The square root of four times a number added to four times the square root of that number is 18. What number is this?

Answer: 9

[hide="Solution"]
We simply set up an equation using what we are given.  Call the number $ x$.  We have:
\[ \sqrt {4x} \plus{} 4\sqrt {x} \equal{} 18\]
We can change the LHS so that we can combine the two terms:
\[ 2\sqrt {x} \plus{} 4\sqrt {x} \equal{} 6\sqrt {x} \equal{} 18\]
Dividing by $ 6$ we simply have:
\[ \sqrt {x} \equal{} 3\]
Squaring our last equation on both sides, we get $ x \equal{} \boxed{9}$, our desired answer.[/hide]

Oh dangg I thought by four times the square root of that number it was $ 4\sqrt{\sqrt{4x}}$, no wonder. :P


calculus
integration
logarithms
limit
trigonometry
algebra
partial fractions

How can we calculate this generalized integral:

$ \int_{0}^{ \plus{} \infty} \frac {x^2}{x^4 \plus{} 1} dx$

Thanks in advance!!

The usual mathematical English for this would be "improper integral," not "generalized integral," which sounds like some very different ideas.

The easiest method is probably the residue theorem from complex analysis.

Here's a very low-level method - the grind it out approach from a first-year calculus course.

Factor the denominator: $ x^4 \plus{} 1 \equal{} (x^2 \minus{} \sqrt {2}x \plus{} 1)(x^2 \plus{} \sqrt {2}x \minus{} 1).$

Perform partial fractions (details not shown):

$ \frac {x^2}{x^4 \plus{} 1} \equal{} \frac1{2\sqrt {2}}\left(\frac {x}{x^2 \minus{} \sqrt {2}x \plus{} 1} \minus{} \frac {x}{x^2 \plus{} \sqrt {2}x \plus{} 1}\right).$

To deal with the integral $ \int\frac {x}{x^2 \minus{} \sqrt {2}x \plus{} 1}\,dx,$ complete the square in the denominator and make the substitution $ u \equal{} x \minus{} \frac1{\sqrt {2}}.$ We wind up with both a logarithm and an arctangent.

Skipping several more steps:

$ \int_0^{\infty}\frac {x^2}{x^4 \plus{} 1}\,dx \equal{} \left.\frac1{2\sqrt {2}}\left(\frac12\ln\left(\frac {x^2 \minus{} \sqrt {2}x \plus{} 1}{x^2 \plus{} \sqrt {2}x \plus{} 1}\right) \plus{} \arctan(\sqrt {2}x \minus{} 1) \plus{} \arctan(\sqrt {2}x \plus{} 1)\right)\right|_0^{\infty}$

$ \equal{} \frac1{2\sqrt {2}}\left(0 \minus{} 0 \plus{} \frac {\pi}2 \plus{} \frac {\pi}2 \minus{} \left( \minus{} \frac {\pi}4\right) \minus{} \frac {\pi}4\right) \equal{} \frac {\pi}{2\sqrt {2}}$

We did know from the beginning that this would converge: it's bounded near zero, and for large $ x,$ we can compare to $ \int_b^{\infty}\frac1{x^2}\,dx,$ which converges.

Cheap numerical estimate: $ \sum_{k\equal{}0}^{30}\frac{(.1k)^2}{(.1k)^4\plus{}1}\cdot(.1)\plus{}\frac13\approxeq 1.1170,$ while $ \frac{\pi}{2\sqrt{2}}\approxeq 1.1107.$

Hence, an answer I can believe in. (I had a wrong answer - twice that - posted at first, and I didn't believe it. The post has now been corrected.)

$ \text{Res}(f,e^{\frac {\pi i}{4}}) \equal{} \lim_{x \to e^{\frac {\pi i}{4}}}\left \{\left ( x \minus{} e^{\frac {\pi i}{4}}\right ) \frac {x^2}{x^4 \plus{} 1}\right \} \equal{} \lim_{x \to e^{\frac {\pi i}{4}}}\frac {x^2}{4x^3} \equal{} \frac {1}{4}e^{\minus{} \frac {\pi i}{4}}$
$ \text{Res}(f,e^{\frac {3 \pi i}{4}}) \equal{} \lim_{x \to e^{\frac {3 \pi i}{4}}}\left \{\left ( x \minus{} e^{\frac {3 \pi i}{4}}\right ) \frac {x^2}{x^4 \plus{} 1}\right \} \equal{} \lim_{x \to e^{\frac {3 \pi i}{4}}}\frac {x^2}{4x^3} \equal{} \frac {1}{4}e^{\minus{} \frac {3 \pi i}{4}}$
$ \oint_{c}\frac {x^2}{x^4 \plus{} 1}dx \equal{} \int_{\minus{}\infty}^{\infty}\frac {x^2}{x^4 \plus{} 1}dx \equal{} 2 \pi i \left \{\text{Res}(f,e^{\frac {\pi i}{4}}) \plus{} \text{Res}(f,e^{\frac {3 \pi i}{4}}) \right \} \equal{} \frac {\pi}{\sqrt{2}}$

Thus, $ \int_{0}^{\infty}\frac {x^2}{x^4 \plus{} 1}dx \equal{} \frac {\pi}{2\sqrt{2}}$.

cf.$ \int_0^\infty \dfrac{1}{x^4\plus{}x^2\plus{}1} dx$ : http://www.mathlinks.ro/viewtopic.php?t=215397

$ \frac{1}{2}\int\frac{x^2\plus{}1\plus{}x^2\minus{}1}{x^4\plus{}1}dx$

$ \equal{}\frac{1}{2}\int\frac{x^2\plus{}1}{x^4\plus{}1}dx\plus{}\frac{1}{2}\int\frac{x^2\minus{}1}{x^4\plus{}1}dx$

$ \equal{}\frac{1}{2}\int\frac{1\plus{}\frac{1}{x^2}}{(x\minus{}\frac{1}{x})^2\plus{}2}dx$+$ \frac{1}{2}\int\frac{1\minus{}\frac{1}{x^2}}{(x\plus{}\frac{1}{x})^2\minus{}2}dx$

$ \equal{}\frac{1}{2\sqrt 2}\tan^{\minus{}1}\left(\frac{x^2\minus{}1}{\sqrt 2x}\right)\plus{}\frac{1}{4\sqrt 2}\ln\left|\frac{x^2\minus{}\sqrt2x\plus{}1}{x^2\plus{}\sqrt2x\plus{}1}\right|$+c


calculus
integration
calculus computations

hello, it is $ \frac{1}{2}\sqrt{\pi}e^\frac{1}{4}$.
Sonnhard.

[quote="sos440"]Evaluate $ \int_{0}^{\infty}e^{-t}\sinh \sqrt{t}\; dt$.[/quote]

Let $ x = \sqrt{t}$ to get,
$ 2\int_{0}^{\infty}e^{-x^{2}}\sinh x \cdot x dx$
So,
$ \int_{0}^{\infty}x\exp\left(-x^{2}+x\right)dx-\int_{0}^{\infty}x\exp \left(-x^{2}-x\right) dx$

Lets look at the first integral cuz the second one is similar,
$ \int_{0}^{\infty}x\exp\left[-\left(x-\frac{1}{2}\right)^{2}+\frac{1}{4}\right] dx = e^{1/4}\int_{0}^{\infty}x\exp \left[-\left(x-\frac{1}{2}\right)^{2}\right]$

Use the substitution $ \xi = x-\frac{1}{2}$ to get (ignore the constant in front),
$ \int_{-1/2}^{\infty}\left( \xi+\frac{1}{2}\right) e^{-\xi^{2}}d\xi = \int_{-1/2}^{\infty}\xi e^{-\xi^{2}}+\frac{1}{2}\left( \int_{-1/2}^{0}e^{-\xi^{2}}d\xi+\int_{0}^{\infty}e^{-\xi^{2}}d\xi\right)$

Now there are three integrals. The first integral is trivial to compute using the substitution $ \mu = \xi^{2}$. The second integral is the problem, but we're not going to compute it, instead give it a name, say $ X$. The last integral is $ \sqrt{\pi}/2$.

Now here is the thing, we only did 1/2 of this problem. There is still another integral to compute in an analogous fashion. Except we will end up with three integrals and the middle one will be $ -X$ so the $ X$ and $ -X$ will cancel and we dont have to worry about non-elementary results.


Let x,y,z are real numbers such that $ x^2\plus{}y^2\plus{}z^2\equal{}2$. Prove that
$ 1\plus{}xyz(x\plus{}y\plus{}z)\ge xy\plus{}yz\plus{}zx$ :D

after normalization and using EMV theorem the problem is X^4 +Y^4 +2(XY)^2 >=2(X^3 Y+Y^3 X)
 (let Z=Min{X,Y,Z})that is correct by using AM_GM two times
 your problem is nice (and easy ):wink:

What do you think with ineq?
(kihe_freety5 )Let $ x;y;z$ be positive real numbers satisfying $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 2$
Prove that $ 1 \plus{} xyz(x \plus{} y \plus{} z) \geq xy \plus{} yz \plus{} xz \plus{} (xyz)^6$

Thank you,M.A.This problem have a nice solution. :D Can you prove it?
And chienthan, your problem is interesting! :)

oke!!!
x²+y²+z²=2; infer that x²+y²≤2 follows that we have xy≤1;
so xz≤1;xy≤1;yz≤1;
consisder (1-xy)(1-yz)(1-zx)≥0;

             => 1+xyz(x+y+z)≥xy+yz+xz + (xyz)²(1)

x²+y²+z²=2  follows that  xyz≤1;(2)
from (1),(2) we have
 1+xyz(x+y+z)≥xy+yz+xz + (xyz)^6
the problem have solved!!

and equal appear when there are 2 value of{x,y,z}=1 and 1 of {x,y,z}=0

Very nice solution, kihe_freety5. I don't believe a problem have interesting solution! And I think this inequality is true for all real numbers x,y,z. :) Congratulation!


geometry
3D geometry
tetrahedron
pyramid
vector
analytic geometry

Use the formula for a pyramid ($V= \frac{1}{3}Bh$), and since all angles are $60^\circ$ you can find the height and the area of the equilateral-triangle base... the volume comes out to [hide] $\frac{s^{3}\sqrt 2}{12}$ if I've done my arithmetic correctly.[/hide]

uhhh??? regular tetrahedron?

you can also use vectors if you are given coordinates

Oh, lol. I didn't see that it wasn't a regular tetrahedron, just any tetrahedron.

But ya, you can still use the formula $V=\frac{1}{3}Bh$, you'll just need to find the area of the base and the distance from the vertex to the plane of the base (i.e. with vectors, as Alex said above ^)

Yeah, and that is why we probably need to use the formula to find the distance form the plane to the point.

[quote="t0rajir0u"]If you're given the coordinates of all of the points or something, it would be easier to do this:

The volume of the tetrahedron $(0, 0, 0), (a, b, c), (d, e, f), (h, i, j)$ is given by

$\frac{1}{6}\cdot \left| \begin{array}{ccc}a & b & c \\ d & e & f \\ g & h & i \end{array}\right|$[/quote]

that is the box product, (a) dot (b cross c), divided by 6 since it is a tetrahedron


very ashamed of score.  This is the worst i got since 5th grade.  31 in total, 17 in sprint.
no one from my team made it to state.  The highest score in our school was 39.  Our team got sixth place.

edit: o yes.  I forgot i live in san jose, calif.

Yeah, the test was way tougher this year.

I'm a coach, but in my chapter, top scores was 38.  My team had 1st, 2nd, 3rd, 4th, 7th, 8th, and 9th.(and one alternate actually had a 37 and would have been 3rd!)  Our team score was 53.5 for 1st.  I think 2nd was like 38.

yeah that's me.  I live in Southern New Hampshire and there are only 9 schools(but 8 teams) at Chapter.  We usually have 4 or 5 kids in top 10(7 of top 9 this year) and have been State Champs the last three years.  The test was harder than usual, but I don't know how a 38 would hold up against the rest of the state.  Last year the kids who made Nats from NH scored 45,42,41,40 at Chapter, so I know our scores are lower than around the country.  But we were 15th at Nats last year!  Good luck to the rest of you.

I did pretty bad.  When I finished, I had like 15 minutes to check, but I didn't do anything.  There was this problem I got wrong I could have brute forced. Got 26 on sprint and a perfect target which is 42.  It kinda sucked in our county cuz like 6 people tied for 4th so I got only Honorable Mention (got a 4th place score tho, stupid tiebreakers).  Yikes! I'm going on to California state....

edit: forgot, live in San Jose, CA

I guess my journey ends here. HariSeldon(guess who) really kicked my but in competition, and I didn't make states because I got 4th and I know I suck at countdown.

Why are you not on the "all star" team?  Do they not take 4 people per chapter, or maybe it is just 3 for yours.  It'd be 3 if it were a smaller chapter.

California now does not reconize the all-star team as a a proper thing b/c every1 gets crammed in at state.

I only needed 1 more place to get into state. :(

36 total, 3rd place, team was 2nd place with 47.  Highest score at chapter was a 40, by the kid from NE who was like 27th or so in nationals last year.  (Don't want to go look it up.)

Not super happy with the score, but at least I still go on to state.

Perfect target, stunk up on sprint, 9/10 on team.

21 on sprint. Perfect on target. 37 total. First place. Countdown wasn't official, so that didn't mess me up.

Team got second. One member got a 33. I had a 37. Other two members didn't write down scores, but got lower than a 32. 9/10 on team.

Hm...

well... 3rd.

42: 26 and 16

got blanked in countdown

team : 1st : 59.50

we had 42, 41, 39, 36 on our team.

scores: (after countdown)
1 - 41
2- 44
3- 42
4- 39

end of news flash...

wow, the scores are really low this year, must've been tough... i remember last year at chapter when i got a 44 and placed 2nd... first was 45 in case you wanted to know

Yep, scores are low because the contest is a lot harder.

The Sprint was hard... when i finished, i thought i had like 20 minutes left... but guess i lost track of time, cause the person said "You have 3 minutes remaining."

yeah, i know laszlo, he's good, but do you go to ICM or w/e?  whats ur name?

of course, if u don't want to tell me, thats fine, i'm just curious

Apparently the MATHCOUNTS team that I was on last year did okay.  One 8th grader got 8th place and her little (grade 6) brother got 4th place.  Granted, this is the northern nevada chapter, which isn't exactly famous for its math skills.  I don't know their scores.

How do they determine what place u r in if countdown is official in your chapter? that doesnt seem fair. countdown should be for fun.
ps any one here from indiana going to state?

[quote="MathFiend"]21 on sprint. Perfect on target. 37 total. First place. [/quote]

wow your guyz chapter's easy. 6th to 9th place here was 42 but there were lots of ties.

Scrambled, do you go to Hyde? im from miller

[i]Please don't mention other peoples' names on the forum without their consent[/i]

[i]-Syntax Error[/i]

[quote="pinkbubblez90"]Scrambled, do you go to Hyde? im from miller[/quote]

yes.

Just in case you cared to know, i think your brother has a picture of me.

it is dark_shadow's brother. i didn't know that tk has a brother.

yes he has a bro and his name is the same as my brother's!!!

[quote="starrynightskies"]scrambled is joseph's brother? i tot he wuz tk's brother.  other pplz chapters ARE easy... lucky...[/quote]

the old dumb siggy i had was to piss joseph off.(incase ur referring to that) i knew he'd get ticked.


trigonometry
polynomial
superior algebra
superior algebra unsolved

Prove that \[ \left[\mathbb{Q}(\sqrt [n] {2}): \mathbb{Q}\right]=n,\forall n\geq 2;  \]
                  \[ \left[\mathbb{Q}(\epsilon): \mathbb{Q}\right]=2;  \]
                  \[ \left[\mathbb{Q}( e^{2\pi i /p}): \mathbb{Q}\right]=p-1;  \]
                  \[ \left[\mathbb{Q}(\cos 20): \mathbb{Q}\right]=3.  \]

    We assume that $p$ is a prime number and $\epsilon=\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}$. :)

1.: $x^n-2$ is irreducible (Eisenstein), thus the result.
2.: See 3. ;)
3.: Let $\zeta_p : = e^{2\pi i /p}$. Then $p(x)=\frac{x^p-1}{x-1}=x^{p-1}+x^{p-2}+...+x^2+x+1$ has the root $\zeta_p$ and is irreducible: look at $p(x+1)$ and use Eisenstein's criterion again.
4.: Either calculate the polynomial. Or: It is $a=cos(30^\circ) = \zeta_{18}+\zeta_{18}^{-1}$ purely real. Thus adjoining $a$ has degree $\frac{\varphi(18)}{2}=3$ (needing well known facts about cyclotomic extensions).


binomial coefficients
algebra unsolved
algebra

Find all pairs of integers, $n$ and $k$, $2 < k < n$, such that the binomial coefficients
\[\binom{n}{k-1}, \binom{n}{k}, \binom{n}{k+1}\]
form an increasing arithmetic series.

[quote="tµtµ"]Condition is $ \binom{n}{k-1} + \binom{n}{k+1} = 2 \binom{n}{k} $

Which simplifies to $ (2k - n)^2 = n+2 $

Solutions are then

$\binom{n^2-2}{\frac {n^2 \pm n}{2} - 2 }$, $\binom{n^2-2}{\frac {n^2 \pm n}{2} - 1}  $ , $\binom{n^2-2}{\frac {n^2 \pm n}{2} }$[/quote]

Just a quick question. How did you find the form of solution from the equation $(2k-n)^2 = n+2$?

 :huh:

Use the formula $ \binom{n}{k}=\frac{n!}{k!(n-k)!}$

how did you get that nice form for k?? wouldn't that involve square roots?


Find the integer that lies between $ \sqrt[3]{\minus{}45}$ and $ \sqrt[3]{\minus{}101}$.

We know that $ (\minus{}4)^3\equal{}\minus{}64$. Also, $ (\minus{}3)^3\equal{}\minus{}27$ and $ (\minus{}5)^3\equal{}\minus{}125$. Thus, the integer in the middle is $ \boxed{\minus{}4}$.


Let $ f(x) \equal{} x \minus{} \frac {1}{x}.$ How many different solutions are there to the equation $ f(f(f(x))) \equal{} 1$?

A. 1
B. 2
C. 3
D. 6
E. 8

[hide="Solution"]
Note that for the x interval (-oo, 0), f(x) is strictly increasing and has range (-oo,oo).
For the x interval (0, oo), again, f(x) is strictly increasing and has range (-oo, oo).

Hence for all a, there is exactly one positive and one negative solution to f(x)=a.

Thus, there are 2^n solutions to f^n(x) = a, and the answer is 2^3 = 8, or [b]E[/b][/hide]

[hide="Another approach"]

Let $ f(x) \equal{} k, f(f(x)) \equal{} f(k),$ and $ f(f(f(x))) \equal{} f(f(k)) \equal{} 1$. Now, $ k \equal{} x \minus{} \frac{1}{x}, \ f(k) \equal{} k \minus{} \frac{1}{k}$ and at last, $ f(f(k)) \equal{} 1 \equal{} f(k) \minus{} \frac{1}{f(k)}$.

Solving for $ k$ gives two values of $ x,$ which are $ \frac{1 \pm \sqrt{5}}{2}$. Now, since the rest of the equations will have two solutions in exact form like this ($ x^2 \minus{} ax \minus{} 1$ where $ a$ is the solution from previous quadratic) and distinct, we can say with confidence that there are $ 2^3 \equal{} 8$ solutions. E. [/hide]


rotation
calculus
integration
calculus computations

Find the exact volume generated when the region under the curve $ y \equal{} (2x \minus{} 3)^{2}$ between $ x \equal{} 0$ and $ x \equal{} 1.5$ is rotated completely about the $ x$-axis.

[I make this $ \frac{243\pi}{10}$.]

The integral is $ \pi\int_{0}^{1.5}(2x\minus{}3)^{4}\,dx$.  Let $ u\equal{}2x\minus{}3\Rightarrow\frac{du}{2}\equal{}dx$.  Then the integral becomes $ \frac{\pi}{2}\int u^{4}\,du\equal{}\frac{\pi u^{5}}{10}\equal{}\frac{\pi (2x\minus{}3)^{5}}{10}\bigg|_{0}^{1.5}$ (I know that I was a bit clumsy with the limits of integration and whatnot).  Evaluating yields $ \frac{243\pi}{10}$.

Thanks, JRav.  (I like to be sure I'm correct as I tend to make silly mistakes.)

Calculators are really amazing tools.  Check your work with one.


[quote="math_mymind"]Hi every one. Please help me solve this problem:
Let a,b,c be real numbers such that:  1/3 <= a,b,c <=3.
Prove that a/(a+b) + b/(b+c) + c/(c+a) >= 7/5[/quote]

$ E(a,b,c)\equal{}\sum\frac{a}{a\plus{}b}\;,a\equal{}\max{a,b,c}$

$ E(a,b,c)\minus{}E(a,b,\sqrt{ab})\equal{}\frac{(\sqrt{a}\minus{}\sqrt{b})(\left(\sqrt{ab}\minus{}c\right)^2}{(\sqrt{a}\minus{}\sqrt{b})(a\plus{}c)(b\plus{}c)}\ge 0$

$ E(a,b,\sqrt{ab})\minus{}\frac{7}{5}\equal{}\frac{(3\minus{}x)((1\minus{}x)^2\plus{}x^2)}{5(x\plus{}1)(x^2\plus{}1)}\ge 0$

which $ x\equal{}\sqrt{\frac{a}{b}}$ then $ 0<t\le 3$

Equality occurs for $ (a,b,c)\equal{}(3,\frac{1}{3},1)$ or any cyclic permutation  :)

Is the method that quangpbc used called Mixing Variable ?


In an electrolytic cell, if we have a copper sulphate or copper chloride solution and platinum electrodes, oxygen is evolved at the anode but in a cell with NaCl as the electrolyte and platinum electrodes, how come chlorine is evolved at the anode and not oxygen?

Another question, in a voltaic cell, how do we decide which is the anode and cathode? Is the one undergoing oxidation always the anode?

In any cell [voltaic (or galvanic) or electrolytic], the cathode is the electrode where reduction occurs and the anode is the electrode where oxidation occurs - which is the same to say that the half reaction with the highest reduction potential occurs in the cathode, and the half reaction with the lowest reduction potential occurs in the anode.

Regarding your first question, I believe the keyword you are looking for is [b][i]overpotential[/i][/b].

i think we are taught the word "overpotential" as selective dischargement which depends on the position of the element discharged in activity series .
i hope i am not wrong

overpotential is not related to position in any series at all :rotfl:  :P

actually its the reason y the series is violated some times
it is the extra potential req 4 release of a gas at electrode :)


This is my first time doing usamts, so I'm not quite sure how everything works...

Do the graders leave comments on how you could have gotten a better score?

Yep. They give brief comments on each of your submissions.

We will provide a numeric score (from 0 to 5) on each problem, along with a written comment.  You will be able to access both your scores and your comments on the USAMTS website.


'Condense' usually has another meaning in here (at least in US). He meant 'simplify' the expression for A with a 'nice' solution.

[hide="Hint"]
Let $ x\equal{}\sqrt{2}$ and $ y\equal{}\sqrt{3}$.  Rewrite the fractions in terms of x and y, then add them using $ \frac{a}{b} \plus{} \frac{c}{d} \equal{} \frac{ad\plus{}bc}{bd}$.  Some stuff will cancel out.
[/hide]

[quote="thanhnam2902"][color=darkblue]Let $ A \equal{} \frac {2 \plus{} \sqrt {3}}{\sqrt {2} \plus{} \sqrt {2 \plus{} \sqrt {3}}} \plus{} \frac {2 \minus{} \sqrt {3}}{\sqrt {2} \minus{} \sqrt {2 \minus{} \sqrt {3}}}$.
I want condense $ A$. Can you give me a nice solution?[/color][/quote]

Start with
\[ \left(\frac {2 \plus{} \sqrt {3}}{\sqrt {2} \plus{} \sqrt {2 \plus{} \sqrt {3}}}\right)\left(\frac{\sqrt {2} \minus{} \sqrt {2 \plus{} \sqrt {3}}}{\sqrt {2} \minus{} \sqrt {2 \plus{} \sqrt {3}}}\right) \plus{} \left(\frac {2 \minus{} \sqrt {3}}{\sqrt {2} \minus{} \sqrt {2 \minus{} \sqrt {3}}}\right)\left(\frac{\sqrt {2} \plus{} \sqrt {2 \minus{} \sqrt {3}}}{\sqrt {2} \plus{} \sqrt {2 \minus{} \sqrt {3}}}\right)\]

Carrying out the multiplications will greatly simplify (condense) the denominators.

It so happens that $ \sqrt {2\pm \sqrt {3}} \equal{} \frac {\sqrt {3} \pm 1}{\sqrt {2}}$. $ A \equal{} \frac {2 \plus{} \sqrt {3}}{\sqrt {2} \plus{} \frac {\sqrt {3} \plus{} 1}{\sqrt {2}}} \plus{} \frac {2 \minus{} \sqrt {3}}{\sqrt {2} \minus{} \frac {\sqrt {3} \minus{} 1}{\sqrt {2}}} \equal{} \frac {2\sqrt {2} \plus{} \sqrt {6}}{3 \plus{} \sqrt {3}} \plus{} \frac {2\sqrt {2} \minus{} \sqrt {6}}{3 \minus{} \sqrt {3}} \equal{} \sqrt {2}$

[quote="azjps"]It so happens that $ \sqrt {2\pm \sqrt {3}} \equal{} \frac {\sqrt {3} \pm 1}{\sqrt {2}}$. $ A \equal{} \frac {2 \plus{} \sqrt {3}}{\sqrt {2} \plus{} \frac {\sqrt {3} \plus{} 1}{\sqrt {2}}} \plus{} \frac {2 \minus{} \sqrt {3}}{\sqrt {2} \minus{} \frac {\sqrt {3} \minus{} 1}{\sqrt {2}}} \equal{} \frac {2\sqrt {2} \plus{} \sqrt {6}}{3 \minus{} \sqrt {3}} \plus{} \frac {2\sqrt {2} \minus{} \sqrt {6}}{3 \plus{} \sqrt {3}} \equal{} \sqrt {2}$[/quote]

Thank you very much.

[quote="azjps"]It so happens that $ \sqrt {2\pm \sqrt {3}} \equal{} \frac {\sqrt {3} \pm 1}{\sqrt {2}}$. $ A \equal{} \frac {2 \plus{} \sqrt {3}}{\sqrt {2} \plus{} \frac {\sqrt {3} \plus{} 1}{\sqrt {2}}} \plus{} \frac {2 \minus{} \sqrt {3}}{\sqrt {2} \minus{} \frac {\sqrt {3} \minus{} 1}{\sqrt {2}}} \equal{} \frac {2\sqrt {2} \plus{} \sqrt {6}}{3 \minus{} \sqrt {3}} \plus{} \frac {2\sqrt {2} \minus{} \sqrt {6}}{3 \plus{} \sqrt {3}} \equal{} \sqrt {2}$[/quote]

[color=darkblue]I like it: $ \sqrt {2\pm \sqrt {3}} \equal{} \frac {\sqrt {3} \pm 1}{\sqrt {2}}$ ? It's a nice solution.[/color]


How do you link a website to your post without the whole name showing up? I mean, I click on the URL thing and put a website but the whole thing will show up like this:

[url]http://www.artofproblemsolving.com/Forum/posting.php?mode=newtopic&f=144[/url]

However, I don't know how to make a hyperlink by clicking on the word this:

Click [b]this[/b] to go to the website.

Thanks in advance!

[code]Click [url=http://www.mathlinks.ro/]this[/url] to go to the website.[/code]

Click [url=http://www.mathlinks.ro/]this[/url] to go to the website.


quadratics
algebra
polynomial
modular arithmetic
IMC
college contests

Let $ f$ be a polynomial of degree 2 with integer coefficients. Suppose that $ f(k)$ is divisible by 5 for every integer $ k$. Prove that all coefficients of $ f$ are divisible by 5.

What easy 1st problem! Even easier than one year ago.
A "direktive" solution:
Let $ f(x)=ax^{2}+bx+c$. Then $ c=f(0)$ is divisible by 5, further $ a+b=f(1)$ and $ 2a+4b=f(2)$ are divisible by 5. Hence $ a$ and $ b$ are divisible by 5.
The same holds true and for every polinomial of degree less than $ p$ in $ \mathbb{Z}_{p}$.

Even faster :P: $ f(\hat 1) = a+b = 0 \pmod 5$ and $ f(-\hat 1) = a-b = 0 \pmod 5$ :)


There are 2 triangles in the 3d space. Let $ n$ be the number of sides of the triangles which, when extended, intersect the other triangle internally. Find all possible values of  $ n$

Hey! Is the question too stupid that you dont bother to type the answer?
I am working on the case when the two triangles intersects, and i think that $ n$ cannot exceed 4.
Actually i am trying to code a 3d engine and need an elegant formula to determine which triangle comes over the other one when projected to a plane. This can be done by finding a plane which separates the two triangles, and it will be convenient if the plane contains an edge of one of the triangle. So i came up with this question.
Please help! Your help is much appreciated.


geometry
IMO 1999
IMO
homothety
circumcircle

Two circles $\Omega_{1}$ and $\Omega_{2}$ touch internally the circle $\Omega$ in M and N and the center of $\Omega_{2}$ is on $\Omega_{1}$. The common chord of the circles $\Omega_{1}$ and $\Omega_{2}$ intersects $\Omega$ in $A$ and $B$. $MA$ and $MB$ intersects $\Omega_{1}$ in $C$ and $D$. Prove that $\Omega_{2}$ is tangent to $CD$.

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

[quote="sprmnt21"]If I remember well, I already posted a solution for this problem.

I don't understand way it isn't here anymore.

Could someone, please, explain what is going on?

Ciao

sprmnt21[/quote]

Sorry, your solution was lost to an attack on the server.  :?  Please be so kind rewrite it.  :)

[color=red][[b]Edit by Darij:[/b] Indeed, (at least) 3 messages from this thread were lost:

[b]IMO ShortList 1999, geometry problem 6
4 orl 86 Sat Nov 27, 2004 3:33 pm
sprmnt21 |>[/b]

4 replies means 5 messages including the original post. The first two messages are here (these are the two posts by Orl), the other 3 were lost.][/color]

{U,V} = o1^o2; E=AN^o2; F=BN^o2;p=DF^CE ; X = centre(o1); Y = centre(o2).
AC*AM = AV*AU = AE*AN ==> MNEC cyclic;
<ACE = <ANM = <ABM = <CDM ==> CE tangent to o1;
<AEC =  <AMN = <ABN = <EFN ==> EC tangent to o2;
G= o2^XY, H = foot(Y) on CX;  YH//CE ==> XH = r1 – r2 = XG ==> CXG == YXH ==> <CGX = 90°
XY symmetry axes of CD and EF   ==> <DGX = 90°.

[color=red]sprmnt21[/color]
you have proved that $CD$ is parallel to $AB$. Can you post the proof?

According to this figure we only need to show that EF||AB, right?
Then you can draw a tangent line throught the point M. And let X be a point on that line. Thus we have <EFM=<EMX(i) and <EMX=<ABM(ii),
so from i and ii we have <EMX=<ABM, similarly <FEM=<BAM.
Note: All typos used in this observation are according the figure.

davron

It’s a nice problem , which with $INVERSION$, will be more nice.

(i changed names of the circles from $\omega$ to $O$)

Let me invert all of the shape(not just a part of it), round point $N$.

$O,O_2$, turn into two parallel lines .

the inversive form of $O_1$, is a circle which goes through the respect of

$N$, to $O_1$, and is also

tangent to $O$.

the inversive form of $AB$, is acircle which goes through the intersection

of $O_1,O_2$, and also through$N$.

and it is known that its radius is equal to the radius of $O_1$.

$AN,BM$ are tangent to the two equal circles , which goes through

$N,M,A,B$.

$C,D$ , are the intersection of circles $AM,BM$ with $O_1$.

Now we should prove that:

The circum circle of triangle $C,D,N$, is tangent to $O_2$.

To prove it , we should say that, $\angle NDT=90$, or the other words

$T,D,B$ should be collinear.

We know that $\angle NMB=90$, then we draw $TB$, and intersect it with

$O_1$ at $R$.

And then we should prove , $RBNM$ is a cyclic quadrilateral. and then

$\angle NRB=\angle NMB=90$.

And it means that: $R=D$.

Then we should prove that: $TR.TB=TM.TN$

Suppose centers of circles $O_1,NAB$ are $V,U$ respectively, and

consider $d$, as the distance of these centers.

And the two circles , intersect at $S,Q$.

And also suppose that the radius of these two circles is $r$.

We have these facts:

$TU=\frac{d}{2}$

$TS^2=r^2-\frac{D^2}{4}$

$TM=r-\frac{d}{2}$

$TN=r+\frac{d}{2}$

$TM.TN=r^2-\frac{D^2}{4}=TS^2$

then we should prove:$TS^2=TR.TB$

if the recent equality holds, triangles $TSR,TSB$ will be similar to each

other.and therefore we should have:

$\angle SBT=\angle RST$.

The places of $SQ$ , in circles $O_1,NAB$ , are equal, and

$\angle SBT$=arc$\frac{SH}{2}$

And $\angle RST$=arc$\frac{QR}{2}$.

We can prove the equality of the arcs $SH,QR$, instead of the equality of

the arcs $QH,SR$,

Now look at the shape, $T$ is the midpoint of $QS$ and $UV$.

We draw two parallel radius from $U,V$. and let them intersects their

circles at $W,X$ respectively.

(NOTE:both of them are in to apposite side)

then we have:
..........................................................................................................................................................................................

$UW$ is parallel to $VX$, and then $\angle TVX=\angle TUW$

$TU=TV$

$WU=VX=r$
..........................................................................................................................................................................................
And conclude is :
..........................................................................................................................................................................................
Triangles $WUT,TVX$ are equal.

$\angle UTW=\angle VTX$

$WT=TX$

…………………………………………………………………………………………………………………………..

then $WX$ goes through$T$.

Now, to prove the equality of the arcs $QH,RS$, we do like this:

Consider a homotechy  with center $T$, and the radius $(-1)$.

$Q$, turns to $S$. and each point $W$, from circle $NAB$, turns to a

point $X$, from circle $O_1$.

Then point $R$, from $O_1$, turns to point $H$ , from $NAB$,and it

means that:

Arcs$QH,RS$ are to homothetic arc, and they are equal to each other:

arc$QH$=arc$RS$.

And every thing is ok.

It was hard for me to find an inversive proof for it.
 ;)

@Ashegh: I have no comments to you, you are really good in Inversion.
I hope that you will teach me using it, too. :-)

davron

uew wellcom dear Darvron... ;)

hmmmm,,,,

i know nothing about geometry,

then how can i teach some thing if i dont know it? :D

[quote="ashegh"]i know nothing about geometry, [/quote]

yes, I think,too.

I have a fairly simple proof using coordinate geometry-

let the center of $ \omega_{2}$ and have radius [i]r[/i].  Let $ \omega{1}$ have radius [i]R[/i].  So the diameter of $ \omega$ is [i]r+2R[/i], and its radius is [i]R+r/2[/i].  Since the circles are symetric about the x-axis, I will just work with the top half of each for now.  With some simplifying, the equations for the top parts of $ \omega_{2}$, $ \omega{1}$, and $ \omega$ are $ y^{2}\equal{}r^{2}\minus{}x^{2}$, $ y^{2}\equal{}R^{2}\minus{}(x\minus{}R)^{2}\equal{}2xR\minus{}x^{2}$, and $ y^{2}\equal{}(R\plus{}r/2)^{2}\minus{}(x\minus{}R\plus{}r/2)^{2}\equal{}(x\plus{}r)(2R\minus{}x)$, respectively.  Now $ \omega_{2}$ and $ \omega_{1}$ intersect at $ r^{2}\minus{}x^{2}\equal{}2xR\minus{}x{2}$, or $ x\equal{}r^{2}/2R$.  Since this is the same for the bottom also, the common chord is a vertical line, and it intersects $ \omega$ at $ x\equal{}r^{2}/2R$, so $ y^{2}\equal{}(r^{2}/2R\plus{}r)(2R\minus{}r^{2}/2R)$, or $ (1\plus{}r/2R)^{2}(r(2R\minus{}r)$.  The sdquare of the line from $ (2R,0)$ to this point is $ y^{2}\equal{}r/(2R\minus{}r)(2R\minus{}x)^{2}$, which meets $ \omega_{1}$ at the point when $ x(2R\minus{}x)\equal{}r/(2R\minus{}r)(2R\minus{}x)^{2}$, or $ 2Rx\minus{}rx\equal{}2Rr\minus{}rx$.  So the x-coordinate of point C is [i]r[/i].  The same can be proven using the same methods for point D.  Thus, CD is a vertical line where $ x\equal{}r$, and thus it is tangent to $ \omega_{2}$ at the point $ (r,0)$.

Let X,Y be the centres of $\Omega_{1}$, $\Omega_{2}$ respectively.

-------
[b]Lemma: the tangent line to $\Omega_{1}$ through C is also tangent to $\Omega_{2}$.[/b]

Let the tangent line to $\Omega_{1}$, $\Omega_{2}$ touch $\Omega_{1}$, $\Omega_{2}$ at C', E respectively. I will show that C' = C.

Let the tangent line to $\Omega_{1}$, $\Omega_{2}$ intersect $\Omega$ at J and K. Let the midpoint of arc JK be A'.
The homothety centered at M that takes $\Omega_{1}$ to $\Omega$ also takes C' to A'. Similarly, the homothety centered at N takes E to A'.
Therefore, lines MC' and NE intersect on the circumcircle of $\Omega$, at A'.

Since the homothety centered at N takes E to A', the measures of arcs EN and A'N are equal. Therefore, ∠KEN = ∠A'MN, meaning MC'EN is cyclic.
Therefore, A' lies on the radical axis of $\Omega_{1}$, $\Omega_{2}$. However, since A' lies on the $\Omega$, this means A' = A.

This means that the lines MC' and NE intersect at A. However, MA intersects $\Omega_{1}$ in only one place other than M, namely C, so C' = C.

-------
The homothety from M taking $\Omega_{1}$ to $\Omega$ also takes C to A and D to B. Therefore, CD is parallel to the radical axis, and since the radical axis is perpendicular to XY, CD is perpendicular to XY.

Let CD intersect XY at L. Let $\Omega_{2}$ intersect XY at L'. Let the foot of the perpendicular from Y to XC be P. Since we have ▲XCL~▲XYP, and XY = XC, we have ▲XCL≅▲XYP.
Therefore, we have XL = XP = XC - YE = XY - L'Y = XL'. Therefore, L = L', and therefore CD is tangent to $\Omega_{2}$.

By homothety $CD || AB$ and, if $MN$ is the common chord of $\Omega_1,\Omega_2$, then $A-M-N-B$ collinear. Easily $M\Omega_1N\Omega_2$ is a kite, so $\Omega_1\Omega_2\bot AB, \therefore\Omega_2$ is the midpoint of the arc $CD$.
Let the tangent at $C$ to $\Omega_1$ intersect the circle $\Omega$ at $E,F$ and $AN$ at $P$. We see that $\angle ANM=\angle ACP$, so $MNPC$ is cycli, ınferring $AC\cdot AM=AP\cdot AN$; but $A$ has equal powers to the cırcles $\Omega_1,\Omega_2$, hence $P$ lies onto $\Omega_2$, i.e. $CP$ is the common tangent to the 2 circles.
Further $\angle PC\Omega_2=\angle BC\Omega_2$ and, $BC$ is symmetrical of $CP$ through $C\Omega_2$, so it is tangent to the cırcle $\Omega_2$ as claımed.

Best regards,
sunken rock

Since $A$ lies on the radical axis of $\Omega_1$ and $\Omega_2$, set $\varrho = \text{pow}(A, \Omega_1) = \text{pow}(A, \Omega_2).$ Then the inversion $\Psi$ with center $A$ and power $\varrho$ fixes $\Omega_1$ and $\Omega_2.$ Hence, $\Psi$ interchanges $\{M, C\}.$ Then since $\Omega$ is tangent to $\Omega_1$ at $M, \Psi(\Omega)$ is the line tangent to $\Omega_1$ at $C.$ However, $\Psi(\Omega)$ is also tangent to $\Omega_2$ because $\Omega$ and $\Omega_2$ are tangent. Therefore, $\Psi(\Omega)$ is a common tangent to $\Omega_1$ and $\Omega_2$ passing through $C.$ Similarly, we can show that there exists a common tangent to $\Omega_1$ and $\Omega_2$ passing through $D.$ Let these two tangents meet at $E$, and denote by $S$ and $T$ the points of tangency of $EC$ and $ED$ with $\Omega_2.$
[asy]
 /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(15cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = 125.90308049860157, xmax = 243.67692567282657, ymin = 88.52559844242623, ymax = 148.12940530451223;  /* image dimensions */

Label laxis; laxis.p = fontsize(10);
xaxis(xmin, xmax, Ticks(laxis, Step = 10., Size = 2, NoZero),EndArrow(6), above = true);
yaxis(ymin, ymax, Ticks(laxis, Step = 10., Size = 2, NoZero),EndArrow(6), above = true); /* draws axes; NoZero hides '0' label */
 /* draw figures */
draw(circle((162.81796556704418,117.30338148064997), 22.272952473938865));
draw(circle((159.77238711367855,125.42492402295828), 13.59914123653413), linetype("2 2") + blue);
draw(circle((173.27868183110525,123.8385301950535), 9.938660703570978), linetype("2 2") + blue);
draw((171.274578812264,137.90848046917992)--(166.26990936151225,95.29955236055868));
draw((154.99740984006004,138.15819675260755)--(171.274578812264,137.90848046917992));
draw((154.99740984006004,138.15819675260755)--(166.26990936151225,95.29955236055868));
draw((164.9357204453974,138.0057281020608)--(209.94995472824428,119.53127330465229), red);
draw((161.88003185707186,111.99010076372485)--(209.94995472824428,119.53127330465229), red);
 /* dots and labels */
dot((162.81796556704418,117.30338148064997),linewidth(3.pt) + dotstyle);
label("$O$", (163.18106279722585,117.91785371634148), NW * labelscalefactor);
dot((154.99740984006004,138.15819675260755),linewidth(3.pt) + dotstyle);
label("$M$", (153.3495070261601,138.80990972985617), NW * labelscalefactor);
dot((159.77238711367855,125.42492402295828),linewidth(3.pt) + dotstyle);
label("$O_1$", (160.00628957948587,126.41805297674208), N * labelscalefactor);
dot((173.27868183110525,123.8385301950535),linewidth(3.pt) + dotstyle);
label("$O_2$", (173.72950284326512,124.47222423038531), E * labelscalefactor);
dot((181.7076627244061,129.10438792965618),linewidth(3.pt) + dotstyle);
label("$N$", (182.12729006438377,129.4904141552001), ESE * 1.5);
dot((171.274578812264,137.90848046917992),linewidth(3.pt) + dotstyle);
label("$A$", (171.6812620576264,138.50267361201037), NE * labelscalefactor);
dot((166.26990936151225,95.29955236055868),linewidth(3.pt) + dotstyle);
label("$B$", (166.2534239756839,93.44137632795909), S * labelscalefactor);
dot((164.9357204453974,138.0057281020608),linewidth(3.pt) + dotstyle);
label("$C$", (164.71724338645487,139.01473380842003), N * 2);
dot((161.88003185707186,111.99010076372485),linewidth(3.pt) + dotstyle);
label("$D$", (161.23523405086908,109.92971465235058), SW * labelscalefactor);
dot((209.94995472824428,119.53127330465229),linewidth(3.pt) + dotstyle);
label("$E$", (210.8026610633255,118.73715003059696), E * labelscalefactor);
dot((177.05220084902166,133.03295953037014),linewidth(3.pt) + dotstyle);
label("$S$", (175.7777436289038,133.8941318443233), ENE * 4);
dot((174.8190141045,114.01995835041882),linewidth(3.pt) + dotstyle);
label("$T$", (174.65121119680254,111.7731313594254), S * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
 /* end of picture */
[/asy]
Let $\Gamma \equiv \odot(ESO_2T)$ be the circle of diameter $\overline{EO_2}.$ Clearly the center $X$ of $\Gamma$ lies on the perpendicular bisector of $\overline{ST}.$ Hence, $O_1, O_2, X$ are collinear, implying that $\Omega_1$ and $\Gamma$ are tangent. Now consider the inversion $\Upsilon$ with center $E$ and power $EC \cdot ES = ED \cdot ET.$ Clearly $\Upsilon$ interchanges $\{C, S\}$ and $\{D, T\}$ and therefore $\Upsilon$ interchanges $\{\Omega_1, \Omega_2\}.$ Moreover, $\Upsilon(\Gamma) \equiv CD.$ Then since $\Omega_1$ and $\Gamma$ are tangent, it follows under inversion that $\Omega_2$ and $CD$ are tangent, as desired.

A nice solution:

Let $O$ be the center of $\Omega_{2}$.Join $NA$ and $NB$. Let $NA\cap \Omega_{2}$ = $G$ and $NB\cap \Omega_{2}$ = $H$. Join $OG$ and $OH$. Also, join $CG$ and $DH$. Let the tangent from $C$ to $\Omega_{2}$ meet it at $X$. Join $DX$. Also, join$GX$, $HX$, $OD$ and $OC$. We shall show that $C$, $X$ and $D$ are collinear.

Now, $\angle OGC$ = $\angle OXC$ = $90$°.
So, $OGCX$ is cyclic.
We have,
$CG$ = $DH$, $OG$ = $OH$ and $\angle OGC$ = $\angle OHD$ = $90$°.
By congruence, $\angle ODH$ = $\angle OCG$...($1$)
Again, $OX$ = $OG$,
or, $\angle OXG$ = $\angle OGX$,
or, $\angle OCG$ = $\angle OCX$...($2$)

($1$) and ($2$) gives,
$\angle ODH$ = $\angle OCX$
But, $\angle ODH$ = $\angle OCF$ (alternate segment theorem)
Thus, $C$, $X$ and $D$ are collinear and hence the result follows.

Let $Z=(a,b), Z_1=(0,0), Z_2=(1,0)$ be the centers of circles $\Omega, \Omega_1, \Omega_2$.

$Z, Z_1, M$ are collinear, and $Z, Z_2, N$ are collinear.

$\angle BMA = \angle DMC$, so $\angle BZA = \angle DZ_1C$.

The triangles $MZ_1C$ and $MZA$ are similar, so $\angle MCZ_1 = \angle MAZ$.

So, $\angle MCD = \angle MAB$.  Since $AB$ is perpendicular to the $x$-axis, that means that $CD$ is perpendicular to the $x$-axis.

Let $P$ be the point where $CD$ intersects the $x$-axis.  So the problem is to show that $|Z_2P| = |Z_2N|$.

$|ZZ_1| + |Z_1Z_2| = \sqrt{a^2+b^2} + 1$, and $ZZ_2 = \sqrt {(a-1)^2 + b^2}$.

So $|Z_2N| = \sqrt{a^2+b^2} + 1 - \sqrt {(a-1)^2 + b^2}$.

Let $A^\prime$ and $B^\prime$ be the points of intersection of $\Omega_1$ and  $\Omega_2$.

By the law of cosines, $|Z_1Q| = \cos{(\angle A^\prime Z_1 Z_2)}
= 1 - \dfrac {|Z_2N|^2} 2 $.

So the distance of $Z$ from the common chord of circles $\Omega_1$ and $\Omega_2$ is $a - 1 + \dfrac {|Z_2N|^2} 2 $.

$|Z_1P| = \dfrac 1 {|ZM|}\left (a - 1 + \dfrac {|Z_2N|^2} 2 \right )$, because triangles $BZA$ and $DZ_1C$ are similar.

So $|Z_2P| = 1 + \dfrac 1 {|ZM|}\left (a - 1 + \dfrac {|Z_2N|^2} 2 \right ) =$

$\displaystyle 1+ \dfrac {2a-2 +|ZM|^2 - 2|ZM||ZZ_2| + |ZZ_2|^2} {2|ZM|} =$

$\displaystyle 1 - |ZZ_2| + \dfrac {|ZM|} 2 + \dfrac {2a-2+(a-1)^2+b^2}
{2(\sqrt{a^2+b^2}+1)} =$

$1 - |ZZ_2| + \dfrac {|ZM|} 2 + \dfrac {a^2+b^2-1}
{2(\sqrt{a^2+b^2}+1)} =$

$1 + \sqrt{a^2+b^2} - |ZZ_2| = |Z_2N|$.

[quote=Tumon2001]A nice solution:

Let $O$ be the center of $\Omega_{2}$.Join $NA$ and $NB$. Let $NA\cap \Omega_{2}$ = $G$ and $NB\cap \Omega_{2}$ = $H$. Join $OG$ and $OH$. Also, join $CG$ and $DH$. Let the tangent from $C$ to $\Omega_{2}$ meet it at $X$. Join $DX$. Also, join$GX$, $HX$, $OD$ and $OC$... [/quote]

What do you mean by "joining" two line segments?

[quote]But, $\angle ODH$ = $\angle OCF$ (alternate segment theorem)
Thus, $C$, $X$ and $D$ are collinear and hence the result follows.[/quote]

And where is the point "F"?

My solution: Let $\Omega_1 \cap \Omega_2 = U,V$ and $NA \cap \Omega_2 = P,NB \cap \Omega_2 = Q$. Also, Let $MN \cap \Omega_1 = R,MN \cap \Omega_2 = S$. Lastly, Let $O_1$ and $O_2$ be the centers of $\Omega_1$ and $\Omega_2$ respectively, and let $O_1O_2 \cap CD = T$.

We denote the tangent at a point $K$ to a circle $\alpha$ by $f_{\alpha}(K)$.
Let $f_{\Omega}(M) \cap f_{\Omega}(N) = X, f_{\Omega_1}(C) \cap f_{\Omega_1}(D) = Y, f_{\Omega_2}(P) \cap f_{\Omega_2}(Q) = Z$.

By Radical Axes Theorem on $\Omega, \Omega_1, \Omega_2$, we get that $MX,NX,UV$ concur
$\Rightarrow X$ lies on $UV$ (i.e. $AB$) $\Rightarrow AMBN$ is harmonic.

$\Rightarrow -1 = (A,B;M,N) \overset{\text{M}}{=} (C,D;M,R)$ and $-1 = (A,B;M,N) \overset{\text{N}}{=} (P,Q;S,N)$

$\Rightarrow CMDR$ and $PNQR$ are harmonic $\Rightarrow Y$ lies on $MR$ and $Z$ lies on $NS$ $\Rightarrow Y,Z,M,N$ are collinear.

Now, By Reim's Theorem, $PQ \parallel AB \parallel CD$ $\Rightarrow PUVQ$ and $CUVD$ are isosceles trapeziums.

And, As $O_1O_2$ is the perpendicular bisector of $UV$, so $O_1O_2$ is the perpendicular bisector of $CD$ and $PQ$ also.

But, $YC = YD$ and $ZP = ZQ$ $\Rightarrow Y,Z$ lie on $O_1O_2$. As $Y,Z$ lie on $MN$ also, thus $Y = Z$, i.e. $CP$ and $DQ$ are common tangents of $\Omega_1$ and $\Omega_2$. This means that $O_2P \perp CZ$ and $O_2Q \perp DZ$.

Also, $\angle ZCO_2 = \angle CDO_2 = \angle DCO_2 \Rightarrow O_2$ lies on the angle bisector of $\angle ZCD$. As $O_2T \perp CD$ and $O_2P \perp CZ$, we get that $O_2P = O_2T$ $\Rightarrow T$ lies on $\Omega_2$ $\Rightarrow CD$ is tangent to $\Omega_2$ at $T$. $\blacksquare$

Invert at $N$. We get the following:

Let $N$  be a point and let $\Omega_2, \Omega$ be lines so that $N$ is on the opposite side of $\Omega$ as $\Omega_2$. Let $O$ be the reflection of $N$ over $\Omega_2$, and let $\Omega_1$ be a circle through $O$ tangent to $\Omega$ at $M$ and intersecting $\Omega_2$ at $X,Y$. Let $\Omega$ meet $(NXY)$ at $A,B$, and let $(MNA), (MNB)$ meet $\Omega_1$ again at $C,D$, Show that $(NCD)$ is tangent to $\Omega_2$.

Note that now $(NXY)$, $(OXY)$ are now reflections over $\Omega_2$. Let $\Omega'$ be the reflection of $\Omega$ over $\Omega_2$. This is still too hard, so we invert again, this time at $X$:

Let $X$ be a point on a fixed line $\Omega_2$. Let $\Omega, \Omega'$ be circles tangent to $\Omega_2$ at $X$ which are reflections of each other over $\Omega_2$. Let $N, O$ be reflections of each other over $\Omega_2$ where $N$ lies inside $\Omega$. Let the $N$-tangent to $\Omega'$ and the $O$-tangent to $\Omega$ meet at $Y$ (then $Y \in \Omega_2$ by symmetry) and let the $O$-tangent be tangent to $\Omega$ at $M$. Let the $N$-tangent to $\Omega'$ intersect $\Omega$ at $A,B$. Let $(MNA), (MNB)$ meet $YM$ again at $C,D$. Show that $(NCD)$ is tangent to $\Omega_2$.

To do this, reflect $A,B,X,M$ over the perpendicular to $\Omega_2$ at $Y$ to get $A', B', X', M'$. Then, let $h$ be the homothety centered at $Y$ taking $M'\to N$. It follows that $h\colon A'\to C$ since \[
\frac{YA'}{YC} = \frac{YA}{YC} = \frac{YM}{YN} = \frac{YM'}{YN}
\] and similarly $h\colon B'\to D$. Say $h\colon X'\to Z$, then $(A'B'X'M')$ is cyclic and tangent to $\Omega_2$ at $X'$, which implies that $(NCDZ)$ is cyclic and tangent to $\Omega_2$ at $Z$ as desired. $\blacksquare$

Let $O,O_1,O_2$ be the center of $\Omega,\Omega_1,\Omega_2$. Then by Reim's theorem, $AB \parallel CD$ implying that $O_1O_2 \perp CD$. 	Let $K,L$ be the midpoint of $AB$ and $CD$. Thus $O_2L \perp CD$.

Consider the homothety of center $M$ sending $\odot O_1$ to $\odot O$. Thus $O_1 \rightarrow O$, $C \rightarrow A$, $D \rightarrow B$, $L \rightarrow K$. Let $Q = MO_2 \cap \odot O$, then $O_2 \rightarrow Q$ and $QK \perp AB$, meaning that $Q,K,O$ are collinear. Thus we have
	\[ \frac{O_2L}{O_2M} = \frac{QK}{QM}. \]
	Consider the tangent line at point $M$ and $N$, by radical axis theorem, they intersect at $T$ which lies on $AB$. Hence $MOKNT$ are concyclic. Thus $\angle MNO = \angle MKO$. Now we prove that $MQ$ bisects the $\angle KMN$. Indeed
	\[ \angle QMN = \angle OMN - \angle OMQ = \angle MNO - \angle MQO = \angle MKO - \angle MQO = \angle KMQ\]
	Therefore, by law of sines
	\[ \frac{QK}{QM} =\frac{\sin \angle KMQ}{\sin \angle MKO} = \frac{\sin \angle QMN}{\sin MNO} = \frac{O_2N}{O_2M}\]
	Therefore $O_2L = O_2N$ implies that $ L \in \odot O_2$ and $\odot O_2$ tangent to $CD$.

This problem is significantly easier if you think of what works well with tangent circles.

[hide=hint]curvilinear config![/hide]

[hide=solution sketch]
Let $XY$ be the chord on $\Omega$ tangent to both circles and let $Z$ be the midpoint of arc $XY$ not containing $M$. By shooting lemma we find that the power of $Z$ to both circles is $ZY^2$, so thus we have that $Z$ is either $A$ or $B$. WLOG, we'll denote it as $A$.

Now, we have that $C$ is just where $\Omega_1$ is tangent to $XY$. By homothety we have that $DC$ is parallel to $AB$, and since $CX$ is tangent to $\Omega_2$ it suffices to prove that $\angle{DCX} = 2\angle{OCX}$, where $O$ is the center of $\Omega_2$.

This last task isn't that hard with the diagram already in hand. Let $AB$ intersect the two circles at $U$ and $V$. If $K$ is the circumcenter of $\Omega_1$, then since $OU = OV$ we have that $OK$ is perpendicular to $UV$. Since $DC$ and $UV$ are parallel, this gives us that $OK$ and $DC$ are in fact perpendicular as well. Now just do a little angle chase to finish.
[/hide]

Let $AN$ and $BN$ intersect $\Omega_{2}$ with center $O$ at $E$ and $F$, respectively. Note that there exists a homothety with center $M$ that takes $\Omega_{1}$ to $\Omega$, thus $\overline{CD}\parallel \overline{AB}$. Similarly, there exists a homothety with center $N$ that takes $\Omega_{2}$ to $\Omega$, thus $\overline{EF}\parallel \overline{AB}$.

Since $AB$, $CM$, and $EN$ concur, by the converse of the radical axis theorem, $CEMN$ is cyclic. Thus $180^{\circ}-\angle ECM = \angle ENM = \angle ABM = \angle CDM$, thus $EC$ is tangent to $\Omega_{1}$. By symmetry, we may claim and prove in the same way that $DFMN$ is cyclic, $FD$ is tangent to $\Omega_{1}$, and $EC$ and $FD$ are tangent to $\Omega_{2}$. Thus $CDEF$ is an isosceles trapezoid, with $O$ lying on the perpendicular bisectors of its bases.

Thus $\angle ECO = \angle CMO = \angle CDO = \angle DOC$. Similarly, $\angle FDO = \angle DMO = \angle DCO =\angle CDO$. Thus $\overline{CD}$ is tangent to $\Omega_{2}$, as desired. $\blacksquare$

Denote by $X,Y$ meet-points of $\Omega_1,$ $\Omega_2.$ Due to homothety $AB$ $\parallel$ $CD,$ and center of $\Omega_2$ is the midpoint of arc $XY$ of $\Omega_1.$
We invert wrt $M$ with power $|MX|\cdot |MY|,$ and reflect over the bisector of angle $XMY.$ Well-known that $\Omega_2$ is fixed.
Note that $\Omega_1,AB$ are swapped, and so the same $A,D$ and $B,C.$ Hence $\Omega,CD$ are swapped and conclusion is trivial.

Six hundred sixty-five.

Let $O,O_1,$ and $O_2$ be the centers of $\Omega,\Omega_1,$ and $\Omega_2.$ Let $\overline{BN}$ and $\overline{CN}$ intersect $\Omega_2$ at $F$ and $E$ so that $C$ and $E$ are on the same side of $\overline{O_1O_2}.$

[b][color=lightblue]Claim:[/color][/b] $\overline{CE}$ is a common external tangent of $\Omega_1$ and $\Omega_2.$
[i]Proof.[/i] By homothety, $\overline{CD}\parallel\overline{AB}\parallel\overline{EF},$ and ny Radical Axis, $BCMN$ is cyclic. Hence, $$\measuredangle(\overline{AE},\overline{CE})=\measuredangle NEC=\measuredangle NMC=\measuredangle NBA=\measuredangle NFE.$$ Similarly, $\measuredangle(\overline{CE},\overline{AC})=\measuredangle CDM.$ $\blacksquare$

[b][color=lightblue]Claim:[/color][/b] $O_2$ is the incenter of $\triangle(\overline{CD},\overline{CE},\overline{DF}).$
[i]Proof.[/i] Notice $$\angle DCO_2=90-\angle CO_2O_1=\tfrac{1}{2}\angle O_2O_1C=\angle O_2CE$$ and similarly $\overline{O_2D}$ bisects $\angle FDC.$ $\blacksquare$

Since $\Omega_2$ is tangent to $\overline{CE},$ it is also tangent to $\overline{CD}.$ $\square$

Somehow this works, okay.

Draw in the points $E = \overline{AN} \cap \Omega_2$ and $F = \overline{BN} \cap \Omega_2$.

[b][color=#f00]Claim.[/b][/color] $\overline{CE}$ is a common external tangent to $\Omega_1$ and $\Omega_2$.

First, notice that for homothety reasons, $\overline{CD} \parallel \overline{AB} \parallel \overline{EF}$ and that $CENM$ is cyclic for radical axis reasons. Now angle chase: $$\measuredangle NEC = \measuredangle NMC = \measuredangle NBA = \measuredangle NFE.$$ The other tangency follows because $$\measuredangle ECD = \measuredangle CEF = \measuredangle ENF = \measuredangle AMB = \measuredangle CMD. \ \blacksquare$$

Thus, it suffices to show that $\overline{CO_2}$ bisects the angle $\angle NCD$. Notice that $\overline{CO_2}$ bisects $\angle PCQ$, so it suffices to show $\angle ECP = \angle QCD$. But this is obviously true as $\angle ECP = \angle CDP = \angle QCD$ by the previous claim.

[quote=v_Enhance]
Now consider the homothety mapping $\triangle WAB$ to $\triangle QCD$. It should map $E$ to a point on line $ME$ which is also on the line through $Q$ perpendicular to $\overline{AB}$; that is, to point $T$. Hence $TCD$ are collinear, and it's immediate that $T$ is the desired tangency point.[/quote]
What if the line $ME$ coincides the perpendicular bisector of CD? Is it an easy casework?

what is this solution

Let $O_1$ and $O_2$ be the centers of $\Omega_1$ and $\Omega_2$ respectively. Observe that $\overline{AB}$ and the tangents to $\Omega$ at $M$ and $N$ concur by radical center, hence $AMBN$ is harmonic, so $\overline{MN}$ passes through the intersection of the tangents to $\Omega$ at $A$ and $B$. By taking a homothety at $M$ sending $\Omega$ to $\Omega_1$, it follows that $\overline{MN}$ also concurs with the tangents to $\Omega_1$ at $C$ and $D$ at some point $X$, which also lies on $\overline{O_1O_2}$. On the other hand, by Monge, the exsimilicenter of $\Omega_1$ and $\Omega_2$ lies on $\overline{MN}$, and since it also lies on $\overline{O_1O_2}$, it must be $X$, hence the tangents to $\Omega_1$ at $C$ and $D$ are also tangent to $\Omega_2$.

Now, it is clear that $O_1CDX$ is cyclic, and that $O_1$ is the midpoint of arc $CD$, hence $\overline{XO_1}$ is the $\angle CXD $-bisector. Therefore, by incenter-excenter, $\Omega_1 \cap \overline{XO_1}=O_2$ is the incenter of $\triangle CXD $, which implies the desired result. $\blacksquare$

[b]Remark:[/b] To top off such a strange solution, I first noticed the concurrency of the tangents at $M$ and $N$ with $\overline{AB}$ because of... [url=https://artofproblemsolving.com/community/c5h1802326p11972613]AIME I 2019/15[/url]. what.

!?!?!?!?!?!

Let $XY$ be the common chord, let $O,O_1,O_2$ be centers, let $C',D'$ be intersections of external tangents with $\Omega_1$ and let $T_C,T_D$ be intersections with $\Omega_2$, also by angle chasing note that $C'D'$ is tangent to $\Omega_2$ at $T$, by Monge also let $MN$, $C'T_C$, and $D'T_D$ intersect at $K$.

Now we need to realize that $KM\cdot KN=KC'\cdot KT_C=KD'\cdot KT_D$.
To see this draw $KM\cap \Omega_1=M'$ and define $N'$ on $\Omega_2$ similarly.

By homothety note
\[\frac{KM}{KM'}=\frac{KN'}{KN}\]
which gives what we need.

This basically implies by radical axis that $MC',XY,T_CN$ and $MD',XY,T_DN$ concur at $A',B'$ respectively.

Now it just suffices to show $MA'NB'$ is cyclic which is trivial by angle chase. The one that works for me is
\[\angle MA'B'=\angle MC'D'=\angle T_DD'B'=\angle MNB'\]
and now this given circle must be tangent to $\Omega_1$ and $\Omega_2$ by homothety and is therefore $\Omega$.

wow that was fun

[quote=Dukejukem]
Let $\Gamma \equiv \odot(ESO_2T)$ be the circle of diameter $\overline{EO_2}.$ Clearly the center $X$ of $\Gamma$ lies on the perpendicular bisector of $\overline{ST}.$ Hence, $O_1, O_2, X$ are collinear, implying that $\Omega_1$ and $\Gamma$ are tangent. Now consider the inversion $\Upsilon$ with center $E$ and power $EC \cdot ES = ED \cdot ET.$ Clearly $\Upsilon$ interchanges $\{C, S\}$ and $\{D, T\}$ and therefore $\Upsilon$ interchanges $\{\Omega_1, \Omega_2\}.$ Moreover, $\Upsilon(\Gamma) \equiv CD.$ Then since $\Omega_1$ and $\Gamma$ are tangent, it follows under inversion that $\Omega_2$ and $CD$ are tangent, as desired.[/quote]

yikes this is really smart what the heck

Let $O_1$ and $O_2$ be centers of $\Omega_1$ and $\Omega_2$. Let the two intersections of $\Omega_1$ and $\Omega_2$ be $P$ and $Q$, with $P$ closer to $A$ than $Q$. Let $NA$ and $NB$ intersect $\Omega_2$ at $E$ and $F$. Let $CE$ and $DF$ intersect at $G$.

[b]Claim 1:[/b] $AB\parallel CD\parallel EF$.
Consider the homothety centered at $M$ that takes $\Omega_1$ to $\Omega$. $C$ is taken to $A$ and $D$ to $B$, so $CD\parallel AB$. $EF\parallel CD$ follows similarly.

[b]Claim 2:[/b] $MCEN$ is cyclic.
By Power of a Point, $AC\cdot AM=AP\cdot AQ=AE\cdot AN$, so the claim is true.

[b]Claim 3:[/b] $EC$ is tangent to both $\Omega_1$ and $\Omega_2$.
We have
\[\angle ACE=\angle ANM=\angle ABM=\angle CDM\]
so $EC$ is tangent to $\Omega_1$, and tangency to $\Omega_2$ follows similarly.

Now, note that $CP=DQ$ because $CD\parallel PQ$, and $O_2Q=O_2P$, so $O_2C=O_2D$. Thus, $\angle ECO_2=\angle O_2CD$ by inscribed angles theorem. In particular, $CO_2$ bisects $\angle DCG$ Similarly, $DO_2$ bisects $\angle CDG$. This implies that $O_2$ is the incenter of $\triangle CDG$. $O_2E$ is the length of the perpendicular from the incenter to one of the sides, so $\Omega_2$ is the incircle of $\triangle CDG$. We are done.


[quote="Little Gauss"]Find the minimal value of k.
∀ $x,y \in R$  $a^{4}+b^{4}+c^{4}+k(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}) \ge 4(ab^{3}+bc^{3}+ca^{3})+(a^{3}b+b^{3}c+c^{3}a)+(a^{2}bc+ab^{2}c+abc^{2})$[/quote]

possibly k=5?
perhaps there is a mr. stefan type identity.

No.
I don't know what is stefan type identity.

[quote="Little Gauss"]Find the minimal value of k.
∀ $x,y \in R$  $a^{4}+b^{4}+c^{4}+k(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}) \ge 4(ab^{3}+bc^{3}+ca^{3})+(a^{3}b+b^{3}c+c^{3}a)+(a^{2}bc+ab^{2}c+abc^{2})$[/quote]
You used $a,b,c$ in the inequality and $x,y$ are the variables. I thing that you wrote something wrong.

It is something like
RHS-LHS=
sigma cyclic(x*a^2+y*b^2+z*c^2+l*ab+m*bc+n*ac)^2

for example, putting in (a^2-2ab+bc+ca)^2>=0
and adding all the cyclic cases
one arrives at (a^2+b^2+c^2)^2>=3(a^3*b+b^3*c+c^3*a)
which is a very strong inequality of Mr. Cartoaje's.
I think yours can be done similarily.


function
limit
integration
probability
probability and stats

take a random variable $ X$ such that $ E[|X|] \equal{} \plus{}\infty$, and suppose that $ (X_i)_{i \geq 1}$ is a sequence of independant random variables with the same law as $ X$. (Example: a Cauchy random variable, density $ \frac{1}{\pi} \frac{1}{1\plus{}x^2}$)
1/ can one say something about the convergence of $ \frac{1}{n} (X_1\plus{}X_2\plus{} \ldots \plus{} X_n)$

2/ take $ \alpha \in (0;1)$ and suppose that $ F(x) \equal{} \alpha$ has a unique solution $ x_{\alpha}$ where $ F(x) \equal{} P[X \leq \alpha]$ is the distribution function of $ X$. Define $ (X_{k(1,n)},X_{k(2,n)}, \ldots, X_{k(n,n)})$ the increasing rearrangement of $ (X_1, X_2, \ldots, X_n)$. Prove that if $ F$ is continuous then $ X_{k([\alpha n], n)}$ converges towards $ x_{\alpha}$

3/ under some good hypothesis, study the convergence in law of $ \sqrt{n} (X_{k([\alpha n], n)} \minus{} x_{\alpha})$

In the 1st question, you can think about weak LLN: if the characteristic function is differentiable in zero, then in holds. Equivalently, $ nP(|X|>n)\to 0, \ n\to \infty$ and $ \lim_{t\to\infty}\int_{\minus{}t}^t xdF(x)$ exists and is finite.

thank you Yustas  :)  Could you tell me more about this weak law of large number ? I have never heard about it ?

[b]Alekk[/b], you can easily prove it yourself: if $ \phi(t)$ is the characteristic function of $ X$ and $ \exists \ \phi'(0)$, the $ \frac {S_n}{n}$ converges in probability. Another exercise is to prove that differentiability of c.f. in 0 is equivalent to the conditions in terms of distribution i wrote above. I also think it can be found in Feller, II in the chapter devoted to characteristic functions.

thank you. But just a stupid question: it is quite straightforward that there is a convergence in law of $ \frac{S_n}{n}$. But how can one prove that the convergnece is actualy in probability ?

Well, the convergence in law to a constant is the same as covergence in probability.


inequalities
calculus
integration
trigonometry
calculus computations

Prove that for all x>0
$ |\int_{x}^{x\plus{}1}{siny^2dy}| \leq \frac{1}{x}$


linear algebra
matrix
induction
combinatorics unsolved
combinatorics

Find all $n$ for whiich in matrix $n$x$n$ can be write numbers ${-1, 0, 1}$ such that sum in every row and every column are different.

using induction prove that for $n=2k$ there exists such matrix and also in the sum of the columns and rows of this matrix,$-n$ doesnt apears and $n$ apears,this is pretty simple
 :D

[quote="zgorkster"]Is this really easy?

I agree it's not hard to use induction to show that every even $n$ has an $n\times n$ matrix satisfying the conditions, and it's seems true that no $n\times n$ matrix can be constructed for odd $n$, but I thought this was an open problem until 10 years ago.  I could be wrong, but I thought this was known as the $(0,-1,1)$ antimagic matrix problem.

Am I way off mark?  Does someone know a simple solution or something about this problem?[/quote]
wow...
I dont think so...
for even $n$ its so easy using induction...
ok I'll post my solution:
assume $n=2k$ as I said above I want to build a $n*n$ matrix such that in the sum of the elements of rows and columns of this matrix,$+2k$ apears and $-2k$ doesnt
ok assume that for $n=2k$ there exists such matrix now I want to show that for $n=2k+2$ there exists such matrix...
ok now name the rows and columns from $1$ to $n+2$ and call the matrix,matrix $A$ now assume the $n*n$ matrix which is at the upside left corner and name it matrix $B$...
by the induction hypothesis we can fill its elements with $-1,0,+1$ such that the sum of rows and columns are pairwise different and $2k$ apears and $-2k$ doesnt...
now put $+1$ in the elements of the column number $n+1$ and $-1$ in the elements of the column number $n+2$
but let the $2$ lowest array of each column be empty...
and put $+1$ in the elements of the row number $n+1$ and $-1$ in the elements of the row number $n+2$
but let the $2$ rightest array of each column be empty...
now assume the rows number $1$ to $n$ I want to show that the sum of these rows are pairwise different...
choose two random rows number $i,j$ such that $1\leq i,j\leq n$ I want to prove that the sum of these two rows are different
we wrote the number $+1$ in the arrays number $(i,n+1)$,$(j,n+1)$ and number $-1$ in the arrays number $(i,n+2)$,$(j,n+2)$ so the sum of row $i$ is equal to it's sum in the matrix $B$ and the sum of row $j$ is equal to it's sum in the matrix $B$ but by the induction hypothesis the sum of row $i$ and the sum of row $j$ in the matrix $B$ were different so the sum of rows $i,j$ are different in the matrix...
and so on we can prove in a similar way that the sum of any rows and any columns numbered with numbers $i,j$ such that $1\leq i,j\leq n$ are different...
now what about rows and columns number $n+1,n+2$??
the answer depends on the $4$ empty arrays at the intersection of these two rows and two columns
fill the arrays number $(n+1,n+1),(n+1,n+2)$ with number $+1$ and the array number $(n+2,n+1)$ with number $0$ and the array number $(n+2,n+2)$ with number $-1$ so we get that the sum of row $n+1$ is equal to $n+2=2k+2$ and the sum of row number $n+2$ is equal to $-n-1=-2k-1$ and the sum of column number $n+1$ is equal to $n+1=2k+1$ and the sum of column number $n+2$ is equal to $-n=-2k$
by the induction  hypothesis at the sum of the rows and the columns of matrix $B$ number $-n=-2k$ doesnt apears so the sum of column number $n+2$ isn't a repeat...
and at the sum of the rows and columns of this matrix number $n+2=2k+2$ apears at the row number $n+1$
and also number $-n-2=-2k-2$ doesn't apear at all...
so the induction is completed succesfully...
and the m atrix $2*2$ is an example for the induction...

any questions?? :D

I agree with your induction babak; at first glance that looks right to me.  However, can you prove that the odd $n$ do not work?  That is what interests me.  I would like to see a proof if you have (or can find) one.


inequalities
real analysis
calculus
calculus computations

Hi. Can someone give me hint on how to prove this?

[b]Question: [/b]If "a < b" and "c > d" then "a-c < b-d"

My answer: "b-a" and "c-d" is positive. $ \therefore$ -d-(-c) means -c < -d. Now here is what I cannot get. b-a is positive. So b-a+(-d-(-c))=b-d+(-a-c)? I am stuck here. Please help. Thank you.

This is [b]very [/b]much the wrong forum, but anyway: you already figured out that the second inequality can be written as $ \minus{}c < \minus{}d$ instead of $ c > d$. Now you have two inequalities with the $ <$ sign going the same way. What happens if you add them up? Can you justify adding together two inequalities?

Oh dear.... Sorry about wrong posting. I will post in the correct forum next time. Thank you for pointing out to me randomgraph. :blush:

So now we have

"a < b" and "-c<-d"

$ \therefore$ When we add them up:

b-a+[-d-(-c)] =b-a-d-(-c) = (b-d)-(a-c)

And thus (a-c)<(b-d)

Do you think I am right, randomgraph?

Er.... not quite. "Adding up two inequalities" means that from

$ A < B$ and

$ C < D$

you can conclude

$ A\plus{}C < B\plus{}D$

(that's just because $ B\minus{}A > 0$ and $ D\minus{}C > 0$ implies $ (B\minus{}A)\plus{}(D\minus{}C) > 0$ or $ (B\plus{}D) \minus{} (A\plus{}C) > 0$). What you've written doesn't make much sense since after "when we add them up" you wrote a couple of equations, with no inequality in sight.

I wouldn't have put that word "very" on the "wrong forum" note - I think it's debatable. But I'll go ahead and move it.

The question of exactly how to answer this depends very much on what assumptions you're starting from.

$ \mathbb{R}$ is an ordered field. Now: what are the axioms of an ordered field (that is, what axioms are you assuming that $ \mathbb{R}$ satisfies)? Symbian, please state these exactly.

(There's more than one possible form of these axioms - we need to see which of the several possible versions Symbian is using before we answer.)

Thank you, randomgraph. This is the starting chapter of my Real Analysis course and its very confusing..... Glad I receive assistance from you all in this forum.

Dear kent, the question which I had posted is an exact extract of the question and there is no leftout part. I am new to this Real Analysis, so do point out should I made any mistake. The word "axiom" is a new word to me. When I made reference, it seem we use a different word. We name them "property". Hope that is the one.  :blush:  Glad this forum have lots of knowledgable guys like you all.

---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
Dear Randomgraph, can I have your view on this? I extract this from an example taught by my friend:

If a<b and c<d, show that a+c<b+d

$ \therefore$ (b-a)-(d-c) >0, thus (b+d)-(a+c) > 0, hence (a+c)<b+d

Do you think we are on the right track and what we did make sense? :maybe:
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

As for my question, I had revised them for your comment. Do let me know if it make sense now.

Question: If $ a < b$ and $ c > d$ then $ a \minus{} c < b \minus{} d$

My Solution: $ c \minus{} d > 0$, thus $ \minus{} d \minus{} ( \minus{} c) > 0$. Hence $ \minus{} c < \minus{} d$

Now the question now become $ a < b$ and $ \minus{} c < \minus{} d$

So;

$ a\plus{}(\minus{}c)<b\plus{}(\minus{}d)$ which certainly satisfy the question. But why? So here is the explanation:

$ b \minus{} a > 0$ and $ \minus{} d \minus{} ( \minus{} c) > 0$ $ \therefore$ $ b \minus{} a \plus{} [ \minus{} d \minus{} ( \minus{} c)] > 0$ Hence $ b \plus{} ( \minus{} d) \minus{} (a \plus{} ( \minus{} c) > 0$. Finally $ a \plus{} ( \minus{} c) < b \plus{} ( \minus{} d)$.


quote="gagarin"]Given a triangle ABC with circumcircle (O) and three sides BC=a;AB=c;CA=b.Denete by $ A_1;B_1 and C_1$ the midpoints of BC;CA and AB  respectively and $ A_2;B_2;C_2$ are the midpoint of the ares BC(wich does not contain A);CA;AB.   Draw the cirdes ($ O_1);(O_2);O_3)$ whose diemeters are $ A_1A_2;B_1B_2;C_1C_2$  respectively.  Prove that:
 $ P_{A/(O_1)} \plus{} P_{B/(O_1)} \plus{} P_{C/(O_1)} \geq \frac {(a \plus{} b \plus{} c)^2}{3}$[/quote]


Find the sum of the series,
(x+1/x) ³   + (x ² + 1/x ²) ³  + ..............+ (xⁿ +1/xⁿ) ³

Expand to get

$ (x^{3}\plus{}x^{6}\plus{}x^{9}\plus{}....\plus{}x^{3n})\plus{}\left(\frac{1}{x^{3}}\plus{}\frac{1}{x^{6}}\plus{}....\plus{}\frac{1}{x^{3n}}\right)\plus{}3(x\plus{}x^{2}\plus{}x^{3}\plus{}....\plus{}x^{n})\plus{}3\left(\frac{1}{x}\plus{}\frac{1}{x^{2}}\plus{}...\plus{}\frac{1}{x^{n}}\right)$.

Now apply summation of G.P for each bracket.


No, String Theory is not the theory that unifies the four major forces, though its intent is to do so and it has yet to accomplish this.
Currently, as I previously posted, String Theory is being refuted in its attempts.
Most people become quite excited when they learn about String Theory, like with the PBS special "The Elegant Universe", but as the host, Brian Greene, says that it is what some believe as being the best attempt at doing so.

Of course, String Theory could also end up being the theory of all theories about the universe, the big T.O.E., the Theory of Everything.

Wait... isn't string theory the theory about everything is made of itsy bitsy strings? Even quarks?

-String Theory in general, there are a couple kinds of String Theory, states that all matter is made up of small vibrating strands or loops of energy or "strings".-

I found this site for the String theory.

[url=http://www.superstringtheory.com/]Click Me![/url]

Yeah that is a pretty good site hunter34.   :)

Well it has to be good, It was the first thing when you googled it.

I'm not saying that it has to be the explanation for the laws of the universe, or even that if a string theory that appears to describe our universe is found, that there isn't an equivalent, non-string theory that yields identical results, but in my opinion, some string theory is going to be the UFT (Unified Field Theory.)

EDIT:
It needs a bit of refinement first.  Quite a bit.  If it does actually work out.

Here's a good website http://www.theory.caltech.edu/people/jhs/strings/string14.html

String theory is a broad idea of an unified theory (that Einstein came painstakingly close to finding). The only standing candidate for that is the M Theory, which is an extension of the string theory and identifies 11 dimensions which may exist according to the Feynman sum over alternate histories. This sum says that a particle takes all paths at the same time (search Buckyball Experiment).

Guys have you read this [url=http://www.amazon.com/The-Trouble-With-Physics-Science/dp/0618551050]book[/url]? Its approved by Peter Shor [url=http://www.amazon.com/review/R2H7GVX4BUQQ68/ref=cm_cr_pr_viewpnt#R2H7GVX4BUQQ68]here[/url]!
And its gonna be hard to see if its right or wrong when there's [url=http://en.wikipedia.org/wiki/String_theory_landscape]10^500[/url] of them

Right now, astrophysics is a huge mess of theories, and it's hard to tell what's what nowadays.
In a book I read by Michio Kaku, he talks about String Theory and later on goes back to the Big Bang, the beginning of time. He shows the separation of the 4 forces (how they were once unified).
Honesly, nothing is proven right now. It's still a huge mess of theories that will hopefully clear up with the young minds of our generation.
It's rumored that if Einstein had thought of a big picture, he might have come up with the unified field theory (he focused on mainly abstract math teh rest of his life, which depresses me).

Well, there is a lot of talking going on about string theory, but accept it or not, it is the only game in the town, that can unify everything.
Although there are numerous theories, I think that M-theory is the only one worth a shot.


calculus
integration
trigonometry
real analysis
real analysis unsolved

Prove

$ \prod_{n\equal{}1}^\infty \left(1\minus{}e^{\minus{}2\pi n}\right)\equal{}\frac{\Gamma\left(\frac14\right)\, e^{\frac{\pi}{12}}}{2\,\pi^{\frac34}}$

Or can you show

$ \prod_{n\equal{}0}^\infty \left(1\plus{}e^{\minus{}(2n\plus{}1)\pi}\right)\equal{}2^{1/4}\, e^{\minus{}\pi/24}$

here, i'll prove $\prod_{n=1}^{\infty }\frac{(1-e^{-2\pi nx})}{(1-e\frac{-2\pi n}{x})}=\frac{e^{\frac{\pi }{12}(x-\frac{1}{x})}}{\sqrt{x}}$.
from easy integral $\int_{0}^{\infty }e^{-ax}\sin (bx) dx=\frac{b}{a^{2}+b^{2}}$ it follows that $\int_{0}^{\infty }\frac{\sin (ax)}{e^{x}-1}dx=\int_{0}^{\infty }\sin (ax)\sum_{n=1}^{\infty }e^{-nx}dx=\sum_{n=1}^{\infty }\frac{a}{n^{2}+a^{2}}=\frac{\pi }{2}+\frac{\pi }{e^{2\pi n}-1}-\frac{1}{2a}$.
therefore after some modification $\int_{0}^{\infty }\frac{ax\cos (2\pi nx)}{e^{2\pi ax}-1}dx=\frac{a}{8\pi ^{2}n^{2}}-\frac{e^{\frac{2\pi n}{a}}}{2a(e^{\frac{2\pi n}{a}}-1)^{2}}$.
Now Poisson summation formula gives
$\frac{1}{2\pi }+2\sum_{n=1}^{\infty }\frac{an}{e^{2\pi an}-1}=\frac{1}{12a}+\frac{a}{12}-\frac{2}{a}\sum_{n=1}^{\infty }\frac{e^{\frac{2\pi n}{a}}}{(e^{\frac{2\pi n}{a}}-1)^{2}}=\frac{1}{12a}+\frac{a}{12}-\frac{2}{a}\sum_{n=1}^{\infty }\frac{n}{e^{\frac{2\pi n}{a}}-1}$.
hence we derived $\sum_{n=1}^{\infty }\frac{an}{(e^{2\pi an}-1)}+\frac{\frac{n}{a}}{(e^{\frac{2\pi n}{a}}-1)}=\frac{1}{24}(a+\frac{1}{a})-\frac{1}{4\pi }$. and integration both side is exactly the same equation  $\prod_{n=1}^{\infty }\frac{(1-e^{-2\pi nx})}{(1-e\frac{-2\pi n}{x})}=\frac{e^{\frac{\pi }{12}(x-\frac{1}{x})}}{\sqrt{x}}$

[quote="Ponclete"]Let $f(x)=\sum_{n=-\infty }^{\infty }x^{n^{2}}$, $g(x)=\sum_{n=-\infty }^{\infty }(-1)^{n}x^{n^{2}}$.
there is amazing identity between f(x),g(x). I'll show two of them.
(1)f²(x)+g²(x)=2f²(x²)
....[/quote]

Hello Ponclete,

Could you give more identities with f and g ?

Or references

Thank you

Moubinool, sorry for late posting.
I think there are many relationship between $f(x)$ and $f(x^{n})$.
Begin to this, I'll prove $\prod_{n=1}^{\infty }(1+x^{2n-1})^{8}-\prod_{n=1}^{\infty }(1-x^{2n-1})^{8}=16x\prod_{n=1}^{\infty }(1+x^{2n})^{8}$.
In my course, we need Jacobi's triple product identity $\prod_{n=1}^{\infty }(1-x^{2n})(1+x^{2n-1}z)(1+\frac{x^{2n-1}}{z})=\sum_{n=-\infty }^{\infty }x^{n^{2}}z^{n}$.
Using this, we can transform f²(x)+g²(x)=2f²(x²) into $\prod_{n=1}^{\infty }(1+x^{2n-1})^{4}+\prod_{n=1}^{\infty }(1-x^{2n-1})^{4}=2\prod_{n=1}^{\infty }(1+x^{2n})^{2}(1+x^{4n-2})^{4}$.
Similary, from obvious identity $\sum_{n=-\infty }^{\infty }x^{n^{2}}+\sum_{n=-\infty }^{\infty }(-1)^{n}x^{n^{2}}=2\sum_{n=-\infty }^{\infty }x^{4n^{2}}$ and $\sum_{n=-\infty }^{\infty }x^{n^{2}}-\sum_{n=-\infty }^{\infty }(-1)^{n}x^{n^{2}}=4x\sum_{n=-\infty }^{\infty }x^{4n^{2}+4n}$,
we get two infinite sereis identity
$\prod_{n=1}^{\infty }(1+x^{2n-1})^{2}+\prod_{n=1}^{\infty }(1-x^{2n-1})^{2}=2\prod_{n=1}^{\infty }(1+x^{2n})(1+x^{4n})(1+x^{8n-4})^{2}$
$\prod_{n=1}^{\infty }(1+x^{2n-1})^{2}-\prod_{n=1}^{\infty }(1-x^{2n-1})^{2}=4x\prod_{n=1}^{\infty }(1+x^{2n})(1+x^{4n})(1+x^{8n})^{2}$,
and by multiplying three identity, proof is over.
This identity shows that there is an algebric relationship between $x^{\frac{1}{24}}\prod_{n=1}^{\infty }(1+x^{n})$ and $x^{\frac{1}{12}}\prod_{n=1}^{\infty }(1+x^{2n})$ and suggests that there might be similar algebric relationship with $x^{\frac{k}{24}}\prod_{n=1}^{\infty }(1+x^{kn})$.
So far I could find when k=2,3,5 but with regard to general identity I have almost nothing.
I hope this writing fulfilled your curiosity.

In your proof the equality

$ \sum_{n=1}^\infty \frac{x^n}{(x^n-1)^2}=\sum_{n=1}^\infty \frac{n}{x^n-1} $ (for $ x>1 $) occured.

Could you prove it, please.


Euler logarithms limit geometry function trigonometry perimeter

You'll get everything back... it will just take a few centuries.

[quote="nutz_for2.718281828"]Some wizard just made everyone forget all mathematics except arithmetic. :(

With only those tools, and ones we will "invent", let's see how far we can go to get all of our math back.

P.S. All formulas are erased, too.

Fun game.  :rotfl:[/quote]

lol sounds fun...how far does arithmetic go? like to powers? etc. lol

Oh, come on.  The wizard left us with arithmetic?  Why didn't he just finish the job?  We can reconstruct arithmetic...

[quote="Lena"]I would like to introduce the fibonacci theorem

$a_{1}=1 , a_{2}= 2$

every next number is the sum of the previous two[/quote]What a stupid sequence, I see no useful application for it. :) And by the way, why that strange name, "Fibonacci"?

i use magic to make all tha math back again

i win

[quote="mathnerd314"][quote="nutz_for2.718281828"]HEY!!! I've got it! Let's use letters for random numbers, like

$A=\frac{1+2+3}{4}$

$B=345827340598345$

$C=0$[/quote]

Good idea! Then we can make expressions like $3x+5$ and set them equal to things like $10$ so

$3x+5=10$

I wonder how we could find x?

Fun fun :rotfl:[/quote]

then using that we discover fractions (part of arithmetic? or no?)

then eventually we have reciprocals.

[quote="lingomaniac88"]I wonder what happens if we added up all the reciprocals of positive integers.  What's $1+\frac12+\frac13+\frac14+\ldots$?  Maybe we could make up some sort of notation for that.[/quote]

$\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...=e$

The Euler Number

[quote="nutz_for2.718281828"]hmm... I KNOW!!!

$\log_{a}c=b$ where $a^{b}=c$[/quote]

then we discover square roots

$\sqrt{4}=2$

if $n$ is really really big , then $\frac{1}{n}$ should be really really small ... but how small ? 0 ? i think we should invent analysis. yeeey, fun  :idea:

I know!
Let's organize proofs into two columns, with the left column being the statements and the right column being the reasons!

Let's say the number of integers is $\aleph_{0}$!

[quote="lingomaniac88"]But I just demonstrated that we cannot divide by 0.  If a false step leads to something true, it does not necessarily mean that the first step cannot be false.  Example:

$2=1$
multiply both sides by 0:
$0=0$

Now, it looks like we're trying to figure out how we can prove things.[/quote]
but you didnt prove that $2\ne0$

[quote="zeb"]Looks like we need to start learning about mathematical rigor. :idea:

Hey, why don't we use a completely different set of axioms this time around?

Like this one:

Axiom $0$: all axioms with even indices greater than two which are not the sum of two odd primes are false.

And this one:

Axiom $1/2$: the set of all axioms is isomorphic to an uncountable cardinal number.

And:

Axiom $\pi$: every true statement can be proved in a finite number of steps, from the axioms.[/quote]
Brilliant!
Let's name those axioms after this amazing mathematician! We can call those the Zeb axioms! man thats cool

[quote="tjhance"]Proof that 2=0

As x gets super small, $\frac{1}{x}$ gets super big

So $\frac{1}{0}=\infty$

Multiply 0 to both sides

$1=0\bullet\infty$
$1=0$

multiply both sides by 2

$2=0$[/quote]
well that proof disbalances our whole world (since you can add/subtract/multiply/divide/take the root of both sides, you can make anything equal anything else) and makes everything equal, i will make 2 rules that disallows that proof.
1. Division by 0 is illegal. the result will be undefined, and undefined will not be a number.
2. Infinity is not a number, just a concept. it can be approached, but never reached. it shall not be defined as $\frac{1}{0}$. If it ever is defined as $\frac{1}{0}$, then the opposite of infinity is equal to infinity.

i cannot justify those rules, so i made up my mind that they are axioms.

[quote="tjhance"]Proof that 2=0

As x gets super small, $\frac{1}{x}$ gets super big

So $\frac{1}{0}=\infty$

Multiply 0 to both sides

$1=0\bullet\infty$
$1=0$

multiply both sides by 2

$2=0$[/quote]

Actually, in some fields that is true. Define $2=1+1$, and take a field with characteristic two, e.g. the field of integers modulo two...

(I don't know what I'm talking about, I just need to make myself feel smart :lol:)

in that case, 2 characteristic 2 would not exist, because it is actually 0 characteristic 2
the arithmetics the wizard left us with states that the remainder has to be less than the divisor.

Ooh, here are two (or zero, if you prefer) more cool self-referential axioms!

Axiom $(1+\sqrt5)/2$: The cardinality of the set of true statements is strictly greater than the cardinality of the set of axioms.

Axiom $-1$: No axiom can be proved from the other axioms in a finite number of steps.

However, I'm worried that eventually my axioms will contradict eachother... maybe we should try to imagine a concrete example of a system that satisfies my axioms, to check that they could make sense?

Axiom $-\frac{1}{2}$: any axiom posted by ProtestanT is true.

Axiom 42: Any axiom proposed by mathnerd314 is false.

Heehee.

:harhar:

since that axiom number is even, it will not be accepted.
it will simply confuse our dimwitted community

Okay:

Axiom 1337: Every axiom proposed by lingomaniac88 is false.

Axiom 1339: Except axiom 1337.
Axiom 1341: And axiom 1339.
Axiom 1343: And axiom 1341.
Axiom 1345: And axiom 1343.
Axiom 1347: And axiom 1345.
Axiom 1349: And axiom 1347.
Axiom 1351: And axiom 1349.
Axiom 1353: And axiom 1351.
$\vdots$
Axiom $2n+1\forall n\in\mathbb{N},n\ge677$: And axiom $2n-1$.
$\vdots$

Okay, I just used up all the odd axioms $\ge 1337$.  Perfect. :dry:

That's fine, we have an uncountable set of axioms.

Somehow I find the idea of an uncountable axiom set appealing (kinda like a Hamel basis or something). I'm pretty sure I'm the only one. It *does* get rid of Goedel's Incompleteness Theorems (I think).

Actually.... could we prove that one exists using Zorn's lemma? One that includes these:

Axiom $a$: every statement is provable or provably false from these axioms.
Axiom $b$: no axiom can be proven from the other axioms.

It should also be able to simulate arithmetic. NBG would be nice.

Of course, constructing these axioms now becomes extremely tricky, mostly because of axiom $b$. For instance, we would have to *not* have an axiom stating that there are uncountably many of them, as this is implied by $a$, $b$, and NBG.

There could also be interesting theorems *about* the axioms themselves. Example: "There exists an axiom stating that Theorem B is true iff there is no axiom that states Theorem A is true." Or: "For any finite subset of axioms containing $a$ and $b$, there are two different sets of axioms containing that finite subset, both of which are consistent, but inconsistent with eachother."


Fulltest 8, Paper1, qn 2:
In the young's double slit apparatus, a parallel beam is incident  making some angle $ \theta$ with the normal to the slit, the fringe width is?
The soln they had said fringe width doesn depend upon the angle made by the inc. beam wid the normal for small angles, but for large angles, how does it ?

[hide="What i thought :"]Suppose the incident beam makes$ \theta$ with the normal to the plane of the slits,
In this case ur total phase difference is $ d\sin\theta \plus{} \frac {y \cdot d}{D} \equal{} n\lambda \mbox {for constructive interefer.}$
Now distance of the nth bright fringe from the centre is$ \frac {n\lambda \cdot D}{d} \minus{} D\sin\theta$
while calculating fringe width wont the $ \sin \theta$ term get cancelled anyway?
D and d hav their usual meaning..[/hide]

does the solution say that it does, coz what u said i think is correct.
i mean if they say that for small angles it is true, its not that they r saying its not true for large angles??


search
linear algebra
matrix
algebra
polynomial
linear algebra unsolved

Let $ A \in M_{m\times n}(F)$ and $ B \in M_{n \times m}(F)$ be matrices. Prove the identity: $ x^n \det(I_{m} x \minus{} AB) \equal{} x^m \det(I_{n} x \minus{} BA)$.

[hide] We are not restricted ourselves to $ F \equal{} \mathbb{R}$.[/hide]

I think that it's not that hard to modify the argument I posted over on the other thread so that it works over any commutative ring. (I also think I mixed up $ n$ and $ m$ in some places over there.)

Let me elaborate.

We will assume that $ A$ is an $ m\times n$ matrix and that $ B$ is an $ n\times m$ matrix. For any square matrix $ M,$ let $ P_{M}(x) \equal{} \det(xI \minus{} M)$ be the characteristic polynomial of $ M.$

Claim: $ x^{n}P_{AB}(x) \equal{} x^{m}P_{BA}(x).$

Let $ U \equal{} \begin{pmatrix}0 & A \\
B & 0\end{pmatrix}.$

Let's compute the characteristic polynomial of the $ (m \plus{} n)\times(m \plus{} n)$ matrix $ U.$

$ \det(xI \minus{} U) \equal{} \det\begin{pmatrix}xI & A \\
B & xI\end{pmatrix}.$

Do row operations here. Our first row operation is to multiply the lower blocks by $ x.$ This multiplies the characteristic polynomial by $ x^n.$ At this point we have

$ x^n\det(xI \minus{} U) \equal{} \det\begin{pmatrix}xI & A \\
xB & x^2I\end{pmatrix}.$

Add to the lower blocks $ \minus{} B$ times the upper blocks. This leaves the determinant unchanged.

$ x^n\det(xI\minus{}U)\equal{}\det\begin{pmatrix}xI & A \\
0 & x^2I \minus{} BA\end{pmatrix}$

$ \equal{} x^{n\plus{}m}\det\left(x^2I \minus{} BA\right) \equal{} x^{n \plus{}m}P_{BA}(x^{2})$

Now, perform the same argument on the matrix

$ V \equal{} \begin{pmatrix}0 & B \\
A & 0\end{pmatrix}$

$ x^mP_{V}(x) \equal{} x^{n \plus{} m}P_{AB}(x^{2}).$

But $ U$ and $ V$ are similar. They can be made similar by a permutaion with the structure

$ \begin{pmatrix}0 & I _n\\
I_m & 0\end{pmatrix}.$

Therefore, $ U$ and $ V$ have the same characteristic polynomial. We have that

$ x^{n\plus{}m}P_U(x)\equal{}x^{n\plus{}2m}P_{BA}(x^2)$
$ x^{n\plus{}m}P_V(x)\equal{}x^{2n\plus{}m}P_{AB}(x^2)$

We conclude that

$ x^{2n}P_{AB}(x^{2}) \equal{} x^{2m}P_{BA}(x^{2}).$

Since these are equal as polynomials, we can replace $ x^2$ by $ x$ to get the desired result.

At no point did I divide by any element; hence this argument is valid over any commutative ring.

[quote="olorin"]Prove this over any commutative ring. :wink:[/quote]
See my second post [url=http://www.mathlinks.ro/viewtopic.php?t=202260]here[/url].

Very nice solution.

My sollution assumes that the conclusion holds in case $ m \equal{} n$. If $ m < n$ then just extends the matrices $ A, B$ to $ A', B'$ so that the extended matrices are in $ M_n$ (fill in the remaining entries with 0). The second task is to expand and watch the result.

Good thing is this is easy to see. Bad thing is that I have to prove the special case $ m\equal{}n$ separately.

[quote="olorin"]Prove this over any commutative ring. :wink:[/quote]

I 've just realized that this fact is a corollary of the original problem on a field.

Let $ m<n$ then for an eigenvalue of $ AB, \lambda$:
$ ABx\equal{}\lambda x\rightarrow BA Bx\equal{}\lambda Bx \rightarrow \lambda$ is an eigenvalue of $ BA$.
Now we know that $ Tr((AB)^k)\equal{}Tr((BA)^k) ,\forall k$, so if $ a_1,a_2,\dots,a_m$ are $ AB$'s eigenvalues and $ b_1,b_2,\dots,b_{n\minus{}m}$ the rest of the eigenvalues of $ BA$ then:
$ \sum b_i^k\equal{}0, \forall k$,
 and from the Newton identity: $ ks_k\equal{}s_{k\minus{}1}p_1\minus{}s_{k\minus{}2}p_2\plus{}\dots\plus{}(\minus{}1)^{k\minus{}1}s_0p_k$ ( where $ s_i$ are Viete sums and $ p_i\equal{}\sum x_k^i$, with $ x_k$ the roots of the polinomial)
we have that $ s_k\equal{}0, \forall k$ so $ b_k\equal{}0, \forall k$.
Then $ f_{BA}\equal{}x^{n\minus{}m}f_{AB}$ and we are done.


Coloring
combinatorial geometry
geometry
Ramsey Theory
IMO
IMO 1984

Given points $O$ and $A$ in the plane. Every point in the plane is colored with one of a finite number of colors. Given a point $X$ in the plane, the circle $C(X)$ has center $O$ and radius $OX+{\angle AOX\over OX}$, where $\angle AOX$ is measured in radians in the range $[0,2\pi)$. Prove that we can find a point $X$, not on $OA$, such that its color appears on the circumference of the circle $C(X)$.

Let $a_{n},\ n\ge 1$ be a sequence of positive reals such that $\left(\sum_{n\ge 1}a_{n}\right)^{2}<2\pi\ (*)$. For each $n\ge 1$, let $\mathcal C_{n}$ be the circle centered at $O$ with radius $\sum_{i=1}^{n}a_{i}$.

Because of $(*)$, we can find points $x_{i,j}\in\mathcal C_{i},\ i,j\ge 1$ such that for all $i,j\ge 1$ we have $\mathcal C(x_{i,j})=\mathcal C_{i+j}$. We now forget about all the other points, and work only with the matrix $M=(x_{i,j})$.

Suppose we use $n$ colors. There must be one, $c_{1}$, which appears infinitely many times on the first row of $M$, in, say, points $x_{1,j_{1}},x_{1,j_{2}},\ldots$. Then $c_{1}$ cannot appear on the lines $j_{k}+1$, $ k\ge 1$. Next, there is a color $c_{2}$ which appears infinitely often among the points $x_{j_{1}+1,j_{k}-j_{1}}$,  $k\ge 2$. But then $c_{2}$ cannot appear on the lines $j_{k}+1$ for such $k$. Repeating this procedure, we reach a stage where we have a row of $M$ (infinitely many actually) on which none of our $n$ colors $c_{1},\ldots,c_{n}$ can appear. This is a contradiction.


Hi  :D  ! $ a,b,c$ are positive and we know that $ \frac{1}{a\plus{}1} \plus{} \frac{1}{b\plus{}1} \plus{} \frac{1}{c\plus{}1} \equal{} 2$.Prove that
$ \frac{1}{4a\plus{}1} \plus{} \frac{1}{4b\plus{}1} \plus{} \frac{1}{4c\plus{}1} \geq 1$

put $ a\equal{} \frac{x}{y\plus{}z}$ and so on.then use Cauchy (i.e. mutiplying the first fraction by $ y\plus{}z$ and ...)

Why we can put  :idea:  $ a\equal{}\frac{x}{y\plus{}z}$ (If we able to do it we should have $ abc\equal{}\frac{xyz}{(x\plus{}y)(x\plus{}z)(y\plus{}z)} \leq \frac{1}{8}$ by AM-GM)

[quote="mahanmath"]Why we can put  :idea:  $ a \equal{} \frac {x}{y \plus{} z}$ (If we able to do it we should have $ abc \equal{} \frac {xyz}{(x \plus{} y)(x \plus{} z)(y \plus{} z)} \leq \frac {1}{8}$ by AM-GM)[/quote]
denote each fraction (in the assumption) by $ k,l,m$ respectively.
since $ k\plus{}l\plus{}m\equal{}2$ there exists obviously positive reals like $ x,y,z$ such that $ \frac{y\plus{}z}{x\plus{}y\plus{}z}\equal{}k$ and so on.
then we get
$ \frac{1}{a\plus{}1}\equal{}\frac{y\plus{}z}{y\plus{}z\plus{}x} \implies a\plus{}1\equal{} \frac{x\plus{}y\plus{}z}{y\plus{}z}\equal{}1\plus{}\frac{x}{y\plus{}z}$.
and the same holds for $ b,c$ so we can put  $ a\equal{}\frac{x}{y\plus{}z}$

[quote="mahanmath"]Hi  :D  ! $ a,b,c$ are positive and we know that $ \frac {1}{a \plus{} 1} \plus{} \frac {1}{b \plus{} 1} \plus{} \frac {1}{c \plus{} 1} \equal{} 2$.Prove that
$ \frac {1}{4a \plus{} 1} \plus{} \frac {1}{4b \plus{} 1} \plus{} \frac {1}{4c \plus{} 1} \geq 1$[/quote]
AM-HM: $ \frac{\frac{1}{4a\plus{}1}\plus{}\frac{1}{3}}{2} \ge \frac{2}{4a\plus{}1\plus{}3} \iff \frac{1}{4a\plus{}1} \ge \frac{1}{a\plus{}1}\minus{}\frac{1}{3}$
So $ LHS \equal{} \sum_{cyc} \frac{1}{4a\plus{}1} \ge \sum_{cyc} \left ( \frac{1}{a\plus{}1}\minus{}\frac{1}{3} \right ) \equal{} 2\minus{}1\equal{}1$.


modular arithmetic
graph theory
combinatorics proposed
combinatorics

There are $n$ football teams in Tumbolia. A championship is to be organized
in which every two teams play exactly one match. All matches are played
on Sundays, and no team plays more than one match in the same day.
Determine the least positive integer $m$ for which it is possible to set up a
championship in $m$ Sundays.

This is very well-known. The question is: what's the minimal number of colors with which we can color the edges of the complete graph $K_n$ s.t. no two adjacent edges share the same color.

The answer is $n-1$ if $n$ is even and $n$ if $n$ is odd. I think there's a proof around here somewhere :?.

Actually I did it "fascinated"  :lol: by Graham-Pollack theorem - dividing $K_n$ into bipartite graphs - though I suppose it is just the same as with colours  :)

I give up.. :); I can't find it on the forum, so here's a proof:

First of all, it's clear that the answer should be at least $n-1$, since there are $n-1$ edges meeting in each vertex of the graph.

We first prove that $n-1$ is not possible for odd $n$: there are at most $\frac{n-1}2$ edges of each color, so, in total, we can have at most $\frac{(n-1)^2}2$ edges, which is absurd.

We now show that we can properly color the edges of a $K_{2n+1}$ with $2n+1$ colors: label the vertices $0,1,\ldots,2n$, and color the edge $ij$ with the color $i+j\pmod{2n+1}$. This is easily seen to be a proper coloring.

Now, in order to show that the edges of $K_{2n}$ can be properly colored with $2n-1$ colors, we do this: we take one vertex out of the picture, and properly color the edges of the remaining $K_{2n-1}$ with $2n-1$ colors as described above. After this, we connect the vertex we took out with the vertex $i,\ i\in\overline{0,2n-2}$ through an edge colored with the color $2i\pmod{2n-2}$.


geometry
geometric transformation
Support
IMO

Here is a serious but although somehow funny question: Who will be first to have IMO2004 problems online ?  :)

John Scholes said that he was too busy to put them on the site..

[quote="Peter VDD"]www.mathlinks.ro :D[/quote]yes of course, we are talking about silver and bronze medals here, not the gold medal (which is of course attributed to us :P).

I know ML is the gold medal winner.  :)  But my poll was to determine the bronze medal winner. Silver medal goes to the Hyacinthos group. Have a look at their yahoo forum: http://groups.yahoo.com/group/Hyacinthos/  :)

bah i've seen ohter web pages having them already, like some personal websites of participants

I won't hope on more than a HM if I were them :D

As for ML the gold medal winner, what do you want? Have we ever been content with less? :P

It is still a tough fight for the bronze medal. But somehow I wonder that neither the Kalva nor the official IMO site is capable of posting it. I think all other science pages have their sites up-to-date for a long time now (except IOI which was not scheduled yet).  :)

[quote="darij grinberg"]Actually there's no chance for any of them to get bronze. The [url=http://www.mathematik-olympiaden.de/IMOs/imo04.pdf]official report of the German team leader Gronau (containing the problems)[/url] is online. And actually, as if this wouldn't be enough, [url=http://www.imo-register.org.uk/2004-report.html]the British[/url] have been quick, too.

  Darij[/quote]

Yeah, I read both documents. But I think these links did not used to be the locations where an average user would look for a report or problems/solutions first. And the document from the German server is written in German and there are not too many people out who know German or translate it with any translation software...

[quote="Peter VDD"]I won't hope on more than a HM if I were them :D[/quote]

I don't think they even deserve HMs... :(  Honestly, how difficult is it to put the questions up? :?

On that note, [url=http://www.srcf.ucam.org/~jsm28/imo-scores/]Joseph Myers' site[/url] is also fashionably late in posting up the results...

[quote="Valiowk"][quote="Peter VDD"]On that note, [url=http://www.srcf.ucam.org/~jsm28/imo-scores/]Joseph Myers' site[/url] is also fashionably late in posting up the results...[/quote][/quote]

I've been waiting for the right one of the organisers to get back from holiday to provide a list of scores on individual problems.  Until then I've been holding off proofreading the names from the printed list without individual problem scores in the hope that an electronic list will be provided that includes names, saving the proofreading.

While waiting, I've added the scores from 1977 (missing names of two Finnish contestants), 1983, 1990 (missing the name of one Mexican contestant), 1991 and 1992, plus corrections and additional names for 1989.  (1977, 1983 and 1990 lack scores on the individual questions.)

[quote="orl"]I see you managed to put up the results I sent you a long time ago.  :)  BTW how far did you come with the scores from the Russian books by Morozova/Petrakov ?[/quote]

I added support for scores without individual question scores when I received an offer of the 1990 scores to add to those for 1977 and 1983 you sent me (and then put up 1983, followed by 1990, followed by 1977).  I haven't done anything about putting up scores where I have neither individual question scores nor a reasonably complete list of names (whether no names, or medallists only as in the Russian books) but do have the individual totals by contestant code.

[quote="jsm28"][quote="Valiowk"][quote="Peter VDD"]On that note, [url=http://www.srcf.ucam.org/~jsm28/imo-scores/]Joseph Myers' site[/url] is also fashionably late in posting up the results...[/quote]

I've been waiting for the right one of the organisers to get back from holiday to provide a list of scores on individual problems.  Until then I've been holding off proofreading the names from the printed list without individual problem scores in the hope that an electronic list will be provided that includes names, saving the proofreading.[/quote][/quote]

I see.  Thanks for letting us know about it! :)

Hi jsm, I like your site of that has IMO scores and IMO reports. :) Thank you for your hard work of putting everything up.

Yes, I have seen it. I like the contents - thanks, Nikolaos, for uploading the photos, that's very nice! -, but I really hate the plugins. I would prefer seeing a normal textbased website without blinking letters and flying elephants (or are they boars?) to a page which takes two minutes to load, causes the internet explorer to hang and displays nonstop java errors...

  Darij

[quote="Anonymous"]Has anyone checked http://www.imo2004.tk ? It's cool...[/quote]

Yeah, it is a nice site. But I do not get why you established a chat room. I looked into it for some times and it was always empty.  :D

I dunno, 2004 seems to be the worldwide year of questions-seem-not-to-get-put-online...

Not for IMO, not for our national contests, not for kalva :P

It's just a bad time I guess :P

[quote="Valentin Vornicu"][quote="Templar"]Judging from the current state of http://www.imo2004.gr, will they ever have it online?[/quote]I was just wondering. You guys DO know that the questions / solutions / comments are online on THIS site since the IMO took place, don't you?

I don't that that it really matters if other sites have or not uploaded the questions, as long as there is one that did. :)

Same goes for the IMO Shortlists / IMO Longists.[/quote]

If you'd read the previous page Valentin, you'll get answer to that ;)

It seems that the official IMO site has won the race...   :)


Let $ ABC$ be an equalateral triangle with sides $ a,b$ and $ c$. Consider a triangle $ PQR$ with sides equal to $ a\plus{}\frac{b}{2}, b\plus{}\frac{c}{2},$ and $ c\plus{}\frac{a}{2}$. Show that $ [PQR]\equal{}\frac{9}{4}[ABC]$, where $ [XYZ]$ represents the area of the triangle $ XYZ$.

Why are you writing $ ABC$ equilateral and again mentioning the sides to be $ a,b,c$? It is $ a \equal{} b \equal{} c \equal{} k$ after all!
Anyway $ [ABC] \equal{} \frac {\sqrt{3}}{4} k^2$
Side length of $ PQR$ is $ (k \plus{} \frac {k}{2}) \equal{} \frac {3k}{2}$
Thus its area=$ [PQR] \equal{} (\frac {3}{2})^2\times[ABC] \equal{} \frac {9}{4}[ABC]$...Done!
It is very very trivial! I wonder if you wanted to mean anything else...'cause it is very easy!
 :)

I know the problem is indeed very easy and I did not mean anything. Moreover mathlinkers must sometimes be given a few easy ones too. Dont u believe that? :lol:

Maybe...but it was rather easy! I wonder if the general case would have been considered... :)


geometry
combinatorics unsolved
combinatorics

You are given $n$ squares, not necessarily all of the same size, which have total area 1. Is it always possible to place them without overlapping in a square of area 2?

I solved this one sometimes ago here...
I will try to find it.

Pierre.

Here :
http://www.mathlinks.ro/Forum/viewtopic.php?p=21783#21783

But, come on Arne, you already proposed it  and I already gave the same answer!

Pierre.

I know I posted it before, but I needed it urgently for something we're working on :) and I couldn't find it anyway :( [I guess you know what I'm talking about]


Unde sunt punctajele de la teste? In mod sigur se stiu deja.
Cine a intrat in lot?
Unde se tine lotul anul asta?

Se tine la Bucuresti. Rezultatele (pentru primele doua teste la seniori) sunt:

Zamorzaev Alexandru (Sasha - R. Moldova) 7 7 7 7 7 7 = 42
Zahariuc Adrian (Dzeta) 7 7 2 7 7 7 = 37
Boreico Iurie (Iura - R. Moldova) 7 7 7 2 7 5 = 35
Kreindler Gabriel (Kappa) 5 7 7 1 7 5 = 32.5
Stefanescu Andrei (Andreis) 7 5.5 7 0 7 5.5 = 32
Michnea Dragos 7 7 0 0 7 7 = 28
Borsenco Ivan (Prowler - R. Moldova) 7 2 0 0 7 7 = 23
Purcaru Alin 7 7 0 0 7 1.5 = 22.5
Ungureanu Andrei 7 7 1 0 0.5 7 = 22.5
Ionescu Dragos 7 0 0 7 7 1 = 22
Banica Dan 7 0.5 0 0 7 7 = 21.5
Dicu Gabriel 7 7 6.5 0 0 1 = 21.5
Dragus Andrei 7 0 0 0 7 7 = 21
Pachitariu Marius (Ikap) 7 0.5 0 0 7 6 = 20.5
Filip Simeon (Dust - R. Moldova) 7 7 0 1 0 5 = 20

Mai atasez si rezultatele finala de la nationala, in format .xls:

[quote="amfulger"]Felicitari celor admisi!
Cei din Republica Moldova sunt parte din lot sau au dat testele fiindca locul la nationala le da dreptul?[/quote]Presupun ca varianta 2. Ideea e ca in fiecare an elevii din R. Moldova dau si ei barajele, si de obicei erau corectate de profesorii lor, deci nu conta prea mult pozitia lor in clasament. Anul asta le-am corectat oricum noi pe toate.