calculus integration trigonometry function symmetry induction LaTeX

Take $ x \equal{} \sin(\frac {\alpha}{2})$, then $ dx \equal{} \frac {1}{2}\cos(\frac {\alpha}{2})d\alpha$, the integration interval becomes $ [0,\pi]$, and the integrand is $ \cos^5(\frac {\alpha}{2})$.  So the integral is
\[ \frac {1}{2}\int_0^\pi\cos^6(\frac {\alpha}{2})d\alpha \equal{} \frac {1}{16}\int_0^\pi(1 \plus{} 3\cos(\alpha) \plus{} 3\cos^2(\alpha) \plus{} \cos^3(\alpha))d\alpha.
\]
Now, over $ [0,\pi]$, $ \cos(\alpha)$ and $ \cos^3(\alpha)$ integrate to 0 (both functions have odd symmetry around $ \frac {\pi}{2}$), while $ \cos^2\alpha \equal{} \frac {1 \plus{} \cos(2\alpha)}{2}$ integrates to $ \frac {\pi}{2}$ since $ 2\alpha$ makes a complete revolution from 0 to $ 2\pi$.  Therefore, the integral is
\[ \frac {\pi}{16} \plus{} \frac {3\pi}{32} \equal{} \frac {5\pi}{32}.
\]

Note from daniel73: edited after kunny pointed out a stupid calculation mistake, originally posted $ \frac {\pi}{16} \plus{} \frac {3\pi}{32} \equal{} \frac {7\pi}{32}$...   :blush:

Observe that :
$ I \equal{} \int_0^1 (1\minus{}x^2)^{\frac{5}{2}} dx \equal{} \int_0^1 \sqrt{1\minus{}x^{\frac{2}{5}}}dx \equal{}5\int_0^1 u^4\sqrt{1 \minus{} u^2}du$ (with $ x^{\frac{2}{5}} \equal{} u^2$)

$ I \equal{} 5\int_0^\frac{\pi}{2}\cos(\theta)\sin^4(\theta)\cos(\theta)d\theta
\equal{}\frac{5}{8}\int_0^\frac{\pi}{2} \sin^2(2\theta)(1 \minus{} \cos(2\theta))d\theta$

$ I \equal{} \frac{5}{8}\int_0^\frac{\pi}{2} (\frac{1}{2} \minus{} \frac{\cos{2\theta}}{4} \minus{} \frac{\cos{4\theta}}{2} \plus{} \frac{\cos{6\theta}}{4})d\theta \equal{}\frac{5\pi}{32}$.

[quote="akech"]Observe that :
$ I \equal{} \int_0^1 (1 \minus{} x^2)^{\frac {5}{2}} dx \equal{} \int_0^1 \sqrt {1 \minus{} x^{\frac {2}{5}}}dx$ [/quote]

 :huh:

I did not understand. Could you please explain?

[hide]
We take $ x \equal{} \sin\alpha$, and we obtain $ dx \equal{} \cos\alpha d\alpha$, $ (1 \minus{} x^2)^\frac {n}{2} \equal{} \cos^n\alpha$, integration interval $ [0,\frac {\pi}{2}]$, or
\[ I_n \equal{} \int_0^\frac {\pi}{2}\cos^{n \plus{} 1}\alpha d\alpha.
\]
Now establish induction using integration by parts for $ n\geq1$, since $ \cos\alpha \equal{} \frac {d\sin\alpha}{d\alpha}$, and since $ \sin\alpha\cos^n\alpha \equal{} 0$ for $ \alpha \equal{} 0,\frac {\pi}{2}$,
\[ I_n \equal{} \minus{} n\int_0^\frac {\pi}{2}\cos^{n \minus{} 1}\alpha( \minus{} \sin\alpha)\sin\alpha d\alpha \equal{} nI_{n \minus{} 2} \minus{} nI_n,
\]
and $ I_n \equal{} \frac {n}{n \plus{} 1}I_{n \minus{} 2}$.  Establish now initial conditions with $ n \equal{} 1$, yielding easily
\[ I_1 \equal{} \int_0^\frac {\pi}{2}\cos^2\alpha d\alpha \equal{} \frac {\pi}{4},
\]
and for $ n \equal{} 0$, yielding trivially
\[ I_0 \equal{} \int_0^1dx \equal{} 1.
\]
Then, if $ n \equal{} 2m$ is even,
\[ I_{2m} \equal{} \frac {2m}{2m \plus{} 1}\frac {2m \minus{} 2}{2m \minus{} 1}\dots\frac {2}{3} \equal{} \frac {2^{2m}(m!)^2}{(2m \plus{} 1)!} \equal{} \frac {2^n\left(\left(\frac {n}{2}\right)!\right)^2}{(n \plus{} 1)!},
\]
and if $ n \equal{} 2m \plus{} 1$ is odd,
\[ I_{2m \plus{} 1} \equal{} \frac {2m \plus{} 1}{2m \plus{} 2}\frac {2m \minus{} 1}{2m}\dots\frac {3}{4}\frac {\pi}{4} \equal{} \frac {(2m \plus{} 2)!}{2^{2m \plus{} 1}\left((m \plus{} 1)!\right)^2}\frac {\pi}{4} \equal{} \frac {\pi(n \plus{} 1)!}{2^{n \plus{} 2}\left(\left(\frac {n \plus{} 1}{2}\right)!\right)^2}
\]
Note that, in this last case, making $ n \equal{} 5$ results in
\[ I_5 \equal{} \frac {\pi6!}{2^73!3!} \equal{} \frac {5\pi}{32},
\]
as expected...

Note: I skipped the trivial induction step used to prove the result for all even $ n$, and for all odd $ n$...
[/hide]

That's right. We can use Walli's formula.

Again,
$ I_n \equal{} \int_0^1 (1 \minus{} x^2)^{\frac {n}{2}} dx \equal{} \int_0^1 \sqrt {(1 \minus{} x^{\frac {2}{n}})} dx \equal{} n\int_0^{\frac {\pi}{2}}\sin^{n \minus{} 1)}(\theta)\cos^2(\theta)d\theta$

After here one can use reduction formula just as daniel73 did.

I have used this: $ \int_0^1 (1 \minus{} x^a)^{\frac {1}{b}} \equal{} \int_0^1 (1 \minus{} x^b)^{\frac {1}{a}} dx$

Suppose neither a nor b is zero and then set $ y^b \equal{} 1\minus{}x^a$.
Then we have:
$ \int_0^1 (1 \minus{} x^a)^{\frac {1}{b}} \equal{} \minus{}\int_0^1 y((1 \minus{} y^b)^{\frac {1}{a}})'dy \equal{} \int_0^1 (1 \minus{} x^b)^{\frac {1}{a}} dx$

i have a doubt

how to post the equations like others have posted on this site  :o

thanks -- J.Y .CHOI  :lol:

This website might be helpful.

http://www.sitmo.com/latex

Other solution

$ \displaystyle\int_{0}^{1} (1\minus{}x^2)^{\dfrac{5}{2}} dx$

let $ x^2\equal{}u$

$ \equal{}\dfrac{1}{2}\displaystyle\int_{0}^{1} (1\minus{}u)^{\dfrac{5}{2}}u^{\dfrac{\minus{}1}{2}} du$

$ \equal{}\dfrac{1}{2} \beta\left(\dfrac{7}{2},\dfrac{1}{2}\right)\equal{}\dfrac{\Gamma\left(\dfrac{7}{2}\right)\Gamma\left(\dfrac{1}{2}\right)}{2\Gamma(4)}\equal{}\dfrac{15\cdot\sqrt{\pi}\cdot\sqrt{\pi}}{2\cdot 8 \cdot 3!}\equal{}\dfrac{5\pi}{32}$

[quote="Srinivasa Ramanujan"]how to post the equations like others have posted on this site[/quote] See http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690 and [[LaTeX]].

A different way ( even though it is circuitous):

$ x\equal{}\cos (u)$
$ dx\equal{}\minus{}sin(u)du$

so we get:
$ \minus{}\int_{0}^{1}sin(u)^6du$

and here we can use the formula:
$ \int \sin (t)^ndt\equal{}\minus{}\frac{1}{n}\cos (t)\sin (t)^{n\minus{}1}\plus{}\frac{n\minus{}1}{n}\int \sin (t)^{n\minus{}2}dt$
three times to get:
$ \minus{}\int_{0}^{\frac{\pi }{2}}\sin (u)^6du\equal{}\left [ \cos (u)(\frac{1}{6}\sin (u)^5\plus{}\frac{5}{24}\sin (u)^3\plus{}\frac{5}{16}\sin (u))\minus{}\frac{5}{16}u \right ]_0^\frac{\pi }{2}$
for the first summand being 0 for upper and lower bound we can simplify it to:
$ \minus{}\int_{0}^{\frac{\pi }{2}}\sin (u)^6du\equal{}\left [ \minus{}\frac{5}{16}u \right ]_0^\frac{\pi }{2}$
and thus $ \int_{0}^{\frac{\pi }{2}}\sin (u)^6du\equal{} \frac{5\pi }{32}$

[quote="isomorphism"][quote="kunny"]He seemed to make a typo.[/quote]

Oh I did not notice it , Thanks kunny.[/quote]

Actually, there was no typo in my calculation as the proof of the claim made was provided later in this thread.


Hello,

This morning my wife and I were thinking about cooking a kind of traditional Costa Rican dish (we currently live in Atlanta, GA, USA) and we went to a couple of supermarkets but we didn't find the main ingredient: the tongue of a cow. It seems like no body here uses what so ever.

That made me think that very often we have in our home countries some special dishes that people from elsewhere consider odd, and even some time (probably by lack of experience) disgusting. For example, some people find disgusting that we cook the tongue of the cow, but I should say that many people find it delicius, both Costa Rican and people from everywhere else.

We also prepare a kind of soup made of one of the stomachs of the cow, that we call "Mondongo". If the person that cooks it knows how to do it, it tastes great, but otherwise is not that delicius.

But the big price of Costa Rican dishes that foreigners often find disgusting is the "Creadilla", that is, the testicles of a bull prepared with a very special sauce. It has a particular flavor, it is not my favorite at all, but it is not disgusting.

Those are types of food that I saw while I was raised, therefore they are not awkward for me as they can be for many of the members of the forum. Nevertheless, I know every country has some kind of dishes that people know are disgusting for foreigners, at least if they ask. So many times they would eat it and enjoy it if they don't know what it is.

For example, something that I considered bizarre to eat was Sushi or Sachime. The idea of raw fish was very disgusting for me, but then a friend took me to a very traditional japanese restaurant in Costa Rica (Goyi) and, with reticence at the begining, I started eating it, and currently, I love most of the Japanese food I have taste (I mean it, I really dislike the fact that it is very expensive here in the US).

Honestly, the only disgusting dish, really repulsive (at least for me) that I have ever taste comes from my own country, and the main ingredients are the flowers of a tree called "itavo" and eggs of some animal, frecuently hen's eggs, but also some time turtle or other kind of eggs. It is not that exotic, is it?, nevertheless, if you ever go to my country and you are offered this dish, eat it carefully. May be you like it, or may be, as I do, you really hate it!

I honestly say, I have taste dishes from several different countries, prepared by people who is native of such countries, and I have to tell you, I have taste things that I really like, others that I didn't like that much, but any way is eatable, but never something that I really find disgusting, even when the ingredients are "exotic' or 'uncommon' for me.

What about in your countries?

Best regards,

I think Chinese sometimes eat cow tongue. Don't they have a mix of cow intestines (lungs, small intestines, etc.) in Chinese restaurants? I don't know, but they taste OK.

We also have shark fins. We eat it as a soup. Quite expensive though.

Well, shark fins I believe have become one of the most common dishes in the wold cousine specialities. Specially because (at least here in the west) people say that they are good for cancer and a good aphrodiciac. I have never eaten them, because as you say, they are quite expensive.

Best,

Duck tongues are used in some chinese dishes. It's very good.

I won't eat about all of the food mentioned here, because I'm a vegetarian.  I don't think any of the food from India sounds disgusting to others (at least the part that I'm from), mainly is vegetarian food, no meat.

Chinese also eat chicken testicles, but I haven't ever eaten one.

Scottish pickled sheep of the same thing. Or Rocky Moutnain oysters.

Have you tryed Harry's?

I know somebody who's vietnamese, and he said that they eat monkey brains.

Some traditional Jewish food is pretty weird and/or disgusting.  Consider kishka (intestines, usually of a cow), and chopped liver, and gefilte fish.  The fish isn't particularly bad, but if you buy jarred gefilte fish it usually has this strange jello like substance there.
A children's book from Germany that I read mentioned pickled pig's knuckles.

don't know about cow tongues... but i've had pig tongues, quite tasty actually

chinese ppl eat everything:
pig knuckles and intestines (basically anything pork related), dog sometimes, abalone, shark fin, snake soup, beef tongue, and i think even horse (im not sure about that one though), and more stuff too that i cant think of right now

I hade durian once and never want to have it ever again

My grandmother used to cook this stuff that was essentially refried bacon.  Called it Fatback.  I have also seen deep fried oreos, candy bars, and twinkies.
I ate a deep fried oreo and thought I was going to have a heart attack.

Gotta love the Southern US!

Actually, I think that here in the Southern US, people cook with too much fat.

I have a couple of theories about it, but think about it. Most of the poeple here in the south worked in agricultural labors with basically no tecnological help during almost two centuries.

That has changed slowly during the last century or so, but think about it, before, you needed to work hard, fisical work a lot of hours a day, and therefore you needed a lot of energy all day long, that is, you needed to have many many calories to burn, and eat fatty food is a quick way to obtain such calories.

The problem, of course, people no longer needs that many calories, therefore the food should be ajusted, but people likes the food they have eatten for generations, and that is, of course, fatty food.

Best,

[quote]Some Koreans (I repeat, some) eat dog.
Of course, nobody eats their pet dogs (at least, I hope not)...
Supposedly dog is one of the healthier types of meat... hmm.[/quote]

many people make fun of my native country( korea ) for eating dogs, and I find it very interesting....

I don't get offended by that because it's a cultural difference....

and I like being different than others...

well, I never actually tasted dogs before but my parents have...

hm....

this is what my formal social studies teacher said, but
on the southern part of Africa, they eat elephant tails...

and I hear some (more ) people in South America eat penguins too....
( penguins live in south america ... VERY sourther part of south america... )

don't make fun of cultural differences.. some of my friends ( american ) eat horse meat..

Yeah, you gotta love the fatty American food (except if you want to lose weight).

Where else can you get a 'heart attack'? (it is a steak with omelet and cheese and bacon all on a huge hoagie loaf :D )

Really, Americans eating horse meat? I can never find here in America, only in Europe.


geometry
trapezoid
symmetry
similar triangles
Pythagorean Theorem

The bases on an isosceles trapezoid are 18 and 30, and its altitude is 8.  Using each leg of the trapezoid as a diameter, semicircles are drawn exterior to the trapezoid.  If the midpoints of those arcs are $ P$ and $ Q,$ compute $ PQ$.

[hide="Sketch with answer"]

Pythagoras yields a radius of 5 for each circle. Use similar triangles on the 3-4-5 triangle and the triangle formed by the line PQ, the leg of the trapezoid and the line from the center of the circle to the midpoint of the arc. You get $ 7 \cdot 2 \plus{} 18 \equal{} 32$

[/hide]

For Fun Geometry Problems 5-8, I do not have an answer to them so I'll post my solution as well. I got same answer as JavaMan did.

[hide="My solution"]
Construct a diagram as attached.

First, note that $ O'E \equal{} O''F \equal{} 5$ since they are all radius of the semicircles.  Now, it's time to note that there are large number of similar triangles, and we can make a good use of them by comparing them to the big triangle (the third vertice isn't name).  It's the triangle that you get if you drop down $ A$ and hit in perpendicularly on $ CD$.  Anyway, by using the similar triangles with $ \triangle AO'E$, we get $ O'I \equal{} \frac{25}{4}$. Note that by symmetry, this means that $ O''J \equal{} \frac{25}{4}$ as well. So, the problem now comes down to finding $ IJ$ and adding them to $ \frac{25}{2}$ will give the length of $ PQ$.

I apologize for not labeling the diagram with $ PQ$.  :roll:

Anyway, by Pythagorean Theorem, $ IE \equal{} \frac{15}{4}$.  Now, use similar triangle again with $ \triangle IDK$ to find that $ IK \equal{} 7$.  Using Pythagorean Theorem on this triangle, we get $ DK \equal{} \frac{21}{4}$.  Again, there are two of them to subtract from 30, or the length of $ CD$.

Once done that, we add the result, which is 19.5, to $ \frac{25}{4} \cdot 2 \equal{} 12.5$ to get 32 as stated above. [/hide]

Yeah, I'm not very good at geometry so this explanation was probably bad and more than needed. But, it still shows that how much powerful thing this similar triangles can be.  :yup:


In triangle ABC, let AD the angle bisector where D lies on segment BC, let M be the midpoint of BC. Let E be the foot of perpendicular from B to line AD, and let line BE intersect line AM at N. Prove that DN is parallel to AB.

I shall rewrite your problem:

[color=blue]In $\triangle{ABC}$, $AD$ is the angle bisector of $\angle{A}$ and $D$ is on $BC$. $M$ is the midpoint of $BC$. $E$ is a point on $AD$. $BE$ meets $AM$ at $N$.
Prove: $BE \perp AD$ if and only if DP // AB.[/color]

Proof: (Your original problem)
Extend BE to meet AC at F. Then use Menelaus on $\triangle{ACM}$ and line $FNB$:
\[\frac{AF}{FC} \cdot \frac{CB}{MB} \cdot \frac{MN}{NA} = 1\]

\[\frac{MN}{NA}=\frac{FC}{2AF}\]

Since $AD$ is the angle bisector and $BF \perp AE$, so $AF=AB$. (Very Important Conclution.)

Now, use $CB=2CM$ and the Angle Bisector Theorem:

$\frac{MD}{DB}=\frac{CD-CM}{BD}=\frac{CD}{BD}-\frac{BC}{2BD}=\frac{CD}{BD}-\frac{CD+BD}{2BD}= \frac{1}{2}(\frac{CD}{BD}-1)$

$=\frac12 (\frac{AC}{AB}-1) = \frac{1}{2} \cdot \frac{AC-AF}{AF} = \frac{CF}{2AF}= \frac{MN}{NA}$

Therefore: PD // AB.

The only if part is a problem of the 58th Moscow Mathmetical Olympiad in 1997. Basically we can use the same method to prove $AF=AB$ to imply $BE \perp AE$.

Shobber


Complete the square to find the center and radius of the circle:

3x^2 + 3y^2 - 6x + 30y + 15 =0

Thanks

Start by dividing by 3:

$ x^{2}\plus{}y^{2}\minus{}2x\plus{}10y\plus{}5 \equal{} 0$.

Now, the $ x^{2}\minus{}2x$ suggests $ (x\minus{}1)^{2}$, and $ y^{2}\plus{}10y$ suggests $ (y\plus{}5)^{2}$, and

$ (x\minus{}1)^{2}\plus{}(y\plus{}5)^{2}\equal{} x^{2}\minus{}2x\plus{}1\plus{}y^{2}\plus{}10y\plus{}25 \equal{} x^{2}\minus{}2x\plus{}y^{2}\plus{}10y\plus{}26$.

This differs from our initial equation by 21, let's add to both sides of it this quantity:

$ x^{2}\plus{}y^{2}\minus{}2x\plus{}10y\plus{}5\plus{}21 \equal{} 21\Rightarrow  (x\minus{}1)^{2}\plus{}(y\plus{}5)^{2}\equal{} 21$.

So the center of our circunference lies on the point (1, -5), and its radius is $ \sqrt{21}$.


induction
least common multiple
algebra unsolved
algebra

Do there exist positive integers  $ a_{1}<a_{2}<.......<a_{100}$  such that for  $ 2\leq k\leq 100$ , the least multiple of  $ a_{k\minus{}1}$  and $ a_{k}$  is greater than the least common multiple of $ a_{k}$ and  $ a_{k\plus{}1}$.

[quote="b555"]Do there exist positive integers  $ a_{1} < a_{2} < ....... < a_{100}$  such that for  $ 2\leq k\leq 100$ , the least multiple of  $ a_{k \minus{} 1}$  and $ a_{k}$  is greater than the least common multiple of $ a_{k}$ and  $ a_{k \plus{} 1}$.[/quote]

Replace $ 100$ by $ n$ and let the proposition be $ P(n)$.

When $ n\equal{}3$ , we can easily find an example , e.g. $ 5<8<12$ . So $ P(3)$ is true.

Assume that $ P(m)$ is true and $ a_1,a_2,a_3, ... ,a_m$ satisfies the condition.

Then let $ p>lcm(a_1,a_2)$ be a prime , we have $ p,(p\plus{}1)a_1,(p\plus{}1)a_2,...,(p\plus{}1)a_m$ is an example to prove $ P(m\plus{}1)$. So $ P(m\plus{}1)$ is also true.

So by induction $ P(n)$ is true for all integer $ n \ge 3$. Put $ n\equal{}100$ , done.


[quote="dogseatcheese"][quote="ch1n353ch3s54a1l"][quote="dogseatcheese"][b]What is[/b] $7 5/7$% [b]of the expression[/b] $17.7 - 2 1/9 + 2.3$ [b] ? Express your answer as a decimal to the nearest hundredth.[/b][/quote]

is it $7\frac{5}{7}$% (as it says in the original text), $\frac{75}{7}$%, or {75}/{7%}?[/quote]

:blush:I am so sorry. I have absolutely no idea how to use latex and so.......I messed up. it is supposed to be $7\frac{5}{7}$% and well let me try to rewrite the problem.

[b]What is[/b] $7\frac{5}{7}$[b]% of the expression[/b] $17.7 - 2\frac{1}{9} + 2.3$[b]? Express your answer as a decimal to the nearest hundredth.

 Please show all of your work. Thanks[/b]  :)[/quote]

[hide]Thanks
$17.7 - 2\frac{1}{9} + 2.3 = 20 - \frac{19}{9} = 17\frac{8}{9}$
so then
$\frac{162}{9} \times \frac{54}{700} = \boxed{\frac{243}{175}}$[/hide]

[hide]
Evaluating the $17.7 - 2\frac{1}{9} + 2.3$ part, we get $17\frac 89=\frac{161}{9}$.
The percent part is equivalent to $\frac{54}{7}\%=\frac{54}{700}=\frac{27}{350}$.
Multiplying together: $\frac{161}{9}\times\frac{27}{350}=1.38$[/hide]

[hide]7 and 5/7 converted to a percent would be infinitely repeating, so we can just convert this into an improper fraction and multiply by 1/100. This gets 54/700, or 27/350. For the expression, rearranging terms gets 20-2 1/9, or 161/9. (161/9)(27/350)=483/350. Divide 483 by 350 and you get [b]1.38[/b].[/hide]

I remember doing this problem last year :lol: .

[quote="dogseatcheese"][b]What is[/b] $7 5/7$% [b]of the expression[/b] $17.7 - 2 1/9 + 2.3$ [b] ? Express your answer as a decimal to the nearest hundredth.[/b][/quote]

[hide]1.66? I added 17.7 and 2.3 first. That is 20. Then Subtracted 2 1/9. That gave me 161/9. Then I multiplied that by 54/7*1/100. That gave me $161/9$*$54/700$ By Cross Canceling, it would be $161*6/700$, which gives 1.38.[/hide]

[hide]Do it on your calculator. Its 1.38. :D [/hide]


what is your question?? this makes no sense
i think the answer is the same number because it has to be at least and at most so < or = as well as > or = so the only thing shared between these two inequalities is the = part. thus they must be equal.

Basically id refer to them as less than or equal to ($\le$) and greater than or equal to ($\ge$). for example less than or equal to x means that all real numbers that are less than x, but also includes x...get it?

OK. first of all please use $\LaTeX$. You're wording is confusing. Secondly, break down what you're talking about with each relation.

if x is atleast y, then the least x can be is y. so it is impossible for $x<y$ so $x\ge y$

if x is atmost y, then the most x can be is y. so it is impossible for $x>y$ so $x\le y$

[quote="Interval"]x  is AT LEAST another number y can be expressed as:

x > or = to y.

AND

x is AT MOST another number y can be expressed:

x < or = y.

My question:

Why is AT LEAST "greater than or equal to" and AT MOST "less than or equal to"?

I don't understand this concept.

Thanks![/quote]

I have at least 3 marbles. what are some possible numbers of marbles I could have?

3, 4, 5, or any number greater than, or equal to,  3

I have at most 10 apples. what are some possible numbers of apples that I could have?

10, 9, 8, or any number less than, or equal to, 10

Hope this helps

This is what I mean:

Express "x at least 7."

My answer: x is greater than or equal to 7.  Why [color=red]GREATER FOR AT LEAST?[/color]

Another sample:

Express "x is AT MOST 6."

I write:

x is less than or equal to 6.  Why [color=orange]LESS THAN FOR AT MOST?[/color]

if x [blank] 6 (or any number) we know either
x<6,
x>6, or
x=6.

the signs for 'less than or equal to' and 'greater than or equal to' are combinations of two those three- we know it's not one of the 3 choices, but of the other two we're not sure which it is. by saying at most, you limit yourself and say taht x can't be greater than that number. by saying at least, you limit yourself saying x can't be less than that number: that number is the least that x can be.

AT LEAST - should not be less than the number

Your age must be at least 18 to blah blah blah... meaning, you have to be 18 or older to blah blah blah

-

AT MOST - means the maximum, the number should not exceed the maximum

The number of students in a class should be at most 20 - it means that the maximum number of students is 20, a class can't have 21 or more students.

-

In your example:

x is at least 7 - it means that x should be no less than 7. :wink:

At this point we can apply a really cool Theorem known as Chen's Lemma
[b]Theorem[/b]: THe minimum of something is less than or equal to its maximum.

Q.E.D.


Putnam calculus geometry induction integration AMC USA(J)MO

Here we go - what did you think?

I'd say it was easier than it is sometime, what with the straightforward calculus problem for A1 and the "what was that?" problem B5. Personally, in real time (while proctoring), I got A1, A2, A3, B1, and B5 - but then, I'm not really the best judge of these things.

B5 was totally outrageous.  Also, whose idea was it to put two combinatorial geometry problems in a row?  Gah.  I thought B was comparable to last year's A (both of which I found tough) and that A was harder than last year's B (which had a couple of real giveaways, if I remember correctly) but still not terrible.  So, a bit harder than last year -- probably more zeros, because there wasn't anything really easy.  I bet scores would have been higher if B5 was numbered B3.

I've made my usual posting to the AMC Putnam archive. The only difference is that the archive front page has changed URLs, to http://www.unl.edu/amc/a-activities/a7-problems/putnamindex.shtml

I'll probably update the solutions a couple of times in the next week or two, as I hear about corrections and additional solutions.

For your solution to A1, that should be "xy plane", right?

I got solutions to A1 and A2 and had the right answer to B3, but apparently not the proof. When I got to B, I knew that I would have no chance at solving any of the calculus problems, so I didn't even attempt them.

Same as last year in difficulty, I guess?

About an hour into the discussion:

Not only does A6 have no replies, it has the least page views so far. It seems to have been genuinely difficult - did everyone just give up on it?

I vote "was genuinely hard."  Everyone around me seemed to draw really nice pictures to illustrate it, though.

B paper was pretty easy, A paper was pretty hard?

I got A1/A3/B2/B3/B5.  A4 was not bad though, and I would have gotten B1 if not for being a total idiot.

I can proudly say I probably got zero.
(probably for being infinitesimally hopeful)

So I am one of the 50% takers.

I submitted A2,A3,B2,B5.

This was my first time, wishing for two-digit next time.

I think that the easy problems (A1 A2 B1 B2) were harder than last years [Its quite hard to beat last year's B1] A1 B1 were still easy to get, nothing much required. A2 B2 had standard arguemnts for questions of that type. Nothing fanciful required.

The medium questions however were easier. A3 was standard, A4 wasn't hard. B3's score might be lowered due to lack of rigour in the combinatorical argument. B4 apparantly wasn't solved by many, but B5 definately made up for it. Seems like the top 25 cutoff would be higher, but thats just my guess. Could potentially hit 80 depending on how well solutions are written up.

The hard questions were really hard though. A6 seems doable only by those who have seen it before. For B6, Kedlaya's numericial solution posted on MAA is almost impossible to guess. A5 was semi-standard if you recognize it in terms of polynomials. B5 bah. I think many people missed out B5 solely from the fact that it was placed so far back. I threw it out when i first saw it, and only looked at it again when i had time to spare.

This putnam seems to favor those who have had lots of experience in competition problem solving. I'm not sure any of them required brillant/creative insights, unlike some other putnam problems. A6 and B6 seems like long ugly computation, with no nice motivation or simplification (I might just be wrong, but the answers really do not suggest a slick solution, unlike last year)

I wasnt too happy with my performance even if i am a freshman.
For A1 i made a crucial mistake and thus my answer was way off. Then A4/A4 (the one about "local maxes") i was going for an induction proof but couldnt guess a formula.
Part B i made some progress on B1. I found the point (-1,-1) but for some reason didnt think to look at the line $x+y=1$ also when i saw B5 i was thinking wow this looks really simple. but decided to spend my time on B2 since usually 5 tends to be harder than 2. o well. at least i was able to get a solution.

in my opinion looking back on the test it really was not as difficult as putnam from previous years. In any case I would be surprised if I got above a 3.  :(

I just bombed on paper $A.$ I felt it was unusually hard - certainly harder than last year when I got at least one problem right on it. I actually left the examination room after two hours - I know this doesn't look good on me, but I just didn't have the energy to proceed for too long. I guess it could also be due to the low phase I am going through these days.

$B1$ was easy for those people who knew exactly what they needed to do. Note that Dr. Titu Andreescu was on the Putnam Committee last year and I believe he is still on it this year too. Now, if any one of you has ever done problems from the first chapter (could be second) of "Mathematical Olympiad Treasures", by Dr. Andreescu, you will see why $B1$ is a giveaway. The equation $x^{3}+y^{3}+3xy = 1$ figures in one of the problem sets.

Sadly, this is the only problem I got right this year, down from almost three last year. I think I will have to give up math.

Lastly, I think this year's Putnam was geared towards students who have been exposed to olympiad style problem solving.

High-school junior here (of course, my first time taking it), and I thought it was about the same level as most of the ones I've seen problems from online. Pretty average as far as Putnams go. No regrets about taking it early, though.

I submitted A1, A2, A4, B2, B3, and a partial solution on B4 (I ran out of time writing it up - hopefully I'll get some partial credit.) So, all in all, hoping I did well!

im my opinion the A test quite harder than i expected.
and the B test, i really disliked seeing B3 and B4. especially B4, since i really dont understand what the question is even asking.
it was a suprise that B5 was so easy, but i didnt even spend anytime on it.
im hoping to get marks from A1/A2/A3/B1/B2 (but probably will get a zero from some of them).

This is the first year I have taken it and it was easier than I expected. Looking at the posted solutions I probably got somewhere in the 50(A1,A4,B1,B2,B3,B4) range so I am happy.

I messed up my computations on A1.  I did the polar substitution, said it was a torus and made an integral but didn't get it to come out right.  On B1 I got it but my solution was horrible since I started writing stuff down, crossed through it, wrote more down, crossed through it etc.  Anyway, I'll probably get between 10 and 20.

I'm definitely going to try harder for next year.  This year I didn't prepare for it very much and during the test I did the first problem on each set and then couldn't get myself to focus on any other problem.  The motivation just wasn't there.  I pretty much had the attitude coming into this year that I didn't have much of a chance of placing in the top 200 or 500 or whatever is posted so I didn't try my hardest.  This, obviously, is the wrong attitude.  Anyway, by next year I'll have had linear algebra, differential equations, and an analysis class so if I can just add in some problem solving on the side I should be in good shape.  This was a good wake-up call to me.

So, as a first-timer, I'm curious: How tough or easy do the graders usually grade the tests? For example, if I had the right general idea for a problem and the proof was correct but not 100% rigorous, will they give me a 9, a 5, or a 1? Also, how horrendous does your handwriting have to be to lose points for it?

Thanks!

[quote="FieryHydra"]Lastly, I think this year's Putnam was geared towards students who have been exposed to olympiad style problem solving.[/quote]
I strongly agree with this and I was pretty annoyed (especially after the A section) because of it. I think people who are familiar with USAMO level number theory and combinatorics considered A2-A4 to be trivial. A1 was just calculation. Most of my friends thought the B section was a lot harder, but since I'm not good at olympiad style problems I thought B was much easier. I wrote actual solutions for A1, A3, B1-4. I made a minor and stupid but possibly fatal mistake on A1 (didn't multiply by 2). I'm going to pessimistically expect about 30 so that I'm unlikely to feel bad about my result (and because I tend to be pessimistic about my own math abilities most of the time).

[quote="Xevarion"]
I'm going to pessimistically expect about 30 so that I'm unlikely to feel bad about my result (and because I tend to be pessimistic about my own math abilities most of the time).[/quote]
I was quite pessimistic about my math problem solving abilities last year after the Putnam; this year, even more so. Earlier I thought I was gonna do a double major in computer science and math, but now I feel I am just not cut out for the latter. I get the feeling that these days one has missed the boat if one hasn't been exposed to olympiad problems from an early age. Several weeks ago, I was reading about Alexander Grothendieck, and his biography gave me hope that all was not lost: you really don't have to be able to do a lot of contest style problems in order to understand and do serious math. But now, all that has changed, and I am even more pessimistic about my math abilities. I am seriously thinking of quitting math and just sticking solely to computer science.

And, as Xevarion pointed out, correctly I think, section B was easier than section A, which was harder than the 2005 section A.

BTW, Xevarion, I think you are extremely good at math! :)

[quote="FieryHydra"]BTW, Xevarion, I think you are extremely good at math! :)[/quote]
 :blush:  :blush: Thanks, haha :D. I will concede that I have studied a lot of analysis for someone my age. But advanced math doesn't always help much on Putnam; in fact, it can hurt -- I didn't get B5 (and wasted time on it) because I tried to kill it with unnecessarily powerful techniques.

I was really hoping that I could use some of my unusually advanced analysis knowledge to destroy one of the hard problems, but unfortunately there weren't any problems where I could use geometric measure theory or ergodic theory...

You are awesome Xevarion!  I hope that one day, I'll be able to understand all this crazy analysis stuff you are always doing.  When I saw B5, I thought to myself, Xevarion will probably destroy that problem with ergodic theory and geometric measure theory or something, it's not even in my league haha.  I just wish the putnam had an afternoon session and an evening session, that's so much more in sync with my sleep schedule.  Thinking about B3 and B4 put me right to sleep.  Overall, putnam was really fun.

[quote="pdiao"]I just wish the putnam had an afternoon session and an evening session, that's so much more in sync with my sleep schedule.[/quote]

If you go to the Budapest Semesters in Mathematics program fall semester, you can take the Putnam in 2 sessions, from 3-6 and 8-11.

I think that many former participants of IMO will score a lot on this putnam since problems A1234 and B12345 dont require some extraordinary effort.  By this I mean that there are no huge steps to do on the solution.

Here is my personal opinion about some problems:
A1 is standard calculus.
A2 is based on the well-known fact that there are very big gaps between prime numbers
A3 is the same to ISL'1998 N1(may be wrong about the year) which uses the idea of periodic sequences.
A4 is double-counting( I admit I couldn't do it on the exam)

B1 needs some concrete observations but is not unnatural to get to them
B2 a simple double use of pigeonhole principle.
B3 is nice combinatorial geometry but an experienced competitor has nowhere to mistake on it.
B4 has an obvious answer, but needs careful proof.
B5 is WTF- just AM-GM

I did A123 and B12345 and hope to get at least 70pts for a H.Mention :)

In my oppinion all putnam fellows will have 90+. I allready know 3 people that have solved 9+ problems and suspect that there are others as good as them.

Btw, does anybody know when the results become available? Thx in advance.

Putnam fell at the end of a ridiculously exhausting week for me, unfortunately.  I ended up oversleeping and missing the first half of the morning session, and I'm still kicking myself for not telling a friend to call me at 7:45 if I wasn't there.

I think I mostly made up for that, though, by being incredibly in the zone for that first hour and a half, more so than I've been in a very long time.  I ended up solving A1, A2, and A4, although I may lose points on A4 because I did the whole thing (solving and writing) in the last 5 minutes.  I think if I had had the full time, I would have had a shot at A3 as well.

B-side went less well.  I came up with a full solution for B1, but couldn't do B2.  I had some sketchy reasoning for B3 (that actually looked a lot like an old USAMO problem), and it probably works.  I mainly BS'ed B4.

So in short, I'm looking at anwhere from high 20's to high 50's, depending on my scores for A4, B3, and B4.  I'd be surprised if I break 50, though.

[quote="Kalle"][quote="Kent Merryfield"]They're very tough. You won't get a 5 on any problem - in fact, the only numbers in use for credit on any problem are 0,1,2,8,9,10. Assume that those answers getting 8 or 9 are really quite correct, there's just a quibble about some detail that should have been accounted for. If you think you're "half right," then maybe that's a 1. If you had some interesting ideas, figure that's probably a zero.[/quote]
Why is this always the case in mathematics competitions? I haven't written the putnam, but in swedish competitions I have been in you would only be awarded 3/7 points for a completely mathematically correct solution, while the other four points would be awarded for extraordinary elegance or something*. I think it's pretentious and makes mathematics into figure skating.

*Note that this is not bitterness, but the actual stated rules.[/quote]

That is not at all equivalent to the Putnam grading -- it is 10 points for a full solution, but if you only get part of the way there, they aren't going to assume you knew how to finish it, so they aren't going to give you 6/10 just because you're 3/5th of the way through the official solution.  The Putnam is stingy with extra credit (including in cases of a right "overall idea" but missing rigor -- something that will happen to a lot of competitors on B3 this year, I expect), but they aren't asking anything more than that you get it exactly right :-)


algebra
polynomial
inequalities unsolved
inequalities

Polynomial $ f(x)\equal{}a_0\plus{}a_1x\plus{}a_2x^2\plus{}\ldots \plus{} a_nx^n$ and $ f(1)\equal{}1 \; f(0,9)\equal{}0 \quad \minus{}0,1 \le f(x) \le 0 \quad \text{ for each  } \; 0\le x \le 0,9$.
 It prove, that  $ n\ge 4$

[quote="aktyw19"]Polynomial $ f(x) \equal{} a_0 \plus{} a_1x \plus{} a_2x^2 \plus{} \ldots \plus{} a_nx^n$ and $ f(1) \equal{} 1 \; f(0,9) \equal{} 0 \quad \minus{} 0,1 \le f(x) \le 0 \quad \text{ for each } \; 0\le x \le 0,9$.
 It prove, that  $ n\ge 4$[/quote]

You mean to say

If $ f(x) \equal{} a_0 \plus{} a_1x \plus{} a_2x^2 \plus{} \cdots \plus{} a_nx^n , f(1) \equal{} 1, f(0.9) \equal{} 0$ and $ \minus{} 0.1 \le f(x) \le 0$ $ \forall x\in[0,0.9]$,prove that  $ n\ge 4$


A sports car passes a police check point at $60\ km/hr$ and then immediately started slowing down uniformly till its speed was $40\ km/hr$.It continues to move at this speed until it was passed by a police car $1\ km$ from the check-point.The police car had started from rest at the same instant as the sports car had passed the check-point.The police car had moved with a constant acceleration until it had passed the sports car.Assuming the time taken by the sports car in slowing down from $60\ km/hr$ to $40\ km/hr$ was equal to the time it had travelled at a consrtant speed before it was passed by the police car,find:-

$a)$time taken by the police car to reach the sports car

$b)$speed of police car at the instant when it passed the sports car

$c)$the time measured from the check-point when the speed of the two were equal.

 :| man,I sure can tell you finding the answers is easier than typing tjhe question.....

I'm not really good at physics... but here goes:

[hide]
1. t = 1/45 hours
2. v = 90 km/hr
3. t = 0.0103 hour[/hide]


geometry
ratio
invariant
parallelogram
analytic geometry

Let ABCD be a parallelogram. Join each vertice with the midpoints of the opposite sides. Determine the area of the formed octagon.

Source:Selection of the team representing Albania in the BO 2004
It seems to be a generalization of problem 5, round 10,International Mathematical Talent Search.

Since area ratios are invariant under stretches and shearings of the plane, we may assume that the parallelogram is a square. Now it isn't too difficult. You could even use coordinates if you want! I think you find that the answer is 1/6 the area of the parallelogram.


For any set A of positive integers for example$\{a_1,a_2,...,a_n\}$,we define the follow set $A*=\{a_i+a_j|1\leq i<j\leq n\}$.
Suppose A and B are different and A*=B*,prove that$|A|=|B|=2^k$for some positive integer $k$. :?:

that $2^k$ smells like number of partitions of a set.
so the problem maybe can be reformulated like this:
if A,B are 2 sets and A*,B* are equal(A*=sets composed with all sums possible between different elements in A ,|A*| also identifies with the number of 2 element subsets in A).
so A,B are the partition sets of sets A* and B* (becuase A*=B*).
again...this is just by intuition,i am aware that what i have written may be false.

this problem is very famous and have been posted before:
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=selfridge&t=29381


Pls Help me with this :
* prove that
sin12.sin48.sin54 = 1/8
all in degrees.(without using calculator. must be derived from product or sum formula)
thx.

hello, use that
$ \sin(12^{\circ})\equal{}\frac{\sqrt{3}}{8}\minus{}\frac{\sqrt{15}}{8}\plus{}\frac{\sqrt{2}}{8}\sqrt{5\plus{}\sqrt{5}}$
$ \sin(48^{\circ})\equal{}\minus{}\frac{\sqrt{3}}{8}\plus{}\frac{\sqrt{15}}{8}\plus{}\frac{\sqrt{2}}{8}\sqrt{5\plus{}\sqrt{5}}$
$ \sin(54^{\circ})\equal{}\frac{1}{4}(1\plus{}\sqrt{5})$
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, use that
$ \sin(12^{\circ}) \equal{} \frac {\sqrt {3}}{8} \minus{} \frac {\sqrt {15}}{8} \plus{} \frac {\sqrt {2}}{8}\sqrt {5 \plus{} \sqrt {5}}$
$ \sin(48^{\circ}) \equal{} \minus{} \frac {\sqrt {3}}{8} \plus{} \frac {\sqrt {15}}{8} \plus{} \frac {\sqrt {2}}{8}\sqrt {5 \plus{} \sqrt {5}}$
$ \sin(54^{\circ}) \equal{} \frac {1}{4}(1 \plus{} \sqrt {5})$
Sonnhard.[/quote]
Can you explain how you got these values?


inequalities
LaTeX
inequalities proposed

[quote="darkdevil"]Let {x_i}, {y_i} real numbers and \frac{x_i}{y_i}=r. Prove that:
(\sum{i=1}{n}{x_i}^2-\sum{i=1}{n}{y_i}^2)(\sum{i=1}{n}{({y_i-r})}) >= (1-r) (\sum{i=1}{n}{x_i}+\sum{i=1}{n}{x_i})^2[/quote]

I think you mean this:

Let ${x_i}, {y_i}$ real numbers and $\frac{x_i}{y_i}=r$. Prove that:
$\left(\sum_{i=1}^{n}x_i^2-\sum_{i=1}^{n}y_i^2\right)\left(\sum_{i=1}^{n}({y_i-r})\right) \ge (1-r) \left(\sum_{i=1}^{n}x_i+\sum_{i=1}^{n}y_i\right)^2$  ;)

[quote="darkdevil"]yeah, that`s what i mean (i`m new to this forum, so don`t know exactly how to use those cmds)[/quote]

For instal LaTeX software to your computer, see [url=http://www.mathlinks.ro/LaTeX/AoPS_L_About.php]here[/url]. And I think the best way how o study LaTeX is watching how other people post...

Good Luck..

Anyway, can I ask you, darkdevil, how did you invented it? Where does it come from?

[quote="darkdevil"]hmmm......something wrong at it......i saw what...so i reviewed it and:

Let $x_i, y_i$ real numbers and $\frac{x_i}{y_i}=r$ Prove that:
$\frac{1+r}{r-1}\left(\sum_{i=1}^{n}x_i^2-\sum_{i=1}^{n}y_i^2\right) \ge \frac{1}{n}\left(\sum_{i=1}^{n}\left(x_i+y_i\right)\right)^2$[/quote]So what? It's just Cauchy :rotfl: $(1+r)^2\sum_{i=1}^{n}y_i^2\ge\frac1n(r+1)^2 (\sum_{i=1}^{n}y_i)^2$

hehehe :)
hmmm...btw.....what`s cauchy ineq? it`s not cbs, right?


Euler
algebra
polynomial
modular arithmetic
algorithm
number theory

find the remainder when $ x^3 \plus{} 3x^2 \plus{} 2x$ is divided by $ x \minus{} 1$.
using number theory

[hide="Not sure if this is what you mean by using number theory"]We have

$ x^3 \plus{} 3x^2 \plus{} 2x$

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

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

Hence the remainder is $ \boxed{6}$.[/hide]

thanx but no i mean by (congruence,mod)

an example.
[hide][quote="binomial_4eva"]1
[hide]Using Euler's theorem $ a^{\varphi (m)} \equiv 1\pmod {m}$
remainder of $ \frac {3^{1874}}{11}$
$ (3,11) \equal{} 1\implies 3^{\varphi (11)} \equiv 1\pmod {11}$
$ {\varphi (11)} \equal{} 10\implies3^{10}\equiv 1\pmod {11}$
$ \implies3^{1874}\equiv(3^{10})^{187}\cdot3^4\equiv1^{187}\cdot3^4\equiv 81\equiv4\pmod{11}$
$ \implies 3^{1874}\equiv4\pmod{11}$
therefore the remainder when $ 3^{1874}$ is divided by $ 11$ is $ \boxed4$[/hide]
i will do the rest soon[/quote][/hide]

Well, this is colse to impossible since the mod ($ x\minus{}1$) is dependant on $ x$ too. Also the exponents are constant which makes Euler-Fermat close to impossible to apply.

[hide="A very straightforward way to solve this problem"]
Set $ x\equal{}1$ in $ f(x) \equal{} (x\minus{}1)\cdot g(x) \plus{} r$ ;)
[/hide]

[quote="bertram"][hide="A very straightforward way to solve this problem"]
Set $ x \equal{} 1$ in $ f(x) \equal{} (x \minus{} 1)\cdot g(x) \plus{} r$ ;)
[/hide][/quote]One theorem: Remainder Theorem

Thank you very much everyone, but i was looking for a solution like that of JBL's, JBL which formula or theorem is that?. @ 100000th user ya remainder theorem is easier to use but i dont know how to use mod.

Which specific claim that I made are you confused about?  (All the proofs given so far are essentially the same, although nayel's calculated more than was necessary (essentially, the $ g(x)$ in bertram's solution).)

[quote="JBL"]bertram, you just said doing something is impossible and then you did it ;)

An equivalent way to phrase this is [hide]the remainder of a polynomial when it's divided by $ x \minus{} 1$ is just that polynomial modulo $ x \minus{} 1$.  Since $ x \equiv 1 \pmod{x \minus{} 1}$, the polynomial $ P(x) \equal{} x^3 \plus{} 3x^2 \plus{} 2x$ is equivalent to $ 1^3 \plus{} 3\cdot 1^2 \plus{} 2 \cdot 1 \equal{} P(1)$ modulo $ x \minus{} 1$.[/hide][/quote]

:( I was kind of fixed on Euler-Fermat.

The basic equation you use is $ x^k\equiv1^k\equiv1\pmod{x\minus{}1}$ and then you just have to add the coefficients.

JBL i get everything you said there, but is there a general formula? :)

The remainder that $ a_nx^n \plus{} a_{n \minus{} 1}x^{n \minus{} 1} \plus{} \dots \plus{} a_1x \plus{} a_0$ yields upon division by $ x \minus{} b$ is $ a_nb^na_{n \minus{} 1}b^{n \minus{} 1} \plus{} \dots \plus{} a_1b \plus{} a_0$. Are you asking for this? It's general enough to me...

Maybe he is looking for something much more general like dividing a polynomial $ P(x)$ by $ Q(x)$ and finding the remainder $ R(x)$...

I'm not sure what you mean by a "general formula," but here are the relevant results.

[b]Division Algorithm:[/b]  Given a polynomial $ A(x)$ and a nonzero, non-constant polynomial $ B(x)$, there exist polynomials $ Q(x), R(x)$ such that

$ A(x) \equal{} B(x) Q(x) \plus{} R(x)$

such that either $ R(x) \equal{} 0$ or $ \text{deg} R < \text{deg} B$.

[b]Corollary:[/b]  When $ B(x) \equal{} x \minus{} r$ for some $ r$, $ R(x) \equal{} P(r)$.

[b]Corollary (Factor Theorem):[/b]  $ x \minus{} r | P(x) \Leftrightarrow P(r) \equal{} 0$.


calculus
integration
modular arithmetic
number theory
relatively prime
combinatorics unsolved
combinatorics

Let $ L$ denote the set of all lattice points of the plane (points with integral coordinates). Show that for any three points $ A,B,C$ of $ L$ there is a fourth point $ D,$ different from $ A,B,C,$ such that the interiors of the segments $ AD,BD,CD$ contain no points of $ L.$ Is the statement true if one considers four points of $ L$ instead of three?

[hide="Solution."]
This can be done with $ 2^d\minus{}1$ points in $ d$ dimensions, but not with $ 2^d$. I'll stick to the $ d\equal{}2$ case but the generalization is trivial.

Observe that the segment between $ (x_1,y_1)$ and $ (x_2,y_2)$ contains no lattice points in its interior iff $ x_1\minus{}x_2$ is relatively prime to $ y_1\minus{}y_2$. Therefore, given $ r$ lattice points points $ A_1, \ldots, A_r$ where $ A_i\equal{}(x_i,y_i)$, in order to find a point $ Z\equal{}(x,y)$ satisfying the requirement of the problem, we need to find integers $ x,y$ such that $ \gcd(x\minus{}x_i,y\minus{}y_i)\equal{}1$ for all $ i$, $ 1 \leq i \leq r$.

If $ r\equal{}4$ (or more), this may be impossible if the $ A_i$ are badly chosen. Indeed, suppose that the $ (x_i,y_i)$ are congruent modulo 2 to $ (0,0)$, $ (0,1)$, $ (1,0)$ and $ (1,1)$. Then whatever point $ Z\equal{}(x,y)$ is chosen, there will be some $ i$, $ 1 \leq i \leq 4$ such that $ x \equiv x_i \pmod{2}$ and $ y \equiv y_i \pmod{2}$. Thus $ x\minus{}x_i$ and $ y\minus{}y_i$ are both divisible by 2, so they're not relatively prime.

If $ r\equal{}3$, we can indeed find such a point, as follows. Let $ (a,b)$ be a pair of residue classes mod 2 that doesn't appear among the $ A_i$. Since there are 4 possible pairs of residue classes and only three points $ A_i$, there must be such a vacant residue class. Similarly, we can find a pair $ (c,d)$ of residue classes mod 3 that is also not represented. Set $ x$ to be any integer congruent to $ a$ modulo 2 and to $ c$ modulo 3.

(The reason we have to worry about the primes 2, 3 is that there are 3 points, and if we choose $ x$ carelessly we may not be able to find any suitable $ y$, since all residue classes $ (x,y)$ mod 2 and 3 may be taken by one of the $ A_i$.)

Having chosen $ x$ as explained, we consider the numbers $ x\minus{}x_i$ for all $ i$. Let $ P$ be the set of all primes dividing any of the numbers $ x\minus{}x_i$. For each $ p \in P$, find some $ y(p)$ such that $ (x\minus{}x_i,y(p)\minus{}y_i)$ is not congruent to $ (0,0)$ mod $ p$ for any $ i\equal{}1,2,3$. This can be done for $ p\equal{}2,3$ by our choice of $ x$, and for $ p \geq 5$ simply because there are $ p$ options for $ y(p)$ and only 3 constraints.

Using the Chinese Remainder Theorem, find a number $ y$ such that $ y \equiv y(p)$ for all $ p \in P$. For this choice of $ y$, the numbers $ (x\minus{}x_i,y\minus{}y_i)$ for $ i\equal{}1,2,3$ are not both divisible by $ p$ for any $ p \in P$. Since all other primes don't divide $ x\minus{}x_i$ at all, those numbers are actually relatively prime, as desired.

[/hide]


search
function
graph theory
combinatorics unsolved
combinatorics

I would like to find the minimum number of vertices needed so that 2 coloring the edges gurantees 2 monochromatic triangles of the same color and prove it.  I think it is ten, but not sure if my reasoning is good enough.  Here is what I have so far, after lots of scratch paper I can draw a complete graph with nine vertices that does not have two monochromatic triangles of the same color.  But for ten vertices I can find one triangle of one color, take it away and im still left with 7 vertices so I can get a second triangle of the other color that leaves 4 vertices and since my graph is complete those four vertices are connected to at least two vertices form my existing triangles so I can get another monochroatic triangle it has to be color one or two so im done.  Am I close?

Maybe i misunderstood something?
But probably the number you are looking for is $ 6$.and it's true by Ramsey theorem...

Yeah 6 vertices guarantees one monochromatic triangle since
R(3,3)=6, but I need two monochromatic triangles and they must be the same color.

It's unclear to me -- do your triangles need to be either edge- or vertex-disjoint?  If not, then 9 suffices: one can show that on 6 vertices actually we always get 2 monochromatic triangles (perhaps sharing an edge, or perhaps of different colors), so finding and removing one triangle gives us our result.  (The result I mentioned has been posted on the forum more than once -- try playing around with the search function for a while to see if you can find it.  Posts by pbornsztein may be extra promising.)  What's the best lower bound you have for this?


In the following number behind the decimal point we write the square numbers one after the other:

0.14916253649...

What is the 100th digit behind the decimal point?

You could just try and list all the square numbers to practice your perfect square memory, but is there a faster way?

[hide]

The first 3 squares take up 3 digits.
The squares from $4^2$ to $9^2$ each take up 2 digits, making 12, and 15 in all.
The squares from $10^2$ to $31^2$ each take up 3 digits, making 66, and 81 in all.
Since 100-81=19=4*4+3, we are looking for the 3rd digit of the 5th square past $31^2$, which is the 3rd digit of $36^2$.
$36^2=1296$

So, our answer is 9
[/hide]

Is there an even easier way?

Edit: mistakenly wrote 5 instead of 6


Prove that the cube of any positive integer greater than 1 can be represented as a difference of the squares
of two positive integers.

There is a general condition for a decomposition into difference of two squares. Blang! Why did you not write it here? ;) :)

The numbers that are not the difference of two squares are $4n+2$.
But Myth, the simple above relation is more funny!  ;)  :)

I know! :) It is cool!

P.S. I don't see you in Yahoo ;)

Blang the relation you gave was simple enough to be verified


message from mathfanatic: I will return your post tomorrow.  Until then, it will sit in a word document on my computer.  I intend to enforce the no-discussion-of-scores-until tomorrow rule as strictly as possible.

Problems, scores, and discussions of difficulty must wait until tomorrow.  Until then I must lock this thread.


[hide]$ \imath^{3}/ \imath^{-1}= \imath^{3-(-1)}= \imath^{4}$
$ \imath^{4}= 1$
Therefore $ \imath^{3}/ \imath^{-1}= 1.$
Any number divided by a number equal to itself is equal to 1.
Therefore $ \imath^{3}= \imath^{-}1.$[/hide]

Yet another edit, still making discoveries here.

[hide="explaination and pattern."]We can write $i^{-1}$ and $i^{3}$:

$\frac{1}{i}$
$i*i*i$

and rationalize of simplify:

$\frac{1}{i}=\frac{i}{i*i}=\frac{i}{-1}=-1$

$i*i*i=-1*i=-i$

Here's a useful pattern:

[u]i | -1 | -i |  1[/u]
1 | 2 | 3 | 4
5 | 6 | 7 |  8
9 |10 |11|12...

The numbers under i represent the powers of i that equal i. The numbers under -1 represent the powers of i that equal -1. The numbers under -i represent the powers of i that equal -i. The numbers under 1 represent the powers of i that equal 1.[/hide]

[hide][quote="XCgirl"]Just going through [u]The Art of Problem Solving: The Basics[/u] and came across the following: Why is i^-1 = i^3?  Thanks in advance for deigning to answer such a simple question![/quote]

i^3=-i

1/i=i/-1=-i
[/hide]


[quote="ondrob"]You have 20 cards with numbers 0,0,1,1,2,2,3,3,...,9,9. Prove that you cannot order them in a line so that between two cards with numbers i will be exactly i cards.[/quote]

What exactly do you mean by the "between two cards with numbers i will be exactly i cards"?

I think he means you can't have like
0 0 1 2 1 3 2 4 5 3 6 7 4 8 5 ... but you run out of numbers?

Between first card with 0 and second card with 0 is 0 cards (so it looks like ...,0,0,...). Between first card with 1 and second card with 1 is exactly 1 card (so it looks like ...,1,something,1,...). Between first card with 2 and second card with 2 there's exactly 2 cards (so it looke like ...,2,something,something,2,...) and so on...

Prove that you cannot order them into a line in such way that all this conditions will be fulfilled.

Is it enough clear now?

see euclid 2004 (a canadian contest) question 10, or Canadian MO problem 5 , 1992 for similar question.

--

Let a_0 .. a_9 and b_0 .. b_9 be integers such that a_i and b_i are values corresponding to the two _positions_ of the cards with value i, where the positions are numbered 1 ... 20, and a_i < b_i.

Suppose we can find such an ordering.  Let the sum of a_i be A, sum of b_i be B.  Clearly A+B = 210.  Now it is given b_i - a_i = i+1, so B-A is clearly 55.  So we can solve to get B = 132.5 , which isnt an integer.  Oops.

I think your solution is all right... :)
Original solution is using coloring. Suppose you have such order and then color first, third, fifth,... card black and other cards let be white. Then when 2 odd cards have to be the same color (even cards have different colors). But there's five pairs of odd cards, so it's easy to see that if this order is proper then there can't be 10 black and 10 white. Contradiction.


ratio
geometry
AMC
AIME
similar triangles

In $\triangle ABC, AB= 425, BC=450,$ and $AC=510.$ An interior point $P$ is then drawn, and segments are drawn through $P$ parallel to the sides of the triangle. If these three segments are of an equal length $d$, find $d$.

There was only one solution posted in the answer topic for this question, and it was a pretty heavy arithmetic solution. I was wondering if there were other ways to do it. Thanks :)

[quote="1=2"]Are you sure its not perpendicular??? If it is, then it would be much easier...

IF it's parallel, do the segments meet the triangle and stop there?[/quote]

No, they're parallel. Each line connects one point on each of two sides so that the line is parallel to the third and that the point is an element of the line.

If they are parallel, then you always think of similar triangles- and the ratio of areas using similar triangles is r^2, where r is the ratio of the side length of those two similar triangles.. ill post a solution later

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=394401#p394401[/url]

my solution is fine...

it is not a lot since everything is symmetrical

Yeah, I guess it's all good then.

[quote="Art of Owna"][quote]Yeah they're parallel, not perpendicular.

In any case, When I tried using similar triangles with area and side ratios, it ends up with a final equation that takes ~2-3 minutes to solve (w/o using calculator since this is AIME question).

So I just wanted to see if anyone had a nice way of working it out.[/quote]

2-3 mins?  Dude, thats very quick-for a number 9-and you way the best method( AIME is 15 questions for 3 hrs-average is 12 mins/problem and thats if you do all of the problmes)[/quote]

I meant that the very last equation you get (after spending time doing the problem) takes 2/3 minutes to solve. It's cool though, thanks anyways.


algebra function domain symmetry LaTeX Asymptote

For the function y = 1/(x-2).

(A) Find Domain

(B) Find Range

(C) Determine any symmetries that are characteristics of the graph.

[hide="I wont get it wrong this time!"]

In $ \text{\LaTeX}$, the funtion is:

$ y=\frac{1}{x-2}$

This graph is the graph of $ y=\frac{1}x$ shifted 2 units to the right. It has an asymptote of $ x=2$.
Domain: $ x\neq2$
Range: R
This graph has point symmetry with respect to the point $ (2,0)$[/hide]

Dear sharkman,

Could you please stop posting homework like problems.  You have already posted multiple topics that ask exactly the same things but with different functions.  Asking us to solve them for you won't help you.  You have to try them yourself too.

-jli


geometry
geometric transformation
reflection
circumcircle
symmetry
geometry unsolved

Let triangle ABC inscribed in a circle (C). d is a line in a plane contained that triangle.
Let $ d_1$ is reflection of d in BC, $ d_2$ is reflection of d in AC,  $ d_3$ is reflection of d in AB.
Prove that: $ d_1$, $ d_2$,  $ d_3$ concurrent in a point M on (C) if only if  d pass through the orthocenter H of ABC.

[quote=ngoclam1088] Prove that: $ d1$, $ d2$ , $ d3$ concurrent in a point M in (C) if only if $ H \in d$ [/quote]

Should it be "M [b] on [/b] (C)" ?

In this case, "reversing" the problem should be useful. Note that the reflection images of $ M$ about the three sides of the triangle are points on $ d$. So: You need to prove that for any point on the circumcircle of triangles $ ABC$, its images after the reflection about the sides of $ ABC$ lie on the line which pass through the orthocenter.

The converse should not be difficult.

Yes! Point M is on the circle (C).But i can't prove that.Could you help me further, please!!

OK!

Suppose that the 3 lines are concurrent at $ M$ on $ (C)$. Let $ N$ and $ P$ be reflected images of $ M$ about the sides $ AB$ and $ AC$. Note that $ N$ and $ P$ are points on $ d$. So it suffices to prove $ N, P$ and $ H$ are collinear.

$ \angle{ANB} \equal{} \angle{AMB} \text{(due to symmetry)}\equal{} \angle{C} \Rightarrow \angle{ANB} \plus{} \angle{AHB} \equal{} \angle{C} \plus{} \angle{AHB} \equal{} \pi \Rightarrow ANBH$ is an inscriptible quadrilateral. Similarly, $ APCH$ is also an inscriptible quadrilateral. These imply $ \angle{AHN} \equal{} \angle{ABN} \equal{} \angle{ABM}$ and $ \angle{AHP} \equal{} \angle{ACP} \equal{} \angle{ACM}$. Then $ \angle{NHP} \equal{} \angle{AHN} \plus{} \angle{AHP} \equal{} \angle{ABM} \plus{} \angle{ACM} \equal{} \pi$. This completes the solution for the first part.

this is Bulgaria 2000 I think:

[b]a line $ d$ is drawn through the orthocenter of acute triangle $ ABC$.prove that the reflections of $ d$ across the sides of the triangle are concurrent.[/b]

[b]Solution.[/b] Because triangle $ ABC$ is acute,its orthocenter $ H$ is inside the triangle.without loss of generality we may assume that $ d$ intersects sides $ AC$ and $ BC$ at $ Q$ and $ P$,respectively.if $ d\parallel AB$,let $ R$ be any point on the reflection of $ d$ across line $ AB$.otherwise,let $ R$ be the intersection of $ d$ and line $ AB$,and assume without loss of generality that $ R$ lies on ray $ BA$.let $ A_1,B_1,C_1$ be the reflections of $ H$ across lines $ BC,CA,AB$,respectively,it is well known that $ A_1,B_1,C_1$ lie on $ (C)$.it suffices to prove that lines $ A_1P,B_1Q,C_1R$ are concurrent.
because lines $ AC$ and $ BC$ are not parallel or perpendicular,lines $ B_1Q$ and $ A_1P$ are not parallel.let $ S$ be the intersection of lines $ A_1P$ and $ B_1Q$.because:

$ \angle SA_1C\plus{}\angle SB_1C\equal{}\angle PA_1C\plus{}\angle QB_1C\equal{}\angle PHC\plus{}\angle QHC\equal{}\pi$

quadrilateral $ SA_1CB_1$ is cyclic,hence,$ S$ is the intersection of line $ B_1Q$ and circle $ (C)$.
likewise,line $ B_1Q$ and $ C_1R$ are not parallel,and their intersection is also the intersection of line $ B_1Q$ and circle $ (C)$.hence,lines $ A_1P,B_1Q,C_1R$ are concurrent at a point on the circumcircle of triangle $ ABC$.


Malo bolj udaren naslov teme, ker slednja pač ni tako - udarna. Gre pa nekako takole:
Danih je N oseb (N liho število). Teh N oseb glasuje. Naj bo res, da je povprečna verjetnost, da neka oseba glasuje za predlog strogo večja od 0,5; t.j. naj bodo p1, ..., pk, ...., pN verjetnosti k-te osebe, da glasuje za predlog, potem: (p1+...+pN)/N>0,5. Pokaži ali ovrži, da iz tega sledi, da je verjetnost da bo predlog izglasovan večji od 0,5. Osebe glasujejo neodvisno!
Intepretacija: Pri majority voting je "pravilni" izid volitev vsaj tako verjeten kot pri naključno izbranem volilcu, ki bi o predlogu sam odločil.
Zadevo lahko posplošite, preoblikujete ...

Nisi napisal, kaj pomeni, da je predlog izglasovan? Je to takrat, ko zanj glasuje vsaj 1 oseba, vsaj eden več kot polovica vseh oseb, natanko eden več kot polovica vseh, vsi?  :?

Če je tako, da mora v prid predlogu, da bo predlog sprejet, glasovati več ljudi kakor proti, potem mislim, da ta teorem ne drži.
Primer:
N = 3
p1 = 0.3001
p2 = 0.3001
p3 = 0.9000
Potem je to, kar ti imenuješ povprečna verjetnost: (0.3001+0.3001+0.9000)/3 > 0.5
Verjetnost, da bo predlog izglasovan, pa je potem približno 0.46813, kar je manj kot 0.5
(Če se seveda pri izračunih nisem zmotil. Verjetnost tega pa je več kot 0.5, zato naj raje
izračun za dane podatke še nekdo preveri. :roll: )


There are 100 boxers, each of them having different strengths, who participate in a tournament. Any of them fights each other only once. Several boxers form a plot. In one of their matches, they hide in their glove a horse shoe. If in a fight, only one of the boxers has a horse shoe hidden, he wins the fight; otherwise, the stronger boxer wins. It is known that there are three boxers who obtained (strictly) more wins than the strongest three boxers. What is the minimum number of plotters ?

[i]Proposed by N. Kalinin[/i]

Let's assume we have $p$ plotters and $u$ usual boxers, ie boxers who are not the three strongest and who do not participate in the plot.
If $p$ is minimum, the plotters should not use horseshoes on normal boxers, because it does not matter how many games they win. So we will assume that all plotters are stronger than usual boxers. In that case, the three strongest boxers and the plotters each get $u$ wins.
Let boxer $s_{1}$ be the strongest boxer, $s_{2}$ the second strongest one, etc.(note that in that case the plotters are boxes $s_{i}$, $4\le i\le p+3$) Then boxer $s_{1}$ beat boxer $s_{2}$ and $s_{3}$ and got $2$ wins among the three strongets boxers; boxer $s_{2}$ got one win, boxer $s_{3}$ got $0$ wins.
Among the $p$ boxers, boxer $s_{4}$ got $p-1$ wins, boxer $s_{5}$ got $p-2$ wins, boxer $s_{6}$ got $p-3$ wins. We want minimum $p$ so the boxers $s_{4},s_{5},s_{6}$ should be the ones getting more wins than boxers $s_{1},s_{2},s_{3}$. Each boxer $s_{4},s_{5},s_{6}$ should use a horseshoe on one of $s_{1},s_{2}, s_{3}$(to make $p$ a minimum) and so will get an extra win, therefore the total number of wins $s_{4}$ will get is $u+(p-1)+1$, $s_{5}$ will have $u+(p-2)+1$ wins, $s_{6}$ will obtain $u+(p-3)+1$ wins.
To minimize the number of wins of $s_{1},s_{2},s_{3}$ each plotter should use a horseshoe on them. To minimmize te maximum number of wins that $s_{1},s_{2},s_{3}$ will get, we need $\frac{p}{3}+1$ plotters using horseshoe on $s_{1}$, $\frac{p}{3}$ plotters using horseshoe on $s_{2}$, $\frac{p}{3}-1$ plotters using horseshoe on $s_{3}$. Then the total number of wins each of $s_{1},s_{2},s_{3}$ will get is $u+1+\frac{2p}{3}$.
Now, we need the minimum number of wins obtained by $s_{4},s_{5},s_{6}$ to be greater than maximum number of wins obtained by $s_{1},s_{2},s_{3}$, si have:
$u+(p-3)+1>u+1+\frac{2p}{3}$
$\frac{p}{3}>3$.
$p>9$. Taking $p=10$, we cannot reach the desired result:need at least $5$ plotters use horseshoe on $s_{1}$, at least $4$ to use horseshoe on $s_{2}$, at least $3$ to use horseshoe on $s_{3}$. Taking $p=11$, we cannot reach the desired result:need at least $5$ plotters use horseshoe on $s_{1}$, at least $4$ to use horseshoe on $s_{2}$, at least $3$ to use horseshoe on $s_{3}$. Finally, $p=12$ works: $5$ plotters use horseshoe on $s_{1}$, $4$ use horseshoe on $s_{2}$, $3$ use horseshoe on $s_{3}$.
So, answer is $12$.


calculus
integration
function
derivative
calculus computations

[1]
Let $ f(x) \equal{} (x \minus{} 1)(x \minus{} 2) ... (x \minus{} n)$

Then find $ \int \frac {f''(x)f(x) \minus{} (f'(x))^{2}}{f^{2}(x)}dx$

[2] The function $ f$ is continious and  have the property

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

Find $ \int^{2a}_{0}f(x)dx$ without integrating $ (1)$

[3] The function $ f$ is continious and  have the property

$ f(f(x)) \equal{} \minus{} x \plus{} 1$

 Then find  $ \int^{1}_{0}f(x)dx$

[quote="giannis"][1]
Let $ f(x) \equal{} (x \minus{} 1)(x \minus{} 2) ... (x \minus{} n)$

Then find $ \int \frac {f''(x)f(x) \minus{} (f'(x))^{2}}{f^{2}(x)}dx$[/quote]

It's easy to see that $ \frac {d}{dx} \frac {f(x)}{f'(x)} \equal{} \frac {f''(x)f(x) \minus{} (f'(x))^{2}}{f^{2}(x)}$.
So we have just to simplify $ \frac {f(x)}{f'(x)}$. Using the generalized rule for the derivative of the product of functions, we find that $ f'(x) \equal{} \sum_{j \equal{} 1}^{n}\prod_{i \neq j}(x \minus{} i)$ and simplifying $ \frac {f(x)}{f'(x)} \equal{} \frac {1}{x \minus{} 1} \plus{} \frac {1}{x \minus{} 2} \plus{} ... \plus{} \frac {1}{x \minus{} n}$.
I'm still trying to solve the other one.

Good night
GM

[quote="giannis"]
[2] The function $ f$ is continious and  have the property

$ f(a \plus{} x) \plus{} f(a \minus{} x) \equal{} x$ $ (1)$

Find $ \int^{2a}_{0}f(x)dx$ without integrating $ (1)$[/quote]

Could you explain, because  $ f(a \plus{} x) \plus{} f(a \minus{} x) \equal{} x$ and $ f(a \minus{} x) \plus{} f(a \plus{}x) \equal{} \minus{}x$  :!:  :?:
 :cool:

[2] ${ \displaystyle \int_0^{2a}}f(x)dx = \int_{ - a}^af(x + a)dx = \int_{ - a}^af( - x + a)dx =$ ${ \frac 12 \int_{ - a}^a(f(x + a) + f( - x + a))}dx$ .. and son ...
[3]$ (x\leftrightarrow f(x + \frac 12)) \Longrightarrow (f(f(f(x + \frac 12))))) = - f(x + \frac 12) + 1$ $ \ \Longrightarrow f(1 - x - \frac 12) = - f(x + \frac 12) + 1 \Longrightarrow f(x + \frac 12) + f(\frac 12 - x) = 1$
Finally (see point [2]) : ${ \displaystyle \int_0^1f(x)dx = \frac 12 \int_{ - \frac 12}^{\frac 12}(f(x + \frac 12) + f( \frac 12 - x))}dx$  .. and son ..
 :cool:


Find the integer n that makes the following equation integer. (99<n<1998)

      (2^n + 2) / n

thanks in advance.

I tried it by deviding n in an even number and an odd number and i failed...sorry it's too hard..

[hide="Answer"]946[/hide]

Will post solution later if noone else does. It wasn't particularly nice.


[quote="Ihatepie"](12x+5)squared[/quote]
[hide="foil suks for squaring a binomial"]$144x^{2}+120x+25$[/hide]

lol i learned foil in pre algebra, im in geometry now

(12x+5)(12x+5)= 144x squared+120x+25

please note that if you are squaring something:

$(x+y)^{2}$

that the answer isn't $x^{2}+y^{2}$. many GT Algebra II students often write that as their answers.

FOIL and the Distributive property are basically the same except that the distributive can be used in problems such as $(x+2y)(x^{2}+7x+42)$ etc.

[quote="funcia"]wait a minute i learned this but what is the method did i just not learn the name or what?[/quote]

If you just read the title you would see it's the foil method.

this is FOIL

[b]F[/b]irst [b]O[/b]uter [b]I[/b]nner [b]L[/b]ast

So say this is the problem

$(x+y)^{2}$

You would write it out so it would be

$(x_{1}+y_{1})(x_{2}+y_{2})$

The F in FOIL indicates that you would multiply the first terms in the problem so you have

$x \cdot x$

and your first term in your answer is $x^{2}$

Outer terms would be multiplying $x_{1}$ and $y_{2}$

Inner terms would be multiplying $y_{1}$ and $x_{2}$

Last terms would be multiplying $y_{1}$ and $y_{2}$

[quote="funcia"]ooops i meant is there a trick or is what i am doing FOIL
(12x + 5)^2

12x^2 + 2(5)(12x) +5^2

is that FOIL?[/quote]

That does work... so yes it is foil

yes it should be; i mentally checked it so i culd be wrong but it looks right to me

Yeah-who would make it to ALg. 2 and think that?

I get FOIL now!!!!!  :blush:  Yes I was taught that in algebra just not by the name.. :)

It has to do with the distributive property.

[hide][quote="Ihatepie"](12x+5)squared[/quote]

(12x+5)(12x+5)
144x^2=F
12x*5=O
5*12x=I
5*5=L
I hope I did that right (i just use a^2+2ab+b^2)
Now add tehm up

144x^2+120x+25[/hide]

[quote="nutz_for2.718281828"]Here's a nice clean formula for $\left(x+y\right)^{2}$:

[hide="in here"]

$\left(x+y\right)^{2}=x^{2}+2xy+y^{2}$[/hide][/quote]

Thanks I learnt something.

[quote="Ihatepie"][quote="Ignite168"][quote="7h3.D3m0n.117"]
that the answer isn't $x^{2}+y^{2}$. many GT Algebra II students often write that as their answers.[/quote]

I highly doubt that's true, please find a citation for that generalization.[/quote]
demon is right[/quote]

umm my math teacher says it is but you probably don't recognize her.

citation: go to a beginning Algebra I(when they are on their polynomials lesson) and ask them. see what they say. (some smart ones should get it)


define $0<a_{n+1}-a_{n}<1989$ and a_n is integer Prove that  there exist p,q is prime st $a_{p}|a_{q}$


Find: $ \frac{1}{2\sqrt{1}\plus{}1\sqrt{2}}\plus{}\frac{1}{3\sqrt{2}\plus{}2\sqrt{3}}\plus{}\frac{1}{4\sqrt{3}\plus{}3\sqrt{4}}\plus{}\cdots\plus{}\frac{1}{(n\plus{}1)\sqrt{n}\plus{}n\sqrt{n\plus{}1}}$ as $ n$ approaches infinity.

For some reason it is $ 1$. I suck at all calculus.

[quote="mewto55555"]Gah i fails at limits.[/quote]

wait, whats a limit???????? :maybe:

Whats a limit? Actually in this case it might not be a limit since the sum converges.

Yeah, I know it approaches 0, but what would it's limit be?

No calculus.  Multiply by the conjugate of the denominator and use induction.
$ \frac{(n\plus{}1)\sqrt{n}\minus{}n\sqrt{n\plus{}1}}{n(n\plus{}1)}\plus{}\frac{(n\plus{}2)\sqrt{n\plus{}1}\minus{}(n\plus{}1)\sqrt{n\plus{}2}}{(n\plus{}1)(n\plus{}2)}\equal{}\frac{(n\plus{}2)\sqrt{n}\minus{}n\sqrt{n\plus{}2}}{n(n\plus{}2)}$
It's not complete, but the terms cancel out as you keep adding them.

First $ (n\plus{}1)\sqrt{n}\plus{}n\sqrt{n\plus{}1}\equal{}\sqrt{n}\sqrt{n\plus{}1}(\sqrt{n\plus{}1}\plus{}\sqrt{n})$
Second rationarization of denominator
Third Telescope.

Good luck! :)

TINY TIM, IT'S LA PROBLEMA.':P'

TU CORRECTA LA GRAMATICA DE ESPANOL, POR PAVOR!!!

actually, it's el problema
lolfish
this post is not spam as it is a public announcement about the masculinity of problema
also, THBAPSO justinahmann

No, problem[b]a[/b] is feminine, so its [b]la[/b] problema. Si?

actually, no
reference:
http://www.spanishdict.com/translate/problema
it's a speecial noun

Yes, even in French, it's [b]le problème[/b].

OOO, that's interesting.  I hope we all got a satisfactory lesson in this world language.  Plus, this has nothing to do with the posted question by tinytim.  The most interesting part  was the title.

GOOD JOB, TINY!

[b][i]This is why the Missouri Forum is so "adored" by the AOPS staff[/i][/b]

dude, MO has the beastliest relationship with the AoPS staff (see math154 sig)

I know, I've seen it, and I know it's very sarcastic... :D

MISSOURI FTW!


Jack can paint his aparment in 12 hours. His wife Cheryl requires 20 hours to do the same job. How long would it take them to complete the job if they worked together?

Jack does $\frac{1}{12}$ in an hour. His wife does $\frac{1}{20}$ in an hour. If they work together, how much do they get done in an hour? [hide="answer"]$\frac{8}{60}$[/hide]

If they work together, how long does it take them?
[hide] We see the rate they do it together, and divide 1 by that rate, or we flip the fractional rate, to get $\frac{60}{8}=\frac{15}{2}=7.5$[/hide]

Billy

[hide="Solution"]If Jack can paint the apartment in 12 hours, then is rate is 1/12, and if Cheryl can paint the apt. in 20 hours, her rate is 1/20 (per hour). If they work together, you just have to add the two => 1/12 + 1/20 = 32/240 = 2/15. If you are dividing into the whole, you now need to divide 1 by 2/15, which is 15/2 or 7.5 [/hide]
-- Pats :!=: lose; same with the Sox... it never does get old, does it.....


I couldn't really think of anywhere else to put this.  So what did people think of it this year?  I left during the algebra round so I didn't get to see who won but from the papers I graded from the team rounds, there were some very impressive scores.

The "Other US Contests and Programs" forum might have been more suitable.

I just noticed this.  JHMT was awesome as usual.  I really enjoyed the Mathematical Explorations Round - I like game theory.  The individual tests seemed a little harder than last year, with more really interesting problems.

Any idea who won?  I got to grade some of the team papers.  Some scores were very good, others were not.

First place team was Quince Orchard High School.  Second place was the Howard Area Homeschoolers Fighting 42's (my team).  We got first place on the Mathematical Explorations Round, and I think one of Langley High School's teams got first on the Puzzle Round.

I was first place on Combinatorics and second on Algebra.  First on Algebra was someone from Mercenberg High School.  First place on Geometry was Sathish Nagappan, the entirety of Quince Orchard's team.


integration
geometry
calculus
inequalities
real analysis
real analysis unsolved

$f(x): [0,1]\rightarrow{[0,1]}$ and $f(0)=0$ and $f(1)=1$ we are also told that $f(x)$ is an increasing continuous function.
prove that

$f\left(\frac{1}{10}\right)+f\left(\frac{2}{10}\right)+\dots+f\left(\frac{9}{10}\right)+f^{-1}\left(\frac{1}{10}\right)+\dots+f^{-1}\left(\frac{9}{10}\right)\leq\frac{99}{10}$

It is better to draw a picture while solving this problem.

Consider the unit square with vertices $(0,0), (0,1), (1,0)$ and $(1,1).$ Then we note that $\int_{0}^{1}f(x)\,dx+\int_{0}^{1}f^{-1}(x)\,dx =$ area of the square = $1.$ This is because the areas represented by the two integrals do not intersect and the sum of their areas is the same as that of the unit square.

Divide the interval $(0,1)$ on $X$ -axis into ten smaller sub-intervals, each of width $\frac1{10}.$ Do the same for the inteval $(0,1)$ on the $Y$ -axis. Then, we note that

$f(\frac1{10})\cdot \frac1{10}+f(\frac2{10})\cdot \frac1{10}+\cdots+f(\frac9{10})\cdot \frac1{10}+A \le \int_{0}^{1}f(x)\,dx$ and
$f^{-1}(\frac1{10})\cdot \frac1{10}+f^{-1}(\frac2{10})\cdot \frac1{10}+\cdots+f^{-1}(\frac9{10})\cdot \frac1{10}+A' \le \int_{0}^{1}f^{-1}(x)\,dx$, where

$A =$ area bounded by $f(x), X$ -axis and the line $x=\frac1{10}$ and
$A' =$ area bounded by $f^{-1}(x), Y$ -axis and the line $y=\frac1{10}.$
Also note that $A+A' = \frac1{10}\cdot\frac1{10}= \frac1{100}.$

Now, adding up the two inequalities above, we get

$\frac1{10}\left( f(\frac1{10})+f(\frac2{10})+\cdots+f(\frac9{10})+f^{-1}(\frac1{10})+f^{-1}(\frac2{10})+\cdots+f^{-1}(\frac9{10})\right)+A+A' \le \int_{0}^{1}f(x)\,dx+\int_{0}^{1}f^{-1}(x)\,dx = 1$

$\Rightarrow f(\frac1{10})+f(\frac2{10})+\cdots+f(\frac9{10})+f^{-1}(\frac1{10})+f^{-1}(\frac2{10})+\cdots+f^{-1}(\frac9{10}) \le 10\left( 1-\frac1{100}\right) = \frac{99}{10}$

[quote="FieryHydra"]
Consider the unit square with vertices $(0,0), (0,1), (1,0)$ and $(1,1).$ Then we note that $\int_{0}^{1}f(x)\,dx+\int_{0}^{1}f^{-1}(x)\,dx =$ area of the square = $1.$ This is because the areas represented by the two integrals do not intersect and the sum of their areas is the same as that of the unit square.
[/quote] how do u know that the 2 areas do not intersect?

[quote="FieryHydra"]
$f(\frac1{10})\cdot \frac1{10}+f(\frac2{10})\cdot \frac1{10}+\cdots+f(\frac9{10})\cdot \frac1{10}+A \le \int_{0}^{1}f(x)\,dx$ and
$f^{-1}(\frac1{10})\cdot \frac1{10}+f^{-1}(\frac2{10})\cdot \frac1{10}+\cdots+f^{-1}(\frac9{10})\cdot \frac1{10}+A' \le \int_{0}^{1}f^{-1}(x)\,dx$
[/quote]
How do u know that you need to add $A$ and $A'$ to both quantities, and then know what $A+A'$ is equal to?


A "modtenacci" sequence is obtained by choosing a 'seed pair' of digits, a and b. The first term of the modtenacci sequence is a, the second is b, and every term thereafter is the units digit of the sum of the two previous terms.

Let the cycle for a pair of digits be the number of terms that appear before the sequence returns to the original seed pair and thus "repeats", including those first terms. For example, the pair 0 and 5 yields the following sequence:
0, 5, 5, 0, 5, 5
which has a cycle of 3 terms. The longest cycle occurs for the seed pair (2,3). How many terms long is that cycle?

I can easily brute force this, but is there a quicker way of doing it?

An expression can be given for the period of the sequence for any modulus, although I forget both the technique and the thread.  It is not that simple.  In other words: brute force is the "quickest" way to do this problem for small moduli.

Edit:  Hmm.  Maybe not.

[hide="The technique"] If you've done the brute force calculations, you'll notice that $ 0$ seems to be appearing with some regularity.  This regularity can be explained as follows:

First, note that $ (2, 3)$ are the third and fourth Fibonacci numbers, so for simplicity we'll deal with periodicity in the Fibonacci numbers $ F_0 \equal{} 0, F_1 \equal{} 1$ taken mod $ 10$ instead.

[b]Lemma:[/b]  $ F_{n\plus{}k} \equiv F_{n\minus{}1} F_k \bmod F_n$.

Proof:  Compute the Fibonacci numbers $ \bmod F_n$.  You'll get $ F_n \equiv 0, F_{n\plus{}1} \equiv F_{n\minus{}1}, F_{n\plus{}2} \equiv F_{n\minus{}1}$, and so forth.  In other words, induction.  (You'll notice this pattern if you brute-force.)

[b]Corollary:[/b]  $ F_n | F_{mn}$.  (Take $ k \equal{} (m\minus{}1)n$ and again work inductively.)  This is why $ 0$ appears regularly.

The problem of finding the period of the Fibonacci sequence $ \bmod m$ first requires finding the smallest Fibonacci number divisible by $ m$.  In our case, this is $ F_{15} \equal{} 610$.

That's the first part.  You'll now notice that after $ 15$ terms, the new terms we get are the original Fibonacci sequence multiplied by $ F_{n\minus{}1} \bmod 10$ (in this case, $ F_{14} \equal{} 377 \equiv 7$).  It should occur to you that after the next $ 15$ terms all that'll happen is that our sequence will get multiplied by $ 7 \bmod 10$ again!

Now we want to find the smallest $ r$ such that $ 7^r \equiv 1 \bmod 10$ (and this is the part that is very hard to do in general).  Here our search is quick:  we get $ 7^4 \equal{} 2401$ so our answer is $ r \equal{} 4$.

Therefore, after at most $ \boxed{60}$ terms we get the original sequence back.  (We need to do a little more reasoning here to be sure that subsequences like $ (2, 3)$ don't appear randomly halfway through and it's more than I want to get into at the moment.) [/hide]

You can take a look at this [url=http://www.mathlinks.ro/weblog_entry.php?t=156030]blog post[/url] on the period of the Fibonacci sequence if interested.

it is not hard to see that multiples of 5 appear every 5th Fib number and multiples of 2 appear every 3rd, so multiples of 10 appear every 15th fib number.
then you know it  has to be a multiple of 15, to create a lower bound...
then i guess you have to brute force it to prove that the term after the multiple of 10 will end in 1,3,7,or 9 and it will repeat when a number that ends in 1 is after a multiple of 10 (or it is about to repeat)


HMMT
Harvard
college
MIT
quadratics
geometric series
algebra

Source: Harvard-MIT Math Tournament - 2002 Guts Round #6

Find all values of x that satisfy
x = 1 - x + x:^2: - x:^3: + x:^4: - x:^5: + ...

Unfortunately, neither of you got the correct answer.

[hide]Using the infinite geometric series, 1/(1+x)=x.  x^2+x=1.  x^2+x-1=0

With the quadratic formula, we get (-1 :pm: sqrt(5))/2.[/hide]

This is what Tare got... Why is this wrong?[/hide]

Go back and check the numbers you got.  That is, verify they work.

Oh, one of them doesn't work...
Hint: [quote="MIT-Harvard students"]Find all values of x that [b]satisfy[/b]...[/quote]

Think when does an infinite series converge?

only (:rt5:-1)/2 is a solution, as x must be <1

I lack to see why bugzpodder's answer is wrong.

Because that is an edited answer.  It used to contain both solutions of the quadratic.

JBL, where in cambridge do you live? What school?

yeah, i was trying to follow the flow of conversation and it wasnt quite making sense because like I said, the second post had the correct answer... at least make a note when you change a wrong answer to the correct one.

The posts all say when they've been edited.

[quote="ComplexZeta"]We should make it a rule that people do not change their solutions after other people have commented. If you make a mistake, just admit it. We all make plenty of mistakes when solving math problems; it's nothing to be ashamed of. Editing solutions to make them correct just confuses people.[/quote]
I agree, though I don't think it's possible for it to be a "rule" since people might disagree and/or break that rule often.

yes JBl it says they have been edited... but how is one to assume that an answer has been changed from incorrect to correct without a note... it could have been a typo that he corrected. An explaination of what was edited would be nice.

[quote="teratomato"]In order to solve the problem you have to be at least as smart as Jello.[/quote]

I guess that proves that jello is smarter than bugzpodder and Tare.   :wink:

Just kidding...

It's interesting that usually when one person gets a question wrong, another person does too (as opposed to the second person noticing the mistake...)

:lol:, I'm even dummer than the pudding bites
Well we all trip over the same gopher hole :lol:, so we get relaxed when we see another person with the same answer so we just post the same answer.

Yeah.. usually wen u edit a post its best to put at the bottom "edited to change blah" or something like that. Saves confusion.

Then again you might as well edit the reply to the edited post...
I guess you shouldn't edit mathematical mistakes at all, unless it's a spelling error or the equivalent.


geometry
rotation
trigonometry
Pythagorean Theorem
trig identities
Law of Cosines

[quote="Go Around the Tree"]Point $P$ is located inside square $ABCD$. If $PA=10$, $PB=6$, and $PC=14$, and $PD=2\sqrt{65}$, find the area of the square.[/quote]

Use the fact that the sides are all equal and the derivation of the British Flag Theorem

Okay, maybe it might not work now that I think about it.

Okay, if we drop perpendiculars to sides we get

$a^{2}+b^{2}=100$
$b^{2}+c^{2}=36$
$c^{2}+d^{2}=196$
$d^{2}+a^{2}=260$.

And, we know that $a+c=b+d=k$, for some real value $k$.

Next we see that the area we are looking for is

$(a+c)(b+d)=ab+ad+cb+cd$.

Yeah, if someone can figure out a way to expand on this they are smart.  I thought it could be expanded from this, but I realized there was a slight error, so there.

Sorry.

would this happen to be a multiple choice question?
i think i got an answer but it could be completely wrong =)

[hide="yay  :lol: "]Rotate about $B$ the point $P$ to the point $P'$. $BP=BP'=6$, and $\angle PBP'=90^\circ$, thus $PP'=6\sqrt{2}$. From the Law of Cosines, $\cos\angle P'PC=\frac{(6\sqrt{2})^{2}+14^{2}-10^{2}}{2\times14\times6\sqrt{2}}=\frac{\sqrt{2}}{2}$, thus $\angle P'PC=45^\circ$. Thus, $\angle BPC=90^\circ$. From Pythagorean Theorem, $[ABCD]=BC^{2}=6^{2}+14^{2}=\boxed{232}$.[/hide]

That is good.  But one question.

I know this is obvious, but how do we know that all three of those interior segments lie within that triangle.


What is the formula for Probability's Binomial Distribution?

[quote="gogators"]http://en.wikipedia.org/wiki/Binomial_distribution

I think this does not belong in this forum.[/quote]

wow. this is the EXACT link i had in mind, with the same EXACT thought
we must be related or something  :wink:


Half-nice, half-bruteforce  :) : Firstly note that there are 4 possibilities for lenght of sides of each triangle, denote it as $1,2,3,4$ (to make notation easier). By $xyz$ ($x,y,z \in \{ 1,2,3,4 \}$) I denote triangle with sidelenghts $x,y,z$. Note that third side is determined by other two. Now show that if one of triangles is $333$ then the other two must have side $3$ (that's brute-force). So futher on we can suppose that no triangle is equilateral. Prescribe $a$ to vertices $A_{1},A_{4},A_{7}$, $b$ to $A_{2},A_{5},A_{8}$ and $c$ to $A_{3},A_{6},A_{9}$. We say that triangle contains $a$ (or $b$,$c$) if one of the vertices has label $a$. Observe that triangle contains $a,b,c$ iff it has two equal sides and is not equilateral. With this we conclude that we can't have two isosceless triangle and one scalene. Also observe that two scalene triangles and one arbitrary always work (two scalenes have at least two sides with the same lenght). So we are left with case when all three triangles are isosceles. Now we note that all isosceles triangles are: $112$, $441$, $224$ (and of course $333$ but we dealt with it earlier). If two of isosceles are congruent we're done, so we have to show that we can't have $112$, $441$, $224$ simultanously, and we use brute force here again :) (not so many cases at it might seem, wlog we can assume that $112$ is $A_{1}A_{2}A_{3}$ and then consider $A_{4}$).

Maybe someone can change parts of my proof where I use brute force into something nicer?   :D

how can it be given in vietnam TST i think it is too easy
in day 2 only phoblem 5 is nice.However it is well-known
and the truth is many contestant know it and have a perfect paper on day 2


Hello,

Given that there are some postings in this forum asking for help with Spanish, I was wondering what do you think, Should we have a "Help me with my Spanish" forum to group all such postings? Please, vote.

Hola,

Dado que hay algunos mensajes en este foro pidiendo ayuda con Español, me preguntaba, ¿Deberíamos tener un foro de "Ayudame con mi Español" para agrupar estos mensajes? Por favor, vota.

Parece que la mayoría de las personas piensa que no deberíamos tener tal foro, pero en realidad pocas personas han votado. ¿Quiere alguno de los votantes decir por qué o por qué no deberiamos tener tal foro?

It seems like most of the people think that we should not have such a forum, but really few people have voted. Does any of the voters want to say why he/she consider that we should or should not have such a forum?

Voté sí pero me di cuenta que mis razones eran demasiado obscenas para expresarlas en este foro :D , ahora lo pienso y no veo por qué deba existir tal foro puesto que existen muchos sitios en la web especializados en el aprendizaje de nuestra lengua. Sin embargo, podríamos tener ayudas en lo que concierne al lenguaje técnico usado en olimpiadas aunque no creo que mucha gente se interese en eso o sí?

Kiubo, Jose, pienso lo mismo que ud, por eso vote que no. A pesar de que estemos en el foro Hispanohablante, estamos aqui para hacer matematicas. Como dices tu, hay muchos lugares en la web q se especializan en la ayuda para el aprendizaje del espanhol, por lo cual no veo una buena razon para que se haga un subforo aca. Uds que piensan?

La verdad, parece que, aunque hubo un voto de diferencia, las pocas personas que opinaron no estan deacuerdo, y en realidad, pues no es tan alarmante la cantidad de post que tenemos por ayuda en español, por lo general me parece de momento innecesario solicitar la creacion del foro. Si vuelve a serlo, pues alli lo volvere a poner como propuesta, por si acaso.

Tambien, el topic pasa a ser normal en lugar de Stiky

[color=blue]No entiendo por qué no puedo votar en la encuesta. Me acuerdo que en una anterior, sí había logrado.

¿Alguien me puede explicar lo que sucede? Muchas gracias.[/color]

Creo que eso es un buen idea.  Sólo sé español por mis clases en la escuela, y me gustaría aprender más en un forum.  (También, podría ayudarme con matemáticas en otras países.)


I think he means Intermediate Geometry, since he mentioned that book before. (Please make your posts more specific though!)
As AIME15 said on the previous thread, the book will come out in the future.

I think he is referring to [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=5003]this book[/url] by Titu Andreescu (past US IMO coach, etc, etc) and Gabriel Dospinescu (who is a very valuable and prolific AoPS/ML community member).


9 bubbles make an "equilateral" triangle. Each of the digits 1-9 are placed in the bubbles such that the sum of every side of the triangle is equal. What is the positive difference between the least and greatest possible sums of a side?

[hide="The non trial and error way"][hide="Are you sure?"]The total value of all the numbers is 1+2+3+...+9 = 45. 3 numbers are going to be used twice when calculating the value of all three sides. The greatest value would be if 9, 8, and 7 were used twice, so we add those to 45 to get 69. 69/3 = 23 is the greatest possible sum of each side. The least possible value of each side would be if 1, 2, and 3 were used twice. We add those to 45 to get 51. In this case, the sum of each side would be 51/3 = 17. The difference between the least and greatest possible sums for each side is 23 - 17 = 6.[/hide][/hide]

WITHOUT LOOKING AT HINT:
[hide]The ones on the corners are the ones you need to worry about, as they're the ones you double count.
Greatest: Corners=9,8,7
Smallest: Corners=1,2,3
Smallest: 4,5,6,7,8,9= 4-9,5-8,6-7=13+4=17
Greatest: 1,2,3,4,5,6= 1-6,2-5,3-4=7+16=23
23-17=6
>>6<<[/hide]
P.S. Tarquin, next time dont post that hint. its too tempting just to read it and post that in their own words. My suggestion: Post the solotion AFTER at least 5 answers (right or wrong)


Let $ x;y;z>0$
we have
$ \sqrt{\frac{a^2}{b^2\plus{}c^2}\plus{} \frac{b^2}{a^2\plus{}c^2}\plus{}\frac{c^2}{b^2\plus{}a^2}\minus{}\frac{3}{2}}\plus{}\frac{ac\plus{}ab\plus{}bc}{a^2\plus{}b^2\plus{}c^2}\ge 1$

[quote="tranvanluan"]Let $ x;y;z > 0$
we have
$ \sqrt {\frac {a^2}{b^2 \plus{} c^2} \plus{} \frac {b^2}{a^2 \plus{} c^2} \plus{} \frac {c^2}{b^2 \plus{} a^2} \minus{} \frac {3}{2}} \plus{} \frac {ac \plus{} ab \plus{} bc}{a^2 \plus{} b^2 \plus{} c^2}\ge 1$[/quote]

[url=http://toanthpt.net/forums/showthread.php?p=48572#post48572]See here for solutions[/url]


Find all functions $ f: \mathbb{R}\to\mathbb{R}$ which are continuous on $ \mathbb{R}$ and satisfy $ 6f(f(x)) \equal{} 2f(x) \plus{} x$ for every $ x\in \mathbb{R}$.

[quote="Mashimaru"]Find every function $ f: \mathbb{R}\mapsto \mathbb{R}$ satisfies: $ 6f(f(x)) \equal{} 2f(x) \plus{} x$ for every $ x\in \mathbb{R}$.[/quote]

Hemmm, beside the two trivial solutions $ f(x)\equal{}\frac{1\pm\sqrt 7}6x$, are you sure we can determine the other or even show that there are no other without any complementary condition (continuity ?, monotonous ?, ...) ?

Since you posted in "proposed and own", it means you have the solution.

I am extremely sorry, Mr.[b]pco[/b], the problem I posted above should have been modified the condition that $ f$ is continuous on $ \mathbb{R}$  :blush:

My outline for the solution is first, we prove that $ f(0) \equal{} 0$, since $ f$ is continuous and injective (from the assertion $ 6f(f(x)) \equal{} 2f(x) \plus{} x$) then $ f$ is monotone. Suppose that $ f$ is increasing, we will first prove that $ f(x) \equal{} \frac{1\plus{}\sqrt{7}}{6}x$ for every $ x>0$ and then for every $ x\in\mathbb{R}$. For this, we will assume in contrast that there exist $ x_0$ such that $ f(x_0) < ax_0$ and point out the contradiction.

Can you explain your idea in a clear way,dear Mashimaru?I didn't understand what is your  [b]a[/b]?