2 B_1 A_2$ are both rhombus. This gives us $XB_2 = B_2 B_1 = C_1 B_2 = C_2 C_1 = C_1 X$, and therefore we conclude tat $\triangle XC_1 B_2$ is equilateral.

[b][color=#f00]Claim.[/color][/b] $\triangle A_1 B_1 C_1$ and $\triangle A_2 B_2 C_2$ are equilateral.
[i]Proof.[/i] Let $A_1 C_2 \cap A_2 B_1 = A^*$. Since $A_1 C_2 \parallel XC_1$ and $A_2 B_1 \parallel XB_2$, we have \[ \angle(A_1 C_2, A_2 B_1) = \angle(XC_1, XB_2) = 60^{\circ}. \]
Therefore we conclude that $\angle C_2 A^* B_1 = 60^{\circ} = \angle C_2 BA_2$, and therefore $C_2 BA^*A_2$ cyclic.
Define $B^*, C^*$ similarly, we could get $A^* A_1 B_1C$, $CA_2 B_2 B^*, AC_1 B_1 B^*, BA_1 C_1 C^*, AB_2 C_2 C^*$ all being cyclic.
This is enough to forces $\angle BA_1 C_2= \angle CB_1 A_2 = \angle AC_1 B_2$, and therefore, we conclude that $\triangle AC_1 B_2, \triangle BA_1 C_2, \triangle CB_1 A_2$ are all congruent. This is enough to forces $\triangle A_1 B_1 C_1, \triangle A_2 B_2 C_2$ are equilateral.
--------------
Now, we conclude that $A_1 C_1 B_2 B_1$ is a kite, and therefore, $A_1 B_2$ bisects $B_1 C_1$. Similarly, $B_1 C_2$ bisects $C_1 A_1$ and $C_1 A_2$ bisects $A_1 B_1$. Hence, $A_1 B_2, B_1 C_2, C_1 A_2$ concur at the centroid of $\triangle A_1 B_1 C_1$.

Excellent problem! Consider what happens when we construct the point $P_A$ such that $B_2B_1A_2P_A$ is a parallelogram. We know that $\overrightarrow{A_2B_1} = \overrightarrow{P_AB_2}$.

Next, observe that $\overrightarrow{A_1A_2} + \overrightarrow{B_1B_2} + \overrightarrow{C_1C_2} = 0$ since they are the same magnitude and are rotations of $120^\circ$ from each other(they form an equilateral triangle).

This implies that $\overrightarrow{C_1B_2}$, $\overrightarrow{B_2P_A}$, and $\overrightarrow{A_1C_2}$ should form an equilateral triangle too, implying that $P_A$ is also the point that makes $C_1C_2A_1P_A$ a parallelogram.

Now since $P_AA_2 = B_2B_1 = A_1A_2$ and so forth, we know that $\triangle A_1A_2P_A$ and $\triangle P_AB_2C_1$ are equilateral. Consequently, $P_A$ is the center of a circle $(B_2C_1A_1A_2)$, and we finish by radical axis applied to $(B_2C_1A_1A_2)$, $(A_1C_2B_1B_2)$, and $(A_2B_1C_1C_2)$.

Notice that $$\overrightarrow{A_1A_2}+\overrightarrow{A_2B_1}+\overrightarrow{B_1B_2}+\overrightarrow{B_2C_1}+\overrightarrow{C_1C_2}+\overrightarrow{C_2A_1}=0$$ as these vectors form hexagon $A_1A_2B_1B_2C_1C_2$. However, $$\overrightarrow{A_1A_2}+\overrightarrow{B_1B_2}+\overrightarrow{C_1C_2}=0$$ as well due to these three vectors having the same magnitude and being rotated $120$ degrees from each other. Therefore $$\overrightarrow{A_2B_1}+\overrightarrow{B_2C_1}+\overrightarrow{C_2A_1}=0$$ which means that these three vectors are also rotated $120$ degrees from each other. This means that $\overrightarrow{B_1C_1}=\overrightarrow{B_1B_2}+\overrightarrow{B_2C_1}$ is rotated $120$ degrees from $\overrightarrow{A_1B_1}=\overrightarrow{A_1A_2}+\overrightarrow{A_2B_1}$. A similar thing applies to $C_1A_1$, and therefore $A_1B_1C_1$ form an equilateral triangle. Then $A_1B_2,B_1C_2,C_1A_2$ are simply the altitudes of $A_1B_1C_1$, and thus concur at the center of $A_1B_1C_1$.

Let $s$ be the side length of hexagon $A_1A_2B_1B_2C_1C_2$. Let $D, E, F$ be the points such that $DC_1C_2$, $EB_1B_2$, and $FA_1A_2$ are equilateral triangles. There are many observations to be made.

[color=#3D85C6][b]Claim: [/b][/color] $DA_2B_1$, $EC_2A_1$, and $FB_2C_1$ are equilateral triangles.

[b][color=#3D85C6]Proof: [/color][/b] I will just prove $FB_2C_1$ is equilateral as the other cases are symmetrical.
Notice that $A_1F \parallel C_2C_1$ and $C_1C_2 = FA_1$, so $FA_1C_2C_1$ is a parallelogram and $A_1C_2 = FC_1 = s$. Also, $A_2F \parallel B_1B_2$ and $A_2F = B_1B_2$, so $FA_2B_1B_2$ is a parallelogram and $A_2B_1 = FB_2 = s$. Therefore $FB_2C_1$ is equilateral as desired.

With this claim, notice that $D, E, F$ are the circumcenters of $(C_1C_2A_2B_1), (B_1B_2C_2A_1), (A_1A_2B_2C_1)$. The result follows from the Radical Axis Theorem on these $3$ circles.

[b][color=#f00]Remarks: [/color][/b] This problem is so beautiful and uses equilateral triangles in such a nice way. It's unfortunate that this is a IMO #1 and definitely should've been at least a 2 or 5.

I think this problem is underrated... anyway :
Let A2B2 and A1B1 meet at X, B1C1 and B2C2 meet at Y, A2C2 and A1C1 meet at Z.

Step1 : CB1XA2 and AB2YC1 and BA1ZC2 are cyclic.
∠CA2B1 + ∠CB1A2 = 120 so ∠A2A1B1 + ∠A2B2B1 = 60 so ∠CB1X + ∠CA2X = 180 so CB1XA2  is cyclic.
we prove the rest with same approach.

Step2 : A2A1C1B2 and A1C2B2B1 and A2C2C1B1 are cyclic.
By simple angle chasing we have XC = XA1 = XB2 so X is center of CA1B2 so ∠XA1B2 = ∠XB2A1 = 30.
with proving that Y and Z are also centers we'll have plenty of 30 angles which they give us what we wanted.

Now by radical axis theorem we have A1B2 and B1C2 and C1A2 are concurrent.
we're Done.

Feels easy for even $P1$.
Let $\angle B_2B_1C_1=\angle B_2C_1B_1=\alpha , \angle C_1B_1C_2=\theta \implies \angle C_1B_2C_2=\angle C_1C_2B_2=60-\alpha, \angle B_2C_2B_1=60-\theta$.

So $\frac{\sin\theta}{\sin(60+\alpha)}=\frac{C_1C_2}{B_1C_2}=\frac{B_1B_2}{B_1C_2}=\frac{\sin(60-\theta)}{\sin(120-\alpha)} \implies \sin(\theta)=\sin(60-\theta) \implies \theta=30$.

 So we get $\angle B_2C_2B_1=\angle B_2A_1B_1=30 \implies B_1B_2A_1C_2$ is cyclic. Similarly $A_1A_2B_2C_1$ and $C_1C_2B_1A_2$ are cyclic. Since $A_1B_2, B_1C_2, C_1A_2$ are radical axises of these $3$ cirlces we get the result.

Hmm wait, is it just me or is this a new solution...
Let the perpendicular-bisector of $\overline{A_1C_1}$ and $\overline{A_2B_2}$ meet at $T$. Then notice that $TC_1 = TA_1$, and $TA_2 = TB_2$. Moreover notice that $\triangle TA_1C_2\equiv TC_1C_2$, $\triangle TA_1A_2\equiv \triangle TC_1B_2$, $\triangle TA_2B_1\equiv \triangle TB_2B_1$, all by SSS. Hence we have that $C_2,T,B_1$ are collinear. Since $C_1A_1, B_2A_2$ both perpendicular to this line, $A_1C_1\parallel A_2B_2$. Do this similar for the other two symmetries, then finish by Desargue's Theorem on perspective triangles.

A very bold solution, not sure if the second paragraph works. If not, simple congruent triangles does the trick.

$A_1A_2B_1B_2C_1C_2$ is a closed polygon, and you can translate $A_1A_2,B_1B_2,C_1C_2$ to form a closed polygon, namely an equilateral triangle. Thus, by vectors you can translate $A_1C_2,C_1B_2,B_1A_2$ to form a closed polygon. Since the lengths are equal this closed polygon is equilateral.

Extend $A_1C_2,C_1B_2,B_1,A_2$ to form another equilateral triangle, and we see that this one and $ABC$ are simply reflections of each other. Note that $A_1B_2,B_1C_2,C_1A_2$ are both possible symmetry lines of the union of the two equilateral triangles, and so they must be concurrent (at the center of gravity).

Solved with [b]pi271828[/b].

We use vectors; note that $\overrightarrow{A_1A_2}+\overrightarrow{B_1B_2}+\overrightarrow{C_1C_2}=0$ which implies $\overrightarrow{A_1C_2}+\overrightarrow{B_1A_2}+\overrightarrow{C_1B_2}=0$ meaning that $A_2B_1,B_2C_1,C_2A_1$ form $60^{\circ}$ angles with each other. Extend $A_1C_2$ and $A_2B_1$ to intersect at $A_3$ and define $B_3$ and $C_3$ similarly. Then note that $\triangle BA_1C_2\cong \triangle A_3A_1A_2\cong \triangle CB_1A_2$ so we easily obtain that $A_1B_1C_1$ and $A_2B_2C_2$ are equilateral. It follows that they share a common circumcenter, implying that all six angle bisectors concur, so we are done by Brianchon's.

[hide=solution synthetically on my own]Anyways this could be bad for me in the future but with a good diagram (and it's not hard to create, this one has no clutter) we can suspect cyclic quads AB_2DC_1 and cyclically, where D,E,F are the intersections of B_1C_1 with B_2C_2, etc. To prove this, it suffices to show that C_1DC_2 is 60 degrees. Indeed, by basic angle chasing we find that for $C_1B_1B_2=x$, $$AC_1B_2=AC_1B_1-B_1C_1B_2=180-60-x-x=120-2x\implies C_1B_2C_2=60-x\implies C_1DB_2=180-x-(60+x)=120,$$ so opposite angles are supplementary. Then$$C_2AD=C_1B_2D=C_1C_2B_2\implies AD=C_2D; \ldots$$ This implies that since D,E,F are circumcenters, $DC_2B_1=\frac{180-2A}{2}=30=FA_1B_2,$ which means that $A_1B_1B_2C_2$ is cyclic and other variants, whence radical axis with radical center that point of concurrence finishes. (We're motivated to find angles after that cyclicity, also a common way of concurrence is radical center so we look for circles of four points since they need two points shared with each of the two other circles, from which it is obvious which points to try to prove, especially with a good diagram.)[/hide]

[hide=sol with vectors, gotten from a hint] I had to get help from someone because I didn't realize you could even say that (all "v"s are vectors) v_1=v_2+v_3 and v_4=v_5+v_6, if v_2 is 120 degree orientation of v_5 and so is v_3 with v_6 then v_1 with v_4 are also 120 degree orientation difference. Again, I think any student with good knowledge of vectors and basic geometry does not need to know any contest prep for this.

We have that $$A_1A_2+A_2B_1+B_1B_2+B_2C_1+C_1C_2+C_2A_1=0$$ and $$A_1A_2+B_1B_2+C_1C_2=0\iff A_2B_1+B_2C_1+C_2A_1=0$$ from them being 120 degrees difference in orientation (idk if rotation describes it accurately, but from here [i]rotation[/i] just means difference in orientation). Then this implies that $B_1C_1=B_1B_2+B_2C_1$ is a 120 degree rotation from $A_1B_1=A_1A_2+A_2B_1$ by comparing each addend respectively. Analogously for $A_1C_1$, we conclude that $A_1B_1C_1$ is an equilateral triangle, whence from kite $A_2B_1C_1A_1$ we get that the diagonals are perpendicular, namely $A_2C_1$ is an (extended) altitude of equilateral triangle, hence all of the diagonals of the hexagon in question concur at the center of this equilateral triangle. $\blacksquare$[/hide]

Notice that $\overrightarrow{A_1A_2} + \overrightarrow{B_1B_2} +\overrightarrow{C_1C_2} = 0$ is the sum of three unit vectors that are $120^\circ$ rotations of each other. It follows that $\overrightarrow{C_2A_1} + \overrightarrow{A_2B_1} + \overrightarrow{B_2C_1} = 0$ is the sum of three unit vectors; these three unit vectors then must form an equilateral triangle. In particular, the lines $\overline{C_2A_1}$, $\overline{C_1B_2}$, $\overline{A_2B_1}$ form an equilateral triangle $\Delta$, say $\delta_A \delta_B \delta_c$.

We can show that $\triangle C_1\delta_C C_2 \cong \triangle A_1BC_2$ and cyclic permutations by AAS congruence. Then $B \to \delta_C$ and $C \to \delta_B$ under reflection about $\overline{B_1C_2}$, and it follows that $\overline{BC} \to \overline{\delta_B \delta_C}$ under this reflection. It then follows that this reflection sends $A_1C_1B_1A_2$ to $C_1C_2B_1B_2$, i.e. $\overline{B_1C_2}$ is an axis of symmetry of the hexagon. It thus passes through the center of mass $O$, and the other two lines are concurrent there similarly.


[quote="shinantori"]Given [b]x+y>0, y+z>0, z+x>0[/b] . Prove that:
 $\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}}<\frac{5}{4}\sqrt{x+y+z}$[/quote]
See here :
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=45024 ;)


AMC
AMC 10
AMC 12
AIME
USA(J)MO
USAMO
geometry

Hi. I'm a freshman about to take the AMC's. My school only offers one test date, so i was wondering whether i should take the 10 or 12. I usually get around 125 on practicing the AMC10 and just over 100 on AMC12 practice questions :blush: . I'd rather take the AMC12, but there's a slight chance i won't make AIME if i do.

It depends what your goal is.
If your goal is to make the AIME and just the AIME, then take the AMC 10.
But if you want to make the USAMO, I think you have a better chance of making it if you take the AMC 12.

Actually I think it would be easier to make the USAMO if you took the 10.  I m pretty sure that the USAMO is calculated exactly the same on the 10 and 12.  So a 130 on the 10 is equivalent to a 130 on the 12.  As for if you just want to make the AIME that can be a little more complicated.  Last year in tenth grade my goal was to just make the AIME.  So I took the 12 and I got into the AIME.  But I think that which one you take depends on your strengths and weaknesses.  If you are strong at geometry go for the 10.  If not you may want to go over the scores you got for the practice tests and use that as a guide.

[quote]Actually I think it would be easier to make the USAMO if you took the 10. I m pretty sure that the USAMO is calculated exactly the same on the 10 and 12. So a 130 on the 10 is equivalent to a 130 on the 12[/quote]

Actually, regardless of taking 10 or 12 when you are a 10th grader or younger, your AMC score counts as an 150 in the USAMO qualification process.

I'm pretty it's easier to qualify for USAMO with a high AMC-12 score than it is with an equal AMC-10 score.  I think you have to get in a higher percentile or something if you qualified through AMC-10, don't remember exactly how it works.

Each correctly answered AIME problem is worht an additional 10 to you AMC score, right?  How many do you need to take the USAMO?

Yes...The USAMO index score is calculated by your AMC 10/12 score plus 10 times your AIME score.   The index needed to get into USAMO varies with the year.  Last year it was 210.

For the last couple of years, the AMC has set two standards for qualifying for the USAMO - a higher standard for everyone, and a lower standard that only applies to 10th graders or younger. The more difficult standard (for everyone) was based on index (AMC+10*AIME), while the easier standard (for 10th graders or younger) was based ONLY on the AIME. I think that was an 8 on the AIME in 2003 and a 7 on the AIME in 2004. In both years, it would have been very, very difficult, maybe impossible, to meet the index-based qualifying standard without also meeting the 10th-grade standard on the AIME. As for how high the qualifying index needed to be: in 2003, it was above 220, while in 2004 it was somewhere near 210.

All of the talk in the above posts about computing the index based on AMC-10 versus AMC-12 is essentially a moot point; if you are a 10th grader or younger,
the AIME, and not the first round test, will determine whether you advance to the USAMO.

On the other hand, if you seriously think you have a chance at the USAMO, you should be exposing yourself to the stiffest challenges you can find, everywhere - and that would mean taking the AMC-12 if you had the choice. Of course, fewer than 1 in 1000 of those who take the first round tests do qualify for the USAMO; most students have little reason to use that to guide their choices.


Hey guys I'm a freshman in a pretty good school, not an Ivy or anything like that though. I'm a physics-math dual major.

Okay so here's a story: My friend's on the Putnam team and he told me about a guy on the team who won a bronze at the IMO a year or two ago. Turns out I sit next to this guy in one of my math classes! WOW, amazing! So I was all excited, I mean I didn't talk to him about it but I'd watch him in class and stuff. To see how amazingly awesome at math he is, and he really is awesome.

Well now the semester's over because I took all my finals. I knew I was in third place grade-wise in the math class, so I decided to race him as fast as I could on the final. Figured it'd be a close match. Well he doesn't know we're racing, and I'm just going as fast as I can. But sometimes I have to stop and think you know. And sometimes I'd look over and he's just blissfully writing, no hurry but no hesitation. Anyway, I finished tied for second with this other guy. But the IMO guy finished in half the time we did, only 45 minutes or something ridiculous.

Yeah, so then I was like, Oh my God, that's really amazing how huge of a math-wiz he is. So I envy him now even though he has no idea about me. He's inspired me to compete in the Putnam... but I've never done any math competition before. Just the basic math route you know, AP Calculus in high school.

Oh and also, my physics hero Dick Feynman competed in the second Putnam as a senior and blew the judges away with his score. So that's another inspiration. I just love seeing huge geniuses and saying, WOW, I'm so dumb, those guys would wipe the floor with me. Anyway, do you guys think by my senior year I'd be able to score at all? I think I could. But more importantly: how do you all recommend I start preparing? Use the AoPS books, past AMC/AIME/USAMO/Putnam exams, yeah? Where do I start with the AoPS books?

Thanks for any tips guys!!!

Able to score at all? Take a look at some of the threads in the [url=http://www.artofproblemsolving.com/Forum/index.php?f=80]Putnam forum[/url] labeled "A1" or "A2" or "B1" or "B2". Those problems are usually the easier ones, to give the normal people a chance to get something right. If you can solve one of those problems completely, you're already in the top half of all competitors in a typical year.

It doesn't matter so much exactly how you prepare; just try to solve problems, and enjoy it. Books or the like are useful for learning techniques that work on whole classes of problems; one that has some explicitly college-level material is Zeitz' Art and Craft of Problem Solving (often abbreviated ACoPS).


[quote="geniusbliss"]Prove that for all positive reals $ a,b,c$,
$ \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a} \ge 3[\frac {a^4 \plus{} b^4 \plus{} c^4}{3}]^\frac {1}{4}$[/quote]
Because of homogeneuinity assume $ a^4 \plus{} b^4 \plus{} c^4 \equal{} 3$. Now we are only required to prove that $ \sum \frac {a^2}{b}\geq 3.$
But, we have that
$ \sum \left[\frac 14\cdot \frac {a^2}{b} \plus{} \frac 14\cdot \frac {a^2}{b} \plus{} \frac 14 \cdot\frac {a^2}{b} \plus{} \frac 14 \cdot\frac {a^2}{b} \plus{} b^4 \plus{} a^4\right] \geq \frac 62 \sum a^4$(AM-GM)
$ \implies \sum \frac {a^2}{b}\geq 3\cdot 3 \minus{} 2\cdot 3 \equal{} 3$.
We are done. Nice inequality :)

[quote="Potla"]
$ \sum \left[\frac 14\cdot \frac {a^2}{b} \plus{} \frac 14\cdot \frac {a^2}{b} \plus{} \frac 14 \cdot\frac {a^2}{b} \plus{} \frac 14 \cdot\frac {a^2}{b} \plus{} b^4 \plus{} a^4\right] \geq \frac 62 \sum a^4$(AM-GM)
  :)[/quote]

I think AM-GM gives $ \frac 62 \sum a^2$ on the RHS  :maybe:
It may not be that easy.

sry i forgot abt this post..
yes potla your solution has a flaw.. the problem is tough and i hope to see an elementary solution...

[quote="geniusbliss"]Prove that for all positive reals $ a,b,c$,
$ \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a} \ge 3[\frac {a^4 \plus{} b^4 \plus{} c^4}{3}]^\frac {1}{4}$[/quote]
By using Holder
\[ (\frac{a^2}{b}\plus{}\frac{b^2}{c}\plus{}\frac{c^2}{a})^2(a^2b^2\plus{}b^2c^2\plus{}c^2a^2)\geq (a^2\plus{}b^2\plus{}c^2)^3\]
It suffice to show
\[ (x\plus{}y\plus{}z)^3\geq 3(xy\plus{}yz\plus{}zx)\sqrt{3(x^2\plus{}y^2\plus{}z^2)}\]
where $ x\equal{}a^2,y\equal{}b^2,z\equal{}c^2$.I think this is easy.By the way,what the maximum $ k$ such that the following one holds:
\[ \frac{a^2}{b}\plus{}\frac{b^2}{c}\plus{}\frac{c^2}{a}\geq 3\sqrt[k]{\frac{a^k\plus{}b^k\plus{}c^k}{3}}\]

[quote="Tourish"][quote="geniusbliss"]Prove that for all positive reals $ a,b,c$,
$ \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a} \ge 3[\frac {a^4 \plus{} b^4 \plus{} c^4}{3}]^\frac {1}{4}$[/quote]
By using Holder
\[ (\frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a})^2(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)\geq (a^2 \plus{} b^2 \plus{} c^2)^3\]
It suffice to show
\[ (x \plus{} y \plus{} z)^3\geq 3(xy \plus{} yz \plus{} zx)\sqrt {3(x^2 \plus{} y^2 \plus{} z^2)}\]
where $ x \equal{} a^2,y \equal{} b^2,z \equal{} c^2$.I think this is easy.[/quote]
Nice proof!
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=38369

[quote="Tourish"][quote="Tourish"]By the way,what the maximum $ k$ such that the following one holds:
\[ \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a}\geq 3\sqrt [k]{\frac {a^k \plus{} b^k \plus{} c^k}{3}}\]
[/quote]
arqady,Can your computer work out this best $ k$ :?:[/quote]
About natural $ k$ see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=149635


calculus
integration
limit
probability
calculus computations

Hello:

Here's an integration problem I am unsure how to tackle. I am well familiar with the usual 'spread of a disease' type problems, but I am unsure how to approach this. Does anyone have any ideas?.

"A New virus has just been declared an epidemic by health officials. Currently, 10,000 people have the disease and it is estimated that t days from now, new cases will be reported at the rate of $R(t)=10te^{-0.01t}$ people per day. If the fraction of victims who still have the virus t days after first contracting is given by $S(t)=e^{-0.015t}$, how many people will be infected by the virus 90 days from now? one year (365 days) from now?"

We want to count all people who are infected at time $T$ ($T=90$, $365$, or whatever).  Out of the $10000$ people infected today, $10000S(T)$ will still have the virus. What about those who got infected between time $0$ and time $T$. Let's focus on a short time interval $[t,t+dt]$. During this period, $R(t)dt$ people got infected, and out of them $S(T-t)R(t)\,dt$ will still have the virus at time $T$. The answer is
\[10000S(T)+\int_{0}^{T}S(T-t)R(t)\,dt \]

The term $10000S(T)$ is an overestimate, since it is based on the (unrealistic) assumption that those $10000$ people all got infected at time $0$. But without any information about $t<0$ this may be the best we can do.

[hide]Let $I(T)$ denote the number of people infected with the disease at time T (T is in days).

$I(T) = 10,000+\int_{0}^{T}R(t)S(T-t) dt$

[hide]This is because during a tiny interval of time $[t_{k}, t_{k+1}]$ of size ${\Delta t}$ there will be $R(t_{k}^{*}){\Delta t}$ people infected. And by time T, a time (T-t) later, there will be $[R(t_{k}^{*}){\Delta t}]S(T-t)$ of those people still infected.

Therefore, the total number of people still infected at time T from those who acquired the disease between t = 0 and t = T will be:

$\lim_{{\Delta t}\to 0}\sum R(t_{k}^{*}) S(T-t){\Delta t}$, which, when added to those who originally had the disease gives the above integral expression.[/hide] [/hide]

mlok makes a good point that the initial 10,000 should be multiplied by S(T) also, as the proportion of them who retain the disease over time should be similar to S.

That $10000S(T)$ is not an overestimate- it's exactly right. The disease rate decays exponentially in those who have already contracted it, and exponential decay has no memory. If two people have the virus at time $t$, you don't need to know when they got it to predict the probability they will still have it at some later time.
If $S$ were anything other than exponential, we would need to know something about the initial infected population to answer the question.

Actual epedemics will start out exponetially and then will slow down as a natural consequence of lack of people who are not infected. The persons  contracting a disease per unit time will actually slow down even though the rate of infection per capita (.01) is constant. The simple exponential model assumes an unlimited supply of non-infected victims but this is hard to attain in actual practice.

Roughly the same thing is to be seen with world population which in theory could be a simple exponetial model but as we know the earth can hold only so many persons

Thanks everyone for their input.

[quote]as we know the earth can hold only so many persons[/quote]

Maybe someone should tell the breeding machines(humans) this.
They multiply like bacteria in a petri dish.


induction
algebra
binomial theorem
Linear Recurrences
pen

[quote="Peter"]An integer sequence $\{a_{n}\}_{n \ge 1}$ is defined by \[a_{0}=0, \; a_{1}=1, \; a_{n+2}=2a_{n+1}+a_{n}\] Show that $2^{k}$ divides $a_{n}$ if and only if $2^{k}$ divides $n$.[/quote]

Let's prove a nice lemma concerning this kind of sequences.
[b]Lemma 1:[/b] If $\{x_{n}\}_{n\ge1}$ is a sequence of real numbers defined by $x_{1}=a,x_{2}=b$ ($a,b\in{R}$) and $x_{n+2}=bx_{n+1}+ax_{n}$
,then $x_{n+m}=x_{n}x_{m+1}+x_{n-1}x_{m}$ for any positive integer $m$.

[i]Proof:[/i] Let's led the proof by induction for $m$.

1.$m=1$ $\Rightarrow$ $x_{n+m}=x_{n}x_{m+1}+x_{n-1}x_{m}\Leftrightarrow x_{n+1}=bx_{n}+ax_{n-1}$

2. Suppose that we have proved the desired relation for all positive
integer $m$s that are less or equal to some $k$.

3. Now we prove it for $m=k+1$.

We have that
\[x_{n+k+1}=bx_{n+k}+ax_{n+k-1}=b(x_{n}x_{k+1}+x_{n-1}x_{k})+a(x_{n}x_{k}+x_{n-1}x_{k-1})\]
but
\[b(x_{n}x_{k+1}+x_{n-1}x_{k})+a(x_{n}x_{k}+x_{n-1}x_{k-1})=x_{n}x_{k+2}+x_{n-1}x_{k+1}\]
$\Rightarrow$ $x_{n+k+1}=x_{n}x_{k+2}+x_{n-1}x_{k+1}.$ So the lemma is proven.

Now let's use this lemma for our sequence:$\{a_{n}\}_{n \ge 1}$.
We get that $a_{n+m}=a_{m}a_{n+1}+a_{m-1}a_{n}$.Putting $m=n$ we find that $a_{2n}=a_{n}(a_{n+1}+a_{n-1})=2a_{n}(a_{n}+a_{n-1}).$
 From the initial property of the sequence we can see that $a_{n}$ and $a_{n-1}$  are  either both odd  or both even. And as we have that $a_{0}=0$ and $a_{1}=1$ $\Rightarrow$ $a_{2n}$ is even and $a_{2n+1}$ is odd for all $n\in{N_{0}}$,then
$a_{n}+a_{n-1}$ is an odd number.So if $2^{k}|a_{2n}\Rightarrow 2^{k}|2a_{n}\Rightarrow 2^{k-1}|a_{n}$.
But considering that $a_{2}=2$ we obtain that $2^{k}|a_{n}\Rightarrow 2^{k}|n$.:wink:

[quote="Peter"]An integer sequence $ \{a_{n}\}_{n \ge 1}$ is defined by
\[ a_{0}=0, \; a_{1}=1, \; a_{n+2}=2a_{n+1}+a_{n}\]
Show that $ 2^{k}$ divides $ a_{n}$ if and only if $ 2^{k}$ divides $ n$.[/quote]

 I really love this problem. It is beautiful and has many solutions. What we need to do is to read $ a_{n}$ (modulo $ 2^{k}$ and) modulo $ 2^{k+1}$. We shall establish that $ a_{2^{k}m}\equiv 2^{k}\; (mod \; 2^{k+1})$ for all [b]odd[/b] $ m \in \mathbf{Z}_{>0}$ and $ k \in \mathbf{Z}_{\geq 0}$. This implies that $ 2^{k}\; \vert \; a_{2^{k}m}$ and $ 2^{k+1}\not\vert \; a_{2^{k}m}$. We need the following simple congruence.

[b][Proposition][/b] Suppose that $ n=2^{k}m \in \mathbf{Z}_{>1}$, where $ m$ is odd. For all odd $ j<n$, we have $ \binom{n}{j}\equiv 0 \; (mod \; 2^{k})$.

It is a particular case of the following claim with $ p=2$:

[b][Claim][/b] Let $ p$ be a prime number and let $ n \in \mathbf{Z}_{>1}$. Write $ n=p^{e}m$, where $ (m, p)=1$. Then,  for all $ j<n$ with $ (j, p)=1$, we obtain
\[ \binom{n}{j}\equiv 0 \; (mod \; p^{e}). \]
[b][Proof of Claim][/b] Since ${ j \; \vert \; j \binom{n}{j}= n \binom{n-1}{j-1}= p^{e}\cdot m \binom{n-1}{j-1}}$ and since  $ (j, p^{e})=1$, we find that $ j \; \vert \; m \binom{n-1}{j-1}$ so that
\[ \binom{n}{j}= p^{e}\cdot \frac{m \binom{n-1}{j-1}}{j}\equiv 0 \; (mod \; p^{e}). \]
In order to read $ a_{2^{k}m}$ modulo $ 2^{k+1}$, we find the [b]combinatorial solution of the linear recurrence[/b]. It's a routine job to get
\[ a_{n}= \frac{1}{2\sqrt{2}}\left( (1+\sqrt{2})^{n}-(1-\sqrt{2})^{n}\right). \]
By the binomial theorem, we obtain
\[ a_{n}= \sum_{j \; \text{: odd}}2^{ \frac{j-1}{2}}\binom{n}{j}= n+\sum_{j>1 \; \text{: odd}}2^{\frac{j-1}{2}}\binom{n}{j}. \]
Suppose that $ n=2^{k}m$ where $ m$ is odd.  Let $ 1<j<n$ be odd. It follows from the above proposition and from $ 2 \; \vert \; 2^{\frac{j-1}{2}}$ that $ 2^{\frac{j-1}{2}}\binom{n}{j}\equiv 0 \; (mod \; 2^{k+1})$. We therefore conclude that
\[ a_{n}\equiv 2^{k}m+\sum_{j>1 \; \text{: odd}}2^{\frac{j-1}{2}}\binom{n}{j}\equiv 2^{k}\; (mod \; 2^{k+1}). \]

[quote="Peter"]An integer sequence $ \{a_{n}\}_{n \ge 1}$ is defined by
\[ a_{0}=0, \; a_{1}=1, \; a_{n+2}=2a_{n+1}+a_{n}\]
Show that $ 2^{k}$ divides $ a_{n}$ if and only if $ 2^{k}$ divides $ n$.[/quote]

 One may modify the method by Tiks to generalize the problem slightly.

 Theorem. Let $ u$ be an integer with $ u \equiv 2 \; (mod \; 4)$. Suppose that the sequence $ \{x_{n}\}_{n \ge 0}$ satisfies the recurrence relation
\[ x_{n+1}= u x_{n}+x_{n-1}, \;\; n \in \mathbb{N}\]
and has the initial condition $ x_{0}=0$ and $ x_{1}=1$. Then,  $ 2^{k}$ divides $ a_{n}$ if and only if $ 2^{k}$ divides $ n$.

 Proof. Since it clearly holds for $ n=0$, we now assume that $ n \geq 1$. It is enough to show the following.

 Claim. For all odd $ m \geq 1$ and $ k \geq 0$, we obtain $ 2^{k}\; \vert\vert \; a_{2^{k}m}$.

 We first work modulo $ 2$. Reading the recurrence relation modulo $ 2$, we find that $ x_{n}\; (mod \; 2)$ is periodic. Indeed, it follows from $ u \equiv 2 \; (mod \; 4)$ that
\[ x_{n+2}\equiv u x_{n+1}+x_{n}\equiv x_{n}\; (mod \; 2). \]
Since $ x_{1}\equiv 1 \; (mod \; 2)$ and $ x_{2}\equiv u \equiv 0 \; (mod \; 2)$, this means that both $ x_{n}$ and $ n$ has the same parity for all $ n \in \mathbb{N}$. Since $ u\equiv 2 \; (mod \; 4)$, we may write
\[ u=2 r \]
for some odd integer $ r$. The trick is to establish the following factorization formula.

 Proposition. Set $ Q_{n}=r x_{n}+x_{n-1}$, where $ n \geq 1$. Then,  for all $ m \geq 1$ and $ k \geq 1$, we obtain
\[ x_{2^{k}m}= 2^{k}x_{m}\prod_{j=0}^{k-1}Q_{2^{j}m}. \]
Proof of claim. Let $ m \geq 1$ be odd and $ k \geq 0$. Our job is to show that  $ 2^{k}\vert x_{2^{k}m}$ and $ 2^{k+1}\not\vert x_{2^{k}m}$.  In the case of $ k=0$, it clearly holds because $ x_{m}$ is odd. Now assume that $ k \geq 1$. By the proposition, we get
\[ x_{2^{k}m}= 2^{k}x_{m}\prod_{j=0}^{k-1}Q_{2^{j}m}\]
We need to show that $ x_{m}\prod_{j=0}^{k-1}Q_{2^{j}m}$ is odd. Recall that $ x_{n}$ and $ n$ has the same parity for all $ n \in \mathbb{N}$. Since $ m$ is odd, $ x_{m}$ is odd. Since $ r$ is odd, we see that
\[ Q_{n}\equiv r x_{n}+x_{n-1}\equiv x_{n}+x_{n-1}\; (mod \; 2) \]
so that $ Q_{n}$ is always odd. Hence, $ \prod_{j=0}^{k-1}Q_{2^{j}m}$ is also odd. This completes the  proof.

 Now, we need to prove the proposition. It comes from the following formula.

 Lemma. For all $ n \in \mathbb{N}$, we have
\[ x_{2n}= 2 x_{n}\left( rx_{n}+x_{n-1}\right) = 2x_{n}Q_{n}. \]
Proof of proposition. Our job is to show that
\[ x_{2^{k}m}= 2^{k}x_{m}\prod_{j=0}^{k-1}Q_{2^{j}m}. \]
By the lemma, $ x_{2m}= 2x_{m}Q_{m}$. Applying the lemma twice, we have $ x_{2^{2}m}= 2^{2}x_{m}Q_{m}Q_{2m}$. Now, it is clear that we need to use the induction on $ k$.

 Proof of lemma. Recall that $ x_{n+1}= u x_{n}+x_{n-1}$,  $ n \in \mathbb{N}$. It can be rewritten as
\[ x_{2n}= 2 x_{n}\left( \frac{u}{2}x_{n}+x_{n-1}\right). \]
or
\[ x_{2n}= x_{n}\left( ux_{n}+2x_{n-1}\right). \]
or
\[ x_{2n}= x_{n}\left( x_{n+1}+x_{n-1}\right). \]
However, this can be deduced by taking $ m=n$ in the following theorem.

 Theorem.  Suppose that the sequence $ \{x_{n}\}_{n \ge 0}$ satisfies the recurrence relation
\[ x_{n+1}= u x_{n}+x_{n-1}, \;\; n \in \mathbb{N}\]
and has the initial condition $ x_{0}=0$ and $ x_{1}=1$. Then, we have
\[ x_{n+m}= x_{m+1}x_{n}+x_{m}x_{n-1}\]
for all $ n \geq 1$ and $ m \geq 0$.

 Proof. This is well-known. For example, we may prove this by induction on $ m$.

[quote="Peter"]An integer sequence $ \{a_{n}\}_{n \ge 1}$ is defined by
\[ a_{0}=0, \; a_{1}=1, \; a_{n+2}=2a_{n+1}+a_{n}\]
Show that $ 2^{k}$ divides $ a_{n}$ if and only if $ 2^{k}$ divides $ n$.[/quote]

  We offer an alternative proof of the previous theorem in the above solution.

 {Theorem.} Let $ u$ be an integer with $ u \equiv 2 \; (mod \; 4)$. Suppose that the  sequence $ \{x_{n}\}_{n \ge 0}$ satisfies the recurrence relation
\[ x_{n+1}= u x_{n}+x_{n-1}, \;\; n \in \mathbb{N}\]
 and has the initial condition $ x_{0}=0$ and $ x_{1}=1$. Then, $ 2^{k}$ divides $ a_{n}$ if and only if $ 2^{k}$ divides $ n$.

 We'll establish the following claim.

 {Claim.} Let $ k \geq 1$ and $ m \geq 0$. Set $ G_{k}(m)=\frac{x_{2^{k}m}}{2^{k}}$. Then, $ G_{k}(m)$ is integer for all $ k$ and $ m$. Furthermore,
\[ G_{k}(m) \equiv \begin{cases}1 \; (mod \; 2) & \left( m \equiv 1 \; (mod \; 2) \right), \\ 2 \; (mod \; 4) & \left( m \equiv 2 \; (mod \; 4) \right), \\ 0 \; (mod \; 4) & \left( m \equiv 0 \; (mod \; 4) \right). \\ \end{cases}\]

 Clearly, this implies that $ 2^{k}\; \vert\vert \; a_{2^{k}m}$ for all odd $ m \geq 1$ and $ k \geq 0$.

 {Proposition.} The sequence $ \{u_{n}\}_{n \ge 1}$ satisfies the recurrence relation
\[ u_{n+1}={u_{n}^{2}-2}, \;\; n \in \mathbb{N}\]
 and has the initial condition $ u_{1}=u^{2}+2$. Then, for all $ k \geq 1$ and $ m \geq 0$, we have
\[ G_{k+1}(m)=\frac{1}{2}G_{k}(2m) \]
and
\[ G_{k}(m+2) = u_{k}G_{k}(m+1)-G_{k}(m). \]

 We first use the proposition to prove the claim and then prove the proposition. Since $ u\equiv 2 \; (mod \; 4)$, we
 write $ u=2 r$ for some odd integer $ r$.

 {Proof of Claim.} Induction on $ k$. First, consider the case $ k=1$. In the previous solution, we showed that both $ x_{n}$ and $ n$ have  the same parity. Hence, we see that $ x_{n}$ is even for all even $ n$. In other words, we find that
 \[ G_{1}(m)=\frac{x_{2m}}{2}\]
 is an integer for all $ m \geq 0$. By the second assertion in the proposition, we get
 \[ G_{1}(m+2) = \left( u^{2}+2 \right) G_{1}(m+1)-G_{1}(m). \]
 Since $ u \equiv 2 \; (mod \; 4)$, this implies that
 \[ G_{1}(m+2) \equiv 2 G_{1}(m+1)-G_{1}(m) \; (mod \; 4). \]
 Since $ G_{1}(0)=\frac{x_{0}}{2}=0 \equiv 0 \; (mod \; 4)$ and $ G_{1}(1)=\frac{x_{2}}{2}=\frac{u}{2}=r \equiv 1 \; (mod \; 2)$, this congruence imply that
\[ G_{1}(m) \equiv \begin{cases}1 \; (mod \; 2) & \left( m \equiv 1 \; (mod \; 2) \right), \\ 2 \; (mod \; 4) & \left( m \equiv 2 \; (mod \; 4) \right), \\ 0 \; (mod \; 4) & \left( m \equiv 0 \; (mod \; 4) \right). \\ \end{cases}\]
 Hence, it is true for $ k=1$. Now, assume that it holds for some $ k \geq 1$. First, we need to show that $ G_{k+1}(m)$ is integer for all $ m \geq 0$. By the inductive hypothesis, we see that $ G_{k}(2m)$ is even. Hence,
 \[ G_{k+1}(m)=\frac{1}{2}G_{k}(2m) \]
 is also an integer for all $ m \geq 0$. It's an easy job to check that $ u_{k}\equiv 2 \; (mod \; 4)$ holds for all $ k \geq 1$. Hence, reading the second assertion in the proposition modulo $ 4$, we obtain
 \[ G_{k+1}(m+2) \equiv 2 G_{k+1}(m+1)-G_{k+1}(m) \; (mod \; 4). \]
 We compute $ G_{k+1}(0)=\frac{x_{0}}{2^{k+1}}= 0$. By the inductive hypothesis, we see that $ G_{k}(2) \equiv 2 \; (mod \; 4)$ so that $ G_{k+1}(1)=\frac{1}{2}G_{k}(2) \equiv 1 \; (mod \; 2)$. Since $ G_{k+1}(0) \equiv 0 \; (mod \; 4)$ and since $ G_{k+1}(1) \equiv 1 \; (mod \; 2)$, one may conclude that
\[ G_{k+1}(m) \equiv \begin{cases}1 \; (mod \; 2) & \left( m \equiv 1 \; (mod \; 2) \right), \\ 2 \; (mod \; 4) & \left( m \equiv 2 \; (mod \; 4) \right), \\ 0 \; (mod \; 4) & \left( m \equiv 0 \; (mod \; 4) \right), \\ \end{cases}\]
 as desired.

 {Proof of Proposition.} The first assertion is easy:
  \[ G_{k+1}(m)=\frac{ x_{2^{k+1}m}}{2^{k+1}}=\frac{1}{2}\cdot \frac{ x_{2^{k}\cdot 2m}}{2^{k}}= \frac{1}{2}G_{k}(2m). \]
  The second assertion is equivalent to
 \[ 2^{k}G_{k}(m+2) = u_{k}2^{k}G_{k}(m+1)-2^{k}G_{k}(m) \]
 or
 \[ x_{2^{k}(m+2)}= u_{k}x_{2^{k}(m+1)}-x_{2^{k}m}. \]
This can be proved by induction on $ k$. We omit the details.

I used 2-adic analysis:

First, one can find the explicit form of $a_n$ as:

$a_n = \frac{(1+\sqrt{2})^n - (1-\sqrt{2})^n}{2\sqrt{2}}$

This formula is equivalent to looking at the integral coefficient of the term with $\sqrt{2}$ in the expansion of $(1+\sqrt{2})^n$.

The problem statement can now be proved with the following, more general statement:

Prove that $v_2(a_n) = v_2(n)$.

To prove this, we use induction. But first let us rewrite $a_i$ to make the 2-adic stuff more workable.

Define $(1+\sqrt{2})^i = b_i + a_i\sqrt{2}$

Note that $b_i$ is odd for all i. This follows directly from the binomial expansion of $(1+\sqrt{2})^i$.

For the base cases, $v_2(a_1) = 0$ and $v_2(a_2) = 1$.

We now carry our induction on $i \leq 2^k \rightarrow i \leq 2^{k+1}$

Suppose that $v_2(a_i) = i$ for all $i \leq 2^k$.

We have that $v_2(a_{2i}) = v_2(2a_i \cdot b_i) = v_2(a_i) + 1 = v_2(i) + 1 = v_2(2i)$
And $v_2(a_{2i+1}) = v_2(2a_i\cdot b_i + b_{2i}) = 0 = v_2(2i+1)$.

So our induction hypothesis holds and the conclusion is implied.


number theory
Diophantine equation
Other problems

[b]Prove that there are infinitely many pentagonal numbers that can be written as a sum of two perfect squares.[/b]

[color=red][b]Note:[/b][/color] The problem was originally posted [url=http://www.mathlinks.ro/viewtopic.php?t=254064]here.[/url]  :wink:

Pretty old problem. Find such pentagonal numbers - which can be represented as a sum of two squares. That is, to prove that such numbers are infinitely many.

So I'm wondering - if I write a formula describing their decision whether it will be sufficient evidence that such numbers are infinitely many?

That is, the following Diophantine equation:

$X^2+Y^2=\frac{Z(3Z\pm1)}{2}$

If we use the equation Pell:  $p^2-2(6k^2-6k+1)s^2=\pm1$

Then using the solutions of this equation can be written solutions of the equation. Sign in and sign Pell alignment of the first alignment should be the same.

$X=kps+2(k-1)s^2$

$Y=(k-1)ps-2ks^2$

$Z=2(2k^2-2k+1)s^2$

more:

$X=p^2+(7k-4)ps+2(2k-1)(3k-2)s^2$

$Y=p^2+(7k-3)ps+2(2k-1)(3k-1)s^2$

$Z=p^2+4(2k-1)ps+4(2k-1)^2s^2$

Can I assume that the problem was solved? Proved - there are infinitely many pentagonal numbers which can be represented as the sum of two squares?

Every pentagonal number can be written as sum of a perfect square and a triangular number:

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

The result follows now from the well-fact that there are [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1426331#p1426331][b]infinitely[/b][/url] many square triangular numbers.

Regards,

And then you still need to look for such number when another term squared. Not better to just solve the equation?
And all your guess gives partial solutions when  $Z=Y$


function
floor function
integration
probability
algebra

Find the value of $ \lfloor\frac{1}{3}\sum_{k\equal{}1}^{2007}\frac{1}{\sqrt{k}}\rfloor$.

I think the following will solve your problem
$ 2\sqrt{n}\minus{}2<1\plus{}\frac{1}{\sqrt{1}}\plus{}\frac{1}{\sqrt{2}}.....\frac{1}{\sqrt{n}}<2\sqrt{n}\minus{}1$

Could you please explain how you arrived at those bounds?

we shall first show that
$ 2\sqrt {m \plus{} 1} \minus{} 2\sqrt {m} < \frac {1}{\sqrt {m}} < 2\sqrt {m} \minus{} 2\sqrt {m \minus{} 1}$

in fact,$ 2\sqrt {m \plus{} 1} \minus{} 2\sqrt {m} \equal{} \frac {2[\sqrt {m \plus{} 1} \minus{} \sqrt {m} ][\sqrt {m \plus{} 1} \plus{} \sqrt {m} ]}{2\sqrt {m \plus{} 1} \plus{} \sqrt {m}} \equal{} \frac {2}{2[\sqrt {m \plus{} 1} \plus{} \sqrt {m}]} < \frac {1}{\sqrt {m}}$
since $ \sqrt {m \plus{} 1} < \sqrt {m}$

note[.]-this doesnot denote greatest integer function , they are simply square brackets

This is only the basic idea......

[hide="The entire Solution :D"]

$ \frac{1}{\sqrt{x}} \equal{} \frac{2}{\sqrt{x}\plus{}\sqrt{x}} < \frac{2}{\sqrt{x} \plus{}\sqrt{x\minus{}1}}$

on rationalizing, this gives

$ \frac{1}{\sqrt{x}} < 2 ( \sqrt{x} \minus{} \sqrt{x\minus{}1})$

Summing it from 1 to n :

$ \frac{1}{\sqrt{1}} < 2 ( \sqrt{1} \minus{} \sqrt{0}) \\
\\
\frac{1}{\sqrt{2}} < 2 ( \sqrt{2} \minus{} \sqrt{1}) \\
\\
\frac{1}{\sqrt{3}} < 2 ( \sqrt{3} \minus{} \sqrt{2}) \\
\\ . \\ .
\\ \frac{1}{\sqrt{n}} < 2 ( \sqrt{n} \minus{} \sqrt{n\minus{}1})$

Alternate terms cancel out

and you are left with

$ \sum_{k\equal{}1}^{n} \frac{1}{\sqrt{k}} < 2 \sqrt{n}$

[b]Similarly proceeding for lower limit[/b]

$ \frac{1}{\sqrt{x}} \equal{} \frac{2}{\sqrt{x}\plus{}\sqrt{x}} > \frac{2}{\sqrt{x} \plus{}\sqrt{x\plus{}1}}$

on rationalizing, this gives

$ \frac{1}{\sqrt{x}} > 2 ( \sqrt{x\plus{}1} \minus{} \sqrt{x})$

Summing it from 1 to n :

$ \frac{1}{\sqrt{1}} > 2 ( \sqrt{2} \minus{} \sqrt{1}) \\
\\
\frac{1}{\sqrt{2}} > 2 ( \sqrt{3} \minus{} \sqrt{2}) \\
\\
\frac{1}{\sqrt{3}} > 2 ( \sqrt{4} \minus{} \sqrt{3}) \\
\\ . \\ .
\\ \frac{1}{\sqrt{n}} > 2 ( \sqrt{n\plus{}1} \minus{} \sqrt{n})$

Alternate terms cancel out

and you are left with

$ \sum_{k\equal{}1}^{n} \frac{1}{\sqrt{k}} > 2 (\sqrt{n\plus{}1} \minus{} 1)$

[b]Coming back to the Question ( n = 2007 )[/b]

$ \frac{2}{3} \ ( \sqrt{2008} \minus{} 1 ) < \frac{1}{3} \ \sum_{k\equal{}1}^{2007} \frac{1}{\sqrt{k}} < \frac{2}{3} \  \sqrt{2007} \\
\\
29.20 <  \frac{1}{3} \ \sum_{k\equal{}1}^{2007} \frac{1}{\sqrt{k}} < 29.8$

The answer =

$ [ \ \frac{1}{3} \ \sum_{k\equal{}1}^{2007} \frac{1}{\sqrt{k}} \ ] \equal{} 29$

[b]Pheww !! [/b] :ninja:

[/hide]

[hide="Alternative nonrigorous method for the less gifted students"]Bound by integrals. You get $ \int_2^{2008} \frac{dn}{\sqrt{n}} \leq \sum_{k \equal{} 1}^{2007} \frac{1}{\sqrt{n}} \leq \int_1^{2007} \frac{dn}{\sqrt{n}}$.  In other words, $ 86.7929989 \leq \sum_{k \equal{} 1}^{2007} \leq 87.5991071$. Thus, $ 28.93 \leq \mbox{what we want} \leq 29.1997024$, so the floor of what we want is 28. We're now going to guess with a decent probability that the number lies in the interval [29, 29.1997024], so its floor will be \mbox{29}. Not the optimal solution, but if I saw this on a competition (and I had no calculators and I don't need to write proofs), this is what I'd do. [/hide]

You'd need a calculator to evaluate $ \int_1^{2007}\frac{dn}{\sqrt{n}}$ though... Unless you could find the exact value and bound it between two integers to find the floor.

We could make a decent estimate for root 2007.

And even with calculators, we didn't manage to bound it between two integers. :P It was guesswork, but if a number lies in [1.95, 2.3], you'd guess that the floor of that number is 2. Which is probably what I'd do in a competition.


calculus
integration
function
geometry
algebra
polynomial
calculus computations

A  sextic funtion $ y \equal{} ax^6 \plus{} bx^5 \plus{} cx^4 \plus{} dx^3 \plus{} ex^2 \plus{} fx \plus{} g\ (a\neq 0)$ touches the line $ y \equal{} px \plus{} q$ at $ x \equal{} \alpha ,\ \beta ,\ \gamma \ (\alpha < \beta < \gamma).$ Find the area of the region bounded by these graphs in terms of $ a,\ \alpha ,\ \beta ,\gamma$.

[quote="Virgil Nicula"][color=darkred]A  polynominal funtion $ y \equal{} ax^{2n} \plus{} \ldots$, where $ a\ne 0$ and $ x\in\mathcal R$,  touches the line $ y \equal{} px \plus{} q$ at $ x_k\ ,\ k\in\overline {1,n}\ .$ Find the area of the region bounded by these graphs in terms of $ a$ and $ x_k\ ,\ k\in\overline {1,n}\ .$[/color][/quote]
[color=darkblue]Kunny, can you generalize this problem to a polynominal function of even degree ?![/color]

Yes, we can, but we seem to have trouble with calculation of Integral for the area.

$ \int_a^b(x \minus{} a)^2(x \minus{} b)^2f(x)\ \mathrm {dx} \equal{} \int_{ \minus{} \frac {b \minus{} a}{2}}^{\frac {b \minus{} a}{2}}\left[x^2 \minus{} \left(\frac {b \minus{} a}{2}\right)^2\right]^2f\left(\frac {a \plus{} b}{2} \minus{} x\right)\ \mathrm {dx} \equal{}$

$ \left(\frac {b \minus{} a}{2}\right)^5\cdot\int_{ \minus{} 1}^1\left(x^2 \minus{} 1\right)^2f\left(\frac {a \plus{} b}{2} \plus{} \frac {a \minus{} b}{2}\cdot x\right)\ \mathrm {dx} \equal{}$

$ \left(\frac {b \minus{} a}{2}\right)^5\cdot\int_0^1\left(x^2 \minus{} 1\right)^2\left[f\left(\frac {a \plus{} b}{2} \plus{} \frac {a \minus{} b}{2}\cdot x\right) \plus{} f\left(\frac {a \plus{} b}{2} \minus{} \frac {a \minus{} b}{2}\cdot x\right)\right]\ \mathrm {dx}\ .$

I"ll use the upper integral, where $ a < b < c$ and $ f(x) \equal{} (x \minus{} c)^2\ :$
$ I(a,b,c) \equal{} \int_a^b(x \minus{} a)^2(x \minus{} b)^2(x \minus{} c)^2\ \mathrm {dx} \equal{} 2\left(\frac {b \minus{} a}{2}\right)^5\cdot\int_0^1\left(x^2 \minus{} 1\right)^2\left[\left(\frac {a \minus{} b}{2}\right)^2x^2 \plus{} \left(\frac {a \plus{} b}{2} \minus{} c\right)^2\right]\ \mathrm {dx}$
In conclusion, coming to the initial notations, the required area is $ a\cdot\left[I(\alpha ,\beta ,\gamma ) \plus{} I(\beta ,\gamma ,\alpha )\right]$ a.s.o. I like it !

[quote="kunny"][color=darkred]A  sextic funtion $ y = ax^6 + bx^5 + cx^4 + dx^3 + ex^2 + fx + g\ (a\neq 0)$ touches the line $ y = px + q$ at $ x = \alpha ,\ \beta ,\ \gamma \ (\alpha < \beta < \gamma).$ Find the area of the region bounded by these graphs in terms of $ a,\ \alpha ,\ \beta ,\gamma$.[/color][/quote]
[color=darkblue][b][u]Proof.[/u][/b] Denote $ I_f(a,b)\equiv\int_a^b(x - a)^2(x - b)^2f(x)\ \mathrm {dx}\ .$ Thus,

$ I_f(a,b) = \int_{ - \frac {b - a}{2}}^{\frac {b - a}{2}}[x^2 - (\frac {b - a}{2})^2]^2f(\frac {a + b}{2} - x)\ \mathrm {dx} =$

$ (\frac {b - a}{2})^5\cdot\int_{ - 1}^1(x^2 - 1)^2f(\frac {a + b}{2} + \frac {a - b}{2}\cdot x)\ \mathrm {dx} =$

$ (\frac {b - a}{2})^5\cdot\int_0^1(x^2 - 1)^2[f(\frac {a + b}{2} + \frac {a - b}{2}\cdot x) + f(\frac {a + b}{2} - \frac {a - b}{2}\cdot x)]\ \mathrm {dx}\ .$

I"ll use the upper integral for the functions $ \|\begin{array}{ccc} u: [a,b]\rightarrow \mathcal R & , & u(x) = (x - c)^2 \\
\ v: [b,c]\rightarrow\mathcal R & , & v(x) = (x - a)^2\end{array}\ :$

$ I_u(a,b) = \int_a^b(x - a)^2(x - b)^2(x - c)^2\ \mathrm {dx} =$

$ 2(\frac {b - a}{2})^5\cdot\int_0^1(x^2 - 1)^2[(\frac {a - b}{2})^2x^2 + (\frac {a + b}{2} - c)^2]\ \mathrm {dx}\implies$

$ I_u(a,b) = \frac {(b - a)^5}{840}\cdot[(b - a)^2 + 7\cdot (a + b - 2c)^2]\ .$ Analogously

$ I_v(b,c) = \frac {(c - b)^5}{840}\cdot[(c - b)^2 + 7\cdot (b + c - 2a)^2]\ .$

Thus, coming to the initial notations, the [b]required area[/b] is $ a\cdot[I_u(\alpha ,\beta ) + I_v(\beta ,\gamma )]$ a.s.o.

[size=134][b]I like very much the your proposed problem ![/b][/size][/color]

Thank you Very much, Virgil Nicula! :)

Here is my solution.

Let $ I(m,\ n) = \int_{\alpha}^{\beta}(x - \alpha)^m(\beta - x)^n\ dx\ (m,\ n = 0,\ 1,\ 2,\ \cdots)$, we have $ I(m,\　n) = \frac {m!n!}{(m + n + 1)!}(\beta - \alpha)^{m + n + 1}$.

The desired area $ S$ is ,

$ S = \int_{\alpha}^{\gamma} (x - \alpha)^2(x - \beta)^2(x - \gamma)^2\ dx$

$ = \int_{\alpha}^{\gamma} (x - \alpha)^2(x - \gamma)^2\{x - \alpha - (\beta - \alpha)\}^2\ dx$

$ = I(4,\ 2) - 2(\beta - \alpha)I(3,\ 2) + (\beta - \alpha)^2I(2,\ 2)$

$ = \frac {2!2!\cdot 6}{7!}(\gamma - \alpha)^5[12(\gamma - \alpha)^2 - 3!\cdot 7(\beta - \alpha)(\gamma - \alpha) + 42(\beta - \alpha)^2]$

$ = \boxed{\frac {1}{210}(2\alpha ^2 + 7\beta ^2 + 2\gamma ^2 - 7\alpha \beta - 7\beta \gamma + 3\gamma \alpha)(\gamma -\alpha)^5}$.


$0<a<1$ and $n \in N$ prove that:
$a^n>a^{n+1}$
i want to see your nice proof. :D

thats a little hard for classroom math. ;)

This is a nice question if you ask which one is greater, and it is classroom math. But proving the inequality is not classroom math, and the mods should move it. I'm interested to see the proof too.  :lol:

[quote="jensen"]ok i will post this question in another forum and i promise i will send the proof here for you.thank both of you.[/quote]

No.  There will be no proofs in the middle school classroom math forum.  Please read the posting guidelines.

I think you ment $n\in\mathbb{R}$ and $0<a<1$, because there are no natural numbers greater than $0$ and less than $1$
$a^n>a^{n+1}$
$a^{n+1}-a^n<0$
$a^n(a-1)<0$
$a^n$ is greater than zero, but $a-1<0$.
End of a proof.

Please be careful about what notation you use when posting in Getting Started or Classroom Math.  This notation alone is too far beyond both, so I am moving this thread to Intermediate.  This is a nice introductory proof exercise, though; thank you for sharing it.