calculus
integration
logarithms
symmetry
trigonometry
calculus computations

Prove that
$ \int_{0}^{\pi/2} ln(tan(x)) dx \equal{} 0$.

[url=http://wangsblog.com/jeffrey/]Source[/url]

[hide="Hints"]One of the tricks you can use is to show that this integral is same as $ \ln \cot$ integral, sum those two, get $ \ln 1$... Another one could be splitting this integral into two ($ (0,\frac {\pi}{4})$ and $ (\frac {\pi}{4},\frac {\pi}{2})$) and showing first is Catalan constant, second one minus Catalan... Or showing that second is the same as $ \ln \cot$ in the bounds of first... There seems to be a variety of ways to prove the statement.[/hide]

Splitting it and "evaluating" each side is clearly worse than looking at symmetry.

Well, I'd never evaluate it in the first place. It's something you know and remember: Catalan constant :) In fact, knowing that helps you not to worry about the converging nature of the integral. Of course, symmetry was my first solution, everything else is just showing-off.

Well having $ \int_{0}^{\frac{\pi }{2}}{\ln (\sin x)\,dx}\equal{}\int_{0}^{\frac{\pi }{2}}{\ln (\cos x)\,dx},$ it's enough to complete the proof.


Show that the vertices of a quadrilateral whose opposite angles add up to $ 180^{\circ}$ lie on a circle.

I don't think that it is Olympiad Section.

Hi Robroobiks.

The question tells us that the opposite angle add up to $ 180^{\circ}$ and we have to deduce that the quadrilateral is cyclic, so the proof only needs to go one way.

(It is easy to show that if a quadrilateral is cyclic, then then oppsite angles add up to $ 180^{\circ}$, by using the fact that the angle at the centre is twice the angle at the circumference.)

From my diagram with $ A, B, C, D$ in cyclic order, I can see that you want $ <CAD \equal{} <CBD$ to get angles on the same arc, but some of the rest of your statements don't fit in with my diagram.

[quote="RobRoobiks"]You could connect AC and BD of quadrilateral ABCD.
Only when <CAD=<CBD can Quad. ABCD be cyclical where these angles corresponding to arc CD
Using Same reasoning, we conclude
<BAD=<BCD
<ABC=<ADC
<ACB=<ADB
then we know in triangle ABC,
<ABC+<ACB+<BAD+<BCD=180
<ADC+<ADB+<BAC=180
<BDC+<BAC=180
These two angles are opposite, and sum to 180 if and only if the quad. is cyclical.

I rarelly do proofs, so thank you AndrewTom for the opportunity. Please give constructive critisims on it.[/quote]
Wait, what do you mean by <BAD=<BCD?

It seems that these are angles on the same arc as RobRoobiks seems to have labelled his quadrilateral in the order $ ABDC$.

[quote="RobRoobiks"]<ABC+<ACB+<BAD+<BCD=180 [/quote]
If the quadrilateral  is $ ABDC$ we should have <CBD instead of <BCD, but it doesn't matter because  this equality is not needed in the two that follow it, and leads to the conclusion.

I have sketched the quadrilateral $ ABDC$ and it seems to me that RobRoobiks should have $ <CAD$ instead of $ <CBD$.

I have the following proof by contradiction.

I have drawn the quadrilateral in the order $ ABCD$ and assumed that, while the opposite angles sum to $ 180^{\circ}$, the points are not concyclic. Without loss of generality, $ D$ is not on the circle through $ A, B$ and $ C$. So we have two cases: either $ D$ is outside this circle or it is inside it. In the first case, the line-segment $ CD$ cuts the circle at $ E$, say and gives a cyclic quadrilateral $ ABCE$ whose opposite angles sum to $ 180^{\circ}$ since it is a cycluic quadrilateral; in the second case, $ CD$ produced meets the circle at $ E$, say, giving a cyclic quadrilateral $ ABCE$ whose opposite angles sum to $ 180^{\circ}$.

This means that if angle $ ABC \equal{} \alpha$, then angles $ AEC$ and $ ADE$ are both equal to $ 180^{\circ} \minus{} \alpha$ and so angle $ DAE$ is $ 0^{\circ}$

Therefore $ D$ and $ E$ are coincident and the vertices $ A, B, C$ and $ D$ are concyclic.


abstract algebra
group theory
superior algebra
superior algebra unsolved

If $G$ is a group and $H$ a subgroup of $G$ and $N \lhd G$ such that $G/N$ and $N$ are abelian.

Prove that there exists a subgroup $D$ of $H$ such that $D \lhd H$ and $H/D$ and $D$ are abelian too.

[hide]
I was thinking to define $D = H \cap N$. But probably I'm totally wrong since I'm missing a proof.
[/hide]

Hi.

$D=H\cap N$  is the natural choice and it works! Since D is a subgroup of N, D is abelien. Since N is normal in G, $H\cap N$ is normal in H. And with the help of one isomorphism theorem we get:
$H/D=H/(H\cap N)\cong HN/N\leq G/N$
So H/D is abelian too.


geometry
power of a point
radical axis
geometry unsolved

Given two circles $O_{1}$ and $O_{2}$ which intersect at two points $A$ and $B$. The segments $X_{1}Y_{1}$ and $X_{2}Y_{2}$ are arbitrary diameters of the circles $O_{1}$ and $O_{2}$, respectively. The line $X_{1}X_{2}$ intersects the line $Y_{1}Y_{2}$ at a point $Z_{1}$; the line $X_{1}Y_{2}$ intersects the line $X_{2}Y_{1}$ at a point $Z_{2}$. Prove that the four points $A$, $B$, $Z_{1}$, $Z_{2}$ are concyclic.

sorry....my solution not very rigourous, but anyways.....

In triangle $X_{1}Z_{1}Y_{2}$, consider the cevians  $Z_{1}Z_{2}$, $Y_{2}x_{2}$, and $X_{1}Y_{1}$. It is well-known, eg. in Coxeter's Geometry Revisited, that the $H$, the orthocenter of triangle $X_{1}Z_{1}Y_{2}$ has equal power to the circles with diameters $Z_{1}Z_{2}$, $Y_{2}x_{2}$, and $X_{1}Y_{1}$.

Repeat for the triangle $Y_{1}Z_{2}Y_{2}$, and cevians $Z_{1}Z_{2}$, $Y_{2}x_{2}$, and $X_{1}Y_{1}$; and we see that the orthocenter of  triangle $Y_{1}Z_{2}Y_{2}$, $H'$, also has equal power to the circles with diameters $Z_{1}Z_{2}$, $Y_{2}x_{2}$, and $X_{1}Y_{1}$. Since $H$ and $H'$ dont coincide, the circles with diameters $Z_{1}Z_{2}$, $Y_{2}x_{2}$, and $X_{1}Y_{1}$ have a common radical axis: AB.

Concluding, the circle wit diameter $Z_{1}Z_{2}$ passes through A and B.

Hope i dont have any flaws  :)

Dear Fang-jh and Mathlinkers,
is there a typo in your problem?
The given proof seems hazardous...
Sincerely
Jean-Louis

Dear Mathlinkers,
can some one precise the origine of this problem and gives the exact wordind of this one like he was given at this competion.
Sincerely
Jean-Louis


analytic geometry
geometry
geometric transformation
reflection

A light ray travelling along the straight line x - 3y + 7 = 0, gets reflected from a spherical mirror whose circular section with the x-y plane is x^2 + y^2 - 2x - 2y - 3 = 0. What is/are the equation(s) of the straight line(s) along which the reflected ray travels?

[hide="solution"]
This might be wrong but it makes it a whole lot easier...
$ x^2 \plus{} y^2 \minus{}2x \minus{} 2y \minus{}3 \equal{} 0 \longrightarrow (x\minus{}1)^2 \plus{} (y\minus{}1)^2 \equal{} 5$

To make this easier we will tanslate the graph to $ x^2 \plus{} y^2 \equal{} 5$
And tanslate the light ray to $ (x\plus{}1) \minus{} 3(y\plus{}1) \plus{}7 \equal{} 0 \Rightarrow x\minus{}3y \plus{}5\equal{}0$

Using these simultaneous equations we can easily see that they meet at point $ (2,1)$ and $ (1,\minus{}2)$
Considering the first point; the normal ray that passes through the origin and $ (2,1)$ is $ y\equal{}\frac{1}{2}x$

Since angle of incidence is equal to angle of reflection the reflected ray will be the original line reflected arround $ y\equal{}\frac{1}{2}x$. So $ (x,y) \rightarrow (y, \frac{1}{2}x)$
$ 2y \minus{} 6x \plus{} 5 \equal{} 0$

Now we have to translate it back to the original position
$ 2(y\minus{}1) \minus{} 6(x\minus{}1) \plus{}5 \equal{} 0$
$ 2y \minus{}6x \minus{}3 \equal{}0$
And that's our answer  :) [/hide]

:(
Thanks for trying but the answer is wrong. The options for this question were:
A) y = 2
B) 3x + y = 7
C) 3x - y = 3
D) x = -1

More than one options may be correct.

~~~~~~~~~~~~~~~~~~~~~~~~~

I think you made a mistake with finding the solutions of the simultaneous equations. Shouldnt it be (-2,1) and (1,2)?


inequalities
LaTeX
logarithms
inequalities solved

i just want to know what is the best solution for the inequality in imo01 and also the easiest to find solution

In my opinion this solution is easier then the official one :
Sorry my first latex :blush:

there is an other solution :
put $x=\frac{1}{a}$;$y=\frac{1}{b}$:$z=\frac{1}{c}$
the inequality becomes $\sum \frac{1}{\sqrt{1+8 \frac{x^2}{yz}}} \geq 1$
We have :
$1+8\frac{x^2}{yz}=1+(2\sqrt[3]{\frac{x^2}{yz}})^3=(1+2\sqrt[3]{\frac{x^2}{yz}})((2\sqrt[3]{\frac{x^2}{yz}})^2-2\sqrt[3]{\frac{x^2}{yz}}+1)$
which means :$\sqrt{1+8 \frac{x^2}{yz}}=\sqrt{(1+2\sqrt[3]{\frac{x^2}{yz}})((2\sqrt[3]{\frac{x^2}{yz}})^2-2\sqrt[3]{\frac{x^2}{yz}}+1)}$
and by using AM-GM the $RHS$ of the last equality is $ \leq 2(\sqrt[3]{\frac{x^2}{yz}})^2+1$
So the inequality is equivalent to :
$\sum \frac{1}{2(\sqrt[3]{\frac{x^2}{yz}})^2+1} \geq 1$
And then we can verify this by clearing the denominators and using AM-GM again.
You can also see the official soution [url=http://imo.wolfram.com/problemset/IMO2001_solution2.html]here[/url]

And here is the proof of general form of inequality: :D

I will $\LaTeX$ it:

Anyways, because it is homogeneous, we may set $a+b+c=1$, and since $f(x)=1/\sqrt{x}$ is convex on $x>0$, we have that

$\sum \frac {a}{\sqrt {a^2+8bc}}\geq \frac {1}{\sqrt{\sum a(a^2+8bc)}}=\frac {1}{\sqrt{\sum a^3+24abc}}$

Now, $(a+b+c)^3=\sum a^3+3\sum a^2b+6abc$, and since $\sum a^2b\geq 6abc$ by AM-GM, we have

$\sum \frac {a}{\sqrt {a^2+8bc}}\geq \frac {1}{\sqrt{\sum a(a^2+8bc)}}=\frac {1}{\sqrt{\sum a^3+24abc}}\geq \frac {1}{\sqrt{(a+b+c)^3}}=1$ which is the desired result.

But blahblahblah you missed a sqrt in the $RHS$

It's funny how ordinary inequalities with stuff in the denominators screams Cauchy, but when there are square roots, it's not so obvious to use Holder. Anyways, using Jensen's for $f(x)=1/\sqrt{x}$ is equivalent to a direct application of Holder's inequality:

\[\left( \sum\frac{a}{\sqrt{a^2+8bc}} \right)^{2/3} \left(\sum a(a^2+8bc)\right)^{1/3} \geq a+b+c,\]

and the result follows.

Holder is generally a statement about convexity (e.g, it is a near-immediate consequence of Young's inequality, which in turn is just a statement of the concavity of the $\log$ function.


Let $ ABCDEF$ be a hexagon with all internal angles equal to $ 120^0$. Show that $ AB\minus{}DE \equal{} CD\minus{}FA \equal{} EF\minus{}BC$.

See http://www.mathlinks.ro/viewtopic.php?t=16758 .

  darij


inequalities
real analysis
real analysis unsolved

Given the convergent series $ \sum a_n$ of positive terms, prove that the series $ \sum\sqrt[n]{a_1\dots a_n}$ must also be convergent.

I'm pretty sure that this problem's been posted here before...

Also, the problem appears as entry 15 or so of [url=http://www.amazon.com/Problem-Seminar-Mathematics-Physical-Sciences/dp/0387907653/ref=sr_1_4?ie=UTF8&s=books&qid=1226188276&sr=1-4]Newman[/url]'s book. :wink:

Maybe Stolz-Cesaro will help you.  :idea:

What is an example where the two series don't have the same limit? Or do they always have the same limit?

Series, not sequence. There's a difference, and this would be false with averages.

I [b]knew[/b] this had been posted [b]her[/b]e before...

Take a look at the following link:

[url=http://www.artofproblemsolving.com/Wiki/index.php/Carleman's_Inequality]*****[/url]

Regards.


Putnam
number theory
least common multiple
inequalities
geometry
college contests

Find the smallest positive integer n such that for every
integer m with 0 < m < 1993, there exists an integer k
for which

m/1993 < k/n < (m + 1)/1994 .

I've got an answer, with a decent amount of reasoning behind it, but how should this be written up on an exam?

Write what you have. We can critique it.

I started by multiplying everything by n, and then subtracting the endpoints.

So we have (m+1)n/1994 - mn/1993 = 1993n - mn / (1993)(1994) = (1993-m)n / (1993)(1994)

so for their to be an integer within that interval for all 0<m<1993, that distance needs to be larger than one.

Start by letting n = lcm (1993,1994), which gives us an interval of > 1 for all m up through 1992.  Unfortunately at m = 1992, we get (1992)(1994) < k < (1993)(1993).  But there are no integers between 3972048 and 3972049, so take n = lcm (1993,1994) + 1.

In general fo integers m&a: (mn / a) < k < (mn + n) / (a + 1).  By taking n = a^2 + a + 1 by subtracting the endpoints we get:

((am+a)(a^2+a+1) - (am + m)(a^2+a+1)) / (a^2 + a)  =  (a-m)(a^2+a+1) / (a^2+a)  =  (a-m) + (a-m) / (a^2 + a)

So for m = a-1, as in the case with 1992 we get 1 + something miniscule, so we get an integer k in between our endpoints for all m < a.

So n = lcm (1993,1994) + 1

- The interval you've identified being larger than $ 1$ is a sufficient, but not a necessary, condition for $ k$ to exist.  Arbitrarily small intervals can contain an integer.  Consequently,
- You've done nothing to show that $ n$ can't be taken to be less than $ 1993 \cdot 1994$.  It's very likely that the correct value is much smaller.

[hide="A useful lemma."]If $ a,b,c,d>0$ and $ \frac{a}{b}<\frac{c}{d},$ then $ \frac{a}{b}<\frac{a\plus{}c}{b\plus{}d}<\frac{c}{d}.$

As a long-time baseball fan, I think of this as the "batting average lemma": if a batter is hitting $ .300,$ which means that he has $ 3n$ hits in $ 10n$ at bats so far, and he then has a game in which he goes 2 for 5 ($ .400$ for the one game), then his new batting average is greater than $ .300$ but less than $ .400.$[/hide]

[quote="jmerry"]Indeed, it is much smaller.
[hide="Correct answer"]$ 3987$[/hide][/quote]

maybe im missing something, but when m = 1992, what is the k if n is what you said.

nvm, i got it, i was using 3997

Great - so you've got a nice answer that works.  Can you now show that no smaller answer is suitable?  :)

This is on the agenda for tomorrow's MATH 491 class.

And the full discussion from that class.

There's nothing at all special about 1993. Let's replace it by $ N$ with $ N\ge 2.$ In the statement of the problem, $ n,N,m,$ and $ k$ will all be positive integers.

The problem is this: find the smallest $ n$ such that for all $ m,$ $ 0<m<N,$ there exists $ k$ such that $ \frac{m}{N}<\frac{k}{n}<\frac{m\plus{}1}{N\plus{}1}.$

Claim: the smallest such $ n$ is $ n\equal{}2N\plus{}1.$ (in the $ N\equal{}1993$ case, that's $ n\equal{}3987.$)

First, the proof that $ n\equal{}2N\plus{}1$ works.

This relies on the "batting average" lemma I referred to above: if $ \frac ab<\frac cd,$ then $ \frac ab<\frac{a\plus{}c}{b\plus{}d}<\frac cd.$ I'll leave the proof of this lemma as an exercise.

By this lemma, $ \frac{m}{N}<\frac{2m\plus{}1}{2N\plus{}1}<\frac{m\plus{}1}{N\plus{}1}.$ This settles the sufficiency of $ n\equal{}2N\plus{}1.$

Now we have to prove that no smaller $ n$ works. The most stringent condition takes place "at the top" for $ m\equal{}N\minus{}1.$ We are looking to see if it is possible to have

$ \frac{N\minus{}1}{N}<\frac kn<\frac{N}{N\plus{}1}$

$ 1\minus{}\frac1N<1\minus{}\frac jn<1\minus{}\frac 1{N\plus{}1}$ where $ j\equal{}k\minus{}n$ is also a positive integer.

$ \frac1{N\plus{}1}<\frac jn<\frac1N.$

If $ n\le N,$ then $ \frac jn\ge\frac1n\ge\frac1N,$ so we cannot satisfy these inequalities.

If $ N<n< 2N\plus{}1,$ which is to say $ N\plus{}1\le n\le 2N,$ then $ \frac1n\le\frac1{N\plus{}1},$ which lies outside the interval, while for $ j\ge 2,$ $ \frac jn\ge\frac 2n\ge\frac2{2N}\equal{}\frac1N$ also lies outside the interval.

So we cannot satisfy the conditions for $ n\le N$ and we cannot satisfy the conditions for $ N<n<2N\plus{}1,$ which makes $ 2N\plus{}1$ the smallest $ n$ for which we can satisfy the conditions.

Here's a geometric picture:  the inequality $ \frac{m}{N} < \frac{k}{n} < \frac{m\plus{}1}{N\plus{}1}$ is satisfied if and only if the point $ (k, n)$ lies between the rays from the origin to the points $ (m, N)$ and $ (m\plus{}1, N\plus{}1)$.  Now, one has

$ \frac{m\plus{}1}{N\plus{}1} \minus{} \frac{m}{N} \equal{} \frac{N\minus{}m}{N(n\plus{}1)}$

so when $ m \equal{} N\minus{}1$ the numerator is $ 1$.  This occurs if and only if the triangle $ (0, 0), (m, N), (m\plus{}1, N\plus{}1)$ has area $ \frac{1}{2}$, so by [url=http://en.wikipedia.org/wiki/Pick%27s_theorem]Pick's theorem[/url] the quadrilateral

$ (0, 0), (m, N), (m\plus{}1, N\plus{}1), (2m\plus{}1, 2N\plus{}1)$

contains no lattice points, hence there are no smaller possible values of $ n$ (since these would correspond to lattice points within the above quadrilateral).

Very pretty, but just one question: if you had a $ (k,n)$ with $ n$ between $ N\plus{}1$ and $ 2N\plus{}1$ so that it was outside the triangle $ (0,0),(N\minus{}1,N),(N,N\plus{}1)$ but still between the rays, why would it have to be in that quadrilateral?

Tricky.  Instead, take the triangle $ (0, 0), (2N\minus{}2, 2N), (2N, 2N\plus{}2)$, which has area $ 2$ and $ 6$ boundary lattice points.


given that:
arctan(x/2)+arctan(2x/3) = pi/4,

what is x?

However, checking, we find that the only solution is $ x \equal{} \frac { \minus{} 7 \plus{} \sqrt {97}}4$ (assuming the original problem was $ \operatorname{Arctan}\left(\frac x2\right) \plus{} \operatorname{Arctan}\left(\frac {2x}3\right) \equal{} \frac\pi4$).

This is because the LHS must be positive, and it is obviously not when the inputs of the $ \operatorname{Arctan}$'s are negative.

hello, you must put your solutions $ x\equal{}\frac{\minus{}7\pm\sqrt{97}}{4}$ in your equation.
Sonnhard.


calculus
integration
trigonometry
calculus computations

Evaluate the integral, \[\int{\sqrt{x^{2}-2x-1}\over x-1}dx.\]


modular arithmetic
number theory
relatively prime
number theory unsolved

Let $ m \equal{} 2007^{2008}$, how many natural numbers n are there such that $ n < m$ and $ n(2n \plus{} 1)(5n \plus{} 2)$ is divisible by $ m$ (which means that $ m \mid n(2n \plus{} 1)(5n \plus{} 2)$) ?

[quote="randomgraph"]What age group is the Vietnam MO targeted at? This seems a bit on the easy side for a MO.[/quote]
It seems that Vietnam MO is vey easy with you  :?: I don't think so. Doing 7 problems in a time 180 minutes is not a simple task. It follows that you are good student and I extremely admire you.

I have a solution, I'll post it soon. What happened to the previous posts?

[quote="randomgraph"](what happened to my post? There were two posts answering the question and both have disappeared. Oh well).

I certainly don't think the Vietnam MO is "very easy", and I would be surprised if it is. Of course I did not know how many problems are involved and what is the alloted time, and I still don't know what age group is involved. I was only commenting on this specific question, which seems relatively easy simply because it has a fairly straightforward and short solution that doesn't involve Fermat's little theorem which is typically a necessary tool for problems of this nature.

Please don't extremely admire me - I have not been a student for many years now.[/quote]
Hey. Orl obviously mistyped, so the question was in fact $ m \mid n(2n \plus{} 1)(5n \plus{} 2)$ and not $ n(2n \plus{} 1)(5n \plus{} 2) \mid m$, as he wrote in the first place..

Well.. It's still pretty much the same.. Notice that $ (2007, (n,2n\plus{}1)) \equal{} (2007, (5n\plus{}2,2n\plus{}1)) \equal{} (2007, (n,5n\plus{}2)) \equal{} 1$, so we can split up the cases: (notice that $ m \equal{} 3^{4016} \cdot 223^{2008}$)
case 1 $ (2007, n)\equal{}2007$: Not possible since $ n < m$
case 2 $ (2007, n)\equal{}9$: Let $ n \equal{} 3^{4016} \cdot k$.
      case 2a $ (2007, 2n\plus{}1)\equal{}223$: Let $ 2n\plus{}1 \equal{} 223^{2008} \cdot j \equal{} 2 \cdot 3^{4016} \cdot k \plus{} 1$ So $ k \equiv \minus{}1 \cdot 2^{\minus{}1} \cdot (3^{\minus{}1})^{4016} \bmod 223^{2008}$
      case 2b $ (2007, 5n\plus{}2)\equal{}223$: Let $ 5n\plus{}2 \equal{} 223^{2008} \cdot j \equal{} 5 \cdot 3^{4016} \cdot k \plus{} 2$ So $ k \equiv \minus{}2 \cdot 5^{\minus{}1} \cdot (3^{\minus{}1})^{4016} \bmod 223^{2008}$
case 3 $ (2007, n)\equal{}1$:
      case 3a $ (2007, 2n\plus{}1) \equal{} 1$: $ 5n \plus{} 2 \equal{} 2007^{2008} \cdot k \Rightarrow n \equal{} \frac{2 \cdot 2007^{2008} \minus{} 2}{5}$.
      case 3a $ (2007, 2n\plus{}1) \equal{} 9$:
      case 3a $ (2007, 2n\plus{}1) \equal{} 223$:
      case 3a $ (2007, 2n\plus{}1) \equal{} 2007$: $ 2n \plus{} 1 \equal{} 2007^{2008} \cdot k \Rightarrow n \equal{} \frac{2007^{2008} \minus{} 1}{2}$

I'll fill out the cases some time :P

I'm not sure I'm able to follow Mathias' case enumeration, so here's my proposed solution.

Let $ a \equal{} 3^{4016}$ and $ b \equal{} 223^{2008}$, so $ m\equal{}ab$ and $ (a,b)\equal{}1$. Since $ n$, $ 2n\plus{}1$ and $ 5n\plus{}2$ are almost relatively prime (we can have $ (n,5n\plus{}2)\equal{}2$), if their product is divisible by $ ab$ then one of them is divisible by $ a$ and one of them (possibly the same one) is divisible by $ b$. So we're trying to simultaneously solve the equations

$ sn\plus{}t \equiv 0 \pmod{a}$
$ un\plus{}v \equiv 0 \pmod{b}$

where $ n$ is the unknown, and each of $ (s, t)$ and $ (u, v)$ is one of $ (1,0)$, $ (2,1)$ and $ (5,2)$. Notice that in all cases $ s$ and $ u$ are relatively prime to both $ a$ and $ b$, so by standard facts from number theory (linear congruences) we have a unique solution to each such system modulo $ ab \equal{} m$. There are 9 combinations for $ (s,t)$ and $ (u,v)$ so basically we get 9 solutions for $ n$.

We do note, however, that when both $ (s,t)\equal{}(1,0)$ and $ (u,v)\equal{}(1,0)$, the solution is obviously $ n \equiv 0 \pmod{m}$. This is perfectly legitimate but the way the problem was phrased it's unclear if $ n\equal{}0$ is allowed ("natural numbers" are sometimes taken to include as sometimes to exclude 0, and I grew up learning that 0 is not a natural number). It is clearly stated that $ n < m$ so $ n\equal{}m$ is not an option.

In all other 8 cases, $ n$ is uniquely defined mod $ m$ and corresponds to a unique number $ 0 < n < m$. We only need to check if any pair of those $ n$s coincide. If this happens then $ a$ or $ b$ divide both $ sn\plus{}t$ and $ un\plus{}v$ for different pairs $ (s,t)$ and $ (u,v)$, which can't happen since these numbers are (almost) relatively prime.

In summary, the number of such $ n$ is 9 or 8 depending on whether $ n\equal{}0$ is allowed.

Note: Contrary to the previous version posted, this version seems to me to be very appropriate for a MO. The previous version still feels way too easy to me.

Ok,the result is $ n\equal{}9$,my solution is used Chinrese Remainder Theorem,But my EL is not good enough to post it here sorry


Calculate:
J=$ \int_0^{\frac{\pi}2} \frac{ sin^2x}{sin x\plus{}2cos x}dx$

$ \int_0^{\frac{\pi }{2}} \frac{\sin ^2(\theta )}{\sin (\theta )\plus{}2 \cos (\theta )} \, d\theta \equal{}\left(\int_0^{\frac{\pi }{2}} \sin (\theta ) \, d\theta \minus{}2 \int_0^{\frac{\pi }{2}} \cos (\theta ) \, d\theta \plus{}4 \int_0^{\frac{\pi }{2}} \frac{\cos ^2(\theta )}{\sin (\theta )\plus{}2 \cos (\theta )} \, d\theta \right)$

$ \left(\underbrace{\int_0^{\frac{\pi }{2}} \frac{1}{\sin (\theta )\plus{}2 \cos (\theta )} \, d\theta }_{\theta \equal{}2 \tan ^{\minus{}1}(z)}\equal{}\underbrace{\int_0^1 \frac{1}{\minus{}z^2\plus{}z\plus{}1} \, dz}_{z\equal{}\frac{1}{2}\minus{}\frac{\sqrt{5} w}{2}}\equal{}\minus{}2 \int_0^{\frac{1}{\sqrt{5}}} \frac{2}{\sqrt{5} \left(w^2\minus{}1\right)} \, dw\equal{}\frac{4 \tanh ^{\minus{}1}\left(\frac{1}{\sqrt{5}}\right)}{\sqrt{5}}\right)$

$ \therefore$

$ \int_0^{\frac{\pi }{2}} \frac{\sin ^2(\theta )}{\sin (\theta )\plus{}2 \cos (\theta )} \, d\theta \equal{}\frac{1}{25} \left(16 \sqrt{5} \tanh ^{\minus{}1}\left(\frac{1}{\sqrt{5}}\right)\minus{}5\right)$


inequalities
trigonometry
function
inequalities proposed

Let $ \Delta ABC$ be an acuted triangle. Prove that
\[ \cos \frac{A}{2} \plus{} \cos\frac{B}{2} \plus{}\cos \frac{C}{2} \ge \frac{4}{\sqrt{3}} (1\plus{} \sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2})\]
:)

Inequality is equivalent whith:$ \sqrt{3}\sum cos\frac{A}{2}\geq 2\sum \cos^2\frac{A}{2}$

 But function $ f(x)\equal{}2cos^2x\minus{}\sqrt{3}cosx$ is concave and follow Jensen inequality


solve for $ x$ the equation :
$ x\plus{} \sqrt{a^2\plus{}x^2} \equal{} \frac{5a^2}{ a^2\plus{}x^2}$

[quote="arcsin"]solve for $ x$ the equation :
$ x \plus{} \sqrt {a^2 \plus{} x^2} \equal{} \frac {5a^2}{ a^2 \plus{} x^2}$[/quote]

I think it should be $ x \plus{} \sqrt {a^2 \plus{} x^2} \equal{} \frac {5a^2}{ \sqrt {a^2 \plus{} x^2}}$
the solution of this one is: $ x\equal{}\dfrac {4a}{3}$

can you post the full solution of this $ x\plus{}\sqrt{a^{2}\plus{}x^{2}}\equal{}\frac{5a^{2}}{\sqrt{a^{2}\plus{}x^{2}}}$

[quote="arcsin"]can you post the full solution of this $ x \plus{} \sqrt {a^{2} \plus{} x^{2}} \equal{} \frac {5a^{2}}{\sqrt {a^{2} \plus{} x^{2}}}$[/quote]

$ x\equal{}\frac{5a^2}{\sqrt{a^2\plus{}x^2}} \minus{} \sqrt{a^2\plus{}x^2}$ ,powering by 2 we get :

$ x^2\equal{} \frac{25a^4}{a^2\plus{}x^2} \minus{} 2*\frac{5a^2}{\sqrt{a^2\plus{}x^2}}*\sqrt{a^2\plus{}x^2} \plus{} a^2\plus{}x^2$

$ x^2\minus{}x^2 \equal{} \frac{25a^4}{a^2\plus{}x^2} \minus{} 10a^2 \plus{} a^2$

$ \frac{25a^4}{a^2\plus{}x^2}\equal{}9a^2$
$ 25a^4\equal{}9a^2(a^2\plus{}x^2)$    divide  by  $ a^2$
$ 25a^2\equal{}9a^2\plus{}9x^2$
$ 16a^2\equal{}9x^2$

$ x\equal{} \frac{4a}{3}$

You forgot negative solution, so it's $ x \equal{} \pm \frac {4}{3}a$.

Actually, its a little more subtle than that.  There is only one answer for $ x$, and it is $ - \frac43|a|$.  The other root is extraneous.  Starting at $ x^2 = \frac {16}{9}a^2$, we can plug back into the original equation:

$ [ \begin{align*} x & = \frac {5a^2}{\sqrt {a^2 + x^2}} - \sqrt {a^2 + x^2} = \frac {5a^2}{\sqrt {a^2 + \frac {16}{9}a^2}} - \sqrt {a^2 + \frac {16}{9}a^2} \\
& = \frac {5a^2}{\sqrt {\frac {25}{9}a^2}} - \sqrt {\frac {25}{9}a^2} = \frac {5a^2}{\frac53|a|} - \frac53|a| = \frac13|a| - \frac53|a| = - \frac43|a| \end{align*}$


complex numbers
number theory unsolved
number theory

$card(S)=+\infty$
because $(x,y,z,u)=(-a^{12},a^{15},-a^{20},a^{30})$ is a solution
but it can have other solutions

[quote="aviateurpilot"]$card(S)=+\infty$
becaue $(x,y,z,u)=(-a^{12},a^{15},-a^{20},a^{30})$ is a solution
but it can have other solutions[/quote]
This is a trivial solution what I also found.
Real part of the  problem is in fact  about "other" solutions what you are pointing at .How to find  and describe them ? :maybe:

Some solutions that are NOT of the type you posted are:
$(-1,6,-11,6),(7,-5,-26,-12)$ etc.
How to show there are infinitely many of these too? :D


If I have $ n$ random variables $ X_1,X_2,\cdots,X_n$, then is there any chance to find \[ Prob(X_1<X_2<X_3<\cdots <X_n)\] regards
bhupala

Of course, but it's complicated in general. Find the joint density and integrate over the appropriate region.

For independent identically distributed continuous variables, it's easy.


[quote="agripa"]\sqrt {x + 2 - x^2 }  How deos have to look this graph function[/quote]

$ y\equal{}\sqrt{x\plus{}2\minus{}x^2}$ $ \Leftrightarrow$ $ (x\minus{}\frac{1}{2})^2\plus{}y^2\equal{}\left(\frac{3}{2}\right)^2$ and $ y\geq 0$

And so graph is the upper half circle with center in $ (\frac{1}{2},0)$ and radius $ R\equal{}\frac{3}{2}$

And next time, use an appropriate forum: This problem is not open, and not olympiad level, either.

I need help in division i don't no where to start with it.If there anyone that can help me please :(  :(


geometry
parallelogram
geometry unsolved

Let ABCD be an quadrilateral, whose opposite sides intersect at 2 points. Two lines through this points intersect quadrilateral's sidelines at 4 points points, which are vertices of an parallelogram P. Prove that sides of P are parallel to the diagonals of the initial quadrilaeral.

Given the quadrilateral $ABCD\ :\ A(-a,0),B(b,0),C(c,e),D(0,d)$.
Choose $P(\lambda,0) \in AB$.

Parallelogram $PQRS\ :\ Q(\frac{ab+b(c+\lambda)-c\lambda}{a+b},\frac{e(b-\lambda)}{a+b}) \in BC, R(\frac{c(b-\lambda)}{a+b},\frac{ad+be+\lambda(d-e)}{a+b}) \in CD, S(\frac{a(\lambda-b)}{a+b},\frac{d(\lambda+a)}{a+b}) \in DA$.

The line $AB$ intersects the line $CD$ in the point $K(\frac{cd}{d-e},0)$.
The line $KS$ intersects the side $BC$ in the point $Q$
if $\lambda=\frac{(a+b)\sqrt{de} \cdot \sqrt{-a[b(d-e)^{2}+cd(e-d)]+bcd(e-d)+c^{2}d^{2}}+a[cd^{2}-b(d-e)^{2}]+bcde}{(d-e)[ae+d(b-c)]}$.
Then, the lines $AD,BC,PR$ are concurrent.

Let P and Q be the intersections of AB and CD, and AD and BC, respectively, and WLOG A lies on segments BP and DQ. Let WXYZ be the parallelogram such that W, X, Y, Z lie on sides AB, BC, CD, DA respectively; then segments WY and XZ bisect each other.

We will now show that WZ || BD. If the lines through Q and P parallel to XZ, WY respectively intersect AB, AD at U, V respectively, then by projecting from P onto AD and from Q onto AB, A, W, B, U and A, Z, D, V are harmonic, and AZ * AU = AW * AV = AP * AQ so AZ / AV = AW / AU, which means AZ / AD = AW / AB, thus WZ || BD.

The other sides can be shown to be parallel to AC and BD in similar ways, by projecting from P and Q onto the other sides of ABCD. Thus, the sides of the parallelogram are indeed parallel to the diagonals of ABCD.

We'll show that, in general, the pairs of opposite sides of $P$ (without the parallelogram restriction) intersect on the diagonals of $ABCD$. The problem is now purely projective, so we could kill it with many things, e.g. mapping $ABCD$ to a rectangle. But for comedy's sake, we'll use projective coordinates.

Let $X$ be $AC\cap BD$. We set up projective coordinates with reference triangle $XYZ$, $A = (1, 1, 1)$, $B = (1,-1,1)$, $C = (-1, 1, 1)$, and $D = (1,1,-1)$. Let the vertices of $P$ on $DA$, $AB$, $BC$, and $CD$ be $Q$, $R$, $S$, and $T$, respectively. Then, suppose that $Q = A + \lambda D = (1 + \lambda, 1 + \lambda, 1-\lambda)$, $R = A + \kappa B = (1 + \kappa, 1 + \kappa, 1 - \kappa)$.

The points on $CD$ are those with $x+z=0$, so it is not hard to see that $T = (1 + \kappa,  1 - \kappa, -1 -\kappa)$. Similarly, $S = (1 + \lambda, -1-\lambda, 1 - \lambda)$.

Now, we want to show that there is a point collinear with each of the pairs $(A, C)$, $(Q, T)$, and $(R, S)$. To this end, observe that
\[(1,1,1) + \lambda \kappa \cdot (-1, 1, 1)  = (1 +\lambda, 1 + \lambda, 1 - \lambda) - \lambda \cdot (1 + \kappa, 1 - \kappa, -1 - \kappa) = (1 + \kappa, 1 - \kappa, 1 + \kappa) - \kappa \cdot (1 + \lambda, -1 - \lambda, 1 - \lambda).\]


Hi guys,

I'm Paolo and this is the second round of the second edition of [i]Oliforum contest[/i]. You can see first round [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=1640043#1640043]here[/url].

I copy below all rules, as always  :wink:

[b]1-[/b] The contest is made of 2 rounds, this is the last one.

[b]2-[/b] Each round is made of some problems (from 3 to 5..), and i will give points 0 to 7: 6 for all correct solutions, 7 for [i]only[/i] elegants and correct solutions (similar to ML contest).

[b]3-[/b] It is tought for high-school student, and also for undergraduate or higher student, but you cannot use non-elementary-theory, otherwise the solution is invalid.

[b]4-[/b] All begins from italian-oliforum and the classification is unique.

[b]5-[/b] The first round starts on 15:00 of 16 september and it ends on 15:00 of 18 seprember, so you have 48 hours :) .
        [b]NB.[/b] All times (date and hours) are according to [i]ROME MERIDIAN (+2GTM), [/i]please attention!

[b]6-[/b] Enrollment is not necessary, it sufficies to send me solution.

[b]7-[/b] Time of arriving solution is not important.

[b]8-[/b] [i]Way of sending solutions[/i].
-          You need to send solutions to both email address: [u]pao_leo88 ML hotmail ML[/u] it and [u]leonettipaolo ML gmail ML com[/u] in a [i]unique pdf file[/i] (so, with attachment)  of a right size;
-          The pdf file must be written with [i]Latex [/i] (or a way that I can understand);
-          You [i]rename the pdf file [/i]as your nickname (the same of yours in ML);
-          You need to write nothing in the email;
-          Try to be [i]clear[/i] in solutions!

{If [i](and only if)[/i] you are not able to create a pdf in latex model, you can send me a email in .doc format with solutions, as latex or in a way that i (and all correctors) can understand, or also with private message to me}.

Later, together with texts, I'll make a list below this topic where i'll write all names of who sent me solutions in right way.

For all doubts and questions, ask me in private messages :wink:

Thanks for the attentions and good luck

[i]Paolo Leonetti[/i]

[b]Problem 1.[/b]
Find all non empty subset $ S$ of $ \mathbb{N}: \equal{} \{0,1,2,\ldots\}$ such that $ 0 \in S$ and exist two function $ h(\cdot): S \times S \to S$ and $ k(\cdot): S \to S$ which respect the following rules:
i) $ k(x) \equal{} h(0,x)$ for all $ x \in S$
ii) $ k(0) \equal{} 0$
iii) $ h(k(x_1),x_2) \equal{} x_1$ for all $ x_1,x_2 \in S$.
[i](Pierfrancesco Carlucci)[/i]

[b]Problem 2.[/b]
Let a convex quadrilateral $ ABCD$ fixed such that $ AB \equal{} BC$, $ \angle ABC \equal{} 80, \angle CDA \equal{} 50$. Define $ E$ the midpoint of $ AC$; show that $ \angle CDE \equal{} \angle BDA$
[i](Paolo Leonetti)[/i]

[b]Problema 3.[/b]
Find all $ (x,y,z) \in \mathbb{Z}^3$ such that $ x^3 \minus{} 5x \equal{} 1728^{y}\cdot 1733^z \minus{} 17$.
[i](Paolo Leonetti)[/i]

[b]Problem 4.[/b]
Let $ a,b,c$ be positive reals; show that $ \displaystyle a \plus{} b \plus{} c \leq \frac {bc}{b \plus{} c} \plus{} \frac {ca}{c \plus{} a} \plus{} \frac {ab}{a \plus{} b} \plus{} \frac {1}{2}\left(\frac {bc}{a} \plus{} \frac {ca}{b} \plus{} \frac {ab}{c}\right)$
[i](Darij Grinberg)[/i]

[b]Problem 5.[/b]
Define the function $ g(\cdot): \mathbb{Z} \to \{0,1\}$ such that $ g(n) \equal{} 0$ if $ n < 0$, and $ g(n) \equal{} 1$ otherwise. Define the function $ f(\cdot): \mathbb{Z} \to \mathbb{Z}$ such that $ f(n) \equal{} n \minus{} 1024g(n \minus{} 1024)$ for all $ n \in \mathbb{Z}$. Define also the sequence of integers $ \{a_i\}_{i \in \mathbb{N}}$ such that $ a_0 \equal{} 1$ e $ a_{n \plus{} 1} \equal{} 2f(a_n) \plus{} \ell$, where $ \ell \equal{} 0$ if $ \displaystyle \prod_{i \equal{} 0}^n{\left(2f(a_n) \plus{} 1 \minus{} a_i\right)} \equal{} 0$, and $ \ell \equal{} 1$ otherwise. How many distinct elements are in the set $ S: \equal{} \{a_0,a_1,\ldots,a_{2009}\}$?
[i](Paolo Leonetti)[/i]

(For all doubts contact me in mp  :wink: )

Users who sent me solutions (which will be corrected as soon as possible):
Ahwingsecretangent
NickNafplio
Giuseppe R
mod_2
exodd
Zhero
azjps
Bugi
Palina.41
TBPL
mavropnevma
Ruby5099991
Maioc92
EUCLA
dario2994
shoki
kn
skytin
elendil
Edit: Inequalities Master,
Edit: grfield

[quote="bboypa"]Users who sent me solutions (which will be corrected as soon as possible):
Ahwingsecretangent
NickNafplio
Giuseppe R
mod_2
exodd
Zhero
azjps
Bugi
Palina.41
TBPL
mavropnevma
Ruby5099991
Maioc92
EUCLA
dario2994
shoki
kn
elendil[/quote]

I am pretty sure that I sent solutions, too..

Oh yes, also you, very sorry!You was the first contestant who sent me solutions, and there were other mails between your.. :P

Hi everyone! Mr Paolo when are you going to post the final results?

Hi! I'm very sorry for the delay, I have some exams to do, so I had no enough time to correct it..
However I hope in the next week to finish it  :(


Here's a problem I just stumbled across in Mathworld:

If x, y and z are real numbers, prove that if |x - y| :le: z :le: x + y, then
|y - z| :le: x :le: y + z and |z - x| :le: y :le: z + x.

I don't think it's very hard, but I'm honestly not really sure.

It stands for "less than or equal to".  In the old days of AoPS (before LaTeX), when you typed :le:, the forum would display a less-than-or-equal-to sign ($\le$).  The forum no longer does that, so many old posts look funny.

Anybody want to try Joel's problem?

[hide]
How about something like this:

$|x-y| \le x+y$

do casework on that and you'll see that $x \ge 0$ and $y \ge 0$, $z \ge 0$ follows.

now think of a triangle with sides x, y and z.

we have inequalities from that: $x+y \ge z$, $x+z \ge y$, $y+z \ge x$.

now as $x+z \ge y$ yields $z \ge y-x$ and
$z+y \ge x$ yields $z \ge x-y$ and as
$|x-y|$ equals either $x-y$ or $y-x$
we have $|x-y| \le z \le x+y$
now it's been shown that the triangle's sides are a solution to the inequality.

now same kind of reasoning goes for the latter $|z-x| \le y \le z+x$ etc.

[/hide]


Prove that there are infinitely many ordered triples of positive integers (x, y, z) such that x3 + y5 = z7.
3,5,7 should be superscript but i don't know how.

well, if you have a solution (x,y,z) the (x*a^35, y*a^21, z*a^15) is also a solution for any integer a, so given a solution you can generate infinitele. now you just have to find a solution but i am too lazy right now to do it


number theory
greatest common divisor
induction
binomial coefficients
number theory solved

If $ n$, $ i$, $ j$ are integers such that $ 0 < i\leq j < n$, prove that $ \gcd\left(\binom{n}{i},\binom{n}{j}\right) > 1$.

[i]Alternative formulation:[/i] If $ 0 < i\leq j < n$ for integers $ i$, $ j$, $ n$, prove that the greatest common divisor of the binomial coefficients $ C\left(n,i\right)$ and $ C\left(n,j\right)$ is greater than $ 1$.

I'll use the notation C(n,i) for the binomial coefficient $ \binom{n}{i}$ (it is also often denoted as nCi or C_n^i).

A try, using the nice Newton identity  :) :

C(n,j) * C(j,i) = C(n,i) * C(n-i,j-i)

If gcd(C(n,i),C(n,j)) = 1 then C(n,i) | C(j,i) which implies C(n,i) <= C(j,i)
which is strange from a combinatoric point of view.

Korrect ?

I have a solution, but I use different notations:

[color=blue][b]Problem.[/b] Let n be a positive integer, and let k and m be positive integers such that $ 0 < m\leq k < n$. Prove that the numbers $ \binom{n}{k}$ and $ \binom{n}{m}$ are not coprime.[/color]

[i]Solution.[/i] Assume, for the sake of contradiction, that the numbers $ \binom{n}{k}$ and $ \binom{n}{m}$ are coprime. We have n - m > 0 (since m < n) and $ 0\leq k \minus{} m < n \minus{} m$ (since $ m\leq k$ and k < n). Thus, the binomial coefficient $ \binom{n \minus{} m}{k \minus{} m}$ is integer and > 0. Note that $ \binom{n \minus{} m}{k \minus{} m} \equal{} \frac {\left(n \minus{} m\right)!}{\left(k \minus{} m\right)!\left(\left(n \minus{} m\right) \minus{} \left(k \minus{} m\right)\right)!} \equal{} \frac {\left(n \minus{} m\right)!}{\left(k \minus{} m\right)!\left(n \minus{} k\right)!}$, so that

$ \binom{n}{k}\cdot\binom{k}{m} \equal{} \frac {n!}{k!\left(n \minus{} k\right)!}\cdot\frac {k!}{m!\left(k \minus{} m\right)!} \equal{} \frac {n!}{m!\left(n \minus{} m\right)!}\cdot\frac {\left(n \minus{} m\right)!}{\left(k \minus{} m\right)!\left(n \minus{} k\right)!}$
$ \equal{} \binom{n}{m}\cdot\binom{n \minus{} m}{k \minus{} m}$.

Hence, the number $ \binom{n}{k}\cdot\binom{k}{m}$ is divisible by $ \binom{n}{m}$. But since we have assumed that the numbers $ \binom{n}{k}$ and $ \binom{n}{m}$ are coprime, this implies that the number $ \binom{k}{m}$ is divisible by $ \binom{n}{m}$. Hence, in particular, $ \binom{k}{m}\geq\binom{n}{m}$ (since $ \binom{k}{m}\neq 0$, as $ 0 < m\leq k$). But since k < n, we have

$ \binom{k}{m} \equal{} \frac {k\cdot\left(k \minus{} 1\right)\cdot ...\cdot\left(k \minus{} m \plus{} 1\right)}{m\cdot\left(m \minus{} 1\right)\cdot ...\cdot 1} < \frac {n\cdot\left(n \minus{} 1\right)\cdot ...\cdot\left(n \minus{} m \plus{} 1\right)}{m\cdot\left(m \minus{} 1\right)\cdot ...\cdot 1} \equal{} \binom{n}{m}$.

Note that here, we have used the constraint that 0 < m (in fact, if m would be 0, then the products in the above two fractions would both have no factors). Thus, we have shown $ \binom{k}{m} < \binom{n}{m}$, what contradicts with $ \binom{k}{m}\geq\binom{n}{m}$. Hence, our assumption that the numbers $ \binom{n}{k}$ and $ \binom{n}{m}$ are coprime must be wrong, and the problem is solved.

This problem appeared in Crux Mathematicorum 1995 as problem 1915, with a reference to

Erdös, Szekeres, [i]Some number theoretic problems on binomial coefficients[/i], Australian Math. Soc. Gazette 5 (1978) pp. 97-99.

  Darij

We first note that the claim can be easily checked to be true for $ n \equal{} 3$. We shall proceed to prove by induction on $ n$, so we assume that $ n > 3$, and that the claim is true for all integers $ n_0$ satisfying $ 3 \leq n_0 < n$.

We write $ {n \choose i}$ and $ {n \choose j}$ as $ \frac {n}{i}{n \minus{} 1 \choose i \minus{} 1}$ and $ \frac {n}{j}{n \minus{} 1 \choose j \minus{} 1}$ respectively. Observe that $ \frac {n}{\gcd (n,i)} \mid {n \choose i}$ and $ \frac {n}{\gcd (n,j)} \mid {n \choose j}$. For the sake of brevity, we denote $ A \equal{} \gcd (n,i), B \equal{} \gcd (n,j)$. If any of $ A,B \equal{} 1$, WLOG say $ A \equal{} 1$, then $ \frac {n}{B} \mid n \Rightarrow \gcd ({n \choose i}, {n \choose j}) \geq \frac {n}{B} > 1$, and we are done.

Thus we can assume $ A,B > 1$. Let $ \gcd (A,B) \equal{} C, A \equal{} CA_0, B \equal{} CB_0$. We note that $ \gcd(A_0, B_0) \equal{} 1$. Thus, we have $ B_0 \mid {n \choose i}$ and $ A_0 \mid {n \choose j}$. Obviously $ CA_0B_0 \mid n$. If $ C > 1$, then since $ \frac {n}{CA_0B_0} \in \mathbb{Z}^ \plus{}$, it follows that $ \frac {n}{A_0B_0} \in \mathbb{Z}^ \plus{}$ and $ \frac {n}{A_0B_0} > 1$. Noting that $ \frac {n}{A_0B_0}$ divides both $ {n \choose i}$ and $ {n \choose j}$, we are done.

We thus can assume that $ C \equal{} 1$. If $ n > A_0B_0$, then $ \frac {n}{A_0B_0} > 1$ divides both $ {n \choose i}$ and $ {n \choose j}$, and we are done. So we can also assume that $ n \equal{} A_0B_0$.

This implies that $ {n \choose i} \equal{} B_0{n \minus{} 1 \choose i \minus{} 1}$ and $ {n \choose j} \equal{} A_0{n \minus{} 1 \choose j \minus{} 1}$. By Induction Hypothesis, there exists an integer $ d > 1$ such that $ \gcd ({n \minus{} 1 \choose i \minus{} 1}, {n \minus{} 1 \choose j \minus{} 1}) \equal{} d$. Thus $ d \mid \gcd({n \choose i}, {n \choose j})$, and we are done.

Suppose on the contrary that ${n\choose i}$ and ${n\choose j}$ are relatively prime.
Consider the identity:
${n\choose i}{n-i\choose j}$=${n\choose j}{n-j\choose i}$
Then ${n-j\choose i}$ is divisible by ${n\choose i}$, impossible.
Observe that using the identity ${n\choose k}={n\choose n-k}$ we may assume that $i\leq n/2$ and that $j\leq n/2$, thus $i+j\leq n$ .

[quote="dblues"]Noting that $ \frac {n}{A_0B_0}$ divides both $ {n \choose i}$ and $ {n \choose j}$[/quote]

Looks wrong (let $i=j$, then $A=B$, so that $A_0=B_0=1$, but $n$ needs not divide ${n\choose i}$).


quadratics
search
algebra
polynomial
Diophantine equation
number theory unsolved
number theory

1) Show that 4xyz - x - y is never a perfect square for x,y,z possitive integers.

2) Solve in possitive integers: $ a^2 \plus{} 1 \equal{} b^2(c^2 \minus{} 4)$

The first one is trivial if z is a quadratic residue modulo a prime divisor of the form 4κ+3 of 4xz - 1 (this prime divisor obviously exists)... maybe we can show this in some way. I have tried to work modulo 4xy also but I cannot find a solution

Can anyone help me?

The problem 1 was posted by moldovan search it in this forum

For the problem 2:
Let $ d_{1,2}\equal{}\frac{2bc\pm\sqrt{((2b)c)^2\minus{}4((2b)^2\plus{}1)(1)}}{2}$.it's easy to understand that $ d_{1,2}$ are  positive integers and also they are the roots of this polynomial:
$ X^2\plus{}(2b)^2\plus{}1\minus{}(2b)(c)X\equal{}0$ it follows that $ \frac{d^2\plus{}(2b)^2\plus{}1}{(2b)d}$ is an integer (equal to $ c$). but it's known that if it is an integer it is equal to 3. So we must have $ c\equal{}3$.The rest is a pell equation that u can easily solve
$ a^2\minus{}5b^2\equal{}\minus{}1$.DONE!


A yeast extract contains all the enzymes required for alcohol production.   The extract is incubated under anaerobic conditions in 1 liter of medium containing: 200 mM glucose, 20 mM ADP, 40 mM ATP, 2 mM NADH, 2 mM NAD+ and 20 mM Pi (inorganic phosphates).  Ethanol production can be summarized by the following equation:

                                                       C6H12O6--------->2 C2H5OH + 2 CO2 + 2ATP
What is the maximum amount of ethanol that can be produced under these conditions?

A. 	2 mM
B. 	20 mM
C. 	40 mM
D. 	200 mM
E. 	400 mM

is it (B) :maybe:

i m not sure 'coz ATP can also cause feedback inhibition...........

Can you write how did you solve problem? :maybe:

does that mean it is correcT?? :o

i just took ADP and Pi as limiting factors

yeah! that is true but i couldnt solve problem by the way that you proposed.can you write more detailed.

only as long as ADP is available, the process can take place

one ADP forms one ATp
so we get maximum amt of ATP synth possible
calculate the coresponding amt of Ethanol :)

good joke!!

i m never going 2 appear :(  :P


algebra
function
domain
calculus
calculus computations

Is it true that for a continuous $f$,
$f$($\limsup{x_{n}}$) = $\limsup f({x_{n}}$)?

No. Even if you require that $\limsup x_{n}$ is in the domain of $f$, there are plenty of things that can go wrong. Try to construct a counterexample.

Hint: It would be true if we added that $f$ is increasing.

could someone give me a more concrete explanation?
I can't really think of any counter examples..
It seems like for both decreasing and increasing $f$, the $f(\limsup x_{n}) = \limsup f(x_{n})$...
Thanks again.

never mind, i think i found the solution.
we can construct $x_{n}$ s.t. it does not converge.
More specifically,

$x_{n}= n$ (for $n$ odd)

and

$x_{n}= \frac{1}{n}$ (for $n$ even)

Then, we have $x_{n}={1, 1/2, 3, 1/4, 5, ...}$

For $f(\limsup x_{n})$, we can clearly see that the $\limsup x_{n}= \infty$
and thus $f(\limsup x_{n})=f(\infty)$

For $\limsup f(x_{n})$, and with $f$ decreasing,
$sup f(x_{n})$ is achieved for the smallest possible $x_{n}$ which is 0 ($lim x_{n}= 0$ for $n$ even).
thus, we have $\limsup f(x_{n})=f(0)$

Since we have $f(\limsup x_{n})=f(\infty)$ and $\limsup f(x_{n})=f(0)$,
we can conclude that it is not necessarily true that $f(\limsup x_{n})=\limsup f(x_{n})$ for $f$ continuous on $R$.

No, this is not a valid counterexample if you are working with a continuous function defined on $\mathbb{R}$. $f(\infty)$ is meaningless. Now, I think we can make your counterexample work if we work with a continuous function defined on the two-point compactification of $\mathbb{R}$ (although I am not quite sure $\limsup$ and $\liminf$ make sense in this context), but certaintly that is not the point of this exercise...

Alternately, make your sequence something like $1,\frac12,2-\frac13,\frac14,2-\frac15,\dots$, alternating between $\frac1n$ and $2-\frac1n$, and use a decreasing function.


geometry
trigonometry
incenter
geometric transformation
homothety
ratio
search

In acute-angled triangle $ABC$, a semicircle with radius $r_a$ is constructed with its base on $BC$ and tangent to the other two sides. $r_b$ and $r_c$ are defined similarly. $r$ is the radius of the incircle of $ABC$. Show that \[ \frac{2}{r} = \frac{1}{r_a} + \frac{1}{r_b} + \frac{1}{r_c}. \]

If $AD$ is the bisector of $\angle BAC$, then $AD=\frac{2bc}{b+c}\cos\frac{\hat A}2$, $D$ is the center
of the semicircle and $\angle DAC=\frac{\hat A}2$, so we have
\[ r_a=AD\cdot\sin\frac{\hat A} 2=\frac{bc}{b+c}\sin\hat A=\frac{2sr}{b+c} \]
where, as usual, $s=\frac{a+b+c}2$.
Consequently
\[ \frac 1{r_a}+\frac 1{r_b}+\frac 1{r_c}=\frac{b+c}{2sr}+\frac{c+a}{2sr}+\frac{a+b}{2sr}=\frac{4s}{2sr}=\frac 2 r \]

Another solution:

Let the angle bisectors of the angles CAB, ABC and BCA intersect the lines BC, CA and AB at the points X, Y and Z, respectively. Also, we know that these angle bisectors all meet at the incenter I of triangle ABC. Then, by the Gergonne theorem (see http://cut-the-knot.com/triangle/Gergonne.shtml ), we have $\frac{AI}{AX}+\frac{BI}{BY}+\frac{CI}{CZ}=2$.

Since the semicircle with radius $r_a$ touches the sides CA and AB of the triangle ABC, its center must lie on the angle bisector of the angle between these two sides, i. e. on the angle bisector of the angle CAB. On the other hand, since this semicircle has its base on the line BC, its center must lie on the line BC. Thus, the center of the semicircle with radius $r_a$ is the point of intersection of the angle bisector of the angle CAB with the line BC, i. e. the point X. Similarly, the centers of the semicircles with radii $r_b$ and $r_c$ are the points Y and Z.

Now, consider the incircle of triangle ABC and the circle with center X and radius $r_a$. The lines CA and AB touch both of these circles; thus, they are the external common tangents of these two circles; consequently, the point of intersection A of these lines CA and AB is the external center of similitude of these two circles. Hence, there exists a homothety with center A which maps the circle with center X and radius $r_a$ to the incircle of triangle ABC.

Now, on the one hand, the ratio of this homothety is $\frac{r}{r_a}$ (in fact, it maps the circle with center X and radius $r_a$ to the incircle of triangle ABC with radius r), but on the other hand, the ratio of this homothety is $\frac{AI}{AX}$ (in fact, it has the center A, and it maps the circle with center X and radius $r_a$ to the incircle of triangle ABC with center I). Hence, $\frac{r}{r_a}=\frac{AI}{AX}$. Similarly, $\frac{r}{r_b}=\frac{BI}{BY}$ and $\frac{r}{r_c}=\frac{CI}{CZ}$. Hence, the equation $\frac{AI}{AX}+\frac{BI}{BY}+\frac{CI}{CZ}=2$ becomes $\frac{r}{r_a}+\frac{r}{r_b}+\frac{r}{r_c}=2$, so that $\frac{1}{r_a} + \frac{1}{r_b} + \frac{1}{r_c} = \frac{2}{r}$, and we are done.

Please forgive me for my style (I'm half asleep)...

  Darij

This problem was created by Bruce Merry, former South African gold medalist :)

By Cauchy we have that $(r_a+r_b+r_c)\left(\frac{1}{r_a}+\frac{1}{r_b}+\frac{1}{r_c}\right)\geq 9$
Bur since $\frac{1}{r_a}+\frac{1}{r_b}+\frac{1}{r_c}=\frac{2}{r}$, we have that
$\frac{9}{2}r \leq r_a+r_b+r_c$

This problem reminds so much (Tobias Pfanner 2004, Bosnia - Herzegovina Mathematical Olympiad 2002, USA International Math Talent Search IMTS 1992, Day 1, Australian-Polish Mathematical Competition APMC 1981)
http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=r_ar_b+tobias&t=55605

So,can we find the best constant $c$ such that $r_a+r_b+r_c\leq cr$

Wonderful solution, Darij!

   You skillfully avoided steering the solution towards the infamous Inradii Problem,

   a fact which will anger Yetti , but it is OK.

   It was also easy to tell that you were partially asleep, because you mentioned

   'the other hand' 3 times.  Was it, by chance, Gergonne's hand you were dreaming

    of?

   Viva Gergonne!

   Thank you.

   M.T.

[quote="Beat"]So,can we find the best constant $c$ such that $r_a+r_b+r_c\leq cr$[/quote]

Nice question. In fact, we have $r_a+r_b+r_c<5r$, and the bound is sharp (equality is obtained for the degenerate triangle with sidelengths 0, 1, 1).

[i]Proof.[/i] We have to show the inequality $r_a+r_b+r_c<5r$. Division by r transforms this into $\frac{r_a}{r}+\frac{r_b}{r}+\frac{r_c}{r}<5$.

Now, since $\frac{r}{r_a}=\frac{AI}{AX}$, we have $\frac{r_a}{r}=\frac{AX}{AI}$. Now, $\frac{AI}{IX}=\frac{b+c}{a}$ (this is a well-known fact; it was proven, e. g., in [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=320163#p320163]http://www.mathlinks.ro/Forum/viewtopic.php?t=50582 post #3[/url], second solution, in the equivalent form $\frac{CI}{IW_c}=\frac{a+b}{c}$). Thus, $\frac{IX}{AI}=\frac{a}{b+c}$, so that $\frac{AX}{AI}=\frac{AI+IX}{AI}=1+\frac{IX}{AI}=1+\frac{a}{b+c}$. Hence, $\frac{r_a}{r}=\frac{AX}{AI}=1+\frac{a}{b+c}$. Similarly, $\frac{r_b}{r}=1+\frac{b}{c+a}$ and $\frac{r_c}{r}=1+\frac{c}{a+b}$. Hence, the inequality in question, $\frac{r_a}{r}+\frac{r_b}{r}+\frac{r_c}{r}<5$, becomes $\left(1+\frac{a}{b+c}\right)+\left(1+\frac{b}{c+a}\right)+\left(1+\frac{c}{a+b}\right)<5$, or, equivalently, $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<2$. But this is a well-known triangle inequality and was proven in http://www.mathlinks.ro/Forum/viewtopic.php?t=42896 . Proof complete.

[quote="armpist"]Wonderful solution, Darij![/quote]

Thank you :oops:

[quote="armpist"]You skillfully avoided steering the solution towards the infamous Inradii Problem, a fact which will anger Yetti , but it is OK.[/quote]

Ehm, it is OK, but I would state it a bit differently: Unlike you, I don't feel obliged to attempt to link every single geometrical problem to the CBT, or to the inradii problem, or any of your other few idee-fixes (though I am fond of interconnections between different problems - but they must really exist and not just be pure speculations).

[quote="armpist"]It was also easy to tell that you were partially asleep, because you mentioned 'the other hand' 3 times.[/quote]

Hmm, I found it only 2 times in my post. However, I am not really famous on this forum for the literary diversity of my posts and their imaginative language, as I am, alas, not a poet, and prefer mathematical correctness.

[quote="armpist"]M.T.[/quote]

Major who (this time)?

[b]PS.[/b] Just wanted to say that I did receive your PM on the Apollonius circles - which are actually not really Apollonius circles, i. e. this is not their main significance, but I'll let you figure out why -, and I liked the result - it seems to be true, yes. What you are good in is making conjectures, as you seem to have a quite precise geometrical intuition. I am thinking about the proof, and I hope I will find it in the next time.

  Darij

Hello Darij,

 Thanks for your unexpectedly kind words. Is it the Holidays Season that

 did the trick of mellowing you down?  Have a good Holiday!

 Anyways, as usual I have a question to ask:

   There is another relationship in this particular triangle    1 / r  -  1 /  Rc  =  1 / Hc

                  r -  incircle,   Rc - semicircle on side c,  Hc - altitude to c

   Naturally, the same for   a    and    b.

  How to prove this without pain?  Is there anything up Gergonne's sleeve to handle

  this algebra?

   Thank you.

   M.T.

This problem is nice and very naturally. [hide="Here is the shortest proof (in my opinion)."][color=darkblue]For the acute $\triangle ABC$ denote the [u]area[/u] $S=[ABC]$ and the [u]semiperimeter[/u]
$p=p(ABC)$. Define the reflection $A'$ of the point $A$ w.r.t. the side-line $BC$.
The deltoid $ABA'C$ is circumscribed to the circle $C(S_a,\rho_a)$, where $S_a\in (BC)$ and
$\widehat {BAS_a}=\widehat {CAS_a}$, i.e. $\frac{S_aB}{S_aC}=\frac cb$ (the ray $[AS_a$ is the bisector of the angle $\widehat {BAC}$).
Thus, $2pr=2S=[ABA'C]=p(ABA'C)\cdot \rho_a=(b+c)\rho_a\Longrightarrow \frac{1}{\rho_a}=\frac{b+c}{2pr}$ a.s.o.
Therefore, $\sum \frac{1}{\rho_a}=\sum \frac{b+c}{2pr}=\frac 2r\Longrightarrow \boxed {\ \frac{1}{\rho_a}+\frac{1}{\rho_b}+\frac{1}{\rho_c}=\frac 2r\ }\ .$[/color][color=darkred] [b][u]That is all ![/u][/b][/color]

[color=red][b]Remark.[/b] $l_a\equiv AS_a=\frac{2bc\cos \frac A2}{b+c},\ \rho_a =l_a\sin \frac A2=\frac{bc\sin A}{b+c}\Longrightarrow \rho_a=\frac{2S}{b+c}\ .$[/color][/hide]

here is another solution:

distances from a point $P$ on the side $AB$ of the triangle $ABC$ to the lines $AC$ and $BC$ are $kh_b$ and $(1-k)h_a$, where $h_a,h_b,h_c$ are the altitudes of the triangle $ABC$, for some $k$.we search $k$ for which $kh_b=(1-k)h_a$ that give us $k=\frac{h_a}{h_a+h_b}$, so the equal distances (that are the same of $r_c$) are $\frac{h_ah_b}{h_a+h_b}$.

so

$\frac{1}{r_a}+\frac{1}{r_b}+\frac{1}{r_c}=$ $\frac{h_b+h_c}{h_bh_c}+\frac{h_a+h_c}{h_ah_c}+\frac{h_a+h_b}{h_ah_b}$ $=\frac{2}{h_a}+\frac{2}{h_b}+\frac{2}{h_c}$

$\frac{2}{h_a}+\frac{2}{h_b}+\frac{2}{h_c}$ $=\frac{2a}{ah_a}+\frac{2b}{bh_b}+\frac{2c}{ch_c}$ $=\frac{a+b+c}{S}=\frac{2}{r}$


linear algebra
matrix
calculus
calculus computations

Using $ \textbf{x}'\equal{}\textbf{P}(t)\textbf{x}$, show that $ \frac{dW}{dt}\equal{}(p_{11}\plus{}p_{22})W$ where $ W$ is the Wronskian of $ \textbf{x}^{(1)}$ and $ \textbf{x}^{(2)}$.  $ \textbf{x}^{(1)}$ and $ \textbf{x}^{(2)}$ are solutions of the first equation.

[quote="JRav"]Using $ \textbf{x}' \equal{} \textbf{P}(t)\textbf{x}$, show that $ \frac {dW}{dt} \equal{} (p_{11} \plus{} p_{22})W$ where $ W$ is the Wronskian of $ \textbf{x}^{(1)}$ and $ \textbf{x}^{(2)}$.  $ \textbf{x}^{(1)}$ and $ \textbf{x}^{(2)}$ are solutions of the first equation.[/quote]

what are $ p_{11}, p_{22}$ ?

[quote="JRav"]The entries of the matrix $ \textbf{P}(t)$.[/quote]

Sorry ! there's some thing I don't understand.

you mean :  $ x \left( t \right) \equal{} \left( {x_{_1} \left( t \right)} ; {x_{_2} \left( t \right)} \right)$ ?  and $ x_{_1} \equal{} x^{(1)} ?$ :maybe:


Let $ ABC$ be a triangle with $ AB \equal{} AC$ . The angle bisectors of $ \angle C AB$ and $ \angle AB C$ meet the sides $ B C$ and $ C A$ at $ D$ and $ E$ , respectively. Let $ K$ be the incentre of triangle $ ADC$. Suppose that $ \angle B E K \equal{} 45^\circ$ . Find all possible values of $ \angle C AB$ .

[i]Jan Vonk, Belgium, Peter Vandendriessche, Belgium and Hojoo Lee, Korea [/i]

If the calculations are ok then the answer is
$ 60^{\circ}$ or $ 90^{\circ}$.

[hide="Solution, I think"]

Compute all angles, and mark S as the incenter of ABC, and mark $ \angle CBE \equal{} x$. $ \angle AES \equal{} 3x,\angle SAE \equal{} 90^\circ \minus{} 2x$

Now, using sine theorem, $ AS: AE \equal{} \sin 3x: \cos x$, but $ AS \equal{} \frac {a}2\cdot\tan x\cdot \frac {1 \plus{} \tan^2 x}{1 \minus{} \tan^2 x}$ and $ AE \equal{} \frac {b^2}{b \plus{} a}$ from angle bisector theorem and sine theorem for other triangles. Also $ \frac a2 \equal{} b\cos 2x$. Now you get a equation of sin x,cos x, tan x. Solve it to get the solutions. I got $ 2\cos^3 x \minus{} 3\cos^2 x \plus{} 1 \equal{} 0$, so $ \cos x \equal{} 1$ or $ \cos x \equal{} \frac { \minus{} 1}2$ I don't know if it's correct, but the problem can be surely done like this

EDIT: I made a mistake, and I didn't use anywhere the angle of 45. But again, we can do the same in triangle SKE, for example, and solve trigonometric equation... [/hide]

Here is an idea of my solution: let I be an incentre of ABC, and IF be perpendicular to AC (F belongs to AC). If F=E, we have an answer 60, if not - then FEIK is cyclic and answer is 90. And it is not difficult to check this two variants.

Let $ K,H$ be the incenter of triangle $ ADC, ABC.AK\cap HE = O$. Put $ AB = AC = x, BC = y$
$ \angle AKH = \angle HEK = 45^o$ then $ HK^2 = HO.HE$
But ${ \frac {HO}{HE} = \frac {AH}{AH + AE}, \frac {HK^2}{HC^2} = \frac {HD^2}(HD + DC)^2}$
We get $ \frac {HC^2.HD^2}{(HD + DC)^2} = \frac {HE^2.AH}{AH + AE}$
$ \Rightarrow \frac {HD^2}{(HD + DC)^2} = \frac {HE^2}{HC^2}.\frac {AH}{AH + AE} = \frac {AE^2}{AB^2}.\frac {AH}{AH + AE}$
$ \Rightarrow \frac {HD}{(HD + DC)^2} = \frac {AE^2}{AB^2}.\frac {AH}{HD(AH + AE)}$
$ \Rightarrow \frac {HD}{(HD + DC)^2} = \frac {AE^2}{AB^2}.\frac {AB}{BD(AH + AE)} = \frac {AE^2}{AB.BD.(AH + AE)}$
$ \Rightarrow AE^2.(HD + DC)^2 = HD.AB.BD.(AH + AE)$
$ \Rightarrow \frac {x^4}{(x + y)^2}.(\frac {1/2y.AD}{1/2y + x} + 1/2y)^2 = \frac {1/2y.AD}{1/2y + x}.x.1/2y.(\frac {x.AD}{1/2y + x} + \frac {x^2}{x + y})$
$ \Rightarrow \frac {x^2}{(x + y)^2}.(\frac {1/2AD}{1/2y + x} + 1/2)^2 = \frac {1/2AD}{1/2y + x}.1/2.(\frac {AD}{1/2y + x} + \frac {x}{x + y})$
Then set $ x = ky$ and note that $ AD = \sqrt {k^2 -1/4}y$ we can easy find $ k$.
$ \Rightarrow \angle A = 60^o$ or $ 90^o$.
PS: Bad proof but easy if we calculate  :blush:

[quote="Peter"]True, but there is a purely synthetical solution. :)[/quote]
[quote="orl"]Solution by the proposers can be found [url=http://ideahitme.wordpress.com/2009/07/16/50th-imo/]here.[/url][/quote]

Like that :D

Let $ D$ be the midpoint of $ BC$; $ I$ be the incenter  of $ \triangle ABC$. $ DK$ meets $ AC$ at
point $ T$.Let $ AB=AC=a$;$ BC=2$. It means $ CD=DB=1$
Firstly, triangle $ IEK$ and triangle $ CDT$ are similar (since $ \angle IED=\angle EDC = 45^o$). Thus $ \frac{IE}{IK}=\frac{CD}{CT}$.
We have:
$ \frac{IK}{ID}=\frac{IC}{ID+1} \rightarrow \frac{1}{IK}=\frac{ID+1}{ID.IC}$
$ \frac{IE}{EC}=\frac{IB}{2}\rightarrow IE=\frac{EC.IC}{2}$ (notice that $ IB=IC)$.
Thus $ \frac{IE}{IK}=(1+\frac{1}{ID})\frac{EC}{2}$
In triangle $ ABC$ : $ \frac{EC}{BC}=\frac{EC}{2}=\frac{AC}{AB+BC}=\frac{a}{a+2}$.
But $ \frac{ID}{DC}=\frac{AD}{AC+DC}=\frac{AD}{a+1}$
Hence $ \frac{IE}{IK}=\frac{(AD+a+1)a}{AD(a+2)}$.
On other hand; $ \frac{CD}{CT}=\frac{AD}{AT}=\frac{1+AD}{a}$.
It implies
\[ \frac{(AD+a+1)a}{AD(a+2}=\frac{1+AD}{a}\]
\[ \leftrightarrow AD.a^2+a^3+a^2=a.AD+a.AD^2+2.AD+2.AD^2\]
\[ \leftrightarrow AD.a^2=a.AD-a+2AD+a^2-2\]
(since $ AD^2=a^2-1$)
\[ \leftrightarrow
AD(a^2-a-2)=a^2-a-2\]
Thus either $ AD=1$ or $ a^2-a-2=0$.
$ AD=1$ implies $ \angle A =90^o$
$ a^2-a-2=0$ implies $ a=2$ thus $ A=60 ^o$.
We have done!.

Simple solution by trigonometry calculation :)

Let $ \alpha = \measuredangle DAC$, and consider triangle $ %Error. "traingel" is a bad command.
EJC$, where $ J$ is the incenter of $ \triangle ACD$.
According to the assumption, this triangle has angles $ \frac{3\alpha}{2}, 45^\circ - \frac{\alpha}{2}, 135^\circ - \alpha$.
From Sine law in $ \triangle BEC$, we have
\[ \frac{EC}{\sin 45^\circ - \alpha/2} = \frac{2R \sin 2 \alpha}{\sin {45^\circ - 3\alpha/2}}.\]
From Sine law in $ \triangle AJC$, we have
\[ \frac{JC}{\alpha/2} = \frac{2R \sin {90^\circ - \alpha}}{\sin 135^\circ}.\]

Finally, in triangle $ \triangle EJC$, we have
\[ \frac{EC}{\sin{135^{\circ} - \alpha}} = \frac{JC}{\sin {3 \alpha / 2}}.\]

After simplification, we have following identity
\[ \frac{\sin {45^\circ + \alpha}}{\sin \alpha} \cdot \frac{\sin {45^\circ + 3\alpha/2}}{\sin {3 \alpha/2}} =
\sqrt{2} \cdot \frac{\sin {45^\circ - \alpha/2}}{\sin {\alpha/2}}.\]

Now, set $ x = \cot \frac{\alpha}{2}$ and using formulas for double and triple angle, it follows
\[ \left(\frac{x^2-1}{2x}+1\right)\left(\frac{3x-x^3}{1-3x^2}+1\right)=2(x-1).\]

This equation is equivalent to $ (x-1) (x^2-4x+1) (x^2-2x-1)=0$.
It follows that $ \cot 2x = \ctg \alpha = \sqrt{3}$ or $ \cot 2x = \ctg \alpha = 1$. This brings us to $ \alpha = 45^\circ$ or $ \alpha = 30^\circ$, and finally $ \measuredangle CAB = 90^\circ$ or $ \measuredangle CAB = 60^\circ$.

Yeah, I too have solved it by trigonometry but it is not an elegant one after all.
I tused the cotangent rule in $ BEC$. That is for a trinagle $ ABC$ divided in the ratio $ m: n$ internally by a cevian making angle $ \theta$ , along with the base angles being $ \alpha$ and $ \beta$ respectively, $ (m\plus{}n)\cot \theta \equal{} m\cot \alpha \plus{} n \cot \beta$. THis laong with the angle bisector theorem got me trhough the sum. TOok some time still.
Yes hojoo lee is having his 8th problem in to the IMO {i think so , probably more}.

[quote="Agr_94_Math"]Yes hojoo lee is having his 8th problem in to the IMO {i think so , probably more}.[/quote]

I think it is problem number 9.

[quote="Valentin Vornicu"]And thus Hojoo manages to squeeze one more IMO problem in :D Good job guys![/quote]

 I really enjoyed our cowork.  Whenever Jan and Peter created breakthroughs on our problems, I was very happy!

Congratulations on 8(9?) th problem on IMO!

Let $\angle ICD = \angle ICE = x$. Simple angle chase yields that $\angle EIK = 2x, \angle KEC  = 135-3x, \angle ECK = x$. It follows that:
$$\frac{IK}{KC} = \frac{\frac{IK}{EK}}{\frac{KC}{EK}} = \frac{\frac{\sin (45)}{\sin (2x)}}{\frac{\sin (135-3x)}{\sin(x)}} = \frac{\sin (45) \sin (x)}{\sin (135-3x) \sin (2x)}$$
We also know that:
$$\frac{IK}{KC} = \frac{ID}{DC} = \tan (x)$$
Hence,
$$\tan (x) = \frac{\sin (45) \sin (x)}{\sin (135-3x) \sin (2x)} \Longleftrightarrow$$ $$\cos (x) \sin (45) = \sin (135-3x)\sin(2x) \Longleftrightarrow$$
$$\sin (45+x) - \sin (x-45) = \cos (135-5x) - \cos (135-x) \Longleftrightarrow$$
$$\sin (45+x) = \cos (135-5x) \Longleftrightarrow$$
From here either $45 - x = 135 - 5x \Longleftrightarrow x = \frac{45}{2}$ or $45+x - 135 + 5x = 90 \Longleftrightarrow 6x = 180 \Longleftrightarrow x =30$. Hence, the possible values for $\angle CAB$ are $60,90$.

Denote by $I$ the incenter of $\triangle ABC$. Notice that both $I$ and $K$ lie on the bisector of $\angle BCA$. Therefore, I, C and K are collinear. Construct point $F$ on segment $AC$ such that $IF$ is a height of $\triangle CIA$. It's easy to see that $\triangle CID$ and $\triangle CIF$ are congruent. In particular, $\angle IDK = \angle IFK = 45^o$, because $DK$ is a bisector of the right angle $\angle CDI$.

Construct a circle passing through $I$ and $K$ such that the chord $IK$ delimits a small arc of $90^o$ facing away from line $AC$. This circle is the geometric locus of all points $P$ with $\angle IPK = 45^o$. Therefore, $E$ lies on this circle, and on the segment $AC$. Since the intersection of a circle and a line is at most two points, there can be at most two possibilities.

We already know that $F$ lies on $AC$ and has $\angle IFC = 45^o$, so either $E=F$ or $E$ is another point.

If $E=F$, then the bisector $AE$ is also a height, which means that $AB=BC$. Therefore, the triangle is equilateral and $\angle CAB = 60^o$.

If $E \neq F$, then the points $I$, $K$, $E$ and $F$ form a cyclic quadrilateral. In particular, $\angle KIF = \angle KEF$. Denote $\angle ICF = \alpha$ and $\angle KEF = \beta$. Summing angles on $\triangle CIF$, we get $\beta = 90^o - \alpha$, and summing on $\triangle CBE$, we get $3\alpha + \beta = 135^o$. Solving this, we get $\alpha = 22,5^o$, which implies $\angle CAB = 90^o$.

Weird problem.

Let $D'$ denote the reflection of $D$ over line $IC$. As $\measuredangle ID'K=\measuredangle KDI=45^\circ=\measuredangle IEK$, point $D'$ lies on $(IEK)$. For now, we assume $D'\ne E$. We deal with the case $D'=E$ later. Note that $D'$ lies on $AC$ because $IC$ bisects $\angle ACD$. Now, observe that
\[\measuredangle AEK=\measuredangle D'EK=\measuredangle D'IK=\measuredangle KID=\measuredangle KIA.\]
Let $I'$ denote the reflection of $I$ over line $AK$. Similarly to before, $I'$ lies on $(AEK)$. Similarly to before, note $I'$ lies on $AKC$, implying either $(AEK)$ is a line, which is absurd, or $I'=E$.

Now, we determine what it means for $I'=E$ or $D'=E$ to hold. In the first case, we get $AI=AE$, so
\[180^\circ-\angle ABE-\angle BAE=\angle AEB=\angle AEI=\angle AIE=\angle DIB=90^\circ-\angle IBD.\]
Since $\angle IBD=\angle ABE$, this yields $\angle BAE=90^\circ$, so $\angle CAB=90^\circ$. Since all steps were reversible, this case works. In the latter case, let $\phi$ denote reflection over $IC$ and note that because $\phi(D)=E$, we must have $\phi(A)=\phi(ID\cap EC)=IE\cap DC=B$. This implies that $\angle BAC=\angle ABC$, so $\angle CAB=60^\circ$. Again, because all steps were reversible, this case works.

In conclusion, we have $\angle ACB\in\{60^\circ,90^\circ\}$.

Let $0<\angle ABC=\alpha<90^{\circ}$. Let $I$ be the incentre of $\triangle ABC$.
Then by the sine law on $\triangle IEK$, $\triangle IDK$, $\triangle AEI$ and the angle bisector theorem on $\triangle ACD$, we get the following angle identity,
$$\sin{(45^{\circ}+\frac{\alpha}{2})}=\sin{(\frac{3\alpha}{2})}\sin{(45^{\circ}+\alpha)}.$$
At the first sight, this equation seems unfeasible to solve by hand, but by some clever algebraic manipulations, we obtain that either $\tan{(\alpha)}=1$ or $\tan{(\alpha)}\cdot \tan{(\frac{\alpha}{2})}=1$. Thus from the first one, we have $\alpha=45^{\circ}$ and from the second one, we have $\alpha=60^{\circ}$. Therefore, we get two possible values of $\angle CAB$, namely $90^{\circ}$ and $60^{\circ}$. Pretty trivial to see those work.

We claim that the answer is $60^{\circ}, 90^{\circ}$. It is clear that these work, so we prove that these are the only angles possible.

Let $I$ be the incenter of $\triangle ABC$. The main idea is to construct $F$, the intouch point on $AC$. Notice that $F$ is also the reflection of $D$ over $CI$, therefore $\angle IFK = \angle IDK = 45^{\circ}$ as $K$ is the incenter of $\triangle ADC$.

[b][color=#f00]Case 1:[/color][/b] $E=F$.

This implies that $B,I,F$ are collinear. Since $IF\perp AC$, $BF\perp AC$. This means that the $B$-angle bisector is also the $B$-altitude, so $\triangle BEA \cong \triangle BEC$, and thus $AB=BC$. However, this directly implies that $\triangle ABC$ is equilateral and thus $\angle BAC = 60^{\circ}$.

[b][color=#f00]Case 2:[/color][/b] $E\neq F$.

As $\angle IEK=\angle BEK = 45^{\circ} = \angle IFK$, $EFIK$ is a cyclic quadrilateral. Then we must have $\angle IKE=180^{\circ}-\angle IFE = 180^{\circ}-90^{\circ} = 90^{\circ}$, so $EK\perp IC$.

Let $x=\angle ACB$. It is clear that $\angle EBC = \frac{1}{2}x$ and $\angle ECB =x$. However, notice that $\angle EKC = 90^{\circ} - \angle ECI = 90^{\circ}-\frac{1}{2}x$, so $\angle BEC = \angle BEK +\angle KEC = 135^{\circ} -\frac{1}{2}x$. Since the angles in triangle $BEC$ must add to $180^{\circ}$, we have $$x+\frac{1}{2}x+(135^{\circ} -\frac{1}{2}x)=180^{\circ}.$$ Solving, we get $x=45^{\circ}$, implying that $\angle BAC = 90^{\circ}$.

Solved with [b]L567[/b].

[hide=Solution]
Use Ratio Lemma and the Angle Bisector Theorem in $\triangle DIC$ and $\triangle IEC$. \\
We get: \\
$$\frac{\sin \alpha}{\cos \alpha} = \frac{DI}{DC} = \frac{IK}{KC} = \frac{\sin \alpha}{\sin 2\alpha} \cdot \frac{\sin 45^{\circ}}{\sin (45^{\circ}+3\alpha)}$$
We therefore have $$\sin (2\alpha)\sin(45^{\circ}+3\alpha) = \cos \alpha \sin 45^{\circ}$$
Then $$\cos(-\alpha-45^{\circ})-\cos(5\alpha+45^{\circ}) = \sin(\alpha+45^{\circ}) - \sin(45^{\circ}-\alpha) = \sin(\alpha+45^{\circ})-\cos(\alpha+45^{\circ})$$
We finally get the equation $$\sin(\alpha+45^{\circ}) = \sin (5\alpha-45^{\circ})$$
and consequently $\alpha \in \frac{\pi}{8},\frac{\pi}{6}$
We get that $\angle CAB \in \{60^{\circ}, 90^{
\circ}\}$. $\blacksquare$[/hide]

Let $I$ be the incenter of triangle $ABC$ and by Quadrilateral Ceva on $CDIE$ $$\frac{\sin45\cdot\sin\angle ECI\cdot\sin45\cdot\sin\angle CID}{\sin\angle KEC\cdot\sin\angle ECI\cdot\sin45\cdot\sin\angle CIE}=1.$$ Let $\angle ABC=2\alpha$ and note that  $\angle CID=90-\alpha,$ $\angle KEC=135-3\alpha,$ and $\angle CIE=2\alpha.$ Hence, \begin{align*}-\cos135&=\sin45\\&=(\sin(135-3\alpha)\cdot\sin(2\alpha))/(\cos\alpha)\\&=2\sin\alpha\cdot\sin(135-3\alpha)\\&=\cos(4\alpha-135)-\cos(135-2\alpha).\end{align*} Therefore, $$\cos(135-2\alpha)=2\cos(135-2\alpha)\cos(2\alpha)$$ and $\angle ABC=60,45$ or $\angle CAB=60,90.$ $\square$

Note that ∠IEK = ∠IDK = 45 so reflecting E across CK would be nice. Let I be incenter of ABC and E' be reflection of E across CK. Note that CI is angle bisector so E' lies on BC. now we have IKDE' is cyclic and we have 2 cases :

1 - E' is D :
It means BE is perpendicular to AC so ABC is regular triangle and ∠CAB = 60.
2 - E' isn't D :
∠EKI = ∠E'KI = ∠E'DI = 90 and ∠IEK = 45 so ∠CIE = 45 so ∠CAB = 90.

we're Done.

:crying: why trig

[img width=50]https://media.discordapp.net/attachments/925784397469331477/946249757964255284/Screen_Shot_2022-02-23_at_8.38.28_PM.png[/img]

We claim that the answer is $60^\circ,90^\circ.$ First, note it's true when $\angle CAB=60^\circ$ or $90^\circ.$ Let $F$ be the incenter of $\triangle ABC$

Now, we reflect $E$ over $CK$ to $E'.$ Note that $\angle FDK=45^\circ=\angle FE'K=45^\circ.$ Given that $D\neq E',$ $FDE'K$ is cyclic. Thus, $\angle FKE'=90^\circ.$ This means that $\angle KEC=180^\circ-\angle AEK=135^\circ-\angle ECB-\angle EBC=90^\circ-\angle ECK$ which implies $45^\circ=\angle FCD+\angle FBC$ so $\angle BAC=90^\circ.$

If $D=E'$ then $EC=CD$ which implies that $AC=BC$ so $\angle BAC=60^\circ$

oops totally didnt already  see solution

The answers are $\angle CAB=\boxed{60^{\circ}}$ and $\angle CAB=\boxed{90^{\circ}}$. Let $I$ be the incenter of $ABC$. Set $\angle CAB=4x$. Angle chasing gives $\angle EIK=90-2x$, $\angle ECK=45-x$, $\angle CEK=3x$, $\angle KEI=45$, $\angle KCD=45-x$, $\angle KID=45+x$. Now
\[\frac{\sin(45-x)}{\sin(3x)}\div \frac{\sin(90-2x)}{\sin(45)}=\frac{EK}{KC}\div \frac{EK}{IK}=\frac{IK}{KC}=\frac{ID}{CD}=\frac{\sin(45-x)}{\sin(45+x)}.\]
So we obtain
\[\frac{1}{2}(\cos(x)-\cos(90+x))=\sin(45)\sin(45+x)=\sin(3x)\sin(90-2x)=\frac{1}{2}(\cos(5x-90)-\cos(90+x))\]
which means that
\[2\sin(3x-45)\sin(2x-45)=\cos(x)-\cos(5x-90)=0\implies x=15, \frac{45}{2}\implies \angle CAB=\boxed{60^{\circ},90^{\circ}}.\]
We are done. $\blacksquare$

bad asdf imagien not covering up the sol so abd

Let $I$ be the incenter of $\triangle ABC$.  There is a homothety at $C$ mapping $I$ to $K$ to $C$, so $I, K, C$ are collinear.  We proceed using [color=blue]trig bash[/color].

Since $K$ is the incenter of $\triangle ADC$, we know by the Angle Bisector Theorem that $$\frac{IK}{KC} = \frac{ID}{DC} = \frac{\sin\angle ICD}{\sin\angle DIC} = \frac{\sin \left(45^{\circ}-\frac A4\right)}{\sin \left(45^{\circ}+\frac A4\right)}.$$Also, by LoS on $\triangle IKE$ and $\triangle EKC$, we have \begin{align*}\frac{IK}{\sin 45^{\circ}} &= \frac{EK}{\sin\left(90^{\circ} - \frac A2\right)} \\ \frac{KC}{\sin\left(\frac{3A}{4}\right)} &= \frac{EK}{\sin\left(45^{\circ} - \frac A4\right)},\end{align*}so\begin{align*}\frac{IK}{KC} = \frac{\frac{\sin 45^{\circ}}{\sin\left(90^{\circ} - \frac A2\right)}}{\frac{\sin\left(\frac{3A}{4}\right)}{\sin\left(45^{\circ} - \frac A4\right)}} &= \frac{\sin\left(45^{\circ} - \frac A4\right)}{\sin\left(45^{\circ} + \frac A4\right)}\\ \sin45^{\circ} \cdot \sin\left(45^{\circ} + \frac A4\right) &= \sin\left(\frac{3A}{4}\right)\cdot\sin\left(90^{\circ} - \frac A2\right) \\ \cos\left(\frac A4\right) - \cos\left(90^{\circ} + \frac A4\right) &= \cos\left(\frac{5A}{4}-90^{\circ}\right) - \cos\left(90^{\circ} + \frac A4\right) \\ \cos\left(\frac A4\right) &= \cos \left(\frac{5A}{4} - 90^{\circ}\right).\end{align*}It's not hard to verify that $\boxed{A = 60^{\circ}, 90^{\circ}}$ are the solutions.  We are done. $\square$

hahahaa i didnt format any of the latex so its jsut a clomp of text  [img]https://artofproblemsolving.com/assets/images/smilies/biggrin.gif[/img]

Let $\angle DAK=\alpha$, so $\angle BAC=4\alpha$. Let $I$ be the incenter of $\triangle ABC$.\\

We have $\angle AIE=\angle DIC=\alpha+45$, so $\angle KEC=2\alpha+\alpha+45-\alpha=3\alpha$. Thus, $$\frac{KC}{IC}=\frac{\sin(90-2\alpha)}{\sin45}\cdot\frac{\sin 3\alpha}{\sin(45-\alpha)}.$$ However, by triangle $\triangle IDC$, we have $$\frac{KC}{IC}=\frac{DC}{ID}=\tan(\alpha+45)$$ Hence, $$\frac{\sin(90-2\alpha)}{\sin45}\cdot\frac{\sin3\alpha}{\sin(45-\alpha)}=\tan(\alpha+45)$$ $$\sin(90-2\alpha)\sin3\alpha=\sin45\sin(45-\alpha)\tan(\alpha+45).$$ However, $\sin(45-\alpha)\tan(45+\alpha)=\sin(45+\alpha)$, so we have $$\sin(90-2\alpha)\sin3\alpha=\sin45\sin(45+\alpha).$$ Multiplying by 2 this becomes $$\cos(90-5\alpha)-\cos(90+\alpha)=\cos\alpha-\cos(\alpha+90)$$ $$\cos(90-5\alpha)=\cos\alpha.$$ This gives $\alpha=\pi/8$ and $\alpha=\pi/12$ as solutions, which correspond to answers of $60$ and $90$ degrees. These clearly work (e.g. by coordbashing), so we are done.

edit: didn't see EK perp CK wow

We claim the answer is 60 or 90 degrees. I got a hint to reflect E over K to get E', and let I the incenter of ABC. Suppose $D\ne E'$. Then IDK=1/2IDC=45=IEK=IE'K, so IDEK is cyclic. Now, let CEE'=CE'E=x. Then $$180-2x=ECB=180-CBE-BEC=180-(ACB/2)-45-x=135-x-(90-x)=45\implies x=\frac{135}2\implies BAC=180-2ACB=180-2(45)=90.$$
On the other hand, if $D=E'$ it follows that CD=CE; since CD is a tangent to the incircle, E lying on AC implies CE is a tangent too; in particular, BE perp. AC whence ABC is equilateral, so BAC=60 degrees.

trig bashing CAN be done but will be omitted here due to bashing sobad

Let $I$ be the incenter of $\triangle ABC$. Since both $CI$ and $KC$ bisect $\angle ACB$, we have that $K$, $I$, and $C$ are collinear. Let $\angle ACI=a$. Notice that since $\triangle ABC$ is isosceles, we have that
\[\angle AIE=\angle BID=\angle DIC=90-\angle DCI=90-a,\]
and that
\[\angle AEI=180-(\angle AIE+\angle AEI)=180-((90-2a)+(90-a))=3a,\]
which in turn gives us that $\angle KEC=180-(45+3a)=135-3a$. \\\\
Now using Law of Sines on $\triangle IEK$ and $\triangle KEC$, we get that
\[\frac{IK}{KC}=\frac{EK*\frac{\sin 45}{\sin 2a}}{EK*\frac{\sin (135-3a)}{\sin a}}=\frac{\sin 45\sin a}{\sin 2a \sin (135-3a)}.\]
However, by Angle Bisector Theorem and Law of Sines on $\triangle IDC$, we get that
\[\frac{IK}{KC}=\frac{ID}{DC}=\frac{\sin a}{\cos a},\]
and setting the two equations equal, we get that
\[\frac{\sin 45\sin a}{\sin 2a \sin (135-3a)}=\frac{\sin a}{\cos a} \iff \sin 45\cos a=\sin 2a\sin (135-3a).\]
By Product-to-Sum, this gives us that
\[\frac{1}{2}(\sin (a+45)+\sin(45-a))=\frac{1}{2}(\cos(135-5a)-\cos(135-a)),\]
however, since $-cos(135-a)=\sin(45-a)$ and $\cos(135-5a)=-\sin(45-5a)$, this gives us
\[\sin(a+45)+\sin(45-5a)=0,\]
which is
\[2\sin(45-2a)\cos(3a)=0,\]
by Sum-to-Product. Since $a<45$, we find that either $a=30$ or $a=\frac{45}{2}$, which gives us that the only possible values for $\angle BAC$ are $60$ and $90$ degrees, finishing the problem.

Let $I$ of the incenter of $\triangle ABC$. Consider the reflection $E'$ of $E$ over $CI$, which we know will land on $BC$. If $D = E'$, the angle bisector $AI$ is an altitude, forcing $\triangle ABC$ to be equilateral, or $\angle CAB = \boxed{60}$.

Otherwise, the angle equality $\angle IDK = \angle IE'K = 45$ tells us $IDE'K$ is cyclic, which induces
\[\angle IKE' = \angle IKE = 90 \implies EKE' \text{ collinear} \implies \angle IKA = \angle EKA = 45\]
\[\implies \triangle EKA \cong \triangle IKA \implies \angle AIE = \angle AEI,\]

from which it's easy to find $\angle B = \angle C = 45$, so $\angle CAB = \boxed{90}$. $\blacksquare$


Sixty-four 4'' $ \times$ 4'' $ \times$ 4'' cubes are stacked to create a larger cube. The faces of the new cube are painted white. What is the total number of faces of the original cubes that remain unpainted?

The number of faces of the original cubes is $ 4\times4\times6\equal{}96$, since it's a $ 4 \text{by} 4$, and there are $ 6$ faces.  Total faces of the original cube is $ 6\times64\equal{}384$.  So we have $ 384\minus{}96\equal{}\boxed{288}$


algebra polynomial quadratics parameterization Vieta induction symmetry

$ax^{2}+bx+c=0$ where does a,b,c come from ?

Depending on what course you're in, $a, b, c$ are usually taken to be reals but can in fact be members of any field.

in this polynomial $2x^{2}+3x+1$ where does 2 ,3, and 1 come from ?

What do you mean by "where do they come from?"

If you mean in terms of the roots, if you say the roots of the quadratic are $r_{1}$ and $r_{2}$, then the polynomial $ax^{2}+bx+c=a(x-r_{1})(x-r_{2})$. Multiplying out and dividing by a, $x^{2}+\frac{b}{a}x+\frac{c}{a}=x^{2}-(r_{1}+r_{2})x+r_{1}r_{2}$. Setting each coefficient equal, we have $r_{1}r_{2}=\frac{c}{a}$ and $-(r_{1}+r_{2})=\frac{b}{a}$.

$a,b,c$ are the parameters to the standard form of a quadratic equation

worthawholebean,yes i mean in terms of roots, thnx for that, but i wanted a general form or way of that for any polynomial, not just for quadratic.is there such a thing?.[/code]

is this vieta's formula ? if yes

 is this formula $(-1)^{i}\frac{a_{n-i}}{a_{n}}$ equal to $-\frac{a_{n-i}}{a_{n}}$ [color=red]always ?[/color] what does $i$ stands for ?

It's not always negative because (for example) $(-1)^{2}=1$. $i$ is the $i$th symmetric sum.

EDIT: Yeah Karth has a point, prove it with induction or go read up a proof of it and you will actually understand it as opposed to people just telling you on the forum.

The general form, called "root form," for a polynomial is $p(x)=k_{1}(x-r_{1})(x-r_{2})(x-r_{3})\cdots(x-r_{n})$, where $k_{1}$ is the first coefficient in standard form, $r_{1}$ through $r_{n}$ are the roots, and $n$ is the degree.

Vieta's formulas are simply the formulas for relating each coefficient to the roots; for example, as I said earlier, the sum of the roots is $-\frac{k_{2}}{k_{1}}$ and the product of the roots is $(-1)^{n}\frac{k_{n}}{k_{1}}$, where $k_{i}$ is the $i^\text{th}$ coefficient and $n$ is the degree. All these can be easily derived by multiplying out the root form.

Dude, go prove Vieta's to understand it. That's the only way, I think. But I could be wrong.

Thanks alot all, also Can we express a,b,c,.. in any other term? other than roots.

The axis of symmetry of a parabola represented by $ax^{2}+bx+c$ is $x=-\frac{b}{2a}$, and the y-intercept is $c$.

So a,b,c can tell the shape of the parabola that the quadratic equation describes.

what about other polynomials ?is it only for quadratic>?
and what other terms can we express a,b,c...

We can always express the coefficients in terms of the roots. For example, the coeffecient of $x^{2}$ in cubic equations is the negative of the sum of the roots multiplied by the coefficient of $x^{3}$(right?).


:huh:  I was a little distracted at school, and I remembered a theorem that said: there is not solution (in integers, obviously) for equation: a4 + b4 = c2. then, i thought: there is solution for equation a2 + b2 = c4 ?  :huh: [/quote]

Try to find one.  Do your integers have to be distinct because if they don't then you have the trivial solutions $ (0,0,0)$ and $ (1,1,1)$.

You mean $ (1, 0, 1)$.

The smallest nontrivial solution is $ 7^2 \plus{} 24^2 \equal{} 5^4$.  In general, let $ (a, b, c)$ be a Pythagorean triple.  Then

$ (2ab)^2 \plus{} (a^2 \minus{} b^2)^2 \equal{} (a^2 \plus{} b^2)^2 \equal{} c^4$.

His last theorem is that there is no solutions for $ x^n\plus{}y^n\equal{}z^n$ for positive integers x,y,z,n and $ n\geq 3$
I know there is a proof  for 3 and 4 but dont think its fully proven yet

So you doubt Wiles' proof¿ Come on...
(By the way, this is rather offtopic)

I don't think he doubts Wiles' proof.  I just don't think he is aware of it.

sorry... I posted the same topic twice, but I´m new and I had a trouble with the computer, then I thought that I did not send it... then, sorry


Gil, have you a solution for this very nice inequality?

[quote="Mircea Lascu"]If $a_1, a_2, a_3, a_4, a_5 \in [1/ \sqrt5,\sqrt5]$, than
$(a_1 - a_2)/(a_2 + a_3) + (a_2 - a_3)/(a_3 + a_4) + (a_3 - a_4)/(a_4 + a_5) + (a_4 - a_5)/(a_5+a_1) + (a_5 - a_1)/(a_1 + a_2) \geq 0.$[/quote]
We'll prove that:
\[\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+e}+\frac{d-e}{e+a}+\frac{e-a}{a+b}\geq0\]
for all $\{a,b,c,d,e\}\subset[\frac{1}{\sqrt5},\sqrt5]$.
$\sum_{cyc}\frac{a-b}{b+c}\geq0\Leftrightarrow\sum_{cyc}\left(\frac{a-b}{b+c}+\frac{2}{3}\right)\geq\frac{10}{3}\Leftrightarrow\sum_{cyc}\frac{3a+2c-b}{b+c}\geq10$.
But $3a+2c-b\geq\frac{5}{\sqrt5}-\sqrt5=0$ and $\sum_{cyc}(3a+2c-b)>0$.
Hence, $\sum_{cyc}\frac{3a+2c-b}{b+c}=\sum_{cyc}\frac{(3a+2c-b)^2}{(3a+2c-b)(b+c)}\geq$
$\geq\frac{\left(\sum\limits_{cyc}(3a+2c-b)\right)^2}{\sum\limits_{cyc}(3a+2c-b)(b+c)}=\frac{16(a+b+c+d+e)^2}{\sum\limits_{cyc}(a^2+4ab+3ac)}$.
Id est, it remains to prove that $\frac{8(a+b+c+d+e)^2}{\sum\limits_{cyc}(a^2+4ab+3ac)}\geq5$.
But $\frac{8(a+b+c+d+e)^2}{\sum\limits_{cyc}(a^2+4ab+3ac)}\geq5\Leftrightarrow\sum_{cyc}(3a^2-4ab+ac)\geq0\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(6a^2-8ab+2ac)\geq0\Leftrightarrow\sum_{cyc}(a^2+4b^2+c^2-4ab-4bc+2ac)\geq0\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(a-2b+c)^2\geq0$. :)

[quote="arqady"]
We'll prove that:
\[\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+a}+\frac{d-e}{e+a}+\frac{e-a}{a+b}\geq0\]
for all $\{a,b,c,d,e\}\subset[\frac{1}{\sqrt5},\sqrt5]$.[/quote]
This is harder: If $a,$ $b,$ $c,$ $d,$ $e \in \left[ \frac{1}{\sqrt{2\sqrt{5}+1}},\, \sqrt{2\sqrt{5}+1}\right],$ then
$\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+a}+\frac{d-e}{e+a}+\frac{e-a}{a+b} \ge 0.$

By the way, Vasc has proposed this as his conjecture: If $a,$ $b,$ $c,$ $d,$ $e$ are positive real numbers, then
$\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+a}+\frac{d-e}{e+a}+\frac{e-a}{a+b} \ge 0.$
I could not prove this conjecture. :(

[quote="can_hang2007"]
This is harder: If $a,$ $b,$ $c,$ $d,$ $e \in \left[ \frac{1}{\sqrt{2\sqrt{5}+1}},\, \sqrt{2\sqrt{5}+1}\right],$ then
$\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+e}+\frac{d-e}{e+a}+\frac{e-a}{a+b} \ge 0.$
[/quote]
Of course! It follows from http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2022233#p2022233
[quote="can_hang2007"]
By the way, Vasc has proposed this as his conjecture: If $a,$ $b,$ $c,$ $d,$ $e$ are positive real numbers, then
$\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+e}+\frac{d-e}{e+a}+\frac{e-a}{a+b} \ge 0.$
I could not prove this conjecture. :([/quote]
I think, the Buffalo way ($b=a+x$, $c=a+y$, $d=a+z$, $e=a+t$, where $a=\min\{a,b,c,d,e\}$) kills it. :wink:


AMC
USA(J)MO
USAMO
function
symmetry
inequalities
inequalities unsolved

Given $ x\plus{}y\plus{}z\equal{}1$ and $ x$, $ y$, $ z>0$

Prove that:\[ x^3\plus{}y^3 \plus{} z^3 \plus{} 6xyz \ge \dfrac{1}{4}\] :D

Let f be the following function: $ f(x,y,z)\equal{}x^3\plus{}y^3\plus{}z^3\minus{}6xyz$. Let x,y,z be non-negative numbers whose sum is 1. WLOG (because of symmetry) $ x \ge y \ge z$. Then-
$ f(x,y,z)\minus{}f(\frac{x\plus{}y}{2},\frac{x\plus{}y}{2},\frac{x\plus{}y}{2})\equal{}\frac{3}{4}(x\minus{}y)^2(x\plus{}y\minus{}2z) \ge 0$
$ \frac{x\plus{}y}{2}\plus{}\frac{x\plus{}y}{2}\plus{}z\equal{}1$, so it is enough to prove the inequality when x=y.

$ f(x,x,1\minus{}2x)\equal{}1\minus{}18x^3\plus{}18x^2\minus{}6x\ge\frac{1}{4}$ iff $ g(x)\equal{}2x\plus{}6x^3\minus{}6x^2 \le \frac{1}{4}$, where $ 0 \le x \le \frac{1}{2}$.
We have: $ (2x\plus{}6x^3\minus{}6x^2)'\equal{}2(3x\minus{}1)^2 \ge 0$, hence, in the interval [0,0.5] g gets its maximum in x=0.5:
$ g(x) \le g(0.5)\equal{}\frac{1}{4}$, Q.E.D

I have a easy solution :roll:

Inequality is equivalent with
\[ \frac {1}{4} \plus{} 3xyz\ge xy \plus{} yz \plus{} zx
\]
Presume $ x\ge y\ge z$, then $ x\ge \frac {1}{3}$ and
\[ xy \plus{} yz \plus{} zx \equal{} yz \plus{} x(y \plus{} z) \equal{} yz \plus{} x(1 \minus{} x)\le 3xyz \plus{} \left(\frac {x \plus{} 1 \minus{} x}{2}\right)^2 \equal{} 3xyz \plus{} \frac {1}{4}
\]

[quote="Nawahd Werdna"]Given $ x \plus{} y \plus{} z \equal{} 1$ and $ x$, $ y$, $ z > 0$

Prove that:
\[ x^3 \plus{} y^3 \plus{} z^3 \plus{} 6xyz \ge\ \frac {1}{4}
\]
:D[/quote]
Put $ a \plus{} b \plus{} c \equal{} p;ab \plus{} bc \plus{} ca \equal{} q;abc \equal{} r$. We have:

$ Ineq \leftrightarrow p^3 \minus{} 3pq \plus{} 3r \plus{} 6r \ge\ \frac {1}{4}$

$ \leftrightarrow 1 \minus{} 3q \plus{} 9r \ge\ \frac {1}{4}$

Case1: $ q \le\ \frac {1}{4}$ and $ r \ge\ 0$

$ \rightarrow 1 \minus{} 3q \plus{} 9r \ge\ \frac {1}{4}$

Case2: $ q \ge\ \frac {1}{4}$ and $ r \ge\ \frac {p(4q \minus{} p^2)}{9} \rightarrow 9r \ge\ 4q \minus{} 1$

$ \rightarrow LHS \ge\ 1 \minus{} 3q \plus{} 4q \minus{} 1\ge\ \frac {1}{4} \leftrightarrow q \ge\ \frac {1}{4}$

We have done! :)

[quote="Nawahd Werdna"]Given $ x \plus{} y \plus{} z \equal{} 1$ and $ x$, $ y$, $ z > 0$

Prove that:
\[ x^3 \plus{} y^3 \plus{} z^3 \plus{} 6xyz \ge \dfrac{1}{4}
\]
:D[/quote]
The inequality is equivalent to:

$ x^3 \plus{} y^3 \plus{} z^3 \plus{} 6xyz \ge \dfrac{(x\plus{}y\plus{}z)^3}{4},$

$ \Leftrightarrow x^3\plus{}y^3\plus{}z^3\plus{}6xyz \geq \sum xy(x\plus{}y),$

which is true since from Schur's inequality we have:

$ x^3 \plus{} y^3 \plus{} z^3 \plus{} 6xyz \geq x^3\plus{}y^3\plus{}z^3\plus{}3xyz \geq \sum xy(x\plus{}y)$


analytic geometry
function
parameterization
calculus
derivative
real analysis
real analysis unsolved

consider jordan curve which has a tangent at every point .does there exist at least one parametric representation of this curve where the coordinates are differentiable functions of parameter and where the derivatives of three coordinates do not vanish simultaneously(or just except a sub set of measure zero)?

What is the exact definition of a "tangent" you want to use here? I'm asking because I'm not sure, for instance, whether we should consider a cusp point as a point where a tangent exists, and there may be some other singularities where the existence of a tangent property may depend on the definition you use.

i meant we have oriented tangent at every point .it seems that i posed vague but the original statement is not correct actually there if we exclude measure zero then there exists approprate parametrization.


How many lists of numbers where the numbers range from 1 to 15 are there such that a term in the list is always greater than the term before it?

I'm not sure if there's a fancy way to do this problem... :huh:

[hide="A fancy way :)"]Do we take one-item lists into account by default? Anyway, such increasing-order lists are nothing else than combinations. (We choose any subset and represent it by any ONE ordering, hence we're free to represent it by an increasing order.) Therefore, if we DON'T take one-item lists into account, our answer is ${15\choose 2}+{15\choose 3}+\dots+{15\choose 15}=2^{15}-{15\choose 0}-{15\choose 1}=2^{15}-16$, and if we DO take one-item lists into account, our answer is ${15\choose 1}+{15\choose 2}+\dots+{15\choose 15}=2^{15}-{15\choose 0}=2^{15}-1$[/hide]

each number can either be chosen or not chosen, given a set of numbers, they can only form one satisfactory seqence, so $2^{15}$, but erasing all of them does not count, so $2^{15}-1$,


$ a,b,c\geq 0$ show that
$ \sum_{cyc} a\sqrt{6a\plus{}21b}\leq \sqrt{8\sum_{cyc} a^3\plus{}30\sum_{cyc} a^2b\plus{}15\sum_{cyc} ab^2\plus{}84abc}$

$ \left(\sum_{cyc} a\sqrt {6a \plus{} 21b}\right)^2\leq\sum_{cyc}\frac{a}{3a\plus{}b\plus{}4c}\sum_{cyc}{a(6a \plus{} 21b)(3a\plus{}b\plus{}4c)}$

$ \leq 8\sum_{cyc} a^3 \plus{} 30\sum_{cyc} a^2b \plus{} 15\sum_{cyc} ab^2 \plus{} 84abc,$

with equality if $ a\equal{}0,b\equal{}1,c\equal{}2.$

Is it the only case, Jichen? I observe that the equality case happens also when $ a\equal{}b\equal{}c >0$

[quote="Rose_joker"]$ a,b,c\geq 0$ show that
$ \sum_{cyc} a\sqrt {6a \plus{} 21b}\leq \sqrt {8\sum_{cyc} a^3 \plus{} 30\sum_{cyc} a^2b \plus{} 15\sum_{cyc} ab^2 \plus{} 84abc}$[/quote]

Dear Rose_joker, what does it mean PrTST?

[quote="Rose_joker"]$ a,b,c\geq 0$ show that
$ \sum_{cyc} a\sqrt {6a \plus{} 21b}\leq \sqrt {8\sum_{cyc} a^3 \plus{} 30\sum_{cyc} a^2b \plus{} 15\sum_{cyc} ab^2 \plus{} 84abc}$[/quote]
Dear friend,
My hint is to note that
\[ 8\sum_{cyc} a^3 \plus{} 30\sum_{cyc} a^2b \plus{} 15\sum_{cyc} ab^2 \plus{} 84abc \equal{}\frac{1}{3}\sum a(6a\plus{}21b)(4a\plus{}b\plus{}4c),\]
and then use the Cauchy Schwarz Inequality.

Anyway, I think PrTST is Peru TST. Am I right? Thanks.

$ \left(1\minus{}\sum_{cyc}\frac {3a}{4a \plus{} b \plus{} 4c}\right)\prod{(a\plus{}4b\plus{}4c)}\equal{}4\sum{a^3}\plus{}12\sum_{cyc}{a^2b}\minus{}15\sum_{cyc}{ab^2}\minus{}3abc$

$ \equal{}9a(a^2\plus{}b^2\plus{}c^2\minus{}bc\minus{}ca\minus{}ab)\plus{}(b\plus{}4c\minus{}5a)(c\plus{}a\minus{}2b)^2\geq0,$

which is clearly true for $ a\equal{}\min\{a,b,c\}.$

[quote="can_hang2007"][quote="Rose_joker"]$ a,b,c\geq 0$ show that
$ \sum_{cyc} a\sqrt {6a \plus{} 21b}\leq \sqrt {8\sum_{cyc} a^3 \plus{} 30\sum_{cyc} a^2b \plus{} 15\sum_{cyc} ab^2 \plus{} 84abc}$[/quote]
Dear friend,
My hint is to note that
\[ 8\sum_{cyc} a^3 \plus{} 30\sum_{cyc} a^2b \plus{} 15\sum_{cyc} ab^2 \plus{} 84abc \equal{} \frac {1}{3}\sum a(6a \plus{} 21b)(4a \plus{} b \plus{} 4c),\]
and then use the Cauchy Schwarz Inequality.

Anyway, I think PrTST is Peru TST. Am I right? Thanks.[/quote]

In Thailand, one should have surmounted about 20 sets of exams in order to win a glorious triumph: a qualification of being IMO representatives. This problem is one of the problems used during the selection. Pr stands for the name of the one who created this set of exam (PrTST = that man's Team selection test)...  :)


[size=150][color=darkblue]Consider the 16 points in the first quadrant of a Cartesian coordinate plane with integer coordinates less than or equal to 4. How many squares can be formed using four of these points as vertices?[/color][/size]
1996 - State - Sprint - Problem#30

I am getting 30 squares ( 16+9+4+1 ). But it is not the correct answer!

Hint: They don't have to have integer side lengths.

16 points make a 4x4 grid. you can make 4.4+3.3+2.2+1.1 = 30 squares from these 16 points. They all have integer side lengths, right?

The answer given was 20

I don't know how they got 20.

3.4 + 2.3+1.2 = 20 though

16 points make a 3x3 grid.
the squares can be rotated, the sides don't have to be parallel to the axis.

[hide="sixteen coordinates such that (1,1)...(4,4)"]

Squares with integer sides :
9 smallest squares such as (1,0) (2,1) (2,2) (1,2)
4 mid sized squares such as (1,1) (3,1) (3,3) (1,3)
1 large square such as (1,1) (4,1) (4,4) (1,4)

Squares with noninteger sides :
4 diamond with ace inside such as (2,1) (3,2) (2,3) (1,2)
2 skewed diamonds such as (3,1) (4,3) (2,4) (1,2)

Total 9+4+1+4+2=20  :P [/hide]

There are n=4 dots per side. $ \frac{16\cdot15}{12}\equal{}\frac{240}{12}\equal{}20$ as desired.

Thanks to all. Some more problems I couldn't solve. I appreciate if anyone solves it.

Five friends Alex, Bill, Charlie, Daniel, and Eddy all own different brands of cars. Their cars are a Ford, a Buick, a Honda, an Audi and a Mercedes. One week they decided to try each others’ cars and they traded cars every day to try all 5 cars. From Monday through Thursday none drove his own car and on Friday everybody drove his own car.
On Monday Daniel drove the Audi.
On Tuesday Charlie drove the Mercedes.
On Wednesday Charlie drove the Honda and Eddy the Mercedes.
On Thursday Alex had the Audi and Daniel had the Buick.
What kind of car does Bill have?
(a) Ford (b) Buick (c) Honda (d) Audi (e) Mercedes

Tommy’s house is next to a road that goes straight to Tommy’s school. A railway runs parallel to that road. When Tommy leaves home at his regular time the moving train just catches him as he reaches school. One day Tommy left 4 minutes 10 seconds later and the train caught him one mile before he reached school. Given that Tommy always bikes at 12 miles per hour, the train always leaves on time, and the train always travels at the same constant speed, which of the following is the closest to the speed of the train?

(a) 50mph (b) 55mph (c) 60mph (d) 65mph (e) 70mph

[quote="Mewto55555"]There are n=4 dots per side. $ \frac {16\cdot15}{12} \equal{} \frac {240}{12} \equal{} 20$ as desired.[/quote]

sorry for my ignorance. Can you please explain this more?

we can inscribe each square into a larger square that is parallel to the axes. Using this, it is not hard to see that the answer is $ 3^2 \cdot 1 \plus{} 2^2 \cdot 2 \plus{} 1^2 \cdot 3$ *, and this pattern holds true for any $ n$ by $ n$ grid of squares (or $ n \plus{} 1$ by $ n \plus{} 1$ lattice points). Using some

[hide="higher level math than mathcounts"]
lagrange interpolation or newton's form or finite differences to express the expression $ \sum_{k \equal{} 1}^n \dfrac{n(n\plus{}1)(2n \plus{} 1)}{6}$* in closed form (is there an easier way not using one of the 3 aforementioned methods? It feels like there should be), we find that the formula for this problem is $ \dfrac{(n \plus{} 1)^2(n \plus{} 2)(n)}{12}$. This is definitely not worth remembering. The pattern might be, though.[/hide]

* post a reply if you want either of these two parts to explain more fully, I am limited in time right now

or if there is easier way to get closed form, im not thinking straight today


For three non negative real numbers $ x$ , $ y$ and $ z$ such that $ xy \plus{} yz \plus{} zx \equal{} 3$ prove inequality:
$ x^{2} \plus{} y^{2} \plus{} z^{2} \plus{} 3xyz\geq 2(xy \plus{} yz \plus{} zx)$

It is well known and also posted in this forum that for all positive reals $ x$ , $ y$ and $ z$ we have:

$ x^{2} \plus{} y^{2} \plus{} z^{2} \plus{} 2xyz \plus{} 1\geq 2(xy \plus{} yz \plus{} zx)$
                                                                     [b] D. Grinberg[/b]

Counterexample: $ x\equal{}y\equal{}z\equal{}\frac{1}{\sqrt{3}}$.
The condition should probably be replaced with $ xy\plus{}yz\plus{}zx\equal{}3$. In this case, the inequality rewrites as $ 3xyz \geq 2(xy\plus{}yz\plus{}zx)\minus{}(x^{2}\plus{}y^{2}\plus{}z^{2})$, which is true by Schur's inequality, since
\[ 3xyz \geq \frac{9xyz}{x\plus{}y\plus{}z} \geq 2(xy\plus{}yz\plus{}zx) \minus{} (x^{2}\plus{}y^{2}\plus{}z^{2}).\]

Oh sorry Thanks [b] pohoatza[/b]

I have edited my post.


Can you prove these ineq without using SOS?Because the solution using SOS is long,so I hope to find another solution
Problem:Prove that if a,b,c≥0,then:
$ \frac{a^4}{a^3\plus{}b^3}\plus{}\frac{b^4}{b^3\plus{}c^3}\plus{}\frac{c^4}{c^3\plus{}a^3}$≥$ \frac{a\plus{}b\plus{}c}{2}$
$ \frac{a^3}{2a^2\plus{}b^2}\plus{}\frac{b^3}{2b^2\plus{}c^2}\plus{}\frac{c^3}{2c^2\plus{}a^2}$≥$ \frac{a\plus{}b\plus{}c}{3}$

[quote="SUPERMAN2"]Can you prove these ineq without using SOS?Because the solution using SOS is long,so I hope to find another solution
Problem:Prove that if a,b,c≥0,then:
$ \frac {a^4}{a^3 \plus{} b^3} \plus{} \frac {b^4}{b^3 \plus{} c^3} \plus{} \frac {c^4}{c^3 \plus{} a^3}$≥$ \frac {a \plus{} b \plus{} c}{2}$
$ \frac {a^3}{2a^2 \plus{} b^2} \plus{} \frac {b^3}{2b^2 \plus{} c^2} \plus{} \frac {c^3}{2c^2 \plus{} a^2}$≥$ \frac {a \plus{} b \plus{} c}{3}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23113
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22937