trigonometry
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algebra
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analytic geometry
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slope

Compute the exact value of $ \sin\ \frac {\pi} 8$ and $ \cos\  \frac {\pi} 8$.

Also, under what condition(s) would the trig. function of a given angle (in radians) to have a closed form?

[hide="Giveaway Hint for Calculations"]Recognize that $ \sin \frac {\theta}{2} \equal{} \sqrt {\frac {1 \minus{} \cos \theta}{2}}$ and $ \cos \frac {\theta}{2} \equal{} \sqrt {\frac {1 \plus{} \cos \theta}{2}}$[/hide]

[quote="Brut3Forc3"]I think that (don't hold me to this) any rational number multiplied by $ \pi$ can be calculated exactly, though some will be extremely ugly.[/quote]Try finding $ \sin\frac {\pi}{7}$

It is possible to express $ \sin \frac{\pi}{7}$ in terms of radicals, but it is (1) messy and (2) requires the use of complex numbers to write such an expression (in as much as certain cases of the cubic formula require the same).

I leave it to the reader as an exercise to demonstrate that

$ P(z) \equal{} z^3 \minus{} z^2 \minus{} 2z \plus{} 1 \equal{} 0$

has $ z \equal{} 2 \cos \frac{\pi}{7}$ as a root.

$ \pi \equal{} 180^{\circ}$.
$ \frac {180}{8}^{\circ} \equal{} \frac {45}{2}^{\circ}$

Let $ 45^{\circ} \equal{} \theta$ and apply the formulas provided above by patterns_34.

Take a look at cyclomatic polynomials; since the general polynomial cannot be solved, you are at the mercy of the particular polynomial.

[hide="Without half-angle formulas"]

The point $ (\cos{\frac{a\plus{}b}{2}},\sin{\frac{a\plus{}b}{2}})$ is equidistant from the points $ (\cos{a},\sin{a})$ and $ (\cos{b},\sin{b})$ on the cartesian plane. The midpoint of the line segment connecting $ (\cos{a},\sin{a})$ and $ (\cos{b},\sin{b})$ is colinear with the origin and the point $ (\cos{\frac{a\plus{}b}{2}},\sin{\frac{a\plus{}b}{2}})$. In this particular problem, we can let $ a\equal{}0$ and $ b\equal{}\frac{\pi}{4}$ to find the values of $ \cos{\frac{\pi}{8}}$ and $ \sin{\frac{\pi}{8}}$.[/hide]

[quote]The point  is equidistant from the points  and  on the cartesian plane.[/quote]Why should this be true?(cosa+cosb)/2=/cos(a+b)/2 right?Same for the y co-ordinates?So for what you wrote to be true it has to be on the perpendicular line through (cosa+cosb)/2.How do you prove it?

Cronus he is using simple geometry. He bisects angle ab and states that this bisector intersects the distinct side of an isosceles triangle at its midpoint. I follow his statements but I'm not quite sure how he's doing it without half angle formulas. The evaluation of $ \sin{\frac{\theta}{2}}$ and $ \cos{\frac{\theta}{2}}$ seems to still need it.

[quote="igiul"]Cronus he is using simple geometry. He bisects angle ab and states that this bisector intersects the distinct side of an isosceles triangle at its midpoint. I follow his statements but I'm not quite sure how he's doing it without half angle formulas. The evaluation of $ \sin{\frac {\theta}{2}}$ and $ \cos{\frac {\theta}{2}}$ seems to still need it.[/quote]

[hide="On solving for the final values"]

Since the points, $ (\cos{\frac {a \plus{} b}{2}},\sin{\frac {a \plus{} b}{2}})$, the origin, and the midpoint of the other two points are colinear, a line passing through any 2 of the points will have the same slope. Also, since our desired point is on the unit circle, it is a distance of 1 unit from the origin.

We now solve the system of equations:

$ x^2 \plus{} y^2 \equal{} 1$ and $ y \equal{} mx$, where $ m$ is the slope of the aforementioned line.[/hide]

Well ok.But on a nonrelevant topic what is the equation for the angle bisector of the angles between 2 straight lines say,
ax+by=c,dx+ey=m.

This may be of interest...

$ \cos 20^\circ \equal{} \frac {1}{2}\cdot\left( \sqrt [3]{\frac {1}{2} \plus{} i\frac {\sqrt {3}}{2}} \plus{} \sqrt [3]{\frac {1}{2} \minus{} i\frac {\sqrt {3}}{2}}\right)$

@Cronus: the angle bisector is equidistant from the two lines. Use the point-line distance formula for an arbitrary point (x,y) and a line, and set that equal to the distance from the point to the other line.

Altheman I think it is easier to do with trig.  Anyway, Cronus, I'd much rather use slope-intercept form.  So lets say our two lines are $ y \equal{} m_1x \plus{} b_1$ and $ y \equal{} m_2x \plus{} b_2$.  Then the angle from the x-axis to each line is $ \tan^{ \minus{} 1}$ of their slopes.  The angle bisector will have the average of these two angles for its slope, which we will denote as $ m$: $ m \equal{} \tan(\frac {\tan^{ \minus{} 1} m_1 \plus{} \tan^{ \minus{} 1} m_2}{2})$ which expands:

$ m \equal{} \frac {\tan((\tan^{ \minus{} 1}m_1)/2) \plus{} \tan((\tan^{ \minus{} 1}m_2)/2)}{1 \minus{} \tan((\tan^{ \minus{} 1}m_1)/2)\tan((\tan^{ \minus{} 1}m_2)/2)}$

pretty messy.  Applying the half angle formula $ \tan(x/2) \equal{} \frac {\tan x \sin x}{\tan x \plus{} \sin x}$, we get $ m \equal{} \frac {\frac {m_1 \sin(\tan^{ \minus{} 1}m_1)}{m_1 \plus{} \sin(\tan^{ \minus{} 1}m_1)} \plus{} \frac {m_2 \sin(\tan^{ \minus{} 1}m_2)}{m_2 \plus{} \sin(\tan^{ \minus{} 1}m_2)}}{1 \minus{} \frac {m_1 \sin(\tan^{ \minus{} 1}m_1)}{m_1 \plus{} \sin(\tan^{ \minus{} 1}m_1)}\frac {m_2 \sin(\tan^{ \minus{} 1}m_2)}{m_2 \plus{} \sin(\tan^{ \minus{} 1}m_2)}}$.

Finally, since $ \sin \tan^{ \minus{} 1} x \equal{} \frac {x}{\sqrt {x^2 \plus{} 1}}$ we have $ m \equal{} \frac {\frac {m_1 \frac {m_1}{\sqrt {m_1^2 \plus{} 1}}}{m_1 \plus{} \frac {m_1}{\sqrt {m_1^2 \plus{} 1}}} \plus{} \frac {m_2 \frac {m_2}{\sqrt {m_2^2 \plus{} 1}}}{m_2 \plus{} \frac {m_2}{\sqrt {m_2^2 \plus{} 1}}}}{1 \minus{} \frac {m_1 \frac {m_1}{\sqrt {m_1^2 \plus{} 1}}}{m_1 \plus{} \frac {m_1}{\sqrt {m_1^2 \plus{} 1}}}\frac {m_2 \frac {m_2}{\sqrt {m_2^2 \plus{} 1}}}{m_2 \plus{} \frac {m_2}{\sqrt {m_2^2 \plus{} 1}}}} \equal{} \frac {\frac {\frac {m_1}{\sqrt {m_1^2 \plus{} 1}}}{1 \plus{} \frac {1}{\sqrt {m_1^2 \plus{} 1}}} \plus{} \frac {\frac {m_2}{\sqrt {m_2^2 \plus{} 1}}}{1 \plus{} \frac {1}{\sqrt {m_2^2 \plus{} 1}}}}{1 \minus{} \frac {\frac {m_1}{\sqrt {m_1^2 \plus{} 1}}}{1 \plus{} \frac {1}{\sqrt {m_1^2 \plus{} 1}}}\frac {\frac {m_2}{\sqrt {m_2^2 \plus{} 1}}}{1 \plus{} \frac {1}{\sqrt {m_2^2 \plus{} 1}}}}$ provided that both $ m_1$ and $ m_2$ are not zero.

So as simple as it gets: $ m\equal{}\frac {\frac {m_1}{\sqrt {m_1^2 \plus{} 1} \plus{} 1} \plus{} \frac {m_2}{\sqrt {m_2^2 \plus{} 1} \plus{} 1}}{1 \minus{} \frac {m_1}{\sqrt {m_1^2 \plus{} 1} \plus{} 1}\frac {m_2}{\sqrt {m_2^2 \plus{} 1} \plus{} 1}}$

In the case that $ m_2 \equal{} 0$ we have $ m \equal{} \tan((\tan^{ \minus{} 1}m_1)/2) \equal{} \frac {\frac {m_1}{\sqrt {m_1^2 \plus{} 1}}}{1 \plus{} \frac {1}{\sqrt {m_1^2 \plus{} 1}}} \equal{} \frac {m_1}{\sqrt {m_1^2 \plus{} 1} \plus{} 1}$.

Note that this method only gives the slope of one angle bisector, dependent on slopes of the lines.  (intersecting lines form 2 pairs of vertical angles, each pair with their own bisector).  The slope of the other bisector is perpendicular to the slope given by this formula. (I think the method Altheman suggests ends up giving both slopes)

Once the slope is found, the intersection of the two lines is then needed since it must be on the bisector.  Then find the intercept with this information.

I'm pretty sure my way is faster...

$ \frac {ax \plus{} by \minus{} c}{\sqrt {a^2 \plus{} b^2}} \equal{} \pm\frac {dx \plus{} ey \minus{} f}{\sqrt {d^2 \plus{} e^2}}$

However, I do think your solution is instructive so it is good too.

[quote="1=2"][quote="atomicwedgie"]$ P(z) \equal{} z^3 \minus{} z^2 \minus{} 2z \plus{} 1 \equal{} 0$[/quote]

How did you get that polynomial?[/quote]

Consider $ \zeta_7 \equal{} e^{2 \pi i/7}$.  Let $ \alpha \equal{} \zeta_7 \plus{} \zeta_7^{\minus{}1} \equal{} 2 \cos \frac{2\pi}{7}$.  Then

$ z^2 \minus{} \alpha z \plus{} 1 \equal{} 0$

has $ \zeta_7$ as a root, because

$ (z^2 \minus{} \zeta_7)(z^2 \minus{} \zeta_7^{\minus{}1}) \equal{} z^2 \minus{} (\zeta_7 \plus{} \zeta_7^{\minus{}1})z \plus{} \zeta_7\zeta_7^{\minus{}1} \equal{} 0$.

Next, consider powers of $ \alpha$:

$ \alpha^2 \equal{} (\zeta_7 \plus{} \zeta_7^{\minus{}1})^2 \equal{} \zeta_7^2 \plus{} 2 \plus{} \zeta_7^{\minus{}2}$

$ \alpha^3 \equal{} \zeta_7^3 \plus{} 3\zeta_7 \plus{} 3\zeta_7^{\minus{}1} \plus{} \zeta_7^{\minus{}3}$.

Then we notice that

$ \alpha^3 \plus{} \alpha^2 \minus{} 2\alpha \minus{} 1 \equal{} \zeta_7^3 \plus{} \zeta_7^2 \plus{} \zeta_7 \plus{} 1 \plus{} \zeta_7^{\minus{}1} \plus{} \zeta_7^{\minus{}2} \plus{} \zeta_7^{\minus{}3}$.

But since $ \zeta_7$ is a primitive 7th root of unity, this is simply equal to zero, being the sum of the 7 roots of $ z^7 \minus{} 1$.  Therefore, we have shown that $ \alpha$ is a root of

$ z^3 \plus{} z^2 \minus{} 2z \minus{} 1$.

A similar principle applies for the other example I provided above.

As an exercise:  What are the other roots of these cubics?

wait a second.. point-line distance formula?  What is that?

I remember a few years ago I tried to do it just by the algebra, using the fact that the bisector is equidistant from the two lines, but that turned into a mess - I think I didn't know when to stop trying to simplify it, but I think I got something similar - after messing around for a few weeks in my spare time.

So what is point-line distance formula?

The formula for distanmce between a point(m,n) and a line (ax+by=c) is:
(am+bn-c)/sqrt(a^2+b^2).

Let's back away from "closed form" to a narrower question: for which angles $ \frac {p}{q}\,\pi$ can you express the cosine and sine in by a formula obtained from integers by addition, subtraction, multiplication, division, and the taking of square roots? Stating it this way does explicitly allow roots to be nested, but only square roots, not cube roots. Hence by these standards, $ \sqrt {\frac {2 \plus{} \sqrt {2}}{2}}$ is fine, but altheman's expression for $ \cos20^{\circ}$ is out of bounds because it involves cube roots (of complex numbers) and atomicwedgie's expression for the trig functions of $ \frac {2\pi}{7}$ is also out of bounds because it requires the irrational roots of a cubic equation.

As it turns out, we can express the trig functions of such an angle in this "square roots only" form if an only if we can construct that angle using compass and straightedge in a standard Euclidean fashion.

The answer is quite a story in its own right. We can construct $ \frac {2\pi}{n}$ if and only if the prime factors of $ n$ are Fermat primes (that is, primes of the form $ 2^{2^k} \plus{} 1$) taken no more than once each, multiplied by an arbitrary power of $ 2.$ (That last provision is easy to explain: we can always bisect angles, and the half-angle formulas given above by Patterns 34 show what is going on algebraically.)

The known Fermat primes: $ 3, 5, 17, 257, 65537.$ The next several Fermat numbers are all composite. It is not known whether there are any others; many would speculate that there aren't.

So you can construct the sine and cosine of $ \frac {2\pi}{n}$ when ...

$ n \equal{} 4,8,16,32,\dots.$ This topic started with $ n \equal{} 16.$
$ n \equal{} 3,6,12,24,\dots.$
$ n \equal{} 5,10,20,40,\dots.$ Finding the closed form for $ \cos72^{\circ}$ is an interesting - and very much do-able - challenge.
$ n \equal{} 15,30,60,120,\dots.$
$ n \equal{} 17,34,68,\dots.$
$ n \equal{} 51,102,204,\dots.$

And so on. On the other hand, even though you can represent the cosine and sine of $ \frac {2\pi}{17}$ using nested square roots, it's rather a large mess.

atomicwedgie's example (which was $ n \equal{} 7$) is not on this list, and altheman's example (which was $ n \equal{} 18,$ which tracks back to $ n \equal{} 9$) is not on this list. You can't do either of those with square roots alone.

I did $ \cos 36^\circ$: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=202083[/url].  $ \cos 72$ is just the double angle formula on $ \cos 36^\circ$.

Kent Merryfield, what if we allow cube roots?  what if we allow any root?  Also, how would you go about finding the values if $ n$ is in one of your last 3 categories? (or is their a general method?)

The regular n-gon is constructible if and only if n is of the form

$ n \equal{} 2^m F_{k_1} F_{k_2} \cdot \ldots \cdot F_{k_r}$,

where $ m$ is a nonnegative integer, and the F's are zero or more distinct Fermat primes.  Consequently, these are precisely the values for which $ \sin \frac {2\pi}{n}$ and $ \cos \frac {2\pi}{n}$ are expressible by square roots.

If we are solely interested in whether the sine and cosine of rational multiples of $ \pi$ are expressible by radicals (permitting n(th) roots), then the answer is yes.  This is a well-known result of the theory of algebraic extensions.

[quote="1=2"][quote="atomicwedgie"]$ P(z) = z^3 - z^2 - 2z + 1 = 0$[/quote]

How did you get that polynomial?[/quote]

If $ \alpha=\frac{\pi}{7}$ then $ \cos(3\alpha)+\cos(4\alpha)=0$. But we know, that

\[ \cos(3\alpha)=4\cos(\alpha)^3-3\cos(\alpha)\\
\cos(4\alpha)=8\, \cos \left( \alpha \right)  ^{4}-8\, \cos \left( \alpha \right) ^{2}+1\]

So, if $ x=\cos(\alpha)$, then

\[ (4x^3-3x )+(8x^4-8x^2+1) = 0\\
(x+1)\,(8\,x^3-4\,x^2-4\,x+1) = 0\]

So, $ x=\cos{\frac{\pi}{7}}$ is the root of the polynomial $ 8\,x^3-4\,x^2-4\,x+1$.

Sergey Markelov

Thanks to all who contributed. :)

I have another question:

Given some angle, it is easy to compute the closed-form of a trig function whose argument is that angle, if it exists.  However, if we are given the closed-form and are asked to find the angle (if it exists) when it is being the argument of a certain trig function, how are we going to do that?  In another word, how do we find the closed-form of an expression with inverse trig function?

For example, if we are given that $ \cos x \equal{} \frac{\sqrt5\minus{}1}4$, then $ x \equal{} \arccos \frac{\sqrt5\minus{}1}4$, but how do we get rid of the inverse cosine function in rewriting x?

@Sergey Markelov: I like your technique. It lets us construct polynomials with $ x\equal{}\frac{\pi}{2k\plus{}1}$ as a root.

We expand $ \cos k x\plus{}\cos (k\plus{}1) x\equal{}0$ and factor as necessary to find the minimal polynomial.

I like your technique! It is quite clever. Is there a simple analogue for sine? (It has something to do with symmetry over the x-axis but I can't quite get it).

I believe that $ \cos(\frac{p}{q}\pi)$ (and $ \sin(\frac{p}{q}\pi)$) are always computable in terms of radicals (not necessarily only square roots), although I am not entirely certain of the proof.

Hamster: someone told me this was a "well-known result of the theory of algebraic extensions" if I remember correctly. (EDIT: wow that was in this thread)

Altheman, try expanding $ \sin (qx) \equal{} 0$ because if $ x \equal{} \frac {p}{q} \pi$ then $ qx$ is a multiple of $ \pi$ and thus has a sine of zero. I don't think this is what you were looking for but it works great for finding polynomials with trig roots.

mathwizarddude

Can't gauruntee this will help... since $ \cos x \equal{} \frac {e^{ix} \plus{} e^{ \minus{} ix}}{2}$.  Let $ y \equal{} \cos x$ and solving the quadratic type we have $ x \equal{} \minus{} i(\ln(y\pm \sqrt {y^2 \minus{} 1}))$.  Of course now $ \minus{} 1 \le y \ge 1$ so in any case where we can't recognize the angle we have a complex number in the square root.  $ ln(A \plus{} Bi) \equal{} a \plus{} bi \implies e^ae^{bi} \equal{} A \plus{} Bi \implies e^a(\cos b \plus{} i\sin b) \equal{} A \plus{} Bi$ hence $ b \equal{} \tan^{ \minus{} 1}B/A$ and $ a \equal{} \ln(A/\cos(\tan^{ \minus{} 1}(B/A))) \equal{} 1/2 \ln(A^2 \plus{} B^2)$ hence $ \ln(A \plus{} Bi) \equal{} 1/2\ln(A^2 \plus{} B^2) \plus{} \tan^{ \minus{} 1} B/A$ but now we are back to using inverse trig... Anyway, we have $ x \equal{} \minus{} i(\ln(y\pm i\sqrt {1 \minus{} y^2})) \equal{} \minus{} i(ln 1 \plus{} i\tan^{ \minus{} 1} (\pm \frac {\sqrt {1 \minus{} y^2}}{y})) \equal{} \tan^{ \minus{} 1} (\pm \frac {\sqrt {1 \minus{} y^2}}{y})$ which shouldn't be a surprise because that's just the inverse tangent of sine divided by cosine.

So another try... since $ \cos (x \plus{} y) \equal{} \cos x \cos y \minus{} \sin x \sin y$ we then have, if $ x \equal{} \cos^{ \minus{} 1}m$ and $ y \equal{} \cos^{ \minus{} 1}n$, then $ x \plus{} y \equal{} \cos^{ \minus{} 1}(mn \minus{} (\pm \sqrt {1 \minus{} m^2})(\pm\sqrt {1 \minus{} n^2}))$ basically we need to find some $ m$ and $ n$ such that $ mn \minus{} (\pm \sqrt {1 \minus{} m^2})(\pm\sqrt {1 \minus{} n^2}) \equal{} \frac {\sqrt {5} \minus{} 1}{4}$ and that we already know $ x$ and $ y$. (if we wanted, we could try using more than two angles)

So none of the general methods are very promising.  But note that $ \cos 72^\circ \equal{} \frac {\sqrt {5} \minus{} 1}{4}$.  This can be obtained from $ \cos 36^\circ \equal{} \frac {1 \plus{} \sqrt {5}}{4}$ which is proved here: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=202083[/url]

The [url=http://mathworld.wolfram.com/ChebyshevPolynomialoftheFirstKind.html]Chebyshev polynomial of the First Kind[/url] $ T_n(x)$ has roots $ \cos \frac {(2k \minus{} 1) \pi}{2n}, k \equal{} 1, 2, ... n$.  The polynomials $ T_n(x) \equal{} \cos \pi \equal{} \minus{} 1$ have roots $ \cos \frac {(2k \minus{} 1) \pi}{n}, k \equal{} 1, 2, ... n$, and the polynomials $ T_n(x) \equal{} \cos 2\pi \equal{} 1$ have roots $ \cos \frac {2k \pi}{n}, k \equal{} 1, 2, ... n$.  All of these values are expressible "in terms of radicals" because

$ \cos \frac {p}{q} \pi \equal{} \frac {\sqrt [q]{ e^{i p \pi} } \plus{} \sqrt [q]{e^{ \minus{} i p \pi } }}{2}$

where $ e^{i p \pi} \equal{} 1, \minus{} 1$ (and this generalizes Altheman's expression of $ \cos 20^{\circ}$).  This tells you little about how to actually compute their values, but the important point is that the Chebyshev polynomials are solvable because they are essentially transformations of the polynomials $ x^n \minus{} 1 \equal{} 0$.

I'm not sure if that works exactly. The square root function outputs the principal root... As far as I know, the most consistent definition is to write $ z\equal{}r\text{cis}\, \theta$ where $ r>0$ and $ \theta\in[0,2\pi)$. From this perspective, the $ q$th root of $ 1$ is just one.

In the example that I used, I computed the third roots of $ \text{cis}\, 60$ and this agrees with the convention for the principal roots of a complex number.

All you did was break up a ninth root into two cube roots.  Similar reasoning works for any $ q$ that can be written as a product of $ d$s such that the $ d^{th}$ roots of unity have simple coefficients.  The real point I was trying to make is that "expressible by radicals" is not necessarily a useful property.  The cubic formula shows that the root of any cubic equation is "expressible by radicals," but when the cube roots involved are of complex numbers, you can't do any better than what I just did anyway.

In other words, do you or do you not agree with the following statement:  the roots of $ x^n \minus{} 1$ are expressible by radicals.

I would agree that they are expressible by radicals, but I don't think that $ \sqrt[n]{1}$ is a valid method of writing them, since $ x^n\minus{}1$ is not irreducible (although maybe it is, since expressible by radicals only requires that the field extension be a subextension of a radical extension - I'm not sure of the accepted definition).

If you want to get technical, the Galois group of the splitting field of $ \Phi_n(x)$ (equivalently, the splitting field of $ x^n \minus{} 1$), a [url=http://en.wikipedia.org/wiki/Cyclotomic_field]cyclotomic field[/url], is (abelian, hence) solvable for every $ n$, and $ \Phi_n(x)$ is irreducible (I believe this is the correct "superior" definition of solvability by radicals).  But see the last paragraph [url=http://en.wikipedia.org/wiki/Root_of_unity#Cyclotomic_polynomials]here[/url], which links to [url=http://cg.cs.uni-bonn.de/personal-pages/weber/publications/pdf/WeberA/WeberKeckeisen99a.pdf]this paper of interest[/url].

[quote="Altheman"]@Sergey Markelov: I like your technique. It lets us construct polynomials with $ x = \frac {\pi}{2k + 1}$ as a root.

We expand $ \cos k x + \cos (k + 1) x = 0$ and factor as necessary to find the minimal polynomial.

I like your technique! It is quite clever. Is there a simple analogue for sine? (It has something to do with symmetry over the x-axis but I can't quite get it).[/quote]

Hello [b]Altheman[/b]!

Thank you for kind words. I love trig very much! Just in case, if we are finding minimal polynomial for $ \cos{\frac{\pi}{n}}$ using the this technology, factor is never necessary if $ n$ is prime, but is always necessary if $ n$ is not prime.

The answer to your question of finding the polynomial, whose root will be $ \sin{\frac{\pi}{n}}$, depends on whether you need the minimal polynomial with integer coefficients, or square roots in the coefficients are OK for you. For example, $ \sin{\frac{\pi}{7}}$ is the root of the polynomial

\[ 64\,{x}^{6}-112\,{x}^{4}+56\,{x}^{2}-7\]

But also it is the root of the polynomial

\[ 8\,{x}^{3}\sqrt {7}+28\,{x}^{2}-7\]

To find the polynomial with integer coefficients, you can notice that if $ x=\frac{\pi}{7}$, then $ \sin{3x}=\sin{4x}$. Now you can use that for any $ x$

\[ \sin{3x} = 3\sin{x} - 4\sin^3{x} \\
\sin{4x} = \cos{x}\, (4\sin{x}-8\sin^3{x})\]

Now the best you can do is to square both identities, and write that is $ t=\sin{x}$, then

\[ \sin^2{3x}-\sin^2{4x} = (3t - 4t^3)^2 - (1-t^2)\,(4t-8t^3)^2 =  \left( 64\,{t}^{6}-112\,{t}^{4}+56\,{t}^{2}-7 \right) {t}^{2}\]

Now what if you want to find minimal polynomial for $ x=\frac{\pi}{7}$ among the polynomial with coefficients can use square roots? In this case you should factor the above polynomial several times under fields with different extentions. And the extention in the next step is $ \sqrt{n}$. For $ x=\frac{\pi}{7}$ the first polynomial is $ 64\,{x}^{6}-112\,{x}^{4}+56\,{x}^{2}-7$. Try to factor it under $ Q[\sqrt{7}]$. You will receive:

\[ 64\,{x}^{6}-112\,{x}^{4}+56\,{x}^{2}-7 =  \left( 8\,{t}^{3}+4\,{t}^{2}\sqrt {7}-\sqrt {7} \right)  \left( 8\,
{t}^{3}-4\,{t}^{2}\sqrt {7}+\sqrt {7} \right)\]

Note that probably in some cases this can give you not the final answer. This is for sure true for $ n=97$, when you can factor not only under $ Q[\sqrt{97}]$, but also under $ Q\left[\sqrt {55193-5604\,\sqrt{97}}\,\right]$, and receive polynomial of degree 24. I expect that $ n=97$ is the first, when the polynomial can be factored more using more square roots (but not 100% sure).

If you like such trig, maybe you will enjoy the following my problems:

\[ \cos{\frac{5\pi}{33}+\cos{\frac{7\pi}{33}-\cos{\frac{10\pi}{33}+\cos{\frac{13\pi}{33}+\cos{\frac{14\pi}{33} = \frac{\sqrt{33}-1}{4}}}}}}\]

\[ \tan{\frac{3\pi}{11}} + 4\sin{\frac{2\pi}{11}} = \sqrt{11}\]

\[ \tan{\frac{\pi}{13}}+\tan{\frac{3\pi}{13}}-\tan{\frac{4\pi}{13}} = -\sqrt{65-18\sqrt{13}}\]

\[ \sqrt[3]{\cos\frac{\pi}{13}+\cos\frac{5\pi}{13}}+
\sqrt[3]{\cos\frac{3\pi}{13}+\cos\frac{11\pi}{13}}+
\sqrt[3]{\cos\frac{7\pi}{13}+\cos\frac{9\pi}{13}}=
\sqrt[3]{\frac{7-3\sqrt[3]{13}}{2}}\]

\[ \sqrt[3]{\cos\frac{\pi}{31}+
\cos\frac{15\pi}{31}+
\cos\frac{23\pi}{31}+
\cos\frac{27\pi}{31}+
\cos\frac{29\pi}{31}}+\\
\sqrt[3]{\cos\frac{3\pi}{31}+
\cos\frac{7\pi}{31}+
\cos\frac{17\pi}{31}+
\cos\frac{19\pi}{31}+
\cos\frac{25\pi}{31}}+\\
\sqrt[3]{\cos\frac{5\pi}{31}+
\cos\frac{9\pi}{31}+
\cos\frac{11\pi}{31}+
\cos\frac{23\pi}{31}+
\cos\frac{21\pi}{31}}=\\
=\sqrt[3]{\frac{3\sqrt[3]{62}-11}{2}}\]

To [b]mathwizarddude[/b]: the answer to your question was found by Jogn Jahnel, who proved the following theorem:

If $ x$ is a quadratic irrationality and $ \frac {\arccos{x}}{\pi}$ is rational then either $ x = \pm\frac {\sqrt {2}}{2}$ or $ x = \pm\frac {\sqrt {3}}{2}$ or $ x = \pm\frac {\sqrt {5}}{4}\pm\frac {1}{4}$.

See page 4 of his work [url=http://www.uni-math.gwdg.de/jahnel/Preprints/cos.pdf]"When is the (co)sine of a rational angle equal to a rational number?"[/url]

I do not expect that anybody will be reading this words since the message is too long, but in case if you are reading this - thank you and greeting from Moscow!  :clap:

Sergey Markelov
markelov@mccme.ru

[quote="Markelov"]
I do not expect that anybody will be reading this words since the message is too long, but in case if you are reading this - thank you and greeting from Moscow!  :clap:

Sergey Markelov
markelov@mccme.ru[/quote]

Please, more!  I really enjoy reading your posts!    :D


geometry
circumcircle
projective geometry
geometry proposed

Quadratilateral  $ABCD$  is inscribed in circle $(O, R)$ . It’ s diagonals intersect at $E$ . Let  $K$   a point  such that  $KA \perp AB$ and  $KD \perp DC$. Prove that points $K , E , O$   lie all on the same line.

  [u]Babis[/u]

[b]Note [/b]

 [i]I moved it from “Unsolved problems”  , because I have  found a solution.[/i]

I note $w$ the circumcircle of the convex quadrilateral $ABCD$ and the second intersections $R,\ P$ of the perpendiculars $KA,\ KD$ with the circle $w$ respectively. We see that $O\in CP\cap BR$ and now we apply the Pascal's theorem to the inscribed hexagon $ARBDPC$:

$K\in AR\cap DP,\ O\in RB\cap PC,\ E\in BD\cap CA\Longrightarrow O\in KE.$

[u]Remark.[/u] I liked this problem and it suggestes me other own compositions with the same  idea.

Perfect !!!

This is also my solution. The problem was proposed ( unsolved)  to EYKLEIDIS  B by N. Dergiades ( he has  solved problems in Darij ' s site , too). Really , a very good application of Pascal theorem. I wonder if we can find a solution without Pascal!

[u]Babis[/u]

Yes, quite trivially:

Let $M$ be the midpoint of $AD$. Then we have that the point $F$ symmetric to $E$ about the line $OM$ is the isogonal conjugate of $K$ with respect to $ADO$ - this is a trivial angle chase. This directly implies that $O,E,K$ are collinear ($\sphericalangle AOE = \sphericalangle DOF = \sphericalangle AOK$).

dear stergiu ,here is a solution with using pascal theorem.i hope it will help u... ;)

let $KA,KD$,intersect the circle in points $H,M$respectivly.

in hexagon $BDMCAH$we have ,the intersection of $HA,DM$,the intersection of $AC,BD$,

and the intersection of $MC,BH$,are colinear.

then we know:$K$,$E$,and the intersection of $BH,CM$ are colinear.

but the intersection of $BH,CM$,is $O$.

because:angle $BAH$is $90$,and $BH$ is diameter.

angle $MDC$is $90$,and $MC$ is diameter.and the intersection of two diameter will be the center.

then we find that:$K,E,O$are colinear.

Nice problem of Mr.  N. Dergiades.

  Thank you.

   M.T.

Many thanks to all of you ! Your solutions are nice!

[u]Babis[/u]


topology
geometry
3D geometry
sphere
advanced fields
advanced fields unsolved

1) find the fundamental group of the following figure:

a circle cutted by two horizantal lines.
i think that the answer is Z*Z*Z but how?

2) show that the torus T=S^1 x S^1 is homeomorphic to a subset of the 3-dimensional sphere S^3?

thx

1) as I understand it (not the circle but the whole disk is cutted by two (open¿) segments, not by lines):
Take a line (a true line, not a segment) parallel to the two given ones lying between them.
Then the whole given space is the union of the upper half and the lower half given by this line. Now when we add some small neighbourhood of each half, we get two subspaces with simply connected intersection.
Thus Van Kampen gives us that the fundamental group is $A*B$ with $A,B$ being those of the halfes.
Now some hard to describe but easy to see deformation brings $A$ into the form of a circle, so the fundamental group of $A$ is $\mathbb{Z}$.
Similar for $B$, thus the total fundamental group is $\mathbb{Z}*\mathbb{Z}$.

2) the torus can be imbedded into the $\mathbb{R}^3$ and the last one can be imbedded into the sphere (the sphere is the compactification of $\mathbb{R}^3$ with $\infty$), the topologies are the same. Or did I missunderstand something¿

So you mean that the answer of the first part will be Z*Z which i think is wrong.
The correct answer is Z*Z*Z so we must split the figure into three pieces not two each is a deformation retract of a cirle.
what do u think?
 all the best and thanks for your help
I

As I understand it (I already said that I don't understand clearly what is meant above), we have a disk with (up to deformation) two holes.
We can go around each of these (or even see the whole thing as the wedge of two circles), giving us $\mathbb{Z} * \mathbb{Z}$.
So please clarify how the figure looks like.


Bonjour, j'ai un question.
Sorry people. I'm not good at French, so I'll express these words in English.

[quote]Let $x,y,z>0$ such that $x+y+z=3$. Find maximum value of:
\[ \frac{x}{y^2}+\frac{y}{z^2}+\frac{z}{x^2}  \]
[/quote]
Why doesn’t the greatest value exist?
It maybe stupid.

[url]http://www.mathlinks.ro/Forum/index.php?f=67[/url]

I think this is a more appropriate place. Why do you post this here?  Not only is this a French speaking community, it is their 'chat space', where they discuss other things than math stuff.

The answer is simple, put : $x=y=t$  and $z=3-2 t$ with $0<t<frac{3}{2}$  you will see it diverges for t on both sides of the interval
There is a minimum value though, which is a very easy result than can be obtained by using AM GM twice!

Pourquoi vous le demandez ici?  Ici on  parle en français et on s'occupe des affaires non mathématiques?
Il n'y a pas une valeur maximale, vous pouvez voir ça avec cette substitution  $x=y=t$  et $z=3-2 t$  $0<t<frac{3}{2}$
Mais il y a une valeur minimale!


hello, to ii)
we get
$ \frac{a\plus{}b\plus{}c}{ab\plus{}bc\plus{}ca}\minus{}\frac{a}{a^2\plus{}ab\plus{}b^2}\minus{}\frac{b}{b^2\plus{}bc\plus{}c^2}\minus{}\frac{c}{c^2\plus{}ca\plus{}a^2}\equal{}$
$ \frac{(a\plus{}b\plus{}c)(a^2b^4\plus{}b^2c^4\plus{}a^4c^2\minus{}ab^3c^2\minus{}a^3b^2c\minus{}a^2bc^3)}{(ab\plus{}bc\plus{}ac)(a^2\plus{}ab\plus{}b^2)(b^2\plus{}bc\plus{}c^2)(c^2\plus{}ac\plus{}a^2)}$
this term is non negative because
$ (ab^2)^2\plus{}(bc^2)^2\plus{}(a^2c)^2\geq ab^3c^2\plus{}a^3b^2c\plus{}a^2bc^3$
after the inequality
$ x^2\plus{}y^2\plus{}z^2\geq xy\plus{}yz\plus{}xz$
Sonnhard.

[quote="nguoivn"]Given $ a, b, c > 0$. Prove that:
i) $ \frac {a \plus{} b \plus{} c}{ab \plus{} bc \plus{} ca} \ge \frac {a}{a^2 \plus{} ab \plus{} bc} \plus{} \frac {b}{b^2 \plus{} bc \plus{} ca} \plus{} \frac {c}{c^2 \plus{} ca \plus{} ab} \ge \frac {9(ab \plus{} bc \plus{} ca)}{(a \plus{} b \plus{} c)^3}$
[/quote]
First we prove $ \frac {a \plus{} b \plus{} c}{ab \plus{} bc \plus{} ca} \ge \frac {a}{a^2 \plus{} ab \plus{} bc} \plus{} \frac {b}{b^2 \plus{} bc \plus{} ca} \plus{} \frac {c}{c^2 \plus{} ca \plus{} ab}$
<=> $ \frac {(a \plus{} b \plus{} c)^2}{ab \plus{} bc \plus{} ca} \ge \frac {a(a \plus{} b \plus{} c)}{a^2 \plus{} ab \plus{} bc} \plus{} \frac {b(a \plus{} b \plus{} c)}{b^2 \plus{} bc \plus{} ca} \plus{} \frac {c(a \plus{} b \plus{} c)}{c^2 \plus{} ca \plus{} ab}$
<=> $ \sum\frac {bc}{a^2 \plus{} ab \plus{} bc} \plus{} \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}\ge 1 \plus{} \sum\frac {ac}{a^2 \plus{} ab \plus{} bc}$
Using CS ineq,we have:
$ \sum\frac {bc}{a^2 \plus{} ab \plus{} bc}\ge 1$
So we must to prove that $ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}\ge\sum\frac {ac}{a^2 \plus{} ab \plus{} bc}$
From CS inequality we get $ \sum\frac {ac(c^2 \plus{} ab \plus{} bc)}{(a^2 \plus{} ab \plus{} bc)(c^2 \plus{} ab \plus{} bc)}\le\sum\frac {ac(c^2 \plus{} ab \plus{} bc)}{(ab \plus{} bc \plus{} ca)^2}$
$ \equal{} \frac {ac^3 \plus{} ba^3 \plus{} cb^3 \plus{} 2abc(a \plus{} b \plus{} c)}{(ab \plus{} bc \plus{} ca)^2}$
So we only need to prove that
 $ (a^2 \plus{} b^2 \plus{} c^2)(ab \plus{} bc \plus{} ca)\ge ac^3 \plus{} ab^3 \plus{} bc^3 \plus{} 2abc(a \plus{} b \plus{} c)$
<=> $ ab^3 \plus{} bc^3 \plus{} ca^3 \ge abc(a \plus{} b \plus{} c)$
<=> $ \frac {b^2}{c} \plus{} \frac {c^2}{a} \plus{} \frac {a^2}{b}\ge a \plus{} b \plus{} c$
  (Obviously true by AM-GM)
I think you has posted the solution for this:
$ \frac {a}{a^2 \plus{} ab \plus{} bc} \plus{} \frac {b}{b^2 \plus{} bc \plus{} ca} \plus{} \frac {c}{c^2 \plus{} ca \plus{} ab} \ge \frac {9(ab \plus{} bc \plus{} ca)}{(a \plus{} b \plus{} c)^3}$

[quote="leedt26"][quote="nguoivn"]Given $ a, b, c > 0$. Prove that:
i) $ \frac {a \plus{} b \plus{} c}{ab \plus{} bc \plus{} ca} \ge \frac {a}{a^2 \plus{} ab \plus{} bc} \plus{} \frac {b}{b^2 \plus{} bc \plus{} ca} \plus{} \frac {c}{c^2 \plus{} ca \plus{} ab} \ge \frac {9(ab \plus{} bc \plus{} ca)}{(a \plus{} b \plus{} c)^3}$
[/quote]
First we prove $ \frac {a \plus{} b \plus{} c}{ab \plus{} bc \plus{} ca} \ge \frac {a}{a^2 \plus{} ab \plus{} bc} \plus{} \frac {b}{b^2 \plus{} bc \plus{} ca} \plus{} \frac {c}{c^2 \plus{} ca \plus{} ab}$
<=> $ \frac {(a \plus{} b \plus{} c)^2}{ab \plus{} bc \plus{} ca} \ge \frac {a(a \plus{} b \plus{} c)}{a^2 \plus{} ab \plus{} bc} \plus{} \frac {b(a \plus{} b \plus{} c)}{b^2 \plus{} bc \plus{} ca} \plus{} \frac {c(a \plus{} b \plus{} c)}{c^2 \plus{} ca \plus{} ab}$
<=> $ \sum\frac {bc}{a^2 \plus{} ab \plus{} bc} \plus{} \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}\ge 1 \plus{} \sum\frac {ac}{a^2 \plus{} ab \plus{} bc}$
Using CS ineq,we have:
$ \sum\frac {bc}{a^2 \plus{} ab \plus{} bc}\ge 1$
So we must to prove that $ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}\ge\sum\frac {ac}{a^2 \plus{} ab \plus{} bc}$
From CS inequality we get $ \sum\frac {ac(c^2 \plus{} ab \plus{} bc)}{(a^2 \plus{} ab \plus{} bc)(c^2 \plus{} ab \plus{} bc)}\le\sum\frac {ac(c^2 \plus{} ab \plus{} bc)}{(ab \plus{} bc \plus{} ca)^2}$
$ \equal{} \frac {ac^3 \plus{} ba^3 \plus{} cb^3 \plus{} 2abc(a \plus{} b \plus{} c)}{(ab \plus{} bc \plus{} ca)^2}$
So we only need to prove that
 $ (a^2 \plus{} b^2 \plus{} c^2)(ab \plus{} bc \plus{} ca)\ge ac^3 \plus{} ab^3 \plus{} bc^3 \plus{} 2abc(a \plus{} b \plus{} c)$
<=> $ ab^3 \plus{} bc^3 \plus{} ca^3 \ge abc(a \plus{} b \plus{} c)$
<=> $ \frac {b^2}{c} \plus{} \frac {c^2}{a} \plus{} \frac {a^2}{b}\ge a \plus{} b \plus{} c$
  (Obviously true by AM-GM)
[/quote]
Ýe, it's also the way I created it (by 2 old results above) :lol:

We also have a simimilar result but very easy:
$ \frac{a}{a^2\plus{}ab\plus{}bc}\plus{}\frac{b}{b^2\plus{}bc\plus{}ca}\plus{}\frac{c}{c^2\plus{}ca\plus{}ab} \ge \frac{a\plus{}b\plus{}c}{a^2\plus{}b^2\plus{}c^2}$  :)

[quote="nguoivn"]We also have a simimilar result but very easy:
$ \frac {a}{a^2 \plus{} ab \plus{} bc} \plus{} \frac {b}{b^2 \plus{} bc \plus{} ca} \plus{} \frac {c}{c^2 \plus{} ca \plus{} ab} \ge \frac {a \plus{} b \plus{} c}{a^2 \plus{} b^2 \plus{} c^2}$  :)[/quote]

By BCS

$ \sum\frac {a}{a^2 \plus{} ab \plus{} bc} \geq \frac {(\sum a)^2}{\sum a^3 \plus{} \sum a^2b \plus{} 3abc} \geq \frac {\sum a}{\sum a^2}$

so it becomes

$ \sum ab^2 \geq 3abc$ true by AM-GM.

[quote="nguoivn"]

$ \frac {a}{a^2 \plus{} ab \plus{} b^2} \plus{} \frac {b}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {c}{c^2 \plus{} ca \plus{} a^2} \ge \frac {a \plus{} b \plus{} c}{a^2 \plus{} b^2 \plus{} c^2}$
:)[/quote]

By BCS

$ \sum \frac{a}{a^2 \plus{} ab \plus{} b^2} \geq \frac{(\sum a)^2}{\sum a^3 \plus{}\sum ab(a\plus{}b)}\equal{} \frac {\sum a}{\sum a^2}$

[quote="manlio"]This is really very nice  :wink:

$ \frac {a \plus{} b \plus{} c}{ab \plus{} bc \plus{} ca} \ge \frac {a}{a^2 \plus{} ab \plus{} b^2} \plus{} \frac {b}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {c}{c^2 \plus{} ca \plus{} a^2}$

Dear nguoivn, have you a nice proof for it without full expansion?

Thank you very much. :)[/quote]
It is equivalent to
$ \frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca} \ge \frac{ab}{b^2\plus{}bc\plus{}c^2}\plus{}\frac{bc}{c^2\plus{}ca\plus{}a^2} \plus{}\frac{ca}{a^2\plus{}ab\plus{}b^2}.$ (1)
The reason is:
$ \frac{ab}{b^2\plus{}bc\plus{}c^2}\plus{}\frac{b}{a\plus{}b\plus{}c} \equal{}\frac{\sum a^2\plus{}\sum ab}{\sum a}\frac{b}{b^2\plus{}bc\plus{}c^2},$
therefore, the inequality (1) is equivalent to
$ \frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca} \plus{}1 \ge \frac{\sum a^2\plus{}\sum ab}{\sum a}\sum \frac{a}{a^2\plus{}ab\plus{}b^2},$
or
$ \frac{a\plus{}b\plus{}c}{ab\plus{}bc\plus{}ca} \ge \sum \frac{a}{a^2\plus{}ab\plus{}b^2}.$

To prove (1), we can use the Cauchy Schwarz Inequality. I will wait for some nice proofs.

Another result: Given $ a, b, c > 0$. Prove that:
\[ \frac{a\plus{}b\plus{}c}{ab\plus{}bc\plus{}ca} \ge \frac{a}{a^2\plus{}bc\plus{}b^2}\plus{}\frac{b}{b^2\plus{}ca\plus{}c^2}\plus{}\frac{c}{c^2\plus{}ab\plus{}a^2} \ge \frac{a\plus{}b\plus{}c}{a^2\plus{}b^2\plus{}c^2}\]
:)

[quote="nguoivn"]Another result: Given $ a, b, c > 0$. Prove that:
\[ \frac {a \plus{} b \plus{} c}{ab \plus{} bc \plus{} ca} \ge \frac{a}{a^2 \plus{} bc \plus{} b^2} \plus{} \frac {b}{b^2 \plus{} ca \plus{} c^2} \plus{} \frac {c}{c^2 \plus{} ab \plus{} a^2} \ge \frac{a \plus{} b \plus{} c}{a^2 \plus{} b^2 \plus{} c^2}\]
:)[/quote]

The RHS inequality is easy. By BCS we have

  $ \sum \frac {a}{a^2 \plus{} bc \plus{} b^2} \geq \frac{(\sum a)^2}{\sum a( a^2\plus{}b^2\plus{}bc)}$

so it suffices to prove

$ (\sum a)(\sum a^2) \geq \sum a^3 \plus{}\sum ab^2 \plus{}3abc$

that is

$ \sum a^2b \geq 3abc$

which is AM-GM. :)

I cannot prove the LHS inequality.  :blush: Can you please give me a hint nguoivn. Thank you very much

[quote="nguoivn"]Another result: Given $ a, b, c > 0$. Prove that:
\[ \frac {a \plus{} b \plus{} c}{ab \plus{} bc \plus{} ca} \ge \frac {a}{a^2 \plus{} bc \plus{} b^2} \plus{} \frac {b}{b^2 \plus{} ca \plus{} c^2} \plus{} \frac {c}{c^2 \plus{} ab \plus{} a^2} \ge \frac {a \plus{} b \plus{} c}{a^2 \plus{} b^2 \plus{} c^2}\]
[/quote]
The right inequality is not hard. Let me prove the left one. By writing the inequality as
$ \sum \left[a \minus{} \frac {a(ab \plus{} bc \plus{} ca)}{a^2 \plus{} bc \plus{} b^2}\right] \ge 0,$
we see that it is equivalent to
$ \sum \frac {a(a^2 \plus{} b^2)}{a^2 \plus{} bc \plus{} b^2} \minus{} \sum \frac {a^2(b \plus{} c)}{a^2 \plus{} bc \plus{} b^2} \ge 0,$
or
$ \sum \frac {a(a^2 \plus{} b^2)}{a^2 \plus{} bc \plus{} b^2} \plus{} \sum \frac {c(b \plus{} c)^2}{a^2 \plus{} bc \plus{} b^2} \ge \sum (b \plus{} c).$
By the Cauchy-Schwarz Inequality, we have
$ \left[ \sum \frac {a(a^2 \plus{} b^2)}{a^2 \plus{} bc \plus{} b^2}\right]\left[ \sum \frac {a(a^2 \plus{} bc \plus{} b^2)}{a^2 \plus{} b^2}\right] \ge (a \plus{} b \plus{} c)^2.$
Furthermore,
$ \sum \frac {a(a^2 \plus{} bc \plus{} b^2)}{a^2 \plus{} b^2} \equal{} \sum a \plus{} \sum \frac {abc}{a^2 \plus{} b^2} \le \sum a \plus{} \sum \frac {c}{2} \equal{} \frac {3}{2}\sum a.$
Combining this with the above inequality, we get
$ \sum \frac {a(a^2 \plus{} b^2)}{a^2 \plus{} bc \plus{} b^2} \ge \frac {2}{3}\sum a.$
Therefore, it is enough to prove that
$ \sum \frac {c(b \plus{} c)^2}{a^2 \plus{} bc \plus{} b^2} \ge \frac {4}{3}\sum a.$
By the Cauchy-Schwarz Inequality,
$ \sum \frac {c(b \plus{} c)^2}{a^2 \plus{} bc \plus{} b^2} \ge \frac {\left[ \sum c(b \plus{} c)\right]^2}{\sum c(a^2 \plus{} bc \plus{} b^2)} ,$
so it is sufficient to check the following inequality
$ 3\left( \sum a^2 \plus{} \sum ab\right)^2 \ge 4\left(\sum a\right) \left( 2\sum ab^2 \plus{} \sum a^2b\right).$
It is easy to see that this inequality is true according to the two known inequalities
$ \left(\sum a^2 \plus{} \sum ab\right)^2 \ge 4\left(\sum a\right)\left(\sum a^2b\right),$
and
$ \left(\sum a^2 \plus{} \sum ab\right)^2 \ge 4\left(\sum a\right)\left( \sum ab^2\right).$
(We can prove these two inequalities by using the AM-GM Inequality)

Thank you very much, my dear friend for your very nice proof. :)

I don't see how to prove  :blush:

$ \sum a^4 \plus{} 2\sum a^3b \geq 2\sum ab^3 \plus{}\sum a^2b^2$

by AM-GM. Sorry.

Thank you very much. :)

[quote="great math"]Here are some things for you, manlio
\[ \sum_{cyc}a^4 \plus{} \sum_{cyc}a^2c^2 \ge 2\sum_{cyc}a^3c\]
and
\[ \sum_{cyc}a^3b \ge \sum_{cyc}(ab)^2\]
[/quote]

You are right, great math, but I don't remember  :blush: how to prove by only AM-GM as can_hang2007 said \[ \sum_{cyc}a^3b \ge \sum_{cyc}(ab)^2\]

Thank you very much

[quote="great math"]Here are some things for you, manlio
\[ \sum_{cyc}a^4 \plus{} \sum_{cyc}a^2c^2 \ge 2\sum_{cyc}a^3c\]
and
\[ \sum_{cyc}a^3b \ge \sum_{cyc}(ab)^2\]
[/quote]
I didn't mean this, great math. Anyways, the second inequality does not always hold (it can only hold for some special case, for example the case $ a,b,c$ are the side-lengths of a triangle is IMO 1983). My proof of the inequality
$ \left( \sum a^2\plus{}\sum ab\right)^2 \ge 4\left(\sum a\right) \left(\sum a^2b\right)$
is as follows:

Assume that $ b$ is the number between $ a$ and $ c.$ Then, by applying the AM-GM Inequality, we get
$ 4\left(\sum a\right) \left(\sum a^2b\right) \le \left( b\sum a\plus{}\frac{\sum a^2b}{b}\right)^2.$
It is thus sufficient to prove the stronger inequality
$ \sum a^2\plus{}\sum ab \ge b\sum a \plus{}\frac{\sum a^2b}{b}.$
This inequality is equivalent to
$ \frac{c(a\minus{}b)(b\minus{}c)}{b} \ge 0,$
which is obviously true according to the assumption of $ b.$

Thank you very much, my dear friend. Now all is clear :)

[quote="leedt26"][quote="can_hang2007"]
$ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \ge \frac {ab}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {bc}{c^2 \plus{} ca \plus{} a^2} \plus{} \frac {ca}{a^2 \plus{} ab \plus{} b^2}.$ (1)
[/quote]
Dear Can , can you post your solution by the Cauchy-Schwarz Inequality for this ineq ?
Thanks.[/quote]
This is my solution to this one:
Rewrite the inequality as
$ \sum {\left[ {{a^2} \minus{} \frac{{ab\left( {ab \plus{} bc \plus{} ca} \right)}}{{{b^2} \plus{} bc \plus{} {c^2}}}} \right]}  \ge 0$
It's equivalent to
$ \sum {\frac{{{a^2}{c^2}}}{{{b^2} \plus{} bc \plus{} {c^2}}}}  \minus{} \sum {\frac{{a{b^2}c}}{{{b^2} \plus{} bc \plus{} {c^2}}}}  \ge 0$
or
$ \sum {\frac{{{a^2}{c^2}}}{{{b^2} \plus{} bc \plus{} {c^2}}}}  \plus{} \sum {\frac{{a{c^2}(b \plus{} c)}}{{{b^2} \plus{} bc \plus{} {c^2}}}}  \ge ab \plus{} bc \plus{} ca$
$ \Leftrightarrow$
 $ \sum {\frac{{a{c^2}}}{{{b^2} \plus{} bc \plus{} {c^2}}}}  \ge \frac{{ab \plus{} bc \plus{} ca}}{{a \plus{} b \plus{} c}}$

Using the Cauchy-Schwarz Ineq we get
$ \sum {\frac{{a{c^2}}}{{{b^2} \plus{} bc \plus{} {c^2}}}}  \ge \frac{{{{\left( {ab \plus{} bc \plus{} ca} \right)}^2}}}{{\sum {a\left( {{b^2} \plus{} bc \plus{} {c^2}} \right)} }} \equal{} \frac{{ab \plus{} bc \plus{} ca}}{{a \plus{} b \plus{} c}}$
Note that $ \sum {a\left( {{b^2} \plus{} bc \plus{} {c^2}} \right)}\equal{}(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)$

[quote="leedt26"][quote="leedt26"][quote="can_hang2007"]
$ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \ge \frac {ab}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {bc}{c^2 \plus{} ca \plus{} a^2} \plus{} \frac {ca}{a^2 \plus{} ab \plus{} b^2}.$ (1)
[/quote]
Dear Can , can you post your solution by the Cauchy-Schwarz Inequality for this ineq ?
Thanks.[/quote]
This is my solution to this one:
Rewrite the inequality as
$ \sum {\left[ {{a^2} \minus{} \frac {{ab\left( {ab \plus{} bc \plus{} ca} \right)}}{{{b^2} \plus{} bc \plus{} {c^2}}}} \right]} \ge 0$
It's equivalent to
$ \sum {\frac {{{a^2}{c^2}}}{{{b^2} \plus{} bc \plus{} {c^2}}}} \minus{} \sum {\frac {{a{b^2}c}}{{{b^2} \plus{} bc \plus{} {c^2}}}} \ge 0$
or
$ \sum {\frac {{{a^2}{c^2}}}{{{b^2} \plus{} bc \plus{} {c^2}}}} \plus{} \sum {\frac {{a{c^2}(b \plus{} c)}}{{{b^2} \plus{} bc \plus{} {c^2}}}} \ge ab \plus{} bc \plus{} ca$
$ \Leftrightarrow$
 $ \sum {\frac {{a{c^2}}}{{{b^2} \plus{} bc \plus{} {c^2}}}} \ge \frac {{ab \plus{} bc \plus{} ca}}{{a \plus{} b \plus{} c}}$

Using the Cauchy-Schwarz Ineq we get
$ \sum {\frac {{a{c^2}}}{{{b^2} \plus{} bc \plus{} {c^2}}}} \ge \frac {{{{\left( {ab \plus{} bc \plus{} ca} \right)}^2}}}{{\sum {a\left( {{b^2} \plus{} bc \plus{} {c^2}} \right)} }} \equal{} \frac {{ab \plus{} bc \plus{} ca}}{{a \plus{} b \plus{} c}}$
Note that $ \sum {a\left( {{b^2} \plus{} bc \plus{} {c^2}} \right)} \equal{} (a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)$[/quote]
Dear leedt26, it's also the old proof of can_hang for it  :)
[quote="nguoivn"]Another result: Given $ a, b, c > 0$. Prove that:
\[ \frac {a \plus{} b \plus{} c}{ab \plus{} bc \plus{} ca} \ge \frac {a}{a^2 \plus{} bc \plus{} b^2} \plus{} \frac {b}{b^2 \plus{} ca \plus{} c^2} \plus{} \frac {c}{c^2 \plus{} ab \plus{} a^2}\]
[/quote]
Dear manlio, it's a special and strong inequality that I haven't found the proof for it  :oops:
Thank you very much, can_hang for your nice proof  :lol:

[quote=nguoivn]Given $ a, b, c > 0$. Prove that:
i) $ \frac {a \plus{} b \plus{} c}{ab \plus{} bc \plus{} ca} \ge \frac {a}{a^2 \plus{} ab \plus{} bc} \plus{} \frac {b}{b^2 \plus{} bc \plus{} ca} \plus{} \frac {c}{c^2 \plus{} ca \plus{} ab} \ge \frac {9(ab \plus{} bc \plus{} ca)}{(a \plus{} b \plus{} c)^3}$

ii) $ \frac {a \plus{} b \plus{} c}{ab \plus{} bc \plus{} ca} \ge \frac {a}{a^2 \plus{} ab \plus{} b^2} \plus{} \frac {b}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {c}{c^2 \plus{} ca \plus{} a^2} \ge \frac{a\plus{}b\plus{}c}{a^2\plus{}b^2\plus{}c^2}$
:)[/quote]

[img]http://s14.sinaimg.cn/middle/006ptkjAzy77H0l4GwR0d&690[/img]


algebra
polynomial
LaTeX
linear algebra
matrix
function
algorithm

What are the most efficient criteria to locate the roots of a complex polynomial ? I know Kakeya theorem, the "criterion" suggested by Rudin's and Cain's proofs of the Fundamental Theorem of algebra, but they complete each other and nothing seems ultimately very good.

What other complementary methods do we have ?

Maybe Sturm's theorem can help: http://planetmath.org/encyclopedia/SturmsTheorem.html
I believe there is a complex version of this somewhere as well..

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=12915]This old topic[/url] has some interesting material, particularly in jmerry's first post.

Warnings: the original problem statement seems to be missing from the topic, and this is mostly from pre-$\LaTeX$ times, hence harder to read.

1. [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=66025]Here[/url] is a recent discussion about that same theorem (from the $\LaTeX$ era).

2. There's also a nice theorem about the eigenvalues of a matrix: [url=http://en.wikipedia.org/wiki/Gershgorin_circle_theorem]Gershgorin circle theorem[/url].
BTW, it is impossible for me to remember that name. It's a good thing that I have a book which has the theorem listed at the index.

@Gershgorin, see also : http://www.mathlinks.ro/Forum/viewtopic.php?t=54738

[quote="julien_santini"]What are the most efficient criteria to locate the roots of a complex polynomial ? [/quote]

Sturm Functions, already mentioned for real polynomials, make root-finding algorithms efficient because they allow the determination of the number of real zeros over a particular interval. However, they do not reveal the actual roots. (Once you know an approximate region for roots, you can focus attention there instead of taking random guesses) At the moment, I do not know if a similar theorem exists for complex polynomials and roots.

Techniques for numerically finding the roots of polynomials include Laguerre's method, calculating the eigenvalues of the polynomial's Companion Matrix, and the Jenkins-Traub Algorithm. Software for calculating the eigenvalues of a polynomial's companion matrix is available that is very robust and gives very accurate results; however, this approach is not as fact or efficient as using a program specifically for finding the roots of polynomials.

I would recommend the Jenkins-Traub Algorithm; it has been around a long time, is well-tested, and is very robust:

[url]http://mathworld.wolfram.com/Jenkins-TraubMethod.html[/url]

For complex polynomials, Fortran source code is posted on the NETLIB site as CPOLY:

[url]http://www.netlib.org/toms/419[/url]

The version for real polynomials is about four times faster (than using the complex program on the same real polynomial), and is posted as RPOLY:

[url]http://www.netlib.org/toms/493[/url]

C++ source code and online root-finders (written in Javascript) are also available.

Regards.

Tom


logarithms
inequalities unsolved
inequalities

[quote="Svejk"]If a,b,c$\in (1,\infty)$ prove that
\[\frac{\log_{ab}^{2}c}{a+b}+\frac{\log_{bc}^{2}a}{b+c}+\frac{\log_{ca}^{2}b}{c+a}\geq \frac{9}{8(a+b+c)}\]
[/quote]
It's stronger: $\sum_{cyc}\frac{\log_{ab}^{2}c}{a+b}\geq\sum_{cyc}\frac{1}{4(a+b)}.$ :wink:


trigonometry
geometry
trapezoid
geometry proposed

Let $A_0A_1A_2A_3A_4A_5$ be a convex hexagon inscribed in a circle. Define the points $A_0'$, $A_2'$, $A_4'$ on the circle, such that
\[ A_0A_0' \parallel A_2A_4, \quad A_2A_2' \parallel A_4A_0, \quad A_4A_4' \parallel A_2A_0 . \]

Let the lines $A_0'A_3$ and $A_2A_4$ intersect in $A_3'$, the lines $A_2'A_5$ and $A_0A_4$ intersect in $A_5'$ and the lines $A_4'A_1$ and $A_0A_2$ intersect in $A_1'$.

Prove that if the lines $A_0A_3$, $A_1A_4$ and $A_2A_5$ are concurrent then the lines $A_0A_3'$, $A_4A_1'$ and $A_2A_5'$ are also concurrent.

[quote="Valentin Vornicu"][quote="mecrazywong"]Just sine law+(trigo form of) Ceva. I believe everyone solved this problem.[/quote]Quite the opposite!  :?[/quote]
Valentin, you mean Ceva doesn't work or someone failed to solve it?

I mean many students failed to solve it.

I don't see any reason unless they haven't heard about the trigo form of ceva :?

[quote="perfect_radio"]can anyone post more details?[/quote]
Just angle chasing+trigo form of Ceva. No other techniques are needed.

i understood that from your previous posts.

to you the solution may seem obvious, but to me it's quite the opposite

Thanks Sailor for explaining this in details :)

[quote="Sailor"]Ok, here are some more datails:

$\frac{A_2A_3'}{A_3'{A_4}}=\frac{\sin{A_2A_3A_3'}}{\sin{A_3A_0A_4}}\frac{\sin{A_3A_4A_2}}{\sin{A_3'A_3A_4}}$.
But $\angle{A_2A_3A_0'}=\angle{A_2A_5A_4'}$ from isosceles trapeziums.
Use the other similar relations and Ceva trigonometric form to complete the solution.[/quote]

From my calculations i get:
$\frac{A_2A_3'}{A_3'{A_4}}=\frac{\sin{A_2A_3A_3'}}{\sin{A_3A_2A_4}}\frac{\sin{A_3A_4A_2}}{\sin{A_3'A_3A_4}}$.

is $A_3A_2A_4=A_3A_0A_4$ ???

i can't get $\angle{A_2A_3A_0'}=\angle{A_2A_5A_4'}$ from isosceles trapeziums

Of course, $\angle{A_3A_2A_4}=\angle{A_3A_0A_4}$ because the equal to half the $arc{A_3A_4}$, but $arc{A_0'A_2}=arc{A_0A4}=arc{A_2A_4'}$ and hence
$\angle{A_2A_3A_0'}=\angle{A_2A_5A_4'}$.

[quote="Sailor"]Ok, here are some more datails:

$\frac{A_2A_3'}{A_3'{A_4}}=\frac{\sin{A_2A_3A_3'}}{\sin{A_3A_0A_4}}\frac{\sin{A_3A_4A_2}}{\sin{A_3'A_3A_4}}$.
But $\angle{A_2A_3A_0'}=\angle{A_2A_5A_4'}$ from isosceles trapeziums.
Use the other similar relations and Ceva trigonometric form to complete the solution.[/quote]
Can write more detail?


What is the sum of the number of faces, vertices and edges in
a triangular prism?

A triangular prism has two triangular bases connected by three quadrilateral faces.

Faces: 5

Vertices: 6

Edges: 9

$ 5 \plus{} 6 \plus{} 9 \equal{} 20$

Who proved V+F=E-2 again?  I forgot who it was.  Sorry.

Euler- It was his polyhedra/polyhedron formula.


advanced fields
advanced fields unsolved

If a is a real number, there is a sequence of rationals {a_n} such that the sequence of real numbers whose n-th term is f(a_n) converges to a.

The definition that I have is that a real number converges (has a limit) to a iff for all e>0, there exists M>0 such that, if n>M, then abs(a - a_n)<e. Then

After that I am stuck. Any thing would help. Thanks

Didn't you say $ a$ was a real number? And do you mean $ f(a_1) \equal{} a$, $ f(a_2) \equal{} a$, $ f(a_3) \equal{} a$, $ \ldots$? In that case, clearly $ f(a_n)$ converges to $ a$. You have to get the question right in your own mind before you can ask it.


If $ a,b,c>0$, then
$ (a\plus{}b)(a\plus{}c)\plus{}(b\plus{}c)(b\plus{}a)\plus{}(c\plus{}a)(c\plus{}b)\geq6\sqrt{abc(a\plus{}b\plus{}c)}$

[quote="lipako"]If $ a,b,c > 0$, then
$ (a \plus{} b)(a \plus{} c) \plus{} (b \plus{} c)(b \plus{} a) \plus{} (c \plus{} a)(c \plus{} b)\geq6\sqrt {abc(a \plus{} b \plus{} c)}$[/quote]

We have
$ (a \plus{} b)(a \plus{} c)\geq2\sqrt{abc(a\plus{}b\plus{}c)}$
by AM-GM; summing this up we obtain the proposed inequality.


triple summation(3 sigmas) of 1/{(3^i )(3^j)(3^K)}where all i,j,k tend to infinity from 0 and i,j,k are not equal (please do not give a computer program.) also will anybody suggest to me how to make the question more readable?

What does "they are not equal" mean? Is it "at least two are distinct" or "all three are distinct"?

So we want
$\sum^{\infty}_{i=0}\sum^{\infty}_{j=0}\sum^{\infty}_{k=0}\frac{1}{2^{i}}\frac{1}{3^{j}}\frac{1}{5^{k}}$.
We just rewrite the sum
$\sum^{\infty}_{i=0}\frac{1}{2^{i}}\sum^{\infty}_{j=0}\frac{1}{3^{j}}\sum^{\infty}_{k=0}\frac{1}{5^{k}}$
and use the geometric sum starting on the right. The reason we can rewrite the sum is because distributing the products gives the top sum.

come here watson .This stuff is not that easy as you think.read it thoroughly. all no.s whose powers have been raised are 3. Well demanded dear farenhajt ;my question provokes it. all are different  at all values eg. if i =1 j=2 k=3 and so on .(else it becomes a question unworthy of being asked in the annals of this esteemed forum. dear watson please teach me how to type those funky characters you have posted.THANK YOU >WAITING IN ANTICIPATION.
As a matter of fact theans is 81/208 .this question is so simple that any indian math. olym. person will solve it in a couple of shakes. i have a method but my palate wants to taste beaux a' la mode

By inclusion-exclusion principle:

$\begin{aligned}S &=\sum_{i}\sum_{j}\sum_{k}{1\over 3^{i}3^{j}3^{k}}-3\sum_{i}\sum_{j}{1\over 3^{i}3^{2j}}+2\sum_{i}{1\over 3^{3i}}\\ &=\left({3\over 2}\right)^{3}-3\cdot{3\over 2}\cdot{9\over 8}+2\cdot{27\over 26}={27\over 8}-{81\over 16}+{27\over 13}={702-1053+432\over 208}={81\over 208}\end{aligned}$

where all summations are taken from $0$ to $\infty$ without any constraints.

thank you . i already knew this method. i was waiting for some sterner stuff.


A unicorn is tethered by a 20-foot silver rope to the base of a magician's cylindrical tower whose radius is 8 feet. The rope is attached to the tower at ground level and to the unicorn at a height of 4 feet. The unicorn has pulled the rope taut, the end of the rope is 4 feet from the nearest point on the tower, and the length of the rope that is touching the tower...

Can someone draw out a diagram to describe this problem or give a description of where the objects are located.


Do there exist any increasing or decreasing functions $ f: R \rightarrow R$such that $ f(f(x)) \equal{} \{\ x \}$ for all $ x \in [0;1]$

_________________________________________
Azerbaijan Land of Fire  :ninja:

[quote="Pirkuliyev Rovsen"]Do there exist any increasing or decreasing functions $ f: R \rightarrow R$such that $ f(f(x)) \equal{} \{\ x \}$ for all $ x \in [0;1]$[/quote]

No : $ f(x)$ strictly monotonous implies $ f(f(x))$ increasing. But this is impossible since $ f(f(\frac 12))\equal{}\frac 12$ while $ f(f(1))\equal{}0$

[quote="prester"][quote="pco"][quote="Pirkuliyev Rovsen"]Do there exist any increasing or decreasing functions $ f: R \rightarrow R$such that $ f(f(x)) \equal{} \{\ x \}$ for all $ x \in [0;1]$[/quote]

No : $ f(x)$ strictly monotonous implies $ f(f(x))$ increasing. But this is impossible since $ f(f(\frac 12)) \equal{} \frac 12$ while $ f(f(1)) \equal{} 0$[/quote]

Is it necessary that $ f(x)$ is [b]strictly[/b] monotonous to prove the assert? We can prove that do not exist any function that is not decreasing or not increasing that satisfies above condition. Am I right?[/quote]

I used strictly because this was the original statement ("Do there exist any increasing or decreasing functions ...") but obviously "strictly" is not necessary to prove there is no such function :)

[quote="Differ"]To correct a few conceptions:

1. The word is monotonic.

2. Decreasing does not mean strictly decreasing. See http://mathworld.wolfram.com/DecreasingFunction.html for a reliable source.[/quote]

1. Ok, thanks

2. In french "decreasing" does not mean strictly decreasing. I thought that in english "decreasing" was used for strictly decreasing and "non increasing" in other cases.
Thanks for your corrections.


Since $ab+bc+ac=1$ and $\sum_{cyc}\tan\frac{A}{2}\tan\frac{B}{2}=1\ \forall\triangle ABC$, make the substitutions $(a,b,c)\mapsto\left(\tan\frac{A}{2},\tan\frac{B}{2},\tan\frac{C}{2}\right),$ where $A,B,C$ are the measures of the angles of $\triangle ABC$.  Then the inequality is equivalent to proving that
\[\sum_{cyc}\cot\left(\frac{\pi-A}{2}\right)\geq \sqrt{3}. \]

[quote="boxedexe"]Since $ab+bc+ac=1$ and $\sum_{cyc}\tan\frac{A}{2}\tan\frac{B}{2}=1\ \forall\triangle ABC$
[/quote]
can uyou or any body else proove it :maybe:

[quote="arqady"]Let $a,$ $b$ and $c$ are positive numbers such that $ab+ac+bc=1.$ Prove that:
$\frac 1{a+b}+\frac 1{b+c}+\frac 1{c+a}\geq \sqrt{3}+\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}.$[/quote]

This is very easy!

${\sum_{cyclic}\frac{1-ab}{a+b}\geq \sqrt r3}$

$\Longleftrightarrow \sum_{cyclic}\frac{c(a+b)}{a+b}\geq \sqrt{3}$

$\Longleftrightarrow a+b+c\geq \sqrt{3}$

$\Longleftrightarrow (a+b+c)^{2}\geq 3(ab+bc+ca)$

$\Longleftrightarrow a^{2}+b^{2}+c^{2}\geq ab+bc+ca$. we are done.

[quote="otman4u"][quote="otman4u"][quote="boxedexe"]Since $ab+bc+ac=1$ and $\sum_{cyc}\tan\frac{A}{2}\tan\frac{B}{2}=1\ \forall\triangle ABC$
[/quote]
can you or any body else proove it :maybe:[/quote]
?[/quote]
steel waiting?

$\tan{\frac{C}{2}}= \tan{\left(\frac{\pi}{2}-\left(\frac{A}{2}+\frac{B}{2}\right)\right)}= \frac{1}{\tan{\left(\frac{A}{2}+\frac{B}{2}\right)}}= \frac{1-\tan{\frac{A}{2}}\tan{\frac{B}{2}}}{\tan{\frac{A}{2}}+\tan{\frac{B}{2}}}$.

Plug that in and you'll find that the identity is indeed true.

The initial inequality, was a problem at the Romanian olympiad, grade IX, 2002.


If: $2sin^{2}\beta cos^{2}\alpha+sin^{2}\alpha sin^{2}\beta-2sin2\alpha sin2\beta \geq 2$

Find all the values of: $cos\alpha+sec2\beta$


geometry
perpendicular bisector
geometry proposed

A new variation In Construction  :D :

Problem: Let $A$ and $B$ be two points and let $\l$ be the lenght of segment $AB$.

Let $R$ be a ruler with lengh $d < l$ .

Construct the Segment $AB$ with

1. a compass and $R$.

2. just $R$ (Its Short ha ? :D )

(1) With a compass and $R$

1) We draw the points $A,B$ (easy, using the compass)

2) We draw the perpendicular bisector $m$ of $AB$ (we can choose a radius for the compass such that the two points $C,D$ which define the line $m$ are too close)

3) We find the midpoint $M$ of $AB$ (we will bring the perpendicular bisector $n$ of $CD$ and then $M \in m \cap n$)

4) If $AM\leq d$, then we connect the points $A,M,B$
If not, then we repeat the steps 1,2,3 for the segment $AM$
Of course the ruler has no zero length, so the procedure will stop after a finite number of steps.

For the second (only the ruler) I don't have an idea!


Following inequality called ''T2's lemma''
$ \sum_{i\equal{}1}^{n}\frac{a_{i}^{2}}{x_{i}}\geq\frac{(\sum_{i\equal{}1}^{n}a_{i})^{2}}{\sum_{i\equal{}1}^{n}x_{i}}\forall a_{i}\in\mathbb{R},\forall x_{i}>0$.

I want know: Why its name is T2's lemma?  :blush:

It's an equivalent form of Cauchy-Schwartz inequality.I know this form under the name "Titu Andreescu's ineq".Where have you found this name ? (T2) .Maybe "T" stands for Titu :).Or maybe if you found this in an informal paper ,the following explanation might be true: "T2" and "Titu" are pronounced the same

Thanks! But i don't think that is Titu Andreescu's ineq because it is well-known and old.  :)

It's called Titu's Lemma because he's pretty fond of using it. He's done an article on this lemma in his book [i]Mathematical Olympiad Treasures[/i] - although he's called it Cauchy's Lemma

Oh, i see! It is ''Cauchy-Schwartz revisited'' p. 7. in that one. Thanks!  :wink:

Well, the name under which I've heard of this inequality (in a lecture) was T-Lemma, and I think that it is derived from Titu's name... It's funny that I was also interested in finding out how this lemma got it's name, few months ago  :)
[/hide]

And actually it has been posted here.
http://www.mathlinks.ro/viewtopic.php?t=15125. :wink:


geometry
rectangle
parallelogram
analytic geometry
graphing lines
slope
trigonometry

We are given a rectangle   $ABCD$  with $DA > AB$ . On side $DA$ we take segment $DE = AB$  and on side  $BA$  segment $BZ = AE$ . If  $BE$ meets $DZ$  at  $K$ , prove that $CK$  is perpendicular to $BE$  .

[u]Babis[/u]

 [i]I put the problem here , because I ask a pure synthetic geometrical solution. [/i]

[b]Lemma:[/b] In triangle $ABD$ (I am using your notation) angle between lines $BE$ and $DZ$ iz $45^o$.

[b]Proof:[/b] Let $a=AB$, $b=DA$, so $EA=b-a$ and $AZ=2a-b$. By Pithagors theorem in $\triangle AZD$ $ZD^2=(2a-b)^2+b^2=2b^2+4a^2-4ab$ simmilari in $\triangle ABE$ $BE^2=(b-a)^2+a^2=b^2+2a^2-2ab$ so as we can se $ZD^2=2BE^2$ and by that triangle with sides length $ZD, BE, BE$ is isosceles and rightangled.

Let's denote point $F$ on $BC$ s.t. $BF=a$. Quadriliteral $EBFD$ is now parallelogram and $DF=BE$. Triangle $BZF$ is congruent with traingle $ABE$ ($AE=BZ$, $BF=a$) so $FZ=BE$. Now look at triangle $ZFD$ his sides lengths are $DZ, BE, BE$ so it's isosceles and rightangled, which implies $\angle FDZ=45^o$ and from that $\angle (DZ, BE)=45^o$ $Q.E.D$

When we have this Lemma we easeily see that quadrilitereal $EKCD$ is cyclyc because angle $ECD$ is $45^o$ too ($DC=DE=a$), so angle $EKC$ is $90^o$. End of proof.

m@re

Thank you M@re! Nice solution ! I think ,too, that the solution must be metric , but I' m not sure!

  babis

Nice solution ! I am a syntetical solution for the case $b>2a$ and the case $a<b\le 2a$. There are and the metrical, trigonometrical or analitical solutions, but here isn't the case.

Dear M@re and Ph -An ! I studied the soluiton by M@re and I see that we can simplify it.

    We take the point $F$ as in the solution. But then triangles $BZF$ and $CFD$ are equal so
  $FZ = FD$  and  $FZ$  is perpendicular to $FD$ ( the last is very easy with angles , since
 $\angle BFZ + \angle CFD = \angle CDF + \angle CFD = 90$ ) .The rest is as in the solution.
 In this way [b]we do not need the pythagorian calcultions , although the inspiration to make them is fantastic ! [/b] !!!

[b] Note[/b]
  As Ph-An said , we can find solutions with other ways , not synthetic. I found a very good solution with analytical geometry , using  the slopes of $DZ$ and $BE$ .Any way , the problem is now well solved and I thank both of you.

[u] Babis[/u]

Let $ABCD,\ AD>AB$ be a rectangle. If the point $E$ belongs to the [b]ray[/b] $[DA$ such that $DE=AB$, the point $Z$ belongs to the [b]ray[/b] $[BA$ such that $BZ=AE$ and $K\in BE\cap DZ$, then $CK\perp BE$ (i.e. the conclusion of proposed problem remains actuallly !).

Hallo!
 Dear Ph - An , thanks for the generalization.

 Does the same argument work again? I hope so! I' ll see it later. I must make a new figure.

   Babis

[quote="stergiu"]
 We take the point $F$ as in the solution. But then triangles $BZF$ and $CFD$ are equal so
  $FZ = FD$  and  $FZ$  is perpendicular to $FD$ ( the last is very easy with angles , since
 $\angle BFZ + \angle CFD = \angle CDF + \angle CFD = 90$ ) .The rest is as in the solution.
 In this way [b]we do not need the pythagorian calcultions , although the inspiration to make them is fantastic ! [/b] !!![/quote]

I saw that but I was to lasy to connect to the net again and edit it :)

[b]P.S.[/b]This problem but only the Lemma was on this year $JBMO$ team selection in my country, and in this year they decided to call 7th contestans which they never had before. Only them, which are about year or two years younger of all the other childrean solved that problem, so I asked how can that b..? Then I remember that Pithagors theorem is learnig for the first time in 7th grade, and all the others must have forgot it and that's why they didn't solve this problem so that's why I used Pithagors when I was solving this problem some time in June few days after that contest... :D

Well,
the most difficult part in a problem is to find the first solution. Then it is easier to improve the solution or to find another , more interesting one ! To this problem I made only  note. The solution  belongs to you and this one with Pythagoras is the best - I like it -  , because only this observation leads as to the solution(.... take  $... F...$).
   [u]Babis[/u]

Apply Menelaos for triangle $ \triangle ABE$ with the transverse line $ ZD$, call $ K\equal{}BE\cap ZD$, and get $ \frac{KB}{KE}\equal{}\frac{b^2\minus{}ab}{2a^2\minus{}ab}$, or $ \frac{KB}{BE}\equal{}\frac{b^2\minus{}ab}{2a^2\plus{}b^2\minus{}2ab} \ ( \ 1 \ )$, but $ BE^2\equal{}2a^2\plus{}b^2\minus{}2ab$, so (1) becomes $ KB \cdot BE \equal{} b^2\minus{}ab \ ( \ 2 \ )$.
Now, take $ F$ on $ BC$ so that $ CDEF$ be a square. As $ BF \equal{} AE \equal{} b\minus{}a$ and $ BC\equal{}b$, the relation (2) may be written as: $ BK\cdot BE \equal{} BF\cdot BC$, i.e. $ K$ belongs to the circle $ (CDEF)$, hence $ CK\perp BE$, q.e.d.

Best regards,
sunken rock

Here is my solution (not synthetic):
We have $ \angle EKD\equal{}\angle AZD\plus{}\angle AEB\minus{}90$. Thus,
$ \tan \angle EKD\equal{} \minus{}\frac{1}{\tan \angle AZD\plus{}\angle AEB}\equal{}\frac{\frac{AE}{AB} \cdot \frac{AB\minus{}AE}{AB\plus{}AE}\minus{}1}{\frac{AE}{AB}\plus{} \frac{AB\minus{}AE}{AB\plus{}AE}}\equal{}1$. Thus, $ \angle EKD\equal{}45\equal{}\angle ECD$, so $ EKCD$ is cyclic, so $ \angle EKC\equal{}90$.
But actually the expression seems to come out to $ \minus{}1$ instead of $ 1$. Can someone show me my mistake? It seems like it should work out but I can't find the mistake...

We draw the lines through the points $ E,\ Z$ and parallel to $ AB,\ AD$ respectively and we denote as $ H$, their point of intersection.

The points $ K,\ H,\ C,$ are collinear as well known ( see the [b]lemma 2[/b] in the post #7# [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=107997][b][color=Blue]Here[/color][/b][/url] ) and let be the points $ F\equiv BC\cap EH$ and $ L\equiv EC\cap BH.$

From the squares $ CDEF,\ BZHF$ we have $ \angle BHF \equal{} \angle EHL \equal{} \angle HEL \equal{} 45^{o}$ $ \Longrightarrow$ $ \angle EHL \plus{} \angle HEL \equal{} 90^{o}$ and so, we have that $ BL\perp CE.$

Because of now $ EF\perp BC,$ we conclude that $ H$ is the orthocenter of the triangle $ \bigtriangleup BCE.$

Hence, we conclude that $ CK\perp BE$ and the proof is completed.

Kostas Vittas.

PS. It is easy to show that the point $ L\equiv EC\cap BH$ is the midpoint of the segment $ CZ$ ( but, it is not necessary for the proof ).

Dear Kostas and Mathlinkers,
this is only a shape of proof... I have in mind.
1. EDCF and BZHF are two squares.
2. The diagonal DF can be seen as the tangent Tf to the circumcircle 1 of BZHF at F
3. Let K' the second intersection of 1 with ZD
4. According to Pascal's theorem applied to Tf BZK'HF, the points    K', H, C are collinear.
5. According to Pascal's theorem applied to Tf BK'ZHF,  the points   B, K', E are collinear.
in consequence, K'=K.
This proof as to be worked...
Sincerely
Jean-Louis

Dear Mathlinkers,
an article concerning "Geometric miniatures on a square" with original proofs has been put on my website
http://perso.orange.fr/jl.ayme    vol. 7 Miniatures géométriques sur un carré p. 7
Sincerely
Jean-Louis


inequalities
integration
trigonometry
calculus
calculus computations

Show that $ \int^{\pi/2}_{\pi/6}\frac{2x}{sinx}dx\ge\frac{2}{9}\pi^2$

$ \frac {\pi}{6}\leq x\leq \frac {\pi}{2}\Longleftrightarrow \sin x\leq \frac {3}{\pi}x.$

$ \therefore \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{2x}{\sin x}\ dx\geq \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{2}{3}\pi \ dx\equal{}\frac{2}{9}{\pi}^2$

O.K. but I will post the detail solution later.

I think the integrand is greater than or equal to 2. So you can use the 'integral of the inequality'


probability
LaTeX
pigeonhole principle
probability and stats

Hey guys!, I have 8 difficult probability questions, the hardest I could find at this level, and I wanna see who can solve them with the best explanations! I actually do not even know some of them!

Here is #1:

A round robin fencing tournament with n contestants takes place; each of the (n choose 2) pairs of fencers fence each other at once. Let 1 < k < n. Show that there is a possible outcome of the tournament so that at least one subset of fencers of size k has the property that they can be lined up in at least
k!/2k-1 ways with each person having beaten the person behind him in line, and explain.

Let me show my work so far on all of these:

ok for this one it is obvious that arranging k people can be done in k! ways, however, not all of these ways ensure the desired outcome. We need to divide out the permutations which do not give the order such that the person in front of another person has beaten that other person. Any idea where the 2^(k-1) comes from?

General advice: don't dump lots of problems at once. That reduces your chances of having any of them responded to. Space them out over time, and wait for the first few to be answered before adding more.

More specific advice: "k!/2k-1" is a ridiculously ambiguous bit of notation. Use $ \text{\LaTeX}$, or at least parentheses and markup like ^ for exponents and _ for subscripts. From your second post, it appears that you intended $ \frac{k!}{2^{k-1}}$ in this case, and there's no way I'd read that without context.

The solution here is very simple: if we choose a sequence of $ k$ fencers and choose the results randomly by flipping coins, that sequence is a chain with each beating the next with probability $ 2^{-k+1}$. Sum over all $ k!$ orders of some $ k$ and the average is $ \frac{k!}{2^{k-1}}$.
Some possibility is at least equal to the average by the pigeonhole principle.

Ok, I am very sorry I had all of this typed up neatly and when I transferred I forgot to edit it. Thank you for your adivce though. Ok, so where did your resultant probability come from (by that I mean 2^(-k+1).) Where do you go to use LATEX?[/code]

$ \text{\LaTeX}$ math mode is enabled on the forum; just put code between dollar signs. You can quote other people's posts to see how it works, or read some of the topics in [url=http://www.artofproblemsolving.com/Forum/index.php?f=123]this subforum[/url].

That probability? We've picked a sequence. There are $ k - 1$ relevant matches. If they all have the right result, it's one of the chains we're interested in.

Ah I see I read it wrong. So, when multiplying by $ k!$ we are in essence taking a single chain, and reocgnizing that there are k! possibilities of this sequence of events happening.


[quote="fleeting_guest"]I don't have a long memory of "before" and "after" on this
board, but: adding more features tends to make a site slower
to load.   Middle schoolers on a home computer may not have
the patience if it takes too long to pull up the threads.[/quote]Loading times are presented in the bottom of each page. For example my loading time when replying is
[ Time: 0.2608s ][ Queries: 12 (0.0195s) ][ GZIP enabled - Debug on ]
The rest is determined by one's internet connection, which we cannot help too much (almost all the graphics is cached, so it's only downloaded once, after that only new data is downloaded - such as posts' text - which adding/removing features does not interfere with).

About the loading speed  (and I will be happy to delete
and repost this elsewhere if off-topic):

I access the AoPS site from several university and other
locations with high-speed connections.  My loading times
are on the order of 0.8sec to several seconds (for the
Forums front page).  The slowdown, and sometimes
out-of-memory and other error messages, seems to
come from the graphics, especially when using Macintosh computers. Also, when going to the Forums front page,
on any of the Mac/PC/Unix systems, first the graphics
loads and then there is a lot of shrinking or expanding
as the site and/or the browser decide which of the
board titles to expand and which to abbreviate.  The
latter step can take quite a long time, during which
the view on the screen changes every half-second.

Maybe I am unlucky and not a representative sample,
but note that this is with modern equipment, fast internet, and
professional (sysadmin) maintenance of the computers,
i.e. probably above-average compared to middle-school
home computer users.   Other sites tend not to
load this slowly.  Do you have statistics on the
distribution of loading times for all users?

The loading times are not user-dependant. They are the times in which our own server loads the information and buildup the page which then serves it to your client. Basically these should not be affected by neither a slow internet connection and / or poor machine settings (though they might be in some detail, because you are sending data back and forth with the server which then is processed, which might make the PGT a little slower).

Can you please post the PGT (page generation times) from the Index page, a forum page (only with topics listing) and a viewtopic page?
Also it seems that you have another problem (an unique one) regarding posting format.

I'm kind of having the same problem as fleeting_guest with the page loading times.  The PGT says [ Time: 0.5898s ][ Queries: 10 (0.1077s) ][ GZIP enabled - Debug on ], but I think that is posted before the Advanced and College forums collapse.  And this is from a Mac too.

Are you guys using the latest Safari / latest Mac OS version?

Also please try what I am suggesting in [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=150254#p150254]the first paragraph of this post[/url] and tell me if this makes the site load faster.

I use firefox on the Mac.

And I tried your suggestion, it loads a lot faster.

Well then, stick with that mode, or change firefox (as I've said numerous times, Ff is far from being a really good quality browser, and it takes [b]too much[/b] memory -> which probably causes slow loading times on certain clients). I recommend you use Safari (if you're on a Mac) or Opera. :)


One statue is 6 times the height of another statue. 60 bars of gold are used in building the first statue. How many are used in the second statue?

This problem is not well defined.
First, it does not say whether the first statue is smaller or bigger. I will assume that the first statue is smaller.
If the height is only thing different(area of the base stays same), we need 60*6=360 bars of gold. But if the statues are similar to each other, we need $60*6^3=12960$ bars of gold.
-------------------------------------------------------------------
EDITED: Sorry, didn't notice the title. 360 is wrong.


[color=purple]Let triangle ABC denote its element asual. Prove that :
      $ l_{a} \plus{} l_{b} \plus{} l_{c}\ge\frac {2r}{R}(4R \plus{} r)\ge\frac {2r}{R}\frac {\sqrt {3}}{2}(a \plus{} b \plus{} c)$
  We have the similar with the 2nd :$ l_{a} \plus{} l_{b} \plus{} l_{c}\ge \sqrt {\frac {2r}{R}}\frac {\sqrt {3}}{2}(a \plus{} b \plus{} c)$
The first is mine if you are interested in this problem try with this one :$ m_{a} \plus{} m_{b} \plus{} m_{c}\ge l_{a} \plus{} l_{b} \plus{} l_{c}\ge3\sqrt {r(4R \plus{} r)}$
[/color]

[color=purple] Here is the gemeralization of one of these problems above :$ xl_{a}\plus{}yl_{b}\plus{}zl_{c}\ge\frac{2r}{R}\sqrt{xy\plus{}yz\plus{}zx}s$
  [/color]


calculus
linear algebra
matrix
induction
ratio
algebra
polynomial

Let $A \in \mathcal M_k \left( \mathbb R \right)$ and $n \in \mathbb{N}_0$ s.t. $A^{n+1} = O_k$. Prove that \[ \det \left( I_k + \sum_{i=1}^n \frac{1}{i!} A^i \right) = 1 . \]

Note that $\left( I_k + \sum_{i=1}^n \frac{1}{i!} A^i \right)=\left( I_k + \sum_{i=1}^{\infty} \frac{1}{i!} A^i \right)=e^A$ since $A^{n+1}=0.$

The desired result is a special case of a much more general theorem: $\det(e^A)= e^{\text{Tr}A}.$

Note that because all of the eigenvalues of a nilpotent matrix must be zero, then the trace of that matrix must be zero.

To prove the main theorem: use Schur's Lemma. We can find a unitary matrix $U$ such that $A=UTU^*$ where $T$ is upper triangular. Without too much trouble we can show that $e^A=Ue^TU^*$ and thus $\det(e^A)=\det(e^T).$

So we're down to showing that for an upper triangular $T,\,\det(e^T)=e^{\text{Tr}T}.$

I'll let you finish that for yourself.

For the first part we use the fact that $A^n = U T^n U^H$. (induction over $n$)

For the second, $e^T$ has on its diagonal the elements $e^{a_{ii}}$, so $\det e^T = e^{\textrm{tr} A}$.

This is a very nice theorem.

But I don't understand a thing: how do we know that $e^A$ exists?

[quote="perfect_radio"]... how do we know that $e^A$ exists?[/quote]
Take any kind of crude norm of $A$ - the sum of the absolute values of all the elements would do. You can then show by induction that each element of $A^i$ is no larger than some constant (which depends on $k$) times $\|A\|^i.$ In other words, the sizes of the elements of $A^i$ can grow at most geometrically fast as $i\to\infty.$ Then by the same argument that shows that the power series for the ordinary exponential has an infiinte radius of convergence, we can show that the series for $e^A$ converges for all complex matrices $A.$

I don't understand how we can conclude that the series convergences.   :blush:

The only way I know to prove the ordinary case is ratio test :(

The solution in the book goes like this:

[quote="Book Solution + A few details from me"]Let $\displaystyle f : \mathbb R \to \mathbb R$, $\displaystyle f \left( x \right) = \det \left( I_k + \frac{x}{1!} A + \frac{x^2}{2!} A^2 + \ldots + \frac{x^n}{n!} A^n \right)$ be a polynomial of degree at most $n$. Let's compute $\displaystyle f(x) f(y)$:

\begin{eqnarray*} f(x) f(y) &=& \det \left( I_k + \frac{x}{1!} A + \ldots + \frac{x^n}{n!} A^n \right) \cdot \det \left( I_k + \frac{y}{1!} A + \ldots + \frac{y^n}{n!} A^n \right) \\ \ &=& \det \left( I_k + \sum_{j=1}^{2n} \left( \sum_{l = \max \left( 0,j-n \right)}^{\min \left( j,n \right)} \frac{x^l y^{j-l}}{l! (j-l)!} \right) A^j \right) \end{eqnarray*}

Since $\displaystyle A^t = 0$ for all naturals $t \geq n+1$, the formula becomes
\begin{eqnarray*} f(x) f(y) &=& \det \left( I_k + \sum_{j=1}^{n} \left( \sum_{l = 0}^j \frac{x^l y^{j-l}}{l! (j-l)!} \right) A^j \right) \\ \ &=& \det \left( I_k + \sum_{j=1}^{n} \frac{(x+y)^j}{j!} A^j \right) \\ \ &=& f(x+y) , \end{eqnarray*}
for all $\displaystyle x,y \in \mathbb R$.

We shall prove that $f$ is a constant polynomial.

For $\displaystyle \left( x,y \right) = \left( x,-x \right)$ we get $\displaystyle f(x) \cdot f(-x) = f(0)$. Thus, we have \[ \displaystyle \lim_{x \to \infty} \left( f(x) \cdot f(-x) \right) = f(0) .  \]

Let's suppose that $f$ is not a constant polynomial. Then
\[ \displaystyle \lim_{x \to \infty} \left( f(x) \cdot f(-x) \right) = \left( \lim_{x \to \infty} f(x) \right) \cdot \left( \lim_{x \to \infty} f(-x) \right) = \left( \pm \infty \right) \cdot \left( \pm \infty \right) = \pm \infty ,  \]
which contradicts what we have obtained earlier.

Hence, $f$ is constant and $\displaystyle f(1) = f(0) = \det \left( I_k \right) = 1$.

On the other hand, $f(1)$ is the determinant we wanted to compute.[/quote]

Any idea how they came up with that trick?

Also, the question of $e^A$ remains...

You don't need such arguments: since $A$ is nilpotent, any polynomial in $A$ which is zero at origin is also nilpotent matrix (obvious) and therefore you need to prove that if $X$ is nilpotent, $I+X$ has determinant 1, which is obvious using eigenvalues.

[quote="perfect_radio"]For the first part we use the fact that $A^n = U T^n U^H$. (induction over $n$)

For the second, $e^T$ has on its diagonal the elements $e^{a_{ii}}$, so $\det e^T = e^{\textrm{tr} A}$.

This is a very nice theorem.

But I don't understand a thing: [/quote]how do we know that $e^A$ exists?[/quote]
Mn(IR) is a Banach space (dim<oo)

I still don't get it... sorry  :(

to harazi: Nice solution.


Given a cube $ABCD-A_{1}B_{1}C_{1}D_{1}$ with side length $a$, find the distance between $AD_{1}$ and $BA_{1}$.

[quote="vishalarul"]What do you mean by your notation $ABCD-A_{1}B_{1}C_{1}D_{1}$?

Does this mean that $A_{1}$ is below $A$, $B_{1}$ is below $B$, etc...?

If so, here's my answer: [hide]$\frac{a\sqrt{2}}{2}$[/hide]
I'll post a solution if I'm right.[/quote]

i dont think this is the answer... cuz the distance is the length of the line segment that is perpendicular to both $AD_{1}$ and $BA_{1}$

[quote="lotrgreengrapes7926"][quote="vishalarul"]If so, here's my answer: [hide]$\frac{a\sqrt{2}}{2}$[/hide]
I'll post a solution if I'm right.[/quote]
I don't think it's that easy. Of course, I could be wrong.
The shortest distance is not half the diagonal of face $ADD_{1}A_{1}$.[/quote]

I am sure it is not the half diagonal of face $ADD_{1}A_{1}$, because that line is only perpendicular to line segment $AD_{1}$, but it is not perpendicular to line $A_{1}B$

[quote="Nerdy_girl_number1"]The distant can be seen as the distance between plate ACD1 and A1BC1,which is one third of line segment B1D.[/quote]

Can you give a proof for that? thanks


geometry
ratio
inequalities
triangle inequality
area of a triangle
geometry solved

IN triangle ABC,K,L,M are variable points on BC,CA,AB and AK,BL and CM meet at P.
Find the position of P when

PK/AP + PL/BP +PM/CP has its least value

It's enough to use $\frac{PK}{AP}=\frac{S_{PBC}}{S_{PCA}+S_{PAB}}$ and the like, where $S_{XYZ}$ is the area of $XYZ$. It's now clear that the minimum is reached when $P$ is the centroid.

[I see the solution was now given by Grobber, but since my solution below solves a slightly extended version of the problem, I post it here.]

I'm pretty busy now but this problem is too nice...

It's not clear to me whether "K,L,M are variable points on BC,CA,AB" means that the points K, L, M must lie on the [i]segments[/i] BC, CA, AB, or they can lie on the [i]extensions[/i] of the segments BC, CA, AB, too. Anyway, I prefer the latter interpretation, since it spices the problem. Actually I will prove the following version of the problem:

[i]Let ABC be a triangle, and K, L, M arbitrary points on the lines BC, CA, AB, respectively, such that the lines AK, BL, CM meet at one point P. Prove that

$\frac{\left| PK\right| }{\left| AP\right| }+\frac{\left| PL\right| }{\left| BP\right| }+\frac{\left| PM\right| }{\left| CP\right| }\geq \frac{3}{2}$,

and equality holds if and only if the point P is the centroid of triangle ABC.[/i]

Hereby, |XY| denotes the [i]non-directed[/i] distance between two points X and Y, whereas XY denotes the [i]directed[/i] distance (we will use directed distances on the lines AK, BL, CM; how exactly we orientate these lines is not important, since we will only consider [i]ratios[/i] of directed distances, but it is important that the orientation is fixed during the solution of the problem).

Here is my [i]solution[/i] of the above problem. At first, I consider the case when the point P lies inside triangle ABC. We will denote by [XYZ] the [i]directed[/i] area of an arbitrary triangle XYZ, and we can WLOG assume that the directed area [ABC] of triangle ABC is positive (i. e., the triangle ABC is directed counterclockwisely). Let x = [PBC], y = [PCA], z = [PAB]. Since [ABC] is positive, and since the point P lies inside triangle ABC, the areas x = [PBC], y = [PCA], z = [PAB] are positive, too.

Now, the area of a triangle equals $\frac12$ $\cdot$ side $\cdot$ corresponding altitude. The triangles PBK and ABP have their B-altitudes equal; hence, the ratio of their areas equals the ratio of their corresponding sides:

$\frac{\left[ PBK\right] }{\left[ ABP\right] }=\frac{PK}{AP}$.

Similarly,

$\frac{\left[ PCK\right] }{\left[ ACP\right] }=\frac{PK}{AP}$.

By the well-known algebraic rule $\left\{ \frac{a}{b}=\frac{x}{y}\wedge \frac{c}{d}=\frac{x}{y}\right\} \Longrightarrow \left\{ \frac{a-c}{b-d}=\frac{x}{y}\right\} $, this implies

$\frac{\left[ PBK\right] -\left[ PCK\right] }{\left[ ABP\right] -\left[ ACP\right] }=\frac{PK}{AP}$,

so that

$\frac{PK}{AP}=\frac{\left[ PBK\right] -\left[ PCK\right] }{\left[ ABP\right] -\left[ ACP\right] }=\frac{\left[ PBK\right] +\left[ PKC\right] }{\left[ ABP\right] +\left[ PCA\right] }=\frac{\left[ PBC\right] }{\left[ PAB\right] +\left[ PCA\right] }=\frac{x}{z+y}=\frac{x}{y+z}$.

Now, since the point P lies inside the triangle ABC, the point P must lie in the interior of the segment AK. Hence, the directed ratio $\frac{PK}{AP}$ is positive, and thus we have $\frac{PK}{AP}=\left| \frac{PK}{AP}\right| =\frac{\left| PK\right| }{\left| AP\right| }$. In other words, we have proved

$\frac{\left| PK\right| }{\left| AP\right| }=\frac{x}{y+z}$.

Similarly, $\frac{\left| PL\right| }{\left| BP\right| }=\frac{y}{z+x}$ and $\frac{\left| PM\right| }{\left| CP\right| }=\frac{z}{x+y}$. This yields

$\frac{\left| PK\right| }{\left| AP\right| }+\frac{\left| PL\right| }{\left| BP\right| }+\frac{\left| PM\right| }{\left| CP\right| }=\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}$.

But by the Nesbitt inequality (which can be applied here since the numbers x, y, z are positive), we have $\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\geq \frac{3}{2}$, with equality if and only if x = y = z. Thus, we have proved that

$\frac{\left| PK\right| }{\left| AP\right| }+\frac{\left| PL\right| }{\left| BP\right| }+\frac{\left| PM\right| }{\left| CP\right| }\geq \frac{3}{2}$,

with equality if and only if x = y = z. Of course, x = y = z is equivalent to [PBC] = [PCA] = [PAB], what is a characteristic property of the centroid of triangle ABC. In other words, we have proved that for every point P lying inside triangle ABC, the inequality $\frac{\left| PK\right| }{\left| AP\right| }+\frac{\left| PL\right| }{\left| BP\right| }+\frac{\left| PM\right| }{\left| CP\right| }\geq \frac{3}{2}$ holds, and equality occurs if and only if the point P is the centroid of triangle ABC.

Now it remains to handle the case when the point P lies outside the triangle ABC. In fact, I will show how to modify the above argument so that it works for an arbitrary point P in the plane of triangle ABC, no matter whether it lies inside or outside of triangle ABC (or on a side).

At first, note that for an arbitrary point P in the plane of triangle ABC, the directed areas x = [PBC], y = [PCA], z = [PAB] are not necessarily all positive.

By the same argument as above, we have

$\frac{PK}{AP}=\frac{x}{y+z}$,

but now, since the sign of the ratio $\frac{PK}{AP}$ can be positive and can be negative as well, we can only conclude

$\frac{\left| PK\right| }{\left| AP\right| }=\left| \frac{PK}{AP}\right| =\left| \frac{x}{y+z}\right| =\frac{\left| x\right| }{\left| y+z\right| }$.

Now, from the trivial inequality $\left| y+z\right| \geq 0$ and the triangle inequality $\left| y+z\right| \leq \left| y\right| +\left| z\right| $, we have $\frac{\left| PK\right| }{\left| AP\right| }=\frac{\left| x\right| }{\left| y+z\right| }\geq \frac{\left| x\right| }{\left| y\right| +\left| z\right| }$. Similarly, $\frac{\left| PL\right| }{\left| BP\right| }\geq \frac{\left| y\right| }{\left| z\right| +\left| x\right| }$ and $\frac{\left| PM\right| }{\left| CP\right| }\geq \frac{\left| z\right| }{\left| x\right| +\left| y\right| }$. Hence, we get the inequality

$\frac{\left| PK\right| }{\left| AP\right| }+\frac{\left| PL\right| }{\left| BP\right| }+\frac{\left| PM\right| }{\left| CP\right| }\geq \frac{\left| x\right| }{\left| y\right| +\left| z\right| }+\frac{\left| y\right| }{\left| z\right| +\left| x\right| }+\frac{\left| z\right| }{\left| x\right| +\left| y\right| }$.

And now, since the numbers |x|, |y|, |z| are positive, we can apply Nesbitt again and get

$\frac{\left| x\right| }{\left| y\right| +\left| z\right| }+\frac{\left| y\right| }{\left| z\right| +\left| x\right| }+\frac{\left| z\right| }{\left| x\right| +\left| y\right| }\geq \frac{3}{2}$.

Hence,

$\frac{\left| PK\right| }{\left| AP\right| }+\frac{\left| PL\right| }{\left| BP\right| }+\frac{\left| PM\right| }{\left| CP\right| }\geq \frac{3}{2}$.

In order to see when equality holds in this inequality, we remember which other inequalities we have used in the proof. In fact, we have used the triangle inequalities $\left| y+z\right| \leq \left| y\right| +\left| z\right| $, $\left| z+x\right| \leq \left| z\right| +\left| x\right| $, $\left| x+y\right| \leq \left| x\right| +\left| y\right| $, and we have used the Nesbitt inequality $\frac{\left| x\right| }{\left| y\right| +\left| z\right| }+\frac{\left| y\right| }{\left| z\right| +\left| x\right| }+\frac{\left| z\right| }{\left| x\right| +\left| y\right| }\geq \frac{3}{2}$. In the triangle inequality, $\left| y+z\right| \leq \left| y\right| +\left| z\right| $, equality holds if and only the numbers y and z have the same sign; in the triangle inequality $\left| z+x\right| \leq \left| z\right| +\left| x\right| $, equality holds if and only if the numbers z and x have the same sign; in the triangle inequality $\left| x+y\right| \leq \left| x\right| +\left| y\right| $, equality holds if and only if the numbers x and y have the same sign; finally, in the Nesbitt inequality $\frac{\left| x\right| }{\left| y\right| +\left| z\right| }+\frac{\left| y\right| }{\left| z\right| +\left| x\right| }+\frac{\left| z\right| }{\left| x\right| +\left| y\right| }\geq \frac{3}{2}$, equality holds if and only if |x| = |y| = |z|. Hence, in the final inequality $\frac{\left| PK\right| }{\left| AP\right| }+\frac{\left| PL\right| }{\left| BP\right| }+\frac{\left| PM\right| }{\left| CP\right| }\geq \frac{3}{2}$, equality holds if and only if all three numbers x, y, z have the same sign and satisfy |x| = |y| = |z|. This is equivalent to x = y = z, what again holds if and only if the point P is the centroid of triangle ABC. Proof complete.

  Darij


[quote="Zamfirmihai"]it`s wrong. try $a=b=c$ or $b,c\to 1$ and $a\to 0$[/quote]
tahts right, sry post edited :blush:
actually i was making this inequality (that you said was wrong) from the one that you see now, but seems i made some Big mistakes...

[quote="sinajackson"]EDIT:
For positive reals $a,b,c$ , prove that:

$\frac{a^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}+\frac{c^{2}}{a^{2}}\geq\frac{b}{a}+\frac{c}{b}+\frac{a}{c}$[/quote]

Denote $x=\frac{a}{b}$, $y=\frac{b}{c}$, $z=\frac{c}{a}$. Now we must prove that $\sum_{cyc}x^{2}\geq \sum_{cyc}\frac{1}{x}$, with $xyz=1$.
But $\sum_{cyc}x^{2}\geq \sum_{cyc}xy=xyz\sum_{cyc}\frac{1}{x}=\sum_{cyc}\frac{1}{x}$


logarithms
limit
function
calculus
derivative
calculus computations

if
$ f(x)\equal{}\frac{\ln{x}}{1\plus{}(\ln{x})^2}$
calculate:
a)$ \lim_{x\rightarrow0^\plus{}}f(x)$
$ \lim_{x\rightarrow\infty}f(x)$
b)maximum and minimum value of $ f(x)$

$ |\ln x|\to\infty$ as either $ x\to0^ \plus{}$ or $ x\to\infty.$ However, since we have the square of that quantity in the denominator but only the quantity itself in the numerator, then the numerator becomes much larger than the denominator.

Hence, $ \lim_{x\to0^ \plus{} }\frac {\ln x}{1 \plus{} (\ln x)^2} \equal{} 0$ and $ \lim_{x\to\infty}\frac {\ln x}{1 \plus{} (\ln x)^2} \equal{} 0.$

We have a continuous function on $ (0,\infty)$ that tends to zero at both ends. Furthermore, this function is negative on $ (0,1)$ and positive on $ (1,\infty).$ Hence we know we will have at least one minimum on $ (0,1)$ and at least one maximum on $ (1,\infty).$

Take the derivative of the function. After some algebraic manipulation (not shown in detail), we get

$ f'(x) \equal{} \frac {1 \minus{} (\ln x)^2}{x\left(1 \plus{} (\ln x)^2\right)^2}.$

This is zero when $ x \equal{} \frac1e$ and when $ x \equal{} e.$ By the previous discussion, we see that we must have a minimum value at $ x \equal{} \frac1e,$ and that minimum value is $ \minus{} \frac12.$ Similarly, we must have a maximum value at $ x \equal{} e,$ and that maximum value is $ \frac12.$

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Here's something we could have done from the start. Let $ u \equal{} \ln x.$ As $ x$ ranges from $ 0$ to $ \infty,$ $ u$ ranges from $ \minus{} \infty$ to $ \infty.$ So our task is to find the limits, maxima, and minima of $ g(u) \equal{} \frac {u}{1 \plus{} u^2}$ on the real line. We would arrive at the same answers as above, and the algebra is a little simpler.

Well, we could rewrite $ f(x)$ as:

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

So, when $ x \rightarrow 0^\plus{}$ then the $ \ln^{\minus{}1} x$ tends to $ 0$ and $ \ln x$ tends to $ \minus{}\infty$ and therefore the function tends to $ 0$. Also, in the other example, the first part tends to $ \plus{}\infty$ and the second so $ 0$ again, the function tends to $ 0$.
Concerning the other part of your question, we need to calculate $ f^{'}(x)$:

$ f^{'}(x)\equal{}\frac{\frac{1}{x}\left(1\plus{}\ln^2 x\right)\minus{}\ln x \left(2\,\frac{\ln x}{x}\right)}{\left(1\plus{}\ln^2 x\right)}\equal{}\frac{1\minus{}\ln^2 x}{x\,\left(1\plus{}\ln^2 x\right)^2}$

So, now solve $ f^{'}(x)\equal{}0$ for $ x$:

$ \frac{1\minus{}\ln^2 x}{x\,\left(1\plus{}\ln^2 x\right)^2}\equal{}0$
$ 1\minus{}\ln^2 x\equal{}0$
$ \pm 1\equal{}\ln x_{1,2}$
$ x_{1,2}\equal{}e^{\pm 1}$

The function has the following values in these points:

$ f(e)\equal{}\frac{1}{2}$
$ f(\frac{1}{e})\equal{}\minus{}\frac{1}{2}$

And this are the maximum and minimum values of the function!

$ x\equal{}e^y$

$ \lim_{x\to 0^\plus{}} \, \frac{\log (x)}{\log ^2(x)\plus{}1}\equal{}\lim_{y\to \minus{}\infty } \, \frac{y}{y^2\plus{}1}\equal{}0$

$ \lim_{x\to \infty } \, \frac{\log (x)}{\log ^2(x)\plus{}1}\equal{}\lim_{y\to \infty } \, \frac{y}{y^2\plus{}1}\equal{}0$

$ \frac{\partial \frac{y}{y^2\plus{}1}}{\partial y}\equal{}0\Longleftrightarrow y^2\equal{}1$

$ \max(f) \equal{} f(e^1) \equal{} \frac{1}{2}$

When searching for max/min, I'd rather change the variable: $ t \equal{} \ln x$. We are to find max/min of $ \frac {t}{1 \plus{} t^2}$ as $ t$ runs over all real numbers (that's what $ \ln x$ does). The derivative is ... and so on. For the first order derivative this is not really different from using the chain rule, but if we use the second derivative test, change of variable works better than the chain rule.

(This is what EulerIntegral did above. And Kent did further above. I'm just reading-challenged.)

No need to use derivative.

b) $ f(1)\equal{}0$, for $ x\neq 1$ and $ x>0,$ we have $ |f(x)|\equal{}\left|\frac{1}{t\plus{}\frac{1}{t}}\right|\equal{}\frac{1}{|t|\plus{}\frac{1}{|t|}}\leq \frac{1}{2\sqrt{|t|\cdot \frac{1}{|t|}}}\equal{}\frac{1}{2}$, yielding $ \minus{}\frac{1}{2}\leq f(x)\leq \frac{1}{2}.$

The equality holds ...., I'm lazy. :P


Hum, I remember it was posted somewhere here. It is in the shortlist of  IMO 2001. It 's really a cool one. :D

Use Cauchy, then telescope.  The means of telescoping are very nice.  And for the record, this has definitely been posted before.


compute the sum

$\sum_{n=1}^{\infty}\frac{n^5}{3^n}$

generalize: provide an algorithm for finding the sum for other values besides 5

[hide="Solution Using Calculus"]
Just use $\sum x^{n}=\frac{1}{1-x}$, take the derivative, and multiply by $x$ each time.

$\implies \sum nx^{n}=\frac{x}{\left(1-x\right)^{2}}$

$\implies \sum n^{2}x^{n}=\frac{-x\left(x+1\right)}{\left(x-1\right)^{3}}$

$\implies \sum n^{3}x^{n}=\frac{x\left(x^{2}+4x+1\right)}{\left(x-1\right)^{4}}$

$\implies \sum n^{4}x^{n}=\frac{-x\left(x^{3}+11x^{2}+11x+1\right)}{\left(x-1\right)^{5}}$

$\implies \sum n^{5}x^{n}=\frac{x\left(x^{4}+26x^{3}+66x^{2}+26x+1\right)}{\left(x-1\right)^{6}}$

So the series you provided is just the case where $x=\frac{1}{3}$.  Therefore

\[ \sum \frac{n^{5}}{3^{n}}=\boxed{\frac{273}{4}}. \][/hide]

Can anyone show a solution without using calculus?

It is somewhat tedious but you can do it.

Here is a simpler example with $\sum_{n=1}^{\infty}\frac{n^3}{3^n}$ instead

$\sum_{n=1}^{\infty}\frac{n^3}{3^n}$
=$\sum_{n=1}^{\infty}\frac{1}{3^n}$ + $\sum_{n=2}^{\infty}\frac{7}{3^n}$+$\sum_{n=3}^{\infty}\frac{19}{3^n}$+...
=$\sum_{n=1}^{\infty}\sum_{k=n}^{\infty}\frac{3n^2-3n+1}{3^k}$
=$\sum_{n=1}^{\infty}(3\sum_{k=n}^{\infty}\frac{n^2}{3^k}-3\sum_{k=n}^{\infty}\frac{n^1}{3^k}+\sum_{k=n}^{\infty}\frac{n^0}{3^k})$

Now you've reduced it to a lower power.  use the same method to find out $\sum_{n=1}^{\infty}\frac{n^2}{3^n}$ and $\sum_{n=1}^{\infty}\frac{n^1}{3^n}$

After you've reduced it all replace all the $\sum_{m=g}^{\infty}\frac{g^0}{3^m}$s with $\frac{g}{1-3^{-1}}$s and simplify it.

Looking back on my previous post, I can generalize most of it except for one of the polynomials.  For example,

\[ n=4\implies x^{3}+11x^{2}+11x+1 \]

\[ n=5\implies x^{4}+26x^{3}+66x^{2}+26x+1 \]

Any ideas on this?

[hide="hint"]let $S_q=\sum_{n=1}^{\infty}\frac{n^q}{3^n}$

what is $3*S_q-S_q$?

( i am not going for $2*S_q$  :P )[/hide]