geometry
trigonometry
incenter
analytic geometry
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geometry proposed

Let the incircle of a triangle ABC have center I and touch the sides BC, CA, AB at the points X, Y, Z. Let the parallels to the lines AX, BY, CZ through the point I meet the lines BC, CA, AB at the points D, E, F. Prove that the point I is the centroid of triangle DEF.

The more solutions the better ;)

  darij

prowler, does c.t.d. mean c(see) touch down?  :D

c.t.d. ~ и т.д. и т.п.   :D
It is like what we have to prove is proven.

Not very hard, but the conclusion is very nice :)
Let $FI\cap DE=P$.
IE/ID=sin<AXC/sin<BEC.
sin<EIP/sin<DIP=sin<YGC/sin<XGC=sin<AXC/sin<BEC.
Since IE/ID=sin<EIP/sin<DIP, we obtain DP=EP, and the problem is solved.

[color=red][Moderator edit: The point G is defined as the point of intersection of the lines AX, BY, CZ.][/color]

[quote="indybar"]Why is IE/ID=sin<AXC/sin<BEC?[/quote]
Sine law.

This problem is very nice. My solution is vectorialy.
I note the intersection M between AI and BC.
From the relations 2pI=aA+bB+cC, (b+c)M=bB+cC, aX=(p-c)B+(p-b)C,
                           DX/DM=IA/IM=(b+c)/a results
2pD=aX+(b+c)M=(p-c)B+(p-b)C+bB+cC=(p+b-c)B+(p+c-b)C ===>
2p.(D+E+F)=3(aA+bB+cC) ===> (D+E+F)/3=I. It is O.K.

Let $ABC$ be a triangle and the points $P(x, y, z),\ R(u, v, w)$ with barycentric coodonates relative to the triangle $ABC$. I note $X,\ Y,\ Z$ which belong to $BC,\ CA,\ AB$ respectively such that $PX\parallel AR,\ PY\parallel BR,\ PZ\parallel CR$. Prove that the point $P$ is the centroid of the triangle $XYZ$ iff there is the relations $\frac{u(v+w)}{x}=\frac{v(w+u)}{y}=\frac{w(u+v)}{z}$.
[i]Remark[/i]. If $P(a,b,c)$ - the incentre and $R(r_a, r_b, r_c)$ - the Gergogne's point, we obtain the propose problem. I believe that the problem is dual: $I\longleftrightarrow\Gamma$.

I add the following relations for to understand the remark (2s=a+b+c):
$\frac {r_a(r_b+r_c)}{a}=\frac{r_b(r_c+r_a)}{b}=\frac{r_c(r_a+r_b)}{c}=s$.

Very nice solutions! Indeed, the two solutions by Levi were exactly the two solutions I had. For what it is worth, let me record the homogeneous barycentric coordinates of the points D, E, F:

D ( 0 : 3a + b - c : 3a + c - b );
E ( 3b + a - c : 0 : 3b + c - a );
F ( 3c + a - b : 3c + b - a : 0 ).

Now I'll comment to some posts.

[quote="marko avila"]Darij; i have an elementary solution to your problem wich i conjured after 5 long hours¡¡¡  i have an important history test tomorrow so i have to study but i promise i will post the solution tomorrow¡¡¡ :)[/quote]

Tomorrow never dies, so could you post your solution now? I would be very interested. Thanks!

[quote="levi"]Let ABC be a triangle and the points P(x, y, z), R(u, v, w) with trilinear coodonates relative to the triangle ABC.[/quote]

The coordinates you using are not [i]trilinear coordinates[/i], but [i]barycentric coordinates[/i].

[quote="levi"]I note X, Y, Z which belong to BC, CA, AB respectively such that PX || AR, PY || BR, PZ || CR. Prove that the point P is the centroid of the triangle XYZ iff there is the relations u(v+w)/x=v(w+u)/y=w(u+v)/z.[/quote]

Strike! This is how I actually [i]found[/i] the problem. If R is the point with homogeneous barycentric coordinates $R\left(u:v:w\right)$ with respect to triangle ABC, then the point with homogeneous barycentric coordinates $\left(u\left(v+w\right):v\left(w+u\right):w\left(u+v\right)\right)$ (this is your point P then) is what I call the [i]isotomcomplement[/i] of the point R with respect to triangle ABC. This is the complement of the isotomic conjugate of the point R with respect to triangle ABC. Over two years ago, I posted [url=http://groups.yahoo.com/group/Hyacinthos/messages/6423?xm=1&m=e&viscount=100]a series of Hyacinthos messages (#6423, #6427, #6431, #6437, #6453)[/url] about isotomcomplements. Actually, to that time, I was quite a beginner, what sometimes is reflected quite clearly in the posts. Anyway, I hope they are halfways interesting.

By the way, try out R = orthocenter. Then, the isotomcomplement of R is P = symmedian point of triangle ABC. Don't you get some familiar result? This was my starting point when I created the problem.

[quote="levi"][i]Remark[/i]. If P(a,b,c) - the incentre and R($r_a, r_b, r_c$) - the Gergogne's point, we obtain the propose problem. I believe that the problem is dual:  the incentre I <----> the Gergogne's point.[/quote]

Hmm, if you mean that R = incenter yields P = Gergonne point, then this is wrong.

[quote="levi"]If we replace I:=$I_a$ and G(Gergogne's point):=N(Nagel's point),
is true the conclusion ???[/quote]

No, but it is true if we replace I by the excenter $I_a$ and the Gergonne point G by the A-exterior Gergonne point $G_a$ (this point is defined as the point of intersection of the lines $AX_a$, $BY_a$, $CZ_a$, where $X_a$, $Y_a$, $Z_a$ are the points where the A-excircle of triangle ABC touches its sidelines BC, CA, AB).

  Darij

[quote="marko avila"]ok darij i will try, it´s just that i dont know   latex thats why i was hoping someone could write it for me. but i will try to post this tomorrow[/quote]
Have a look at http://www.mathlinks.ro/Forum/viewtopic.php?t=5261

See [url=http://www.artofproblemsolving.com/blog/39266][color=red][b]here[/b][/color][/url] my proof of my above extension.

Dear Mathlinkers,
why no synthetic proof has appeared until to day?
Sincerely
Jean-Louis


Prove that there exist infinitely many even positive integers $k$ such that for every prime $p$ the number $p^2+k$ is composite.

We can take k such that it's even, it's equal to 2 (mod 3) and 9+k is composite, for example 5/9+k=3^2+k. This is because p^2=1 (mod 3) for any p different from 3 and we also consider the case p=3. Infinitely many such k exist from the Chinese remainder thm.

If we take $k=4a^2$,we can factorize $p^2+k$ with Sophie Germain.(a is an arbitrary number)

Unfortunately, you're wrong. Even if you take $k=4a^4$, the intended Sophie Germain factorization looks like $p^2 + k = p^2 + 4a^4 + 4a^2p - 4a^2p = (p+2a^2)^2 - (2a)^2 p$, but $p$ is no square. This idea would only work for $p^4+k$ ...

We choose k to be odd, then obviously for odd p (p^2)+k is composite. Let k=n-4 where n is an odd composite, this k satisfies the condition when p=2 also. Since there are infinitely many such n's there exists infinitely many such k too.

Have you not noticed it is clearly specified $k$ must be[b] even[/b]? since for $k$ odd it is so trivial ...

But the problem statement requires $k$ to be [u]even[/u].

It can easily be seen that $k=30a-4$ (for positive integers $a$) works.

Sorry I hadn't noticed that k has to be even.

For every natural $t>0$ $k=66t+2$ works .
There is three cases:
1)If $p=2$ then $p^2+k=66t+4$ which is obviously an even integer bigger that two!
2)If $p=3$ then $p^2+k=66t+11$ again is divisible by eleven and bigger than it!
3)If $p>3$ then trivially $p^2\equiv 1\ (mod3)$ so $p^2+k$ divisible by three!

I think for $k=26+30*a$  for a natural $a$ we get $k+4=30*(1+a)$ and $k+9=5*(7+6*a)$ and for $p\ge5$ we have $p^2\equiv{1}(mod 3)$ so
$p^2+k\equiv{2+1}\equiv{0(}mod 3)$. $\Box$


Prove that if there are infinitely many twin primes, that the sum of their inverses converge.

They converge trivially if there are only finitely many, thus the assumption is not needed ;)
The result itself is called Brun's Theorem, and the constant it converges to is called after Brun, too.
This is too complicated for Pre-Olympiad, and too ugly for the Number Theory forum... You better look it up in a book.

[quote="Anaxerzia"][quote="Math^_^"]This problem is certainly too hard for Pre-Olympiad :) , it is one of the unsolved problems in Mathematics. :!:
I have found a post in Mathlink about Brun's theorem http://www.mathlinks.ro/viewtopic.php?t=158579[/quote]

He isn't asking you to prove there are infinite twin primes, only that if there are, that the sum of their inverses converges.[/quote]

Oh I have mis-read the question, my apology. :fool:


Hey guys I think my book is wrong on this one. Anyways heres the problem hope someone can help.

A ferry is used to transport passenger vehicles between Cap- Rouge and Cap-Espoir, a distance of 8 km. Its direction with respect to Cap Rouge is North 35 Degree East. The current of the river that the ferry crosses is 3.5 km/hr to the west. So as to land in Cap Espoir in 1 hour, what speed and in what direction should the captain set out in the ferry?

I found the speed to be 6.64km/hr, the book sais 10.41 km/hr, who is right?

Shouldn't this be moved to intro physics?

Draw a right angled triangle with the directions accordingly. all you have to do is add/subtract the velocity components in the direction or against the boat, whichever applies. You sure will need a calculator!


algebra polynomial calculus topology search function

Hi,

I know someone who is doing research on Groups of Polynomial Growth. That person's background is say, about AP Calculus BC. So, can you please tell what are the books/articles/researches that that person needs to explore/read in order to understand and finally do a research on this subject matter? Please give a list of books/etc. And, any tips?

Oh btw, go to http://en.wikipedia.org/wiki/Gromov%27s_theorem_on_groups_of_polynomial_growth for more info.

First, he or she must learn a fair amount of group theory. Perhaps a book like [u]Topics in Algebra[/u] by I.N. Herstein would suffice. They should also get an intuitive understanding of relative rates of growth as they are used in Analysis. Any book that explains Big-O and little-o notation might be helpful. I know that Kent Merryfield wrote a handout on it that can found on the forum somewhere. Also they should learn about Metric Spaces. I find the explanation of metric spaces in topology books to be better than those found in analysis books. So maybe the book [u]Elementary Topology[/u] by Gemignani. But mostly, it will require a good knowledge of group theory.

Basically, that person just needs to read Topics in Algebra and Elementary Topology?

In order to understand group theory, does one needs to study abstract algebra first? And, is abstract algebra a pre-req for the book Topics in Algebra?

How can that person gets an intuitive understanding of relative rates of growth? Any suggestions on books, articles, etc?

Do you need to real the whole Topology book or just a portion on metric spaces?

Kent Merryfield, if you are reading this, can you please post the handout here? Or anyone knows where the handout is located? If you do, please link it.

The search function at the top of the forum knows ...

Since you are not that busy, I suggest you do it yourself...

For God's sake, if you are really busy, you will try to look at ways that you can do to short-cut a process. That's the way of any successful people in this world (I am not saying I am or I am not here btw!)!

Anyway, I have gotten the handout.

Are there any differential geometer here? Any suggestions/ideas?

Thank you!


inequalities
algebra
polynomial
function
logarithms
inequalities open

Let $f(x)$ be polynomial of degree $n \in N$. Is the following inequalities valid for all $n$:

$f(x)f(y) \geq f(\sqrt{xy})^2$
...

$f(x_1)...f(x_k) \geq f(\sqrt[k]{x_1...x_k})^k$     where  $k=2,3,...,n$;        $x_1,...,x_k$ are arbitrary numbers

Or for what values of $(n,k)$ are these inequalities true ?

P.S. I know that for $(2,2)$ it is true  :D

i think you can look for some articles about geometric convexity written by chinese.

The proposed is very similar to Holder's inequality - following directly from it if all coefficients are non-negative. Suppose $f$ can have negative coefficients. Then the claim is false for odd $n$, since multiplication by negative numbers flips the inequality sign.

[quote="Megus"]Siuhochung: can you supply any links ? Or just mention some interesting papers/authours ?[/quote]

http://zgbdsyjxz.nease.net/jht/jhths.htm
but it's in chinese...

I don't speak Chinese so...  :(

I've tried to find something on Google but failed.

Siuchochung: can you tell approximately what's contained in these Chinese articles? if something is useful to give a solution to mentioned inequalities can you please translate a little? I would be very thankful

Well, basically, geometric convexity means that for any functions f, if
$af(x) + (1-a) f(y) \ge f(x^a y^{1-a})$, then it is called geometric convex. It is analogus to the schur-convex functions we usually consider.
Thus, it is also enough to consider $f(x)+f(y) \ge 2f(\sqrt{xy}$ to show it is geometric convex.
We can also check that if $f''(e^x) \ge 0$, then f is geometric convex.
Next, for your question, as long as it satisfies $f''(e^x)$, it is geometric convex, and thus the desired inequality holds by the jensen's inequality for geometric convex functions. (Basically you can move all things from schur-convex to geometric convex)
Specificly in one article there, it provided a way to check whether a degree-three polynomial is geometric convex. I did not read it though.

[quote="Megus"]So if I understand correctly to prove these proposed inequalities it is enough to show that:

$(\ln f(e^x))'' \geq 0$, where $f$ is some polynomial.

Or am I mistaken?[/quote]
i think that is one of the ways but i guess there is something better, and I think you mean
$f(e^x)'' \geq 0$ or $(\ln f(x))'' \geq 0$

I meant $(\ln f(e^x))'' \geq 0$ because we want to have product instead of the sum, don't we?


calculus
integration
trigonometry
limit
calculus computations

Compute $ \int_0^{\infty} \frac{e^{\minus{}x} \sin \, (x)}{x} \, dx$

Let $ I(s) \equal{} \int_{0}^{\infty} \frac{\sin x}{x} e^{\minus{}sx} \, dx$ and calculate $ I'(s)$. Then you'll know that $ I(1) \equal{} \frac{\pi}{4}$.

$ \int_0^{\infty} \frac{\sin x}{x} e^{\minus{}x}\,dx \equal{} \lim_{s \to 1} \int_0^{\infty} \frac{\sin x}{x} e^{\minus{}sx}\,dx$

$ \equal{} \lim_{s \to 1} \mathcal{L} \left(\frac{\sin x}{x}\right) \equal{} \lim_{s \to 1} \arctan \left(\frac{1}{s}\right) \equal{} \arctan 1 \equal{} \frac{\pi}{4}$.

@Carcul: What does the symbol $ \mathcal{L}$ mean?

That's something i haven't read up.. Thanks!  :lol:


Stan drove 300 miles in 5 hours, 20 minutes. Next, he drove
360 miles in 6 hours, 40 minutes. What was Stan's average speed in
miles per hour for the total trip?

Average speed is the total distance travelled over total time taken.

5 hours 20 minutes = 16/3 hours
6 hours 40 minutes = 20/3 hours

$ \frac{300 \plus{} 360}{\frac{16}{3} \plus{} \frac{20}{3}} \equal{} \frac{660}{12} \equal{} \boxed{55}$ miles per hour

You can also add 5hrs 20mins to 6hrs 40 mins for a total time of 12hrs, while 300 mi. + 360 mi = 660 miles of total travel. His average speed would then be 660/12 = 55 mph


Anyone with the solution to the attached question...?

a) i) the guy is walking away from home at a pretty slow rate  (around 3 km/hour), then he stops (presumably to get his directions?) for 5 minutes, then walks/runs home at a brisk pace (around  18 and 2/3 km and hour), then gets tired and walks the remaining way at around 3 km/hour.
ii)around...57.5 minutes?

b) i) i'm assuming you mean gradient as in slope, so 1/20. this means for every 20 minutes, he walks 1 km, or 3 km/hour.
ii) i shall assume he is walking, because not many other modes of transportation are that slow.

c) in segment BC, we can see that the man is not going any distance relative to his goal, so he is either moving in a circle or staying still. because we have a straight road, he is standing still. My explanation is that the man realizes he is not going the right way and asks for directions.

d) i) the slope is around 14/45. In this example it is negative because the distance from the man to his destination is decreasing. This slope corresponds to a speed of 18 and 2/3 km per hour.
ii) i assume the worker is either running, or is riding a bike.

e) i) in section DE, the man either for tired of running or dismounted his bike.
ii) the average speed is 8 km/42.5 minutes or 192/17 km/hour. somwheres around 11.2941 km/hour.

f)....i leave that to you.

Where did you get this problem (mathspro)?


geometry
geometric transformation
reflection
ratio
projective geometry
geometry solved

Given a circle, a diameter, and a point on the circle (not on the diameter). Construct, using an unmarked straight edge only, a line through the point, perpendicular to the diameter.

Let $AB$ be the diameter, and $C$ the other point. Take a point $M$ on the circle (different from the three). Construct the line throuygh $M$ which is the harmonic conjugate of $MC$ wrt $MA,MB$ (harmonic conjugates are easy to construct with the straightedge alone). This line will cut the circle precisely in $M$ (of course :)) and the reflection $D$ of $C$ in $AB$. The line $CD$ is the line we're looking for.

How do you define harmonic conjugates for line segments? I thought that harmonic conjugates is something like: $A$, $B$, $C$, $D$ on a line such that $CA/CB = DA/DB$.

Not for line segments, but for lines passing through the same point ($M$ in our case). If you cut four concurrent lines with another line $d$, the cross-ratio of the four lines is defined as the cross-ratio of the four intersection points of the concurrent lines with $d$. It can be shown that it doesn't matter what line $d$ you use: you always get the same cross-ratio, so you can take this cross-ratio to define the quadruple of concurrent lines.

Let $AB$ be the circle diameter and $D$ the given point on the circle different from the points $A, B$. Pick an arbitrary point $E$ on the circle different from $A, B, D$, for example, between $A$ and $D$ (this is not absolutely necessary). Construct lines $AD, BE$ intersecting at a point $H$ and lines $AE, BD$ intersecting at a point $C$.  Since $AE \perp BC$ and $BD \perp AC$ are 2 altitudes of  the $\triangle ABC$ $\Longrightarrow$ $H$ is its orthocenter. Line $CH \perp AB$ is the 3rd altitude. Let $CH$ intersect the circle at points $K, K'$. Let $DK'$ intersect $AB$ at a point $M$. Let $KM$ intersect the circle again at a point $D'$. Since $D'$ is a reflection of $D$ in $AB$ $\Longrightarrow$ $DD' \perp AB$.


It is not a homework... I am practicing on this problem  :lol:

I don't have any idea what this is about, but if you'd like to make a nicer diagram, try using the code tags.  When you use it, everything will get spaced out -- each character is given exactly the same width (including spaces).  So, for example:

[code]   *
  * *
 * * *
* * * *[/code]

Hello!   I will just have to find it out... about the answer...

By the way, I want to know... does this forum allows homework questions?... Can that person show his solutions, whether he's right or wrong?

No. This forum is not for going over homework problems.

If you are doing your homework, as a student in the other recent thread here was, and you come across an idea (such as taking a cross-product) that you don't understand, it is perfectly reasonable to come here and ask people to help explain how to take a cross-product to you.  Also, if you've done one step that you are unsure of, it is reasonable to ask if that single step was correct.  However, to go and either post whole problems from your homework and ask people to do them, or to post whole solutions and ask people to correct them, is unacceptable.  In order to make the line clear, it is probably best to do two things: (1) state what the problem is and where it came from.  People will help you if you present an idea from class that you are having difficulty with or a step of which you are unsure.  I personally get very annoyed when people put up homework problems without identifying them as such, but much less annoyed when people provide sufficient explanation that I don't feel I'm doing their work for them.  (2) Write down, on the homework you hand in, that you received help from individuals in an online mathematics forum.

Hello Joel...

thank you for that... so that I'll know what to do next time...

on my problem on operations research... it was not that of a big problem since it is possible to have different possible starting basic feasible solution... anyways, it is not optimal yet..

Thank you again... I ask that because it's my first time here. :)

[code]

 () < I pwn you
-|-
 /\
 [/code]

I think JBL has made an important point in this thread, which is that it is GENERALLY helpful for people to indicate the source of a problem when they first post a problem. Now the forum software includes a source field for each post, and it is a very good idea to use the source field, perhaps with additional explanation in the body of a post, to indicate where a problem comes from. That helps people helping you give better answers too, so it's a good idea all around.


Let $ a,$ $ b$ and $ c$ are positive numbers such that $ \frac{1}{\sqrt a}\plus{}\frac{1}{\sqrt b}\plus{}\frac{1}{\sqrt c}\equal{}3.$ Prove that
\[ \frac{1}{a\plus{}b}\plus{}\frac{1}{a\plus{}c}\plus{}\frac{1}{b\plus{}c}\leq\frac{3}{2}\]

[hide="remark"] This inequality is not true already for positive $ a,$ $ b$ and $ c$ such that $ \frac{1}{\sqrt[3] a}\plus{}\frac{1}{\sqrt[3] b}\plus{}\frac{1}{\sqrt[3] c}\equal{}3.$ :wink: [/hide]

Here is my proof
Let $ \frac {1}{\sqrt a} \equal{} x$ then $ a \equal{} \frac {1}{x^2}$ $ y,z$ is similar to  So $ x \plus{} y \plus{} z \equal{} 3$
The ineq can be rewritten in this form
$ \sum \frac {x^2y^2}{x^2 \plus{} y^2} \leq\ \frac {3}{2}$
Then use AMGM we have $ x^2 \plus{} y^2\geq\ 2xy$
AND $ (xy \plus{} yz \plus{} zx)\leq\ 3$ The proof is completed
Is my proof right

[quote="arqady"]Let $ a,$ $ b$ and $ c$ are positive numbers such that $ \frac {1}{\sqrt a} \plus{} \frac {1}{\sqrt b} \plus{} \frac {1}{\sqrt c} \equal{} 3.$ Prove that\[ \frac {1}{a \plus{} b} \plus{} \frac {1}{a \plus{} c} \plus{} \frac {1}{b \plus{} c}\leq\frac {3}{2}
\][/quote]$ \frac {1}{a \plus{} b} \plus{} \frac {1}{a \plus{} c} \plus{} \frac {1}{b \plus{} c}\leq\frac {3}{2}$
$ \Longleftrightarrow \frac {2}{a \plus{} b} \plus{} \frac {2}{a \plus{} c} \plus{} \frac {2}{b \plus{} c}\leq\frac {1}{3}\left(\frac {1}{\sqrt a} \plus{} \frac {1}{\sqrt b} \plus{} \frac {1}{\sqrt c}\right)^2$
$ \Longleftrightarrow \frac {6}{a \plus{} b} \plus{} \frac {6}{a \plus{} c} \plus{} \frac {6}{b \plus{} c}\leq \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\plus{}2\left(\frac{1}{\sqrt{ab}}\plus{}\frac{1}{\sqrt{bc}}\plus{}\frac{1}{\sqrt{ca}}\right)$
It is sufficient to show that $ \frac{6}{a\plus{}b}\leq \frac{1}{2}\left(\frac{1}{a}\plus{}\frac{1}{b}\right)\plus{}\frac{2}{\sqrt{ab}}$, which is equivalent to $ 12\leq \frac{(a\plus{}b)^2}{ab}\plus{}4\left(\frac{a\plus{}b}{\sqrt{ab}}\right)$, true because $ 4\leq \frac{(a\plus{}b)^2}{ab}$ and $ 2\leq \frac{a\plus{}b}{\sqrt{ab}}$.


Find the greatest positive real number $ k$ such that the inequality below holds for any positive real numbers $ a,b,c$:
\[ \frac ab \plus{} \frac bc \plus{} \frac ca \minus{} 3 \geq k \left( \frac a{b \plus{} c} \plus{} \frac b{c \plus{} a} \plus{} \frac c{a \plus{} b} \minus{} \frac 32 \right).
\]

$ k \equal{} 2$
We shall first prove that $ k\leq2$
Substitute $ a \equal{} 1, b \equal{} 1, c \equal{} n$, for some real number $ n$
Then the original inequality is equivalent to
$ n \plus{} \frac {1}{n} \minus{} 2\geq k(\frac {n}{2} \plus{} \frac {2}{n \plus{} 1} \minus{} \frac {3}{2}$)
Since the LHS = $ (2 \plus{} \frac {2}{n})(\frac {n}{2} \plus{} \frac {2}{n \plus{} 1} \minus{} \frac {3}{2})$
So if $ LHS\geq RHS$, $ 2 \plus{} \frac {2}{n}\geq k$, as n tends to infinity, $ k\leq2$.

Now we shall prove that $ k \equal{} 2$ is suitable.
Set $ k \equal{} 2$ in the original inequality.
Then it is equivalent to prove $ \frac {a}{2b} \plus{} \frac {b}{2c} \plus{} \frac {c}{2a} \geq \frac {a}{b \plus{} c} \plus{} \frac {b}{a \plus{} c} \plus{} \frac {c}{a \plus{} b}$
After expansion, it is equivalent to $ a^3b^3\plus{}b^3c^3\plus{}c^3a^3\plus{}a^2b^4\plus{}b^2c^4\plus{}c^2a^4 \geq 3a^2b^2c^2 \plus{} a^4bc \plus{}b^4ca \plus{} c^4ab$
Since $ a^3b^3\plus{}b^3c^3\plus{}c^3a^3 \geq 3a^2b^2c^2$ by AM-GM inequality and $ a^2b^4\plus{}b^2c^4\plus{}c^2a^4 \geq a^4bc \plus{}b^4ca \plus{} c^4ab$ by Muirhead inequality, so the original inequality $ \frac {a}{2b} \plus{} \frac {b}{2c} \plus{} \frac {c}{2a} \geq \frac {a}{b \plus{} c} \plus{} \frac {b}{a \plus{} c} \plus{} \frac {c}{a \plus{} b}$ is true. Thus $ k \equal{} 2$.

[u][b]SOLUTION:[/b][/u]
Without loss of generality, suppose that $ c = max(a,b,c)$.
   Observe that
$ \frac {a}{b} + \frac {b}{c} + \frac {c}{a}-3 = \frac {(a - b)^2}{ab} + \frac {(c - a)(c - b )}{ac} .$

$ \frac {a}{b + c} + \frac {b}{c + a} + \frac {c}{b + a} - \frac {3}{2} = \frac {(a - b)^2}{(a + c)(b + c)} + \frac {(a + b + 2c)(c - a)(c - b)}{2(a + b)(b + c)(c + a)} .$

   Thus we have the equivalent inequality:
$ (\frac {1}{ab} - \frac {k}{(c - a)(c - b)})(a - b)^2 - (\frac {1}{ac} - \frac {k(a + b + 2c)}{2(a + b)(b + c)(c + a)})(c - a)(c - b) .$               $ (*)$

   We will show that $ 1$ is the greatest value of the positive real number $ k$ such that this inequality holds for any positive real numbers $ a, b, c$.
   Indeed if $ k = 1$, Because $ c = max(a,b,c)$, we have
$ \frac {1}{ab} - \frac {1}{(a + c)(b + c)} \ge 0 ;$

$ \frac {1}{ac} - \frac {a+b+2c}{2(a + b)(b + c)(c + a)} \ge 0 ;$

$ (c - a)(c - b) \ge 0 .$

   Therefore, the inequality $ (*)$ is true.

   Finally, we must show that if $ k = m + 1 > 1$  (Let $ m$ be a positive real), then there exists a triple $ (a,b,c)$ such that the inequality $ (*)$ is not true.
   Consider the function
$ f(c) = (\frac {1}{ab} - \frac {k}{(c - a)(c - b)})(a - b)^2 - (\frac {1}{ac} - \frac {k(a + b + 2c)}{2(a + b)(b + c)(c + a)})(c - a)(c - b)$

$ = (\frac {1}{ab} - \frac {m + 1}{(c - a)(c - b)})(a - b)^2 - (\frac {2(b - ma)c^2 + (a + b)((1 - m)a + 2b)c + 2ab(a + b)}{2(a + b)(b + c)(c + a)})(c - a)(c - b).$

   We choose $ a$ and $ b$ such that $ b < ma$. Hence
$ \displaystyle \lim_{c\rightarrow + \infty}{(\frac {1}{ab} - \frac {k}{(c - a)(c - b)})(a - b)^2}) = \frac {(a - b)^2}{ab}$
$ \displaystyle \lim_{c\rightarrow + \infty}{(\frac {1}{ab} - \frac {m + 1}{(c - a)(c - b)})(a - b)^2 - (\frac {2(b - ma)c^2 + (a + b)((1 - m)a + 2b)c + 2ab(a + b)}{2(a + b)(b + c)(c + a)})(c - a)(c - b)) = - \infty.}$
   Thus $ \displaystyle \lim_{c\rightarrow + \infty}f(c) = - \infty$. Now we choose $ c$ such that it is great enough to $ f(c) < 0$. Consequently, there exists a triple $ (a,b,c)$ such that the inequality $ (*)$ is not true.

  [i][b] In conclude, we can finally write the full answer to the problem:  $ k = 1$.[/b][/i]

Here is my solution :)
Let $ a\equal{}t^3,b\equal{}t,c\equal{}1$ $ (t>0)$ then the inequality becomes
\[ \frac{(t\minus{}1)^2(t^3\plus{}3t^2\plus{}2t\plus{}1)}{t^3} \ge k\cdot \frac{(t\minus{}1)^2(2t^6\plus{}2t^5\plus{}3t^4\plus{}2t^3\plus{}2t^2\plus{}t\plus{}2)}{2t(t^3\plus{}1)(t^2\plus{}1)}\]
\[ \Leftrightarrow k \le \frac{2(t^3\plus{}1)(t^2\plus{}1)(t^3\plus{}3t^2\plus{}2t\plus{}1)}{t^2(2t^6\plus{}2t^5\plus{}3t^4\plus{}2t^3\plus{}2t^2\plus{}t\plus{}2)}\]
Now, let $ t\rightarrow \plus{}\infty$, we get $ k \le 1$. We will show that $ 1$ is the value which we find, that is
\[ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a} \minus{}3 \ge \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b} \minus{}\frac{3}{2}\]
\[ \Leftrightarrow \frac{ca}{b(b\plus{}c)}\plus{}\frac{ab}{c(c\plus{}a)}\plus{}\frac{bc}{a(a\plus{}b)} \ge \frac{3}{2}\]
By the Cauchy Schwarz Inequality, we get
\[ \frac{ca}{b(b\plus{}c)}\plus{}\frac{ab}{c(c\plus{}a)}\plus{}\frac{bc}{a(a\plus{}b)} \ge  \frac{(ab\plus{}bc\plus{}ca)^2}{abc(b\plus{}c)\plus{}abc(c\plus{}a)\plus{}abc(a\plus{}b)} \equal{}\frac{(ab\plus{}bc\plus{}ca)^2}{2abc(a\plus{}b\plus{}c)}\]
Moreover, it is easy to show that
\[ (ab\plus{}bc\plus{}ca)^2 \ge 3abc(a\plus{}b\plus{}c)\]
And so, we get the conclusion
\[ k_{\max}\equal{}1\]
We have done.
:)

Let $ f(a,b,c,k)\equal{}LHS\minus{}RHS$.
Then $ 2a^{2}(1\plus{}a)b(a\plus{}b)(a^{2}\plus{}b)f(1,\frac{1}{a},\frac{b}{a^{2}},k)$
$ \equal{}(2\minus{}2b(k\minus{}1))a^{7}\plus{}(2\plus{}2b^{2}\plus{}b(k\minus{}2))a^{6}\plus{}b(1\plus{}b)(k\minus{}2)a^{5}\plus{}2b(1\minus{}3b\plus{}b^{3}\minus{}k)a^{4}\plus{}b^{2}(1\plus{}b)(k\minus{}2)a^{3}\plus{}b^{3}(k\minus{}2)a^{2}\plus{}2b^{3}(1\plus{}b\minus{}bk)a\plus{}2b^{4}$.
If $ k>1$, we can choose $ b$ large enough so that $ 2\minus{}2b(k\minus{}1)<0$. Then the coefficient of $ a^{7}$ become negative, so if we choose $ a$ large enough, $ f(1,\frac{1}{a},\frac{b}{a^{2}})<0$.

For $ k\equal{}1$, $ 2abc(a\plus{}b)(b\plus{}c)(c\plus{}a)f(a,b,c,1)\equal{}2\sum_{cyc}a^{3}b^{3}\plus{}2\sum_{cyc}a^{2}b^{4}\minus{}\sum_{sym}a^{3}b^{2}c\minus{}6a^{2}b^{2}c^{2}$. This is positive since
$ 2\sum_{cyc}a^{3}b^{3}\equal{}\sum_{sym}a^{3}b^{3}\ge \sum_{sym}a^{3}b^{2}c$ (Muirhead Inequality)
$ 2\sum_{cyc}a^{2}b^{4}\ge 6a^{2}b^{2}c^{2}$ (AM-GM)

Therefore minimum $ k$ is $ 1$, equality occurs iff $ a\equal{}b\equal{}c$.

[quote=Ilthigore]We must find the greatest $ k$ such that
\[ f_k(a,b,c)\equal{}\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a} \minus{} 3 \minus{} k\left(\frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b}\minus{}\frac{3}{2}\right) \geq 0\]
I adopt the (slightly modified) notation that $ O(x)$ is any function for which there exist real constants $ C,D$ such that $ |O(x)|\leq Cx$ for all $ x\geq D$. In particular, any monomial $ ax^r$ with $ a,r$ real and $ r\leq 1$ (including zero) is $ O(x)$, and $ O(x)\plus{}O(x)\equal{}O(x)$ (adding their constants).

I claim that $ k\equal{}1$ is the solution.

Firstly, I show that $ k\leq 1$. Consider letting $ a\equal{}x^2,b\equal{}1,c\equal{}\frac{1}{x}$. Then we obtain
\[ f_k(x^2,1,\frac{1}{x})\equal{}x^2\plus{}x\plus{}\frac{1}{x^3} \minus{} 3 \minus{} k\left(\frac{x^3}{1\plus{}x}\plus{}\frac{x}{1\plus{}x^3}\plus{}\frac{1}{x^3\plus{}x}\minus{}\frac{3}{2}\right)\]
\[ \equal{} x^2 \minus{} \frac{kx^3}{1\plus{}x} \plus{} O(x) \equal{} x^2 \minus{} \frac{k(x^3\plus{}1)}{1\plus{}x} \plus{} \frac{k}{x\plus{}1} \plus{} O(x) \equal{} x^2\minus{}k(x^2\minus{}x\plus{}1) \plus{} O(x)\]
This is $ (1\minus{}k)x^2\plus{}O(x)$, so for $ x$ sufficiently large and $ k>1$, this will become negative. Thus $ k\leq 1$ as required.

Now, it remains to demonstrate that \[ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\geq \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b}\plus{}\frac{3}{2}\][/quote]
$$\Leftrightarrow \frac{ac}{b(b+c)}+\frac{cb}{a(a+b)}+\frac{ba}{c(c+a)}\geq \frac{3}{2}$$
By C-S inequality: $$LHS\geq \frac{(bc+ca+ab)^{2}}{2abc(a+b+c)}\geq \frac{3}{2}$$
[url]https://artofproblemsolving.com/community/c4h2054805p14768263[/url]
[url]https://artofproblemsolving.com/community/c6h583302p3448303[/url]


Online Math Open
ratio
algebra
polynomial
Vieta

[color=blue][u][b]Πρώτα απ' όλα ας ευχηθόύμε σε όλους τους υποψηφίους[/b][/u][/color]
[size=150][color=red][b]ΚΑΛΗ ΕΠΙΤΥΧΙΑ Τύχη Ἀγαθὴ.[/b][/color][/size]
[size=150][b]Σε αυτό το topic ας βάλουμε τα θέματα λύσεις και τις εντύπωσεις μας[/b][/size]

Πλέον εύχομαι

[b][color=blue][size=150]ΚΑΛΑ ΑΠΟΤΕΛΕΣΜΑΤΑ[/size][/color][/b]

κι από εμένα σε όλους τους διαγωνιζόμενους!

Αλέξανδρος

Καλα αποτελεσματα σε ολα τα παιδια του φορουμ!

Μηπως εχουν αναρτηθει πουθενα τα θεματα?

ΚΑΛΑ ΑΠΟΤΕΛΕΣΜΑΤΑ ΚΑΙ ΑΠΟ ΕΜΕΝΑ ΣΤΑ ΠΑΙΔΙΑ.....
Περιμενω να ανεβασετε τα θεματα....

[quote="Eukleidis"]Τα θέματα δεν έχουν αναρτηθεί επίσημα.  Περιμένω απο τις 10 να ανρτηθουν και ακομά τίποτα. ¨Οταν όμως τα ανεβάσουν θα τα ανεβασω[/quote]

Η ΕΜΕ μπορει να καθυστερισει αρκετα να τα ανεβασει, οποτε ας περιμενουμε κανενα διαγωνιζομενο να τα βαλει

Έχεις δίκιο φίλε μου. Και στον αρχιμήδη τα βαλαν Δευτερα τα θέματα

Τι γινεται ρε παιδια???Τοσο δυσκολο ειναι να βαλει καποιος τα θεματα????

δες εδω . Το ιδιο ισχυει και για φετος  :wink:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=198376

Τα θέματα ήταν εύκολα.

Έλυσα 3 (δεν έκανα τη γεωμετρία) (sry re Nick... :( )

Καλά αποτελέσματα σε όλους. :wink:

Ειχε συνδιαστικη?

[quote="bilot4"][quote="filosofimenos"]Ειχε συνδιαστικη?[/quote]
Απ' όλα είχε! Συνδιαστική, γεωμετρία, θεωρία αριθμών και άλγεβρα.

Καλά αποτελέσματα σε όλους![/quote]

Εγω εμαθα την θεωρια αριθμων :P
Δεν την ποσταρω αφου απαγορευεται αλλα πρεπει να πω οτι ηταν απιστευτα trivial θεμα που μπορουσε να λυθει και σε 30 seconds (εμενα 1 min μου πηρε)... δεν νομιζω να ηταν θεμα για πιο πανω απο Ευκλειδη.

Aνισοτητα υπηρχε :huh:  :?:

οσο για την ανισότητα, εμμέσως....
και η συνδιαστική αμα την έβαζες κάτω ήτν επιλύσιμη...σε σχέση με πέρυσι...τεσπα, ασ περιμένουμε επισήμως...

[quote="NickNafplio"]Παιδια στη συνδιαστικη εβγαλα ακεραιο μερος $ \frac { \minus{} n}{2}$ αλλα χρησημοποιω ελαχιστη συνδιαστικη... το περισσοτερο ειναι Αλγεβρα και μελετη συναρτησης ωστε να βρω την ελαχιστη ακεραια τιμη της... ο Ηλιας νομιζω μου ειπε οτι την ελυσε μονο με συνδιαστικη, εκανε κανενας αλλος καμια τετοια λυση.[/quote]

Φαινεται να εχουμε κανει την ιδια λυση. Ξεκινας εστω μ μαυρα π πρασινα και κ κοκκινα, βγαζεις τις σχεσεις της συνδιαστικης (πολυ απλες) και κατασκευαζεις τη συναρτηση σου που θες να ελαχιστοποιησεις. Nick, προσεχε ομως γιατι με εναν πολυ απλο αλγεβρικο μετασχηματισμο και λαμβνοντας υποψην οτι τα μ,κ,π ειναι φυσικη γλιτωνεις τη μελετη και ξεχωριζεις την περιπτωση αρτιου και περιττου ευκολα.

Ευκλείδης γιατί ισχύει $ x,y,z>\equal{}1$???

Αν θεωρήσουμε το πολυώνυμο

$ P(x)\equal{}x^3\minus{}sx^2\plus{}sx\minus{}1$

όπου $ s$ το άθροισμα είναι προφανές από τους τύπους του Vieta ότι αν έχει τρεις πραγματικές ρίζες αυτές θα είναι οι $ x,y,z$.

Με παραγοντοποίηση βγαίνει ότι μία από αυτές είναι το $ 1$ οπότε οι άλλες δύο που έχουν γινόμενο $ 1$ έχουν μέγιστο άθροισμα όταν είναι ίσες με $ 1$ (δες το και με AM-GM)

Κάθε τριάδα του τύπου $ (1,k,1/k)$ είναι δεκτή, όχι υποχρεωτικά οι άσσοι. :wink:

[quote="Protonios"]Ευκλείδης γιατί ισχύει $ x,y,z > \equal{} 1$???

Αν θεωρήσουμε το πολυώνυμο

$ P(x) \equal{} x^3 \minus{} sx^2 \plus{} sx \minus{} 1$

όπου $ s$ το άθροισμα είναι προφανές από τους τύπους του Vieta ότι αν έχει τρεις πραγματικές ρίζες αυτές θα είναι οι $ x,y,z$.

Με παραγοντοποίηση βγαίνει ότι μία από αυτές είναι το $ 1$ οπότε οι άλλες δύο που έχουν γινόμενο $ 1$ έχουν μέγιστο άθροισμα όταν είναι ίσες με $ 1$ (δες το και με AM-GM)

Κάθε τριάδα του τύπου $ (1,k,1/k)$ είναι δεκτή, όχι υποχρεωτικά οι άσσοι. :wink:[/quote]
Με ελάχιστο άθροισμα θέλουμε. Σε παραπέμπω στον προκκριματικό μικρών 4ο θέμα. Μπορεί να κάνω και λάθος. :oops:

Η λύση μου για το 4ο θέμα των μικρών
Λόγω της συνθήκης $ xyz \equal{} 1$ :
$ x \plus{} y \plus{} z \equal{} xy \plus{} yz \plus{} zx \equal{} \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z}$
Απο τη γνωστή ανισότητα:
$ (x \plus{} y \plus{} z)(\frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z})\geq 9$ έπεται ότι $ x \plus{} y \plus{} z \geq3$
Εφόσον ψάχνουμε το ελάχιστο άθροισμα, αρκεί τώρα να λύσουμε το σύστημα:
$ x \plus{} y \plus{} z \equal{} 3$
$ xyz \equal{} 1$
$ xy \plus{} yz \plus{} zx \equal{} 3$
το οποίο έχει λύση την τριάδα $ (1, 1, 1)$

Επίσης πολλά συγχαρητήρια στα παιδιά που είναι στην ομάδα αλλά και σε όσους έφτασαν ως τον Προκριματικό.

[size=150][color=blue]Συγχαρητήρια σε όλα τα παιδιά που διαγωνίστηκαν ![/color][/size]

 Στα μέλη της εθνικής μας ομάδας εύχομαι να έχουν τη μέγιστη  απόδοση και καλά μετάλια !

 Σε όσους δεν κατάφεραν φέτος να μπουν στην ομάδα και δεν είναι Γ΄λυκείου , θέλω να τους  πω να βάλουν τα δυνατά τους και να αξιοποιήσουν το καλοκαίρι που έρχεται , ώστε να διεκδικήσουν με περισσότερες αξιώσεις μια θέση την ερχόμενη χρονιά ! Ήδη η μέχρι τώρα πορεία τους είναι μεγάλη επιτυχία, κάτι που πρέπει να τους δίνει δύναμη και πίστη !

[size=150][color=red] Καλή Ανάσταση  !!![/color][/size]

Οι άσσοι έχουν το ελάχιστο άθροισμα όντως αλλά εσύ είπες ότι όλοι είναι μεγαλύτεροι του 1 που δεν ισχύει γιατί αλλιώς οι μόνες λύσεις του συστήματος θα ήταν άσσοι. :wink:

Για το 4 των μικρων νομιζω οτι αρκει να πουμε οτι $ x\plus{}y\plus{}z \geq 3$ με ισον μονο για $ x\equal{}y\equal{}z\equal{}1$, οποτε η ζητουμενη τιμη ειναι 3 αρκει οι αριθμοι αυτοι να επαληθευουν και τη 1η εξισωση... που πραγματι την επαληθευουν

Με το 3ο ασχοληθηκα λιγακι και αν δεν μου ξεφυγε κατι εβγαλα οτι ο αριθμος ειναι ισος με $ a \plus{} c \minus{} b$ που ειναι ακεραιος...

Στο 2ο απο την καθετοτητα των Ρ.Αξονων με τις διακεντρους και τη χρηση μερικων εγγραψημων βγαζουμε ισες τις απεναντι πλευρες ανα 2 ως τοξα που βαινουν σε ισες γωνιες σε ισους κυκλους

Μπράβο στην ομάδα κι από εμένα.  :wink:

Τώρα, για το 4ο των μικρών, πήγα αρχικά στο $ x(y\minus{}1) \plus{} y(z\minus{}1) \plus{} z(x\minus{}1) \equal{} 0$ για να ξεχωρίσω τις περιπτώσεις 1) όλοι ισούνται με 1 και το άθροισμα είναι 3 και 2) ένας από αυτούς ισούται με 1 και οι άλλοι δύο είναι αντίστροφοι, οπότε για υ = 1 βγαίνει $ z(\frac{1}{z} \minus{} 1) \equal{} \minus{}z \plus{} 1 \Rightarrow z \equal{} z$ άρα ισχύει για όλους τους αριθμούς στο R, αρκεί ο ένας να είναι αντίστροφος του άλλου. Δεν έχω αποδείξει όμως γιατί είναι το ελάχιστο άθροισμα 3, το οποίο ισχύει για χ = υ =ζ = 1.  :|

[quote="Asynartisies"]Μπράβο στην ομάδα κι από εμένα.  :wink:

Τώρα, για το 4ο των μικρών, πήγα αρχικά στο $ x(y \minus{} 1) \plus{} y(z \minus{} 1) \plus{} z(x \minus{} 1) \equal{} 0$ για να ξεχωρίσω τις περιπτώσεις 1) όλοι ισούνται με 1 και το άθροισμα είναι 3 και 2) ένας από αυτούς ισούται με 1 και οι άλλοι δύο είναι αντίστροφοι, οπότε για υ = 1 βγαίνει $ z(\frac {1}{z} \minus{} 1) \equal{} \minus{} z \plus{} 1 \Rightarrow z \equal{} z$ άρα ισχύει για όλους τους αριθμούς στο R, αρκεί ο ένας να είναι αντίστροφος του άλλου. Δεν έχω αποδείξει όμως γιατί είναι το ελάχιστο άθροισμα 3, το οποίο ισχύει για χ = υ =ζ = 1.  :|[/quote]

Σου παραθέτω την εξής γνωστή ανισότητα:
\[ a \plus{} \frac {1} {a} \geqslant 2
\]
για κάθε
\[ \alpha  \in \mathbb{R}_ \plus{} ^*\]
με ισότητα για α=1

[quote="NickNafplio"]Για το 4 των μικρων νομιζω οτι αρκει να πουμε οτι $ x \plus{} y \plus{} z \geq 3$ με ισον μονο για $ x \equal{} y \equal{} z \equal{} 1$, οποτε η ζητουμενη τιμη ειναι 3 αρκει οι αριθμοι αυτοι να επαληθευουν και τη 1η εξισωση... που πραγματι την επαληθευουν

Με το 3ο ασχοληθηκα λιγακι και αν δεν μου ξεφυγε κατι εβγαλα οτι ο αριθμος ειναι ισος με $ a \plus{} c \minus{} b$ που ειναι ακεραιος...

Στο 2ο απο την καθετοτητα των Ρ.Αξονων με τις διακεντρους και τη χρηση μερικων εγγραψημων βγαζουμε ισες τις απεναντι πλευρες ανα 2 ως τοξα που βαινουν σε ισες γωνιες σε ισους κυκλους[/quote]

Ε και το πρωτο , για λογους πληροτητας, modulo6 βγαινει αμεσως.  :)

[quote="Asynartisies"]Never mind...το απέδειξα αφότου έγραψα την ισότητα ως 1 + χ + 1/χ >= 3 και κατέληξα στο (χ-1)^2 >= 0, αποδεικνύοντας κι αυτό που είπες, Eukleidis. Thanks, πάντως.[/quote]
Τίποτα ευχαρίστηση μου

Εχω την εντύπωση οτι το 2ο θέμα λυνόταν εύκολα μεμιγαδικούς
Αν ξεκινήσουμε απο την $ BA_1\equal{}GA_1\equal{}CA_1$ κι την τρέψουμε σε μιγαδική μορφή  μετά υψώνοντας στο τετράγωνο καταλήγουμε σε ενα πρωτοβάθμιο σύστημα για το $ z_{A_1}$
Οι υπολοιπες σχέσεις προκύπτουν κυκλικά (άρα χωρίς πολλές πράξεις) και το ζητούμενο είναι να δειχθεί ότι $ z_{A_1}\plus{}z_{B_1}\plus{}z_{C_1}\equal{}0$ που προκύπτει και αυτό χωρίς πολλές πράξεις μια που οι παρονομαστές στα $ z_{A_1},z_{B_1},z_{C_1}$ είναι κοινοί $ Re(ab\plus{}bc\plus{}ca)$εδώ χρησιμοποιούμε μια μικρή ιδιότητα οριζουσων
Το τελικο αποτέλεσμα βγαίνει και αυτό εύκολα με την θεώρηση οτι ο κύκλος ABC είναι ο μοναδιαίος

[color=darkred][size=150]ΕΠΙΣΗΜΑ ΠΛΕΟΝ ΟΙ ΛΥΣΕΙΣ ΤΟΥ ΠΡΟΚΡΙΜΑΤΙΚΟΥ ΜΙΚΡΩΝ ΚΑΙ ΜΕΓΑΛΩΝ:[/size][/color]
Για μικρούς: http://www.hms.gr/images/stories/prokrimatikos_2009/JUNIORS2009SOLUTIONS.pdf
Για μεγάλους: http://www.hms.gr/images/stories/prokrimatikos_2009/SENIORS2009SOLUTIONS.pdf

Και να ευχηθούμε σε ολούς που διαγωνιστηκαν στη Μεσογειάδα
[color=blue]ΚΑΛΗ ΕΠΙΤΥΧΙΑ[/color]

Γράφω αναλυτικά την λύση με τους μιγαδικούς για το 2ο γιατί στο προηγούμενο post είχα ένα μικρό λαθάκι το οποίο δεν επηρέαζε βεβαια την λύση αλλά...

Έστω a,b,c,g οι αντίστοιχοι μιγαδικοί των κουφών και του βαρύκεντρου του ΑΒC
ΧΒΤΓ θεωρώ ότι ο κὐκλος ABC είναι ο μοναδιαίος οπότε $ \bar{a}\equal{}1/a...$
Αν z ο μιγαδικός που έχει εικόνα το $ A_1$ τότε
$ |z\minus{}b|^2\equal{}|z\minus{}c|^2\equal{}|z\minus{}g|^2$ Λαμβάνοντας υπ όψη τα προηγούμενα και ότι $ 3g\equal{}a\plus{}b\plus{}c$ έχουμε
$ z(1/c\minus{}1/b)\plus{}\bar{z}(c\minus{}b)\equal{}0$
$ z(1/a\plus{}1/c\minus{}2/b)\plus{}\bar{z}(a\plus{}c\minus{}2b)\equal{}3(|g|^2\minus{}1)$
Η ορίζουσα D του συστήματος είναι..$ D\equal{}(a/c\minus{}c/a)\plus{}(b/a\minus{}a/b)\plus{}(c/b\minus{}b/c)$ κυκλικά συμμετρική άρα κοινή για τα w,v ,μιγαδικών, που έχουν αντίστοιχα εικόνες τα $ B_1,C_1$
H $ D_z\equal{}3(|g|^2 \minus{}1)(b\minus{}c)$
τότε $ \frac{z\plus{}w\plus{}v}{3}\equal{}\frac {(|g|^2 \minus{}1)(b\minus{}c\plus{}c\minus{}a\plus{}a\minus{}b)}{(a/c\minus{}c/a)\plus{}(b/a\minus{}a/b)\plus{}(c/b\minus{}b/c)}\equal{}0$
που σημαίνει ότι βαρύκεντρο του $ A_1B_1C_1$ είναι το περίκεντρο του $ ABC$


Multiply 10.5 by 1.05 and write the product as a common fraction.

I think you can write the answer as "11 1/40", too.

The question specifically states that it needs to be written as the product  [b]as a common fraction[/b]as a common fraction

You have a mixed number, which is different from a mixed number which you wrote.

Jongy did have the correct answer. A common fraction COULD have the numerator larger than the denominator.

-Sky


find the biggest real number  $ k$ such that for all real numbers $ a$ ,$ b$ and $ c$ : $ a\plus{}b\plus{}c \leq k$  we find  $ abc \leq k$
from moroccan TST 2002

i think you want say the least k,if it's yes :it suffice to have $ (abc)^{1/3} < \equal{} k^{1/3} < \equal{} \frac {a \plus{} b \plus{} c}{3} < \equal{} k/3$
so $ k >\equal{} 2\sqrt3$,the least  k is $ 2\sqrt3$

Should be $ a,b,c > 0$
If not let
$ a,b\to\ \minus{} \infty$
$ c \equal{} 1$
Then it is not true.


Let a,b,c real positive numbers. Prove that

 \sqrt (a 2 +1) + \sqrt (b 2 +1) + \sqrt (c 2 +1) \geq  \sqrt (6(a+b+c))

No doubt, Manlio really loves inequalities  :)

f(x) = \sqrt(x+1)
f''(x) = 1/((x^2+1)^(3/2)) > 0 so f (strictly) convex.

1/3  \sum f(a) >= f(\sum a / 3) by Jensen

x+y>= 2xy, so   \sqrt ((a+b+c) + 9) >= \sqrt (6(a+b+c))


Let $ a\equal{}\pi/2008$. Find the smallest positive integer $ n$ such that
\[ 2[\cos(a)\sin(a)\plus{}\cos(4a)\sin(2a)\plus{}\cos(9a)\sin(3a)\plus{}\cdots\plus{}\cos(n^2a)\sin(na)]\] is an integer.

[hide="Solution"]
Distributing the $ 2$ and using the product to sum identity, we have that $ 2\cos(k^2a)\sin(ka) = sin((k)(k + 1)a) - sin((k - 1)(k)a)$, so the series looks like it telescopes. If you don't remember all the product-sum identities you can pretty easily derive them on the spot (look at the product, which looks like it comes from one term of the sine of a sum). We can verify this:
\begin{eqnarray*} \sum_{k = 1}^n 2\cos(k^2a)\sin(ka) & = & \sum_{k = 1}^n sin((k)(k + 1)a) - sin((k - 1)(k)a) \\
& = & \sum_{k = 1}^n \sin((k)(k + 1)a) - \sum_{k = 1}^n \sin((k - 1)(k)a) \\
& = & \sum_{k = 1}^n \sin((k)(k + 1)a) - \sum_{k = 0}^{n - 1} \sin((k)(k + 1)a) \\
& = & \sum_{k = 1}^n \sin((k)(k + 1)a) - \sum_{k = 1}^{n - 1} \sin((k)(k + 1)a) - \sin(0) \\
& = & \sin(n(n + 1)a) \\
& = & \sin\left(\frac {n(n + 1)}{2008}\pi\right) \end{eqnarray*}
$ \sin(\pi x)$ is an integer for any $ x$ which is a multiple of $ \frac {1}{2}$, so $ n(n + 1)\equiv0\pmod{1004}$. Since $ n$ or $ n + 1$ must therefore be divisible by $ 251$, we check $ n = 250$, which doesn't work, and $ n = 251$, which does, so the smallest $ n$ which works is $ \boxed{251}$.
[/hide]

[quote="SorcererofDM"]OMFFFFFFFFFFFFFGGGGGGGGGGGGGGGGGGGGG!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
I blind guessed the freaking problem and got it right!!!!!!!!!!!!!!!! OMFFFFFFGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG!!!!!!!!!!!!!
(well not completely, i was going for the largest prime factor of 2008)!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!![/quote]

 :rotfl: Grats man.

Damn. Made a silly mistake...wrote 250 instead...

Ugghhh

[hide="I thought this was pretty straightforward:"]
Product to Sum formula (had to derive it, but you probably should know how to if you don't) $ 2\cos a\sin b \equal{} \sin(a\plus{}b)\minus{}\sin(a\minus{}b)$

So $ 2\cos k^2a\sin ka \equal{} \sin(ak(k\plus{}1)\minus{}\sin(ak(k\minus{}1)$.

Oooh, telescoping sum $ \sum^n_{k\equal{}1} [\sin(ak(k\plus{}1)\minus{}\sin(ak(k\minus{}1)] \equal{} \sin(an(n\plus{}1))\minus{}\sin(a*1*0)\equal{}\sin(an(n\plus{}1))$...

...is an integer, or $ an(n\plus{}1)$ is a multiple of $ \pi/2$.

$ \frac{\pi n(n\plus{}1)}{2008}\equal{}\frac{k\pi}{2} \implies n(n\plus{}1)\equal{}1004k\equal{}2^2251k$.

So 251 is prime, and divides either n or n+1, while the other is divisible by 4: 252 is divisible by 4, so $ n\equal{}\boxed{251}$.[/hide]

[quote="SorcererofDM"]OMFFFFFFFFFFFFFGGGGGGGGGGGGGGGGGGGGG!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
I blind guessed the freaking problem and got it right!!!!!!!!!!!!!!!! OMFFFFFFGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG!!!!!!!!!!!!!
(well not completely, i was going for the largest prime factor of 2008)!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!![/quote]
Eh... Lucky.  I guess 502 because I thought it would have something to do with factors of 2008.  Oh well...

[quote="pianoforte"]
So 251 is prime, and divides either n or n+1, [/quote]

Assuming you haven't memorized all of the primes between 1 and 251, how did you know that 251 was prime so easily? It took me quite a while to figure out that 251 divides n or n+1... Am I missing something?

Are those separate questions?

1)  You only have to test for primes up to $ \sqrt{251} < 16$.  If a number is composite, it can be written in the form $ n \equal{} ab$ for some $ a, b$, and $ \min(a, b) \le \sqrt{n}$.

2)  A number is prime if and only if it has the property that $ p | xy \implies p | x \text{ or } p | y$.  Can you prove it?

[quote="t0rajir0u"]Are those separate questions?

1)  You only have to test for primes up to $ \sqrt {251} < 16$.  If a number is composite, it can be written in the form $ n \equal{} ab$ for some $ a, b$, and $ \min(a, b) \le \sqrt {n}$.

2)  A number is prime if and only if it has the property that $ p | xy \implies p | x \text{ or } p | y$.  Can you prove it?[/quote]

I remembered 2. But 1 I had forgotten. So you conclude 251 is prime after testing that none of 2,3,5,7,11,13 divide it. That would still take me like 2 minutes but I see now thanks. :)

EDIT: ok more like 35 seconds

[hide="Faster than that"] Ruling out $ 2, 3, 5, 11$ is easy (are you familiar with the relevant primality test for $ 11$?).  That leaves $ 7$ and $ 13$.  But

$ 251 \equal{} 260 \minus{} 9 \equiv \minus{}9 \bmod 13$
$ 251 \equal{} 210 \plus{} 41 \equiv 41 \bmod 7$ [/hide]

[quote="t0rajir0u"][hide=Faster than that]  (are you familiar with the relevant primality test for $ 11$?).  [/quote]

Yes. A number is divisible by 11 iff its representation in decimals $ a_0 ... a_n$ has the property that $ a_0 \minus{} a_1 \plus{} ... \plus{} ( \minus{} 1)^{n}a_n \equiv 0 (\mod 11)$. While were on the topic,  a number is divisible by 3 (9) iff the sum of its digits is divisible by 3 (9). :)

Come on, you guys should have the prime factorization of 2008 memorized, right? You shouldn't have to test anything.

p6: By Product-to-Sum, we can simplify to $\sin(2a)-\sin(0)+\sin(6a)-\sin(2a)+\ldots + \sin(na(n+1)) - \sin(na(n-1)) = \sin(na(n+1))$. For this to be an integer, $n(n+1)$ must have a factor of $1004 = 2^2 \cdot 251$. Thus one of $n, n+1$ is a multiple of $251$. The smallest possibility is at $n = \boxed{251}$, which indeed works since $2^2|252$.


These two spinners are divided into thirds and quarters, respectively. If each of these spinners
is spun once, what is the probability that the product of the results of the two spins will be an
even number? Express your answer as a common fraction.

problem no 28  please see the figure in  the following pdf link

http://www.mnspe.org/mathcounts/2007tests/510-07_C_Countdown.pdf

answer is 2/3

We get an even product from $ (a, b)$ when at least one of $ a$ and $ b$ is even, that is $ (a, b) \equal{} (3, 6), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8), (5, 6)$ or $ (5, 8)$; $ 8$ possibilities out of $ 12$: with probability $ \frac{8}{12} \equal{} \frac{3}{4}$.

We look for the ways in which we can get an odd product, then use complementary counting. This will happen if both numbers are odd. There is a $ \frac {2}{3}$ chance of the first spinner resulting in an odd number. There is a $ \frac {1}{2}$ chance of the second spinner resulting in an odd number.
Therefore,
               $ \frac {2}{3}\times{\frac {1}{2}} \equal{} \frac {1}{3} \implies 1 \minus{} \frac {1}{3} \equal{} \boxed{\frac {2}{3}}$.

@ AndrewTom $ \frac {8}{12} \equal{} \frac{2}{3}$  :wink:


probability
search
logarithms
number theory proposed
number theory

It is quite discouraging that no one can show that the exponent of a prime $p$ in $2^{p-1}-1$ does not go to infinity  :!: ... But here is a reason to believe that $p^{3}$ does not divide $2^{p-1}-1$ for any $p$, or at least for all but finitely many $p$: one would "expect" the Fermat quotient $(2^{p-1}-1)/p$ to take every residue $\mod{p^{2}}$ with equal probability, so that the expected probability for the residue $0$ should be $1/p^{2}$; hence, the expectation for the number of primes $p \leq x$ with $p^{3}\mid 2^{p-1}-1$ should be about $\sum_{p \leq x}1/p^{2}< 1$!

Assuming this conjecture (that $p^{3}\nmid 2^{p-1}-1$ for large enough $p$) -- and only it -- find the density of the set $\{n \mid 2^{n}-1 \text{ cubefree}\}$.

[Edit: "It is quite discouraging" means here that this is a famous and tantalizing unsolved problem. ]

Let me tell you something that you won't like, Vess: it is quite discouraging that people like you make such assertions. Yes, we are all stupid compared to you, your Highness and yes, we cannot prove that. But, your Excellence, should I recall you that usually problems on forums like mathlinks (or at least what used to be mathlinks) were discussed and discussed again, that everyone participated with some idea, advice, hint or something like that. And now? For how many times did I revive the topic with your bloody p-adic estimation? How many times did I ask you to help us on some topics, that really interest us? Yet, how many answers? If all you want is to post challenging but beautiful problems that no one can solve just because you feel in this way that you are extraordinary, then so be it, but if all you do is to post problems, saying how trivial, easy, elementary they are and how people on mathlinks are stupid for not solving them, then you can solve them alone, since I'm quite sick of this attitude. Look closer and you will see I was among those who tried every time to revive your topics, to come with approaches, even though not finished or stupid. I thought they deserved attention, now I see that all you want is completely different. Too bad, they are nice however...

Dear Harazi, I am sorry. I have never wanted to sound arrogant, nor I thought I did. In this case, saying "it is quite discouraging" meant that no one in the world has proven as simple and natural assertion as this (that, say, there are infinitely many primes with $p^{100}$ not dividing $2^{p-1}-1$), that this problem is so formidable. This is a famous unsolved problem, not an "easy" or "trivial" etc. This was figuratively speaking. I could have said "I am quite discouraged that I can never hope to show that [...]" -- same thing. Yes, I don't like that attitude, but I asked for it  :blush:  I owe you an apology  :blush:  for leaving the topic with the p-adic valuation without a response... I was going to respond, you can believe me. I needed to spend a lot more time on this problem to fix a crucial mistake that I found. This is why I delayed -- I only had a sketch idea when I wrote this. I humbly ask for an apology on this -- I will return on this topic at some point. For now, I retract the stated estimate. However, I can at once show an ineffective such result (that there are only finitely many [...] -- that is, without an estimate).  :oops:

About other topics: could you please remind me? I have always tried to relate with what I can...

I am sorry about that psychological analysis that you just made about me. It may seem like this, I don't know. If it does, then -- my sincere apologies. . I realize it may be insulting to call a nice problem "trivial" -- I have never done it with any "bad" intention. By the way, I haven't done that recently, as far as I can recall... I'm not nearly such megalomaniac, you just don't know me. But I like that you wrote what you wrote, and will take the critique from it. About my "purpose" to pose new and new problems on a given theme -- well, there is not any purpose at all, I never thought I might offend someone, I just liked to post incidentally some things that I have thought about and, in case other people become interested, to talk about them further. For instance on this problem, I don't know many things on it -- mine is a simple observation that I have shared, and I am sure, other people will be able to come up with much more valuable observations. And of course, I have long valued all your comments and approaches on my problems (and not only), and by the way, I have read many of your topics even though I rarely reply (whenever I go to this forum, I never browse or search topic-wise, but look at the latest posts by, e.g., harazi, grobber, Myth, Peter Scholze, Ravi B,...) It's only that, as you see, only occasionally post on the forum, and can't contribute as extensively as I'd like. I understand your critique, and will try to be more helpful/responseful in the future.

Thanks! I'll respond soon [edit: at some point... sorry, once again] about the p-adic estimation. My mistake there was that I posted something that was in a premature stage. I make mistakes far too often...  :oops:

Peace :)

No problem, Vess. I have to apologize for my exagerated reaction. It's interesting, because after a few days of terrible struggling  with that marvelous problem on p-adic estimation, I told mathmanman that I believe that any solution that exists must be wrong:the result was simply too beautiful. Unfortunately, I have no idea whether such an effective estimation exists or not, but this largely exceeds my poor capacity. Please, accept my apologizes for the above words and please continue posting beautiful problems, but let's discuss them more.

:)

Again, I'm sorry about those $p$-adic stuff. For $k=2$ at least, there are effective estimates, and these come from the $p$-adic Baker theory of linear forms in logarithms of algebraic numbers (e.g. van der Poorten's estimate in the paper solving the Nathanson conjecture that you attached). I believed I had a surprisingly simple proof of the "estimates" that I wrote, but alas, once I tried to expand it I see that it fails in general :(. It works, e.g., when all $a_{i}\in \{-1,1\}$, and in a couple of other cases -- but this excludes the interesting cases... Later (at some point), I will think about what else can be rescued of the idea, and will post something in that thread and on that topic.

On the current [in this thread] problem, I have to make yet another apology and correction. Now that I can't, for some reason, edit the original post anymore:

Assuming that $p^{3}\nmid 2^{p-1}-1$ for [b] all[/b] $p$ (which IMHO is quite likely to be the case, for the reason stated in the first post), give an explicit (absolute) $c>0$, depending only on this conjecture, such that the density of $n$ with $2^{n}-1$ cubefree is $> c$. (BTW: if $p^{3}\nmid 2^{p-1}-1$ for all $p$, then it is immediately clear that $2^{n}-1$ is cubefree infinitely often -- for example, when $n$ is prime, then $2^{n}-1$ is cubefree; the problem I propose is to show that the set of those $n$ with $2^{n}-1$ cubefree has in fact positive density, and to give an explicit lower bound for that density). I don't claim to be able to even give -- even conjecturaly -- the exact density...

How does one approach problems like this about density? I can find a couple of infinite classes of solutions, but I have no idea how to relate prime factorizations to density.

I can show, under the hypothesis that $p^{3}\nmid 2^{p-1}-1$ for all $p$, that the density of $n$ with $2^{n}-1$ cubefree is at least $0.7$. The idea is hinted here: http://www.mathlinks.ro/Forum/viewtopic.php?t=125386 . I will post a proof if anyone is interested, but only after the problem has stayed for sufficiently long time.

Ok, let's settle this! Suppose that $2^{n}-1$ is not cube-free. Thus, there exists $p>2$ prime such that $p^{3}|2^{n}-1$. Claim: $p$ divides $n$. Otherwise the order of $2$ mod $p^{3}$ will divide $n$ and $p^{2}(p-1)$, thus also $p-1$ and so $p^{3}|2^{p-1}-1$, wrong (HOWEVER, we assume this, which is probably intractably difficult). Thus $n=kp$ et $p^{3}|2^{kp}-1$. Now, it follows that $p|2^{k}-1$ and standard arguments show that we must have $p^{2}|2^{k}-1$ (because $1+2^{k}+...+2^{k(p-1)}$). Now, using the same argument, we observe that either $p$ divides $k$ or $p^{2}|2^{p-1}-1$.
 In any case, either $p ord_{2}(p)$ divides $n$ or $p^{2}$ divides $n$. Moreover, we know that if $p^{2}$ does not divide $n$, then $p^{2}|2^{p-1}-1$. Let $p_{1},p_{2},...$ les nombres premiers of last type (probably there are infinitely many, but I have no idea) and let $A_{i}$ be the set of $n$ for which $p_{i}ord_{2}(p_{i})$ divides $n$. Also, for any odd prime $p$ let $B_{p}$ the set of multiples of $p^{2}$. Now, take $x$ and look at those $n$ for which $2^{n}-1$ is not cube free and $n\leq x$. They are all in the union of all $A_{i}$ for $i\leq 2^{x}-1$ or in the union of all $B_{p}$ for $p\leq 2^{x}-1$. In any case, their cardinality is at most $x(U+V)$ where $U$ is the sum of the inverses of all squares of odd primes and $V$ is the sum of $\frac{1}{p ord_{2}(p)}$ for those $p$ with $p^{2}|2^{p-1}-1$. It means that the set of those $n$ with $2^{n}-1$ cube free has density at least $1-(1/9+1/25+1/49+...)-(1/1093+1/3000+...)$. I used the fact that the only $p$ with $p^{2}| 2^{p-1}-1$ and $p<10^{6}$ are $1093$ and something near 3000 (I always forget them). Now, using the fact that $ord_{2}(p)>log_{2}(p)$, we can easily bound the sum of $\frac{1}{p ord_{2}(p)}$ starting with let's say $10^{6}$. This will give an estimation of error at most $10^{-3}$ if my computations are not extremely wrong.  Thus we can say that the density is at least $0,995-(1/9+1/25+1/49+...)$, which is surely quite large (I think it's greater than $0,7$, but we have to compute a few terms from the sum since the sum of $\frac{1}{k log^{2}k}$ starting from $k_{0}$ is about $\frac{1}{ln k}$. Anyway, modulo probably many stupid computations, this should be the idea.

[quote=nbdaaa]For such a nice conjecture as this problem, I just want to remind these two famous estimations about the congruence of $2^{p-1}-1$ in modulo $p^2$ and $p^3$:
\[\begin{array}{l}
{2^{p - 1}} \equiv 1 - \frac{p}{2}{H_{\frac{{p - 1}}{2}}} \pmod {p^2}\\
{2^{p - 1}} \equiv 1 - \frac{p}{2}{H_{\frac{{p - 1}}{2}}} + \frac{{{p^2}}}{8}H_{\frac{{p - 1}}{2}}^2 \pmod {p^3}
\end{array}\]
For $\[{H_{\frac{{p - 1}}{2}}} = 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{{\frac{{p - 1}}{2}}}\]$[/quote]

15 year old post bumping , really ?

[quote=TheMathCruncher_007]

15 year old post bumping , really ?[/quote]

15 years old but the post is still meaningful though  :maybe:


inequalities geometry perimeter inradius trigonometry

Can anyone help me in solving the following problems please?
1.if [b][i]a,b,c[/i][/b] are the three sides of a triangle with area F, prove that:
         [b][i]ab+bc+ca[u]<[/u]4.(3^1/2).F [/b][/i]

2. if [b][i]a,b,c[/b][/i] are the three sides of a triangle with area A then prove that
    [b][i]      (abc)^2 [u]>[/u]4A/(3^1/2)    [/b][/i]

3. if s is the semi-perimeter of a triangle with inradius r, prove that :
             [b][i]   s^2 [u]>[/u] 27.r^2  [/b][/i]

[hide="3"]

Substitute $ r$ with $ \frac{A}{s}$, where $ A$ is the area of the triangle

$ s^2\ge 27.\frac{A^2}{s^2}$
$ s^4\ge27. s(s\minus{}a)(s\minus{}b)(s\minus{}c)$
The inequality above becomes
$ (a\plus{}b\plus{}c)^3\ge27(a\plus{}b\minus{}c)(a\plus{}c\minus{}b)(b\plus{}c\minus{}a)$

So we need to prove the inequality above
Let $ a\equal{}x\plus{}y,b\equal{}x\plus{}z,c\equal{}y\plus{}z$

Then the inequality changes to
$ (x\plus{}y\plus{}z)^3\ge27xyz$
Which is true by $ AM\minus{}GM$
QED

[/hide]

Anyone please help me with the other two problems!!

     Thanks.

I think for no 2, the problem is wrong
May we try $ a \equal{} b \equal{} c \equal{} \frac {1}{2}$ ?
If we may, we'll get

$ LHS \equal{} \frac1{64}$
$ RHS \equal{} \frac1{4}$

$ LHS < RHS$

1; We have $ A = \frac{ab sin C}{2}
     \Leftrightarrow ab = \frac{2A}{sin C}
\Rightarrow \sum ab = 2A(\sum \frac{1}{sin C})$
Since f(x) = 1/sin(x) is convex and $ A+B+C =\pi$, we have( Jensen);
$ \sum \frac{1}{sin C} \geq \frac{3}{sin \pi /3} = 2 \sqrt{3}$ , which means that $ 2A(\sum \frac{1}{sin C}) \geq  4 \sqrt{3} A$
So if Im not wrong, ou have reversed the inequality sign.

Sorry but I've typed the inequality sign wrong in both the cases. For#1, it will be [u]>[/u] ; and for #2, it will be [u]<[/u]  :blush: !

Maybe I had not slept well the night before......

Pro2 should be
$ (abc)^2 \ge (\frac {{4A}}{{\sqrt 3 }})^3$(1)
And solution is
we have
$ 4A \equal{} \frac {{abc}}{R}$
(1) equiv
$ \ 3\sqrt 3 R^3 \ge abc$
$ \frac {{3\sqrt 3 }}{8} \ge \sin A\sin B\sin C$ which is obvious.

[quote="TRAN THAI HUNG"]Pro2 should be
$ (abc)^2 \ge (\frac {{4A}}{{\sqrt 3 }})^3$(1)
And solution is
we have
$ 4A \equal{} \frac {{abc}}{R}$
(1) equiv
$ \frac {{3\sqrt 3 }}{8}R^3 \ge abc$
$ \frac {{3\sqrt 3 }}{8} \ge \sin A\sin B\sin C$ which is obvious.[/quote]

I wonder how can you get (2) equvi. to  (1)

$ \boxed{\frac {{3\sqrt 3 }}{8}R^3 \ge abc}$.......(1)

$ \boxed{\frac {{3\sqrt 3 }}{8} \ge \sin A\sin B\sin C}$....(2)

For number 2

We know $ A=\frac{1}{2} ab\sin(C) \longrightarrow \frac{2A}{\sin(C)} = ab \longrightarrow (abc)^2 = \frac{8A^3}{\sin(A)\sin(B)\sin(C)}$

So we need to find the maximum of ${ \sin(A)\sin(B)\sin(C)}$ given $ A+B+C = \pi$

From AM-GM and jensen this can be shown to be $ \left(\frac{\sqrt{3}}{2}\right)^3$

$ \Rightarrow (abc)^2 \ge \left(\frac{4A}{\sqrt{3}}\right)^3$

Thank Vinskman,I have edited it :blush: .


Let $ AH_a$ and $ BH_b$ be the altitudes of triangle $ ABC$. Points $ P$ and $ Q$ are the projections of $ H_a$ to $ AB$ and $ AC$. Prove that line $ PQ$ bisects segment $ H_aH_b$.

[i]Author: A.Akopjan, K.Savenkov[/i]

OK,Let us prove HaQHbK is a rectangle : that is to prove :HaK is perpendicular to BHb;just look at the picture we can get :∠QAHa=∠QPHa=∠KBHa;That is P,K,Ha,B are concyclic;∠ BPHa=∠BKHa=90;  obviously HaQHbK is already a rectangle!...Q-E-D.  :blush:  :maybe:

My solution is similar to yours CCmath  :blush: I saw the same rectangle

[geogebra]d213c4c58195d17e2fc4ade2a985b4611bd6bdef[/geogebra]


inequalities
function
algebra unsolved
algebra

Let $f: R\to R$ a function, such that $\frac{f(x)+f(y)}{2}\geq f(\frac{x+y}{2})$, for any $x,y\in R$. Is it true that f is convex? (if yes, show a demonstration). You can consider f-continuous, differentiable, or even twice differentiable.

Graphically, the midpoint of any secant line lies above the function of the midpoint of the two endpoints of the secant line, which is essentially the visual definition of a convex function.

However, that is not very rigorous.

The inequality becomes $f(A)-f(M)\geq f(M)-f(B)$ where $A>B$ and $M$ is the midpoint of the segment $AB$ on the $x$-axis. Then, take two points $A>B \in \rr$ and $n$ points $B=C_0,C_1,...,C_n=A$ that break up the segment $AB$ into $n$ equal parts. We then have that $C_i$ is the midpoint of $C_{i-1}C_{i+1}$ and therefore the inequality gives us

\[ f(A)-f(C_{n-1})\geq f(C_{n-1})-f(C_{n-2})\geq...\geq f(C_1)-f(B) \]

Dividing this by $A-C_{n-1}=C_1-B=\dsp \frac{A-B}n$ gives us

\[ \dsp \frac{f(A)-f(C_{n-1})}{A-C_{n-1}}\geq \dsp \frac{f(C_1)-f(B)}{C_1-B} \]

Letting $n$ go to infinity will make $C_{n-1}$ go to $A$ and $C_1$ go to $B$ and therefore

\[ f'(A)\geq f'(B) \]

Since this works for all $A\geq B$ we need $f''\geq 0$ q.e.d.

As you see, I assumed $f$ to be twice differentiable.


Eleven pencils cost as much as three pens. If seven pens cost $ \$ 9.24$, what is the cost, in cents, of one pencil?

Thus, three pens cost (3/7)9.24 = 11 pencils. 1 pencil costs (3/77)9.24=[b]36[/b] cents


Euler
logarithms
function
number theory
number theory unsolved

let $F(n)=\frac{1+\phi(1)+\phi(2)+...+\phi(n)}{n^{2}}$ where $\phi$ is eulers totient function. Show as n tends to infinity $F(n)$ tends to $3/(\pi)^{2}$.

Myth posted a solution about one and a half year ago.

It is obviosly, because $n^{2}F(n)$ is number Faries rational $\{\frac{a}{b}|0\le a<b\le n, \ (a,b)=1 \}$.

So what? Why is it so obvious? I think this is the way we count Farey sequences.

Guys, if you take mr. Panaitopol's course or read his book, this is just a simple consequence of Mertens theorem which states that

              \[\sum_{k=1}^{n}\varphi(k)=\frac{3}{\pi^{2}}+\mathcal O(n\ln n).\]

 We can also rewrite the above equality as

 \[\sum_{k=1}^{n}\varphi(k)=\frac{3}{\pi^{2}}+F(n),\]

  where $F: \mathbb{N}^{*}\ro\mathbb{R}$ is a function such that $\frac{|F(n)|}{n\ln n}\leq M$.

 The problem stated is just a simple consequence. :wink:

You forget $n^{2}$ and F(n) used for another value.
\[\sum_{k=1}^{n}\varphi(k)=\frac{3n^{2}}{\pi^{2}}+\mathcal O(n\ln n). \]
We can also rewrite the above equality as
\[\sum_{k=1}^{n}\varphi(k)=\frac{3n^{2}}{\pi^{2}}+R(n), \]

Always a pleasure to hear your comments, Cezar. However, some small remarks: believe me, one doesn't need to take Mr Panaitopol's course or book in order to prove that (this does not mean I don't have a huge respect for him), ANY decent book on analytic number theory should do the job. Secondly, all you did was practically to restate the problem (ok, with a big O there, just to make it fun...), and not to give a solution. The order of magnitude you presented follows from the relation:
  \[\frac{1}{n}\sum_{k=1}^{n}{\varphi(k)}=\frac{n}{2}\sum_{k=1}^{n}{\frac{ f(k)}{k^{2}}+\frac{1}{2}\cdot\sum_{k=1}^{n}{{\frac{f(k)(1-2\{n/k\})}{k}}+\frac{1}{2n}\sum_{k=1}^{n}{f(k)\{n/k\}(1-\{n/k\})}}}\]
   The formula follows easily using the fact that $\varphi(n)=n\cdot\sum_{k|n}{\frac{f(k)}{k}}$.
  Here $f$ is Moebius's function and $\{\}$ is the fractional part. All you need now is that $\sum_{k=1}^{n}{\frac{f(k)}{k^{2}}}=\frac{6}{\pi^{2}}+O(1/n)$.


which is larger ? Tell me answers (1)to(5) :P

$(1)$  $e^{\pi},\pi ^e$

$(2)$  $e^{\pi^{\pi ^e}},{\pi}^{e^{e^{\pi}}}$

$(3)$  $\pi ^{\sqrt e},e^{\sqrt{\pi}}$

$(4)$  $(\sqrt{6})^{2\sqrt 2},(2\sqrt{2})^{\sqrt 6}$

$(5)$   $e^{\pi},21$

(5) is an extra probrem,have nothing to do with other problem.
Maybe (1)(2)(3) are far more easier than (4)(5) :lol:

If you need in(4),use $\frac{\sqrt{6}+2\sqrt{2}}{2}<e ,$  $e<2\sqrt{2} \cdots (\ast)$ with no explanation :)

Answer to question 1:
Let $f(x)=\frac{lnx}{x}$ for $x > 0$.
Then $f'(x)=\frac{1-lnx}{x^2}$.
Note that $f'(x) > 0$ for $0 < x < e$, $f'(x)= 0$ when $x = e$ and $f'(x) < 0$ for $x > e$.
Therefore, $f(x)$ attains absolute maximum at $x = e$.
In particular, we have $f(\pi) <f(e)$.
$\frac{ln\pi}{\pi}<\frac{ln e }{e}$ implies ${\pi}^e < e^{\pi}$.

clarification: on all the other problems 1-5, does $a^{b^{c}}$ = $a^{(b^c)}$ or does it equal ${{(a^b)}^c}$

[quote="porte"]which is larger ?
$(4)$  $(\sqrt{6})^{2\sqrt 2},(2\sqrt{2})^{\sqrt 6}$[/quote]

Just taking Napier's logarithm both sides, and I don't know what do you want to ask here...?

[quote="porte"]Right answer,luimichael  :D
(2)(3) can solve the same way,but I think (4) cannot solve using only $f(x)=\frac{\log{x}}{x}$,need another function ;)[/quote]


do you have obvious trend ? pls share opinion  :(  :(


I'm not sure if this is the right section for this but...I'm a junior in highschool and am now picking a project that I have to present next year as a senior.  I decided I want to do something math related--particularly, I want to learn some number theory as I now know relatively little.  For the project I will probably have to apply the things i learn to something that I can easily present to average non-math teachers.  Does anyone have ideas on a NT application i could do?  Also, recommendations on introduction number theory texts would be appreciated.

NT books, Number Theory by George Andrews
and the book by Hardy.

I like

John Stillwell's Elements of Number Theory

Gareth Jones's Elementary Number Theory

and

R. P. Burn's A Pathway into Number Theory

as beginning number theory texts.

Encryption might make a good application of NT to present to the uninitiated.  Most people have experienced submitting their credit card and other private information to a "secure" web server.  Explaining how this works, perhaps with other applications of encryption might make a respectable project.

There are many interesting issues associated with encryption that might be explored as well.  For many years the US government was opposed to commercial sale and use of the most sophisticated encryption technologies, reasoning that wiretaps used in criminal investigations might be rendered useless against these technologies.  Others argued that their need for secure communication and ability to generate globally competitive products and services hinged upon these sophisitcated encryption technologies.  Relating such social debates to the underlying mathematics might keep your audience interested.


let a,b,c and d be odd integers such that a<b<c<d and ad=bc prove that if   a+d=2^k   ;   b+c=2^m
for some integers m and k then a=1
(IMO 1984)


In the Fibonacci seqeunce $F_n=F_{n-2} + F_{n-1}$ where $F_1=F_2=1$.  Prove that the term, $F_n$, will always divide the term $F_{mn}$

[hide="Solution"]
Take the series $\mod F_n$.  Thus we see $F_n \equiv 0, F_{n+1} \equiv F_{n-1}, F_{n+2} \equiv F_{n-1}, F_{n+3} \equiv 2F_{n-1}$, and generally $F_{n+k} \equiv F_k F_{n-1}$.  At $n + k = mn$ we have $F_{mn} \equiv F_{(m-1)n} F_{n-1}$, so starting from $m = 2$ and incrementing by 1, we have a proof by induction.  Q.E.D.

(Note that I am taking $\mod F_n$ implicitly the whole time.  That last line is not an explicit formula for $F_{mn}$.)
[/hide]


Let $x,y,z$ such that $x^{2}+\frac{y^{2}}{4}+\frac{z^{2}}{9}= 1$. What is the maximum and the minimum of $x^{2}+y^{2}+z^{2}$?

I think the problem can't solve because the assumtion and the conclusion is same then it is impposible to analyze the relation between of them


geometry
inequalities
calculus
number theory

You may have seen the following: [url]http://www.artofproblemsolving.com/Resources/AoPS_R_A_DiscreteMath.php[/url].

To quote the last bit of it,

[quote]Many students, especially bright and motivated students, find algebra, geometry, and even calculus dull and uninspiring. Rarely is this the case with most discrete math topics. When we ask students what the favorite topic is, most respond either "combinatorics" or "number theory." (When we ask them what their least favorite topic is, the overwhelming response is "geometry.") Simply put, most students find discrete math more fun than algebra or geometry.[/quote]

Well, let's get some concrete evidence.

I think geometry is my favorite, and I find like combinatorics as well, but it's probably my worst. I'm not as fond of algebra and number theory, but it looks like I'm better at those than at geometry.  :|

You missed some areas

What about number theory?

That's odd, somehow it didn't end up on the poll. Bleargh.

Personally, my favorite is Number Theory, but I like Inequalities also, which I'm guessing would be apart of Algebra...

I will have to go with Geometry!

I guess the reason that students find Calculus or Algebra 'boring' in a crude sense because a lot of it is really just fundamental things that you remember (i.e. there seem to lack a creative side to it). Like a system of simultaneous equations may be solved quite easily using standard procedures while a combinatoric or number theory problem may require a stroke of inspiration, 'Ah, what if we look at the parity of the numbers!' or 'Maybe, I should colour some points red and others blue!' and things like that.

That's just my opinion.

Well I would put number theory, but seeing as it isn't a choice, I'm going with combinatorics since they usually involve something clever...

After that my choice would be geometry.

I like basic algebram but I also like geo :)


Do anyone have any suggestions on any books to read to help prepare for the Avogadro chemistry contest? Also, I have a chemsitry question on one of the practice contests. The question is ...

      Which of these elements form the most basic oxide?
The choices are A) Bismith   B) Antimoney   C)  Arsenic   D)  Phosphorus  E) Nitrogen.  The answer is A but I do not know why. How can you tell which is the most basic oxide? Can anyone please help me. Thanks.

In general, metal oxides form basic solutions in water ($\text{Li}_2\text{O} + \text{H}_2\text{O} \longrightarrow 2\text{LiOH}$), and nonmetal oxides tend to form acidic solutions ($\text{SO}_3 + \text{H}_2\text{O} \longrightarrow \text{H}_2\text{SO}_4$).  $\text{Bi}$ is the most metallic element listed, and so will form an acidic solution.

Thanks for your help. But I have a question. How come you said that metal oxides form basic solutions in water and then you said that Bi is the most metallic and forms an acidic solution so shouldn't it form a basic solution?

:blush: Oops, sorry.  Yes, bismuth oxide will for a [i]basic[/i] solution because it's the most metallic.


inequalities
AMC
USA(J)MO
USAMO
inequalities proposed

Let a, b, c be the sides of a triangle ABC.  Let $m_a, m_b, m_c$ be the medians and S be the area. Prove that:

$a^2m_a^2 + b^2m_b^2+c^2m_c^2 \ge 12 S^2$.

I'm just getting started with inequalities...so please let me know if I have made a mistake. Also, any general advice for working with inequalities would be appreciated.

Note that:

[tex]m_a^2 = \frac{1}{4}(2b^2+2c^2-a^2)[/tex]

...and similarly for [tex]m_b[/tex] and [tex]m_c[/tex]. The LHS then equals:

[tex]\displaystyle \sum_{cyc} \frac{a^2}{4}(2b^2 + 2c^2 - a^2) = \sum_{cyc} \frac{a^2b^2}{2} + \frac{a^2c^2}{2} - \frac{a^4}{4} = a^2b^2 + b^2c^2 + c^2a^2 - \frac{a^4}{4} - \frac{b^4}{4} - \frac{c^4}{4}[/tex].

Now, note that [tex]S^2 = (s)(s-a)(s-b)(s-c) = \frac{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}{16}[/tex].

Multiplying both sides by 16 and expanding the RHS yields:

[tex]16(a^2b^2 + b^2c^2 + c^2a^2) - 4(a^4 + b^4 + c^4) \geq 24(a^2b^2 + b^2c^2 + c^2a^2) - 12(a^4 + b^4 + c^4)[/tex].

[tex]8(a^4 + b^4 + c^4) - 8(a^2b^2 + b^2c^2 + c^2a^2) \geq 0[/tex]

[tex]a^4 + b^4 + c^4 \geq a^2b^2 + b^2c^2 + c^2a^2[/tex]

...which is true by Rearrangement!  :lol:  We have equality when [tex]a=b=c[/tex].

Is this proof correct?

I have a solution for this one:
first we have:
[tex] b^2+c^2=2m_a^2+a^2/2[/tex]
then[tex] 2a^m_a^2=a^2b^2+c^2a^2-a^4/2[/tex]
so [tex] T=a^2m_a^+b^2m_b^2+c^2m_c^2=a^2b^2+b^2c^2+c^2a^2- \frac{a^4+b^4+c^4}{4}[/tex]
but [tex] 16S^2=2(a^2b^2+b^2c^2+c^2a^2)-(a^4+b^4+c^4)[/tex]
then we want to prove
T \ge 12S^2 \leftrightarrow a^4+b^4+c^4 \ge a^2b^2+b^2c^2+c^^2a^2.
wich is clairly right. :lol:

[quote="JSRosen3"]Is this proof correct?[/quote]

It is correct! Good job for a beginner!

I actually created this problem from:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=2897

and the fact that $a^2+b^2+c^2 \leq 9R^2$.

then later I realized that it can be derived from:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15558

by taking $x=a^2$, etc and M the centroid.

[quote="JSRosen3"]Thanks!  ;)

Another question...what is the approximate difficulty of this question relative to others? For instance, is it USAMO level difficulty, or an easier national olympiad level difficulty, or easier? I have little to no frame of reference for this.[/quote]

This turns out to be an easy one. Probably at the same difficult level of IMO at 60s.

Examples:

1. http://www.kalva.demon.co.uk/imo/isoln/isoln612.html

2. http://www.kalva.demon.co.uk/imo/isoln/isoln642.html

Probably comparable with #1 of 2004 USAMO.


Apply power mean to reduce it to

$\frac{ \sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}}{2}\le \sqrt{\left( \frac{a+b}{2}\right) \left( \frac{\frac{1}{a}+\frac{1}{b}}{2}\right) }$,

which is Cauchy.

[quote="Svejk"]If $a,b>0$ prove that:
\[\sqrt[3]{\frac{a}{b}}+\sqrt[3]{\frac{b}{a}}\leq \sqrt[3]{2(a+b)(\frac{1}{a}+\frac{1}{b})}\]
[/quote]

Don't double-post! You posted at [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=122017[/url] already.


counting
distinguishability
combinatorics unsolved
combinatorics

[quote="soccer022483"][b]If anyone can answer any of the following I would really appreciate it:[/b]

[color=darkred]1) 4 runners are involved in a footrace. If ties are allowed (even 3 and 4-way ties), in how many ways can the race finish?[/color]
[color=orange]
2) A group of m*n people are to be placed into m teams, each having n players. If each team is distinguishable in some way (maybe they're named), find the number of ways that this can be accomplished.[/color]

[color=darkblue]3) Akari, Kris, Anna, Lahrri, Mylanda, and Elaine are standing in line at a cafeteria. Anna and Elaine are twins and are indistinguishable from the point of view of a third party. From this person's point of view how many visually different ways can this group of people line up?[/color]

[color=green]4) 2*n people are at a party. They pair off into n pairs of 2 to make conversation. In how many ways can this be accomplished?[/color][/quote]

I'll answer 4 for you (1 has a lot of casework, 3 is really easy, and 2 can be figured out easily if you understand 4)

The number of pairs is clearly

$C^{2n}_2\cdot C^{2n-2}_2\cdot C^{2n-4}_2\cdot \ldots \cdot C^{2}_2$

Expanding all the coefficients, we see that everything in the numerator drops out except for the $(2n)!$ and we have $n$ factors of $2$ on the bottom.  So the answer is just

$\frac {(2n)!}{2^n}$

This obviously offers another way of looking at it.

Write all $2n$ of the players names in a row.  There are $(2n)!$ ways to do this.  Now, taking the first two names gives us a pair, and so on.  However, we have to divide by $2^n$ because it doesn't matter if we interchange two consecutive names - they give us the same pair.

Hi Matt!
So maybe right answer is $\frac{(2n)!}{2^nn!}=(2n-1)!!$...  ;)  :D

Of course, we divide by $n!$ because the order of the pairs does not matter.  :blush:

Hey guys I just wanna say thanks for the help that you gave.


trigonometry
ARML
function
sine double angle

How do you find out the value of i^i, where i is the sqrt(-1)?

And,
how would you use complex trig to figure out the exact value of cos20*cos40*cos80?

thank you very much in advance

The second problem I think I've done before.  Was it an ARML tryout problem?

Anyways, how you do it is:
cos(30-10)*cos(30+10)*cos(90-10)
=(cos30cos10+sin30sin10)(cos30cos10-sin30sin10)(cos90cos10+sin90sin10)  --> the first two are (a+b)(a-b), the second you can simplify b/c cos90=0 and sin90=1

=[(cos30cos10)^2-(sin30sin10)^2](sin10) -->(cos30)^2=3/4, (sin30)^2=1/4

=[3/4(cos10)^2-1/4(sin10)^2](sin10) -->(cos10)^2=1-(sin10)^2

=[3/4-3/4(sin10)^2-1/4(sin10)^2](sin10) -->simplify

=3/4(sin10)-(sin10)^3 -->factor out 1/4

=1/4(3sin10-4(sin10)^3) -->realize that sin(3x)=3sinx-4(sinx)^3

=1/4(sin30)

=1/8

For some reason, I feel like I've left out a negative somewhere.  You should check my work, but that's the gist of how to do the problem.

$ i \equal{} e^{\dfrac{i\pi}{2}}$
$ i^i \equal{} e^{ \minus{} \dfrac{\pi}{2}}$
--
x=cos20cos40cos80
x sin20=sin20cos20cos40cos80
sine double angle formula, stuff cancels like dominos
2x sin 20=sin40cos40cos80
4xsin20=sin80cos80
8x sin20=sin160
x=sin160/(8sin20)
sin a= sin 180-a
x=1/8

from what i see, i dont see any missing negatives. i just worked it out and its fine as is!

funtwo, could you explain the first one a little more? i dont quite get it and am very new to this

[quote="funtwo"][quote="crazypianist1116"]That works but...

[hide]Let $ n \equal{} \cos {20} \cos{40} \cos{80}$
$ \sin {20} n \equal{} \sin {20} \cos {20} \cos{40} \cos{80}$
$ \sin {20} n \equal{} \frac {\sin {40} \cos{40} \cos{80}}{2}$
$ \sin {20} n \equal{} \frac {\sin {80} \cos{80}}{4}$
$ \sin {20} n \equal{} \frac {\sin {160} }{8}$
$ \sin {20} n \equal{} \frac {\sin {20} }{8}$
$ n \equal{} \frac {1}{8}$[/hide][/quote]
thats the same thing. T.T[/quote]

You posted as I was posting. But you did edit after I posted  :roll:

[hide="1"]Because $ \cos\frac{\pi}{2}\plus{}i\sin\frac{\pi}{2}$ is obviously equal to $ i$, by Euler's formula, $ i\equal{}e^{i\frac{\pi}{2}}$.

Then $ i^i\equal{}e^{i^2\frac{\pi}{2}}\equal{}e^{\minus{}\frac{\pi}{2}}$.
[/hide]

People have posted nice solutions to #2, but I don't know how you'd do it with complex trig. (Complex trig can be useful when you're, for example, summing trigonometric functions, because you can look at real/imaginary parts and apply De Moivre; but with products of cosines you're probably better off doing what they did).

cosX * cos 60-X * cos 60+x = 1/4 cox 3X
same as for sine


Hello. I need some help starting some of these problems.

1. The fitness bug is out for a daily run. It runs at a speed of 1 cm/sec for the first centimeter, 2 cm/sec for the second centimeter, 3 cm/sec for the third centimeter, and so on until the sixth centimeter, when it runs at 6 cm/sec. What is the average speed for the entire 6 centimeters.

2. A bug lives on a corner of a cube and is allowed to travel only on the edge of the cube. In how many ways can the bug visit each of the other seven corners once and only once, returning to its home corner at the end of the trip?

Thanks.

Perhaps I am oversimplifying the problem, but for number one, isn't it just the average of all the numbers?

So, you can either add up all of the numbers and divide by 6:
1+2+3+4+5+6= 21

21/6

7/2

3.5 cm/sec.

Or...you could use Gauss Summation:
[N(N+1)]/2

In other words:
6(7)/2

=21

Then divide by six again...
Hope that helps.

yankeeboy, your solution is incorrect.

[hide="My Solution to number 1"]
The first centimeter takes $ 1$ second for the bug to complete, the second centimeter $ \frac {1}{2}$ of a second, the third centimeter $ \frac {1}{3}$ of a second , and so on, so that the sixth takes $ \frac {1}{6}$ of a second.  So the total time the fitness bug runs for is
\[ 1 \plus{} \frac {1}{2} \plus{} \frac {1}{3} \plus{} \frac {1}{4} \plus{} \frac {1}{5} \plus{} \frac {1}{6} \equal{} 2.45 \text{ seconds}\]
So the average speed for the entire $ 6$ centimeters is $ \frac {6}{2.45} \equal{} \frac {120}{49}\approx 2.45$ cm/sec.[/hide]


How many 4-inch by 4-inch squares can be cut from a rectangular piece of leather measuring 4 feet by two-thirds of a yard?

Well, the area of 1 of these squares is 4*4 = 16 square inches.
The area of the whole piece of leather is 4 feet * 2/3(3 feet) = 4ft * 2ft = 48in * 24in = 1152.
1152/16 = [b]72[/b].

The area of 1 square is $ 4\cdot4 \equal{} 16$ square inches

The area of the whole piece of leather is $ 4\times \frac {2}{3}\cdot 3$

$ \equal{} 4\cdot 2 \equal{} 48 \cdot 24 \equal{} 1152$

$ \frac {1152}{16} \equal{} 72$

EDIT: beaten.

First, we figure out the area of the large rectangular piece of leather. To do so, we use the conversion factor: $$\frac{1 \mbox{ yard}}{3 \mbox{ feet}} = 1.$$
Hence, 4 feet by two-thirds of a yard is: $$4 \mbox{ feet} \times \frac{2}{3} \mbox{ yard} \times \frac{3 \mbox{ feet}}{1 \mbox{ yard}} = 8 \mbox{ feet}^2$$
Now, recall that $1 \mbox{ foot} = 12 \mbox{ inches} \implies 1 \mbox{ foot}^2 = 144 \mbox{ inches}^2$. So, $$8 \mbox {feet}^2 \times \frac{144 \mbox{ inches}^2}{1 \mbox{ foot}^2} = 1152 \mbox{ inches}^2$$
It turns out that both $4 \mbox{ feet} = 48 \mbox{ inches}$ and $\frac{2}{3} \mbox{ yard} = 24 \mbox{ inches}$ are divisible by $4$. Hence, we can figure out how many $4 \mbox{ inch} \times 4 \mbox{ inch}$ squares can be cut from the rectangular piece of leather just by dividing $$\frac{1152 \mbox{ inches}^2}{16 \mbox{ inches}^2} = \boxed{72\text{ pieces.}}$$

4in.*4in.=16 sqaure inches.

One yard is 3 feet. 2/3 of a yard is 2 feet.
48*24/16=72
the answer is 72


trigonometry
LaTeX
calculus
integration
irrational number
calculus computations

[quote="math92"]The region bounded by the x-axis and the graph of y = cos² x from x = 0 to x = 2π is revolved about the x-axis. Find the volume of the resulting solid. Express your answer as an EXACT, RATIONAL NUMBER.[/quote]math92, do you know $\LaTeX$ so you can put $\pi$ instead of a primitive symbol like this one----->π<---------? BTW, this looks like a typical calculus question and the answer is an [b]irrational number[/b]...
Explain your behaviour, math92  :huh:

Umm $\pi\int_{0}^{2\pi} \cos^4 xdx$ where did you get this problem?

[hide]
The volume of this solid will be 4 times that of a solid made by revolving $y=\cos x$ from $x=0$ to $x=\pi /2$ about the x-axis, which has volume $\pi \int_0^\frac{\pi}{2} \cos^4 x\ dx$.

So if $V$ is the volume of the original solid, then $V=4\pi \int_0^\frac{\pi}{2} \cos^4\ dx$. Reversing this integral gives $V=4\pi \int_0^\frac{\pi}{2} \cos^4 \left( \frac{\pi}{2}-x \right) \ dx=4\pi \int_0^\frac{\pi}{2} \sin^4 x\ dx$.

Adding these two integrals together and dividing by 2, $V=2\pi \int_0^\frac{\pi}{2} \cos^4 x+\sin^4 x\ dx$. Simplifying the expression that is being integrated, $\cos^4 x+\sin^4 x=1-2\cos^2 x\sin^2 x=1-\frac{\sin^2 2x}{2}=\frac{3+\cos 4x}{4}$.

Evaluating this integral, $V=\frac{\pi}{2}\int_0^\frac{\pi}{2} 3+\cos 4x\ dx=\frac{\pi}{2} \left[ 3x+\frac{\sin 4x}{4} \right]_0^\frac{\pi}{2}=\boxed{\frac{3\pi^2}{4}}$.
[/hide]
So much for exact and rational.

Moved from Pre Olympiad.

By the way, that answer looks exact to me, Jason  :? :P

[quote="math92"]i got this from a website that had math problems...[/quote]Can you post the URL so that I can see how 'smart' it was to ask for EXACT and RATIONAL?


analytic geometry
graphing lines
slope
function
conics

Okay, so there was a thread a while back about how many points are needed to uniquely define a conic.

It was mentioned that if there are 6 variable coefficients in an equation, only 5 points are needed because the entire equation can be divided by a coefficient in the equation that is known. I don't understand this. Why is one coefficient always known?

Like say you have the equation $Ax+By+C = 0$.

There are 3 unknowns, but only 2 points are needed because we can transform it to 2 unknowns by dividing by C, which yields:

$\frac{A}{C}x+\frac{B}{C}y+1 = 0$

But the problem arises when I enter 2 linear points, like (a, b), (na, ba) such as (1, 2), (2, 4)

Then I get:

$A'+2B'+1 = 0$
$2A'+4B'+1 = 0$

And there is no solution because it's the same equation only different constant value???? I am very confused.

Common logic shows that the slope is obvious 2 with a y-intercept of 0. However, how do I solve for these using the technique above? The technique above shows it is unsolvable?

**EDIT: Okay, so hypothetically, if C is 0, then dividing by it would result in an unsolvable system?

So is this why if one is determining a function based on points, then a function with n coefficients based on m variables would need n-1 points consisting of $(x_{1}, x_{2}, ..., x_{n})$?

I guess you can't if it was like that. But yeah, a conic can be unique determined with only 5 points because you can divide the equation by the leading coefficient. But I guess this only works when the leading coefficient is not zero.

Or I guess you can try dividing the whole equation by B or A instead of C.

Yeah, it only works (+ makes sense) when you divide it by a nonzero coefficient.


In triangle $ ABC$, $ AB \equal{} 5$, $ BC \equal{} 4$, and $ CA \equal{} 3$.

[asy]defaultpen(1);

pair C=(0,0), A = (0,3), B = (4,0);
draw(A--B--C--cycle);

label("\(A\)",A,N);
label("\(B\)",B,E);
label("\(C\)",C,SW);[/asy]

Point $ P$ is randomly selected inside triangle $ ABC$.  What is the probability that $ P$ is closer to $ C$ than it is to $ A$ and $ B$?

[asy]defaultpen(1);
pair C=(0,0), A = (0,3), B = (4,0), M=(0,1.5), N=(2,0), O=(2,1.5);
draw(A--B--C--cycle); draw(N--O--M--cycle);
 label("\(M\)",M,W);
label("\(N\)",N,S);
label("\(O\)",O,NE);
label("\(A\)",A,N);
label("\(B\)",B,E);
label("\(C\)",C,SW);[/asy]
$ MO$ is the perp. bis. of $ AC$ so all points on $ MO$ are equidistant from $ A$ and $ C$, and likewise for $ ON$.
So, anything in Rectangle $ MONC$ is closer to C than A or B.
Since $ \triangle AMO \cong \triangle ONB \cong \triangle NOM \cong \triangle MCN$, The area of $ MONC$ is half of $ \triangle ABC$ so it's $ \frac12$


[b]1.[/b] Sa se arate ca daca $ z \equal{} cos\alpha \plus{} isin\alpha$ atunci:
a)$ cosn\alpha \equal{} \frac {z^{2n} \plus{} 1}{2z^n}$, $ sinn\alpha \equal{} \frac {z^{2n} \minus{} 1}{2iz^n}$
b) Calculati produsul $ P \equal{} {cos20^o}\cdot{cos40^o}\cdot{cos80^o}$
[i]Gurzau Ileana, Cehu Silvaniei[/i]

[b]2.[/b] Fie $ a,b\in\mathbb{R}, a^2\ne{b}$. Care din numerele $ log_{2a}^2\left(a \plus{} \sqrt {b}\right)$ si $ log_{2a}\left(\sqrt {b}\right)$ este mai mare?
[i]Flonta Stelian, Simleu Silvaniei[/i]

[b]3.[/b]Fie $ z\in\mathbb{C}$astfel incat $ \mid2z \plus{} 3\mid \equal{} \mid3z \plus{} 4\mid$. Aratati ca $ \mid{z}\mid\in\left[1,\frac {7}{5}\right]$
[i]Vlaicu Liviu, Zalau[/i]

3. Appolonius's circle!

$ |2z \plus{} 3| \equal{} \left|3z \plus{} \frac {3}{4}\right|\Longleftrightarrow 2\left|z \plus{} \frac {3}{2}\right| \equal{} 3\left|z \plus{} \frac {1}{4}\right|$ gives a circle, then consider the distance from the origin.

$ \therefore \overline{OD}\leq |z|\leq \overline{OC}$, yielding $ \frac{7}{5}\leq |z|\leq 1.$

Regretatul poet  T. Tudor Topan , nu mi-a fost doar un profesor , mi-a fost ca un al doilea tata. Si nu exagerez cu  nimic cand spun asta .Omul acesta reusit sa ma transforme . In clasa a 5-a , imi puneam fratele mai mic sa declare ca nu sunt acasa cand suna domnul profesor sa ma cheme la dansul acasa  pentru a rezolva probleme mai interesante de olimpiada . Odata cu inaintarea mea in varsta , am inceput sa realizez cate eforturi depunea domnul profesor pentru mine si alti elevi.A fost un om de  sacrificiu, nu avea o familie . Familia dansului,eram noi elevii care il vizitam aproape in fiecare zi.Statea la dispozitia noastra,pana la ore tarzii noaptea.Imaginati-va o garsoniera $ 5 \times 4$ cu o masa in mijloc la care stateau in jur de 10 elevi,pregatindu-se pentru olimpiada.Dansul traia odata cu noi,se bucura la fiecare reusita a noastra ,suferea  alaturi de noi la fiecare "esec" .Nimic nu-l facea mai fericit decat faptul ca-si vedea elevii progresand.
In clasa a 8-a , am ajuns sa-l iubim . Toata clasa astepta cu drag orele de matematica , chiar si cei mai slabi elevi, acestia fiind constienti ca profu',desi ii ironiza, le dorea doar binele.
     Am ajuns la final, Testarea Nationala 2006 .Aceasta a trecut foarte usor,pentru multi dintre elevii dansului.Zecile de probleme facute la clasa si la consultatiile de la ora 7 dimineata,alaturi de raposatul profesor si-au spus cuvantul. Imi aduc aminte ca miercurea , chiar dupa examenul de matematica am mers la dansul acasa. Se simtea rau , foarte rau, dar eu credeam ca e doar o gripa puternica .
Am incercat sa-i arat subiectul de la Testare ,sa-i explic cum am rezolvat problemele dar nu se putea concentra , ii tremura pixul in mana.A fost foarte ciudat sa-ti vezi mentorul,care in mintea mea avea o imagine de om foarte puternic, neputincios ,rapus de boala.
S-a intins pe pat , vazand ca nu poate fi atent la problema si mi-a zis : " - Va fi 10 , Catalin? - Da , domnule profesor ? - Da' colegii , cum s-au descurcat ? - Majoritatea au stiut ." Atunci a schitat un zambet de multumire si mi-a spus ca vorbim mai tarziu . Pe la ora 22:00 , tot miercuri m-a sunat sa-mi ureze succes la proba de Geografie.Avea un ton ciudat,imi vorbea parca pentru ultima oara.
Eu tot il intrebam "Va simtiti rau,d-le profesor?" , dar dansul evita raspunsul la intrebare.In noaptea aceea , nu am dormit prea bine gandindu-ma la telefonul primit .Am incercat sa-l sun la 7 dimineata , dar nu mi-a raspuns nimeni.
       Cand am iesit de la proba de geografie  ma astepta tatal meu cu o veste ingrozitoare, pe care desi parca am presimtit-o din noaptea aceea , nu puteam sa o cred. Vestea mortii lui a venit ca un fulger asupra noastra.Parca si-a indeplinit misiunea, ne-a indrumat pe noi 4 ani si apoi s-a dus fara sa apuce sa vada roadele muncii sale.
      Acum imi lipsesc orele dansului , glumele , imi lipsesc vorbele sale de duh , imi lipsesc serile petrecute impreuna rezolvand probleme sau doar povestind , imi lipseste ceea ce a fost profesorul si mentorul meu [b][url=http://www.agonia.ro/index.php/author/0019341/index.html]Teodor Topan[/url][/b] .Dumnezeu sa-l odihneasca in pace.
 Puteti vedea o parte din poeziile dansului , din volumul "Triunghiul cu stele" , click [url=http://www.agonia.ro/index.php/author/0019341/index.html]aici[/url] [url=http://www.agonia.ro/index.php/author/0019341/index.html]http://www.agonia.ro/index.php/author/0019341/index.html[/url]

[color=darkred]$ 3\blacktriangleright$ Fie $ z\in\mathbb{C}$ astfel incat $ \mid 2z + 3\mid = \mid 3z + 4\mid$. Aratati ca $ \mid {z}\mid\ \in[1,\frac {7}{5}]\ .$[/color]

[color=darkblue][b][u]Demonstratie.[/u][/b] Folosim relatiile $ \{\begin{array}{c}
\mid z\mid^2=z\cdot\overline z\\\\
2\cdot\mathcal Re\ z=z+\overline z\\\\
\pm\ \mathcal Re\ z\le\ \mid z\mid\end{array}$. Astfel, $ \mid 2z + 3\mid = \mid 3z + 4\mid$ $ \Longleftrightarrow$

$ (2z+3)(2\overline z+3)=(3z+4)(3\overline z+4)$ $ \Longleftrightarrow$ $ 5\mid z\mid^2+6(z+\overline z)+7=0$ $ \Longleftrightarrow$

$ 5\mid z\mid^2+12\cdot\mathcal Re\ z+7=0$ $ \Longrightarrow$ $ 5\mid z\mid^2+7=-12\cdot\mathcal Re\ z\le 12\mid z\mid$ $ \Longrightarrow$

$ 5\mid z\mid^2-12\mid z\mid +7\le 0$ $ \Longleftrightarrow$ $ \mid {z}\mid\ \in[1,\frac {7}{5}]\ .$[/color]

[color=darkred]$ 3\blacktriangleright$ Fie $ z\in\mathbb{C}$ astfel incat $ \mid 2z + 3\mid = \mid 3z + 4\mid$. Aratati ca $ \mid {z}\mid\ \in[1,\frac {7}{5}]\ .$[/color]

[color=darkblue][b][u]Metoda I.[/u][/b] Folosim relatiile $ \{\begin{array}{c} \mid z\mid^2 = z\cdot\overline z \\
 \\
2\cdot\mathcal Re\ z = z + \overline z \\
 \\
\pm\ \mathcal Re\ z\le\ \mid z\mid\end{array}$. Astfel, $ \mid 2z + 3\mid = \mid 3z + 4\mid$ $ \Longleftrightarrow$

$ (2z + 3)(2\overline z + 3) = (3z + 4)(3\overline z + 4)$ $ \Longleftrightarrow$ $ 5\mid z\mid^2 + 6(z + \overline z) + 7 = 0$ $ \Longleftrightarrow$

$ 5\mid z\mid^2 + 12\cdot\mathcal Re\ z + 7 = 0$ $ \Longrightarrow$ $ 5\mid z\mid^2 + 7 = - 12\cdot\mathcal Re\ z\le 12\mid z\mid$ $ \Longrightarrow$

$ 5\mid z\mid^2 - 12\mid z\mid + 7\le 0$ $ \Longleftrightarrow$ $ \mid {z}\mid\ \in[1,\frac {7}{5}]\ .$

[b]Metoda II[/b] (similara cu [b]Kunny[/b]'s) Definim punctele fixe $ A( - \frac 32)$, $ B( - \frac 43)$ si punctul mobil $ M(z)$. Observam ca $ |z| = OM$, unde $ O$ este originea planului complex. Atunci $ \mid 2z + 3\mid = \mid 3z + 4\mid$ $ \Longleftrightarrow$ $ 2\cdot|z - ( - \frac 32)| = 3\cdot|z - ( - \frac 43)|$ $ \Longleftrightarrow$ $ 2\cdot MA = 3\cdot MB$, adica $ \frac {MA}{MB} = \frac 32\ .$ Asadar punctul $ M$ apartine cercului lui Appolonius $ w$ asociat segmentului $ [AB]$ si raportului $ \frac 32\ .$ Notam $ \{M_1,M_2\} = w\cap Ox$, unde $ M_1\in (AB)$, adica $ \frac {M_1A}{M_1B} = \frac {M_2A}{M_2B} = \frac 32\ .$ Se arata usor ca $ M_1( - \frac 75)$ si $ M_2( - 1)\ .$ Intrucat $ OM_2\le OM\le OM_1$ se obtine $ |z|\in[1,\frac 75]\ .$

[b]Problema propusa[/b] (din cartea "[u]Exercitii si probleme de Algebra -  Nita, Nastasescu, etc[/u]").[/color]

[quote][color=darkred]Sa se determine $ z\in\mathcal C^*$ de modul maxim (minim) pentru care $ |z + \frac 1z| = k$ (constant).[/color][/quote]


At Pizza Perfect, Ron and Harold make pizza crusts.  When they work separately Ron finishes the job of making 100 crusts 1.2 hours before Harold finishes the same job.  When they work together they finish making 100 crusts in 1.8 hours.  How many hours, to the nearest tenth of an hour, does it take Ron working alone to make 100 crusts?

[hide]let x be amount of time it takes ron to finish . x+1.2 for harold. so 1/x+1/x+1.2=1/1.8 there for (x+x+1.2)1.8=x^2+1.2x therefore x^2-2.5x-2.16=0 so x=(2.5+ :sqrt: 14.89)/2 so x is around 3.2[/hide]

I'm not sure whether it's right or not but I think the solution is
0.9.... If it's wrong, post the right answer, please... I think I'm pretty close though...

Theres not much help in saying whether a particular answer is correct or not - post your solution, and maybe we can see whats wrong/right with your proof.
If you think logically about what the question says,  they take 1.8 hours both working together, so how could one person by themself take less time?

I think you're right but I don't know the answer, then...
Do you know the answer?
Post your answer please...

No, you're missing the point -- post your full solution (how you got your answer, whether it's right or wrong).  Then, we can see either (a) that it's correct or (b) [b]where[/b] you went wrong, which is far more important than whether or not you're right.  Don't just post answers.


[quote="DonkeyKong"]15. Seventy-two 50-cent pieces are lined side-by-side on a table. Joe replaces every second 50-cent piece with a quarter and leaves. Jane replaces every third remaining 50-cent piece with a dime and leaves. Jim replaces every fourth remaining 50-cent piece with a nickel and leaves. What is the number of dollars in the amount of money left on the table? Express your answer to the nearest cents.

Are there a shortcut on this kind of problem except bruce forcing it?[/quote]

[hide="My Solution"]
Ok so in the beginning, you have $ 72$ $ 50$ cent pieces, which is $ \$36$
But you take away $ 36$ $ 50$ cent pieces.
So now you have $ \$18$
Then you add you add $ 36$ dimes.
Which is $ \$3.60$
So now you have $ \$21.60$
But then, you take away $ 12$ $ 50$ cent pieces and $ 12$ dimes.
So you subtract $ \$7.20$
But then, you add $ 24$ nickels.
Which is $ \$1.20$
Which leaves you with $ \$15.60$
But then you take away $ 6$ $ 50$ cent pieces, $ 6$ dimes, and $ 6$ nickels.
Which is $ 3.90$
And then you add $ 18$ pennies
So you end up with $ \$11.88$
I'm pretty sure this is right.... :P
[/hide]