If $ x \minus{} y \equal{} 6$ and $ x \plus{} y \equal{} 12$, what is the value of $ y$?

so we know x is 6 larger than y.
So we just test values where x+y=12
and we see when x=9, y=3
and 9-3=6
y=3.

Alternatively, adding the two equations gives us $ 2x = 18 \implies x = 9$, so $ y = \boxed{3}$.


quadratics
calculus
calculus computations

have this problem, it does not look difficult, but I do not seem to be able to bring n to one side to find N s.t. $ x_{n} \plus{}1$ < $ \frac{1}{10^2}$ for all n > N.

$ x_{n}\equal{}\frac{6\minus{}n^2}{n^2 \plus{} 5*n}$

$ \frac{6\minus{}n^2}{n^2 \plus{} 5*n} \plus{} 1$ < $ \frac{1}{10^2}$

$ \frac{6\plus{} 5*n}{n^2 \plus{} 5*n}$ < $ \frac{1}{10^2}$

$ \frac{(6/(5*n)) \plus{} 1}{(n/5) \plus{}1}$ < $ \frac{1}{10^2}$

would appreciate some help. thanx.

hello, multiplying  your term by $ 10^2$ we get
$ 5\cdot10^2n\plus{}6\cdot10^2<n^2\plus{}5n \Leftrightarrow 0<n^2\minus{}495n\minus{}600$, now you can solve this with the formual for a quadratic equation.
Sonnhard.


calculus
derivative
calculus computations

Let $ f(x)$ be a thrice continuously differentiable function.Prove or disprove that between  two consecutive roots of $ f'''(x) \equal{} 0$ there exist atmost $ 4$ roots of $ f(x) \equal{} 0$.

Note that $ f(x)\equal{}0$ is third time differentiable, then $ f'''(x)\equal{}0$ for all $ x$. Now choose $ x_0\equal{}1$ and $ x_1\equal{}2$, then $ f'''(1)\equal{}f'''(2)\equal{}0$ but $ f(x)\equal{}0$ in all points between $ 1$ and $ 2$ because $ f(x)\equal{}0$ for all $ x$ then we have more than $ 4$ roots of $ f$ in the interval.

According to Rolle's theorem, between two consecutive zeros of $ f'''(x)$ there is atmost one zero of $ f''(x)$. But this means that there is, atmost, one possible point of inflexion, which means that there is a maximum of two zeros of $ f'(x)$, which implies that there are at most 3 zeros of $ f(x)$. Therefore, the given theorem is false.


Is there a spot on this forum for elementary number theory questions? More specifically questions on following  logic in proofs.

http://www.artofproblemsolving.com/Forum/index.php?f=6

For posting olympiad level elementary number theory problems.


logarithms
trigonometry
function
calculus
algebra unsolved
algebra

Let $k$ be a positive integer. Let $\log{(\cos{x})_k}=\log{(\cos{(\log{(\cos{\cdots})})}}$ with $k$ times. Find all $x$ when $\frac{2\pi}{3}<|x|<\frac{9\pi}{2} $ such that the sum(k=1 to n) of $k\log{(\cos{x})_k}$ is maximal

http://www.math.umn.edu/~garrett/a08/Graph.html
use this to make the function graph and see for yourself.
you can use calculus also but i didnt do that

But the given condition is $\frac{2\pi}{3}<|x|<\frac{9\pi}{2} $

ขอเมล์มั่งค้าบ อยากคุยกับพี่แต่อ่านไม่รุเรื่อง


Δεν ξέρω πως θα σας φανεί απλώς μου φάνηκε ωραία μιας και είναι η πρώτη μου τέτοιου είδους...Λοιπόν....

Έστω ομάδα G με τις ακόλουθες ιδιότητες:
i) Για κάθε χ,ψ στη G , είναι (χψ)^2 = (ψχ)^2
ii) Στη G δεν υπάρχουν στοιχεία τάξης 2.
Να αποδείξετε ότι η G είναι Αβελιανή (abelian)

Παιδιά επειδή τη μεταφράζω από τα αγγλικά τη δίνω και στα αγγλικά μήπως βρει κανένας κανα μεταφραστικό λάθος...

"Let G be a group with the following two properties:

(i)   For all x, y in G, (xy)^2 = (yx)^2,
(ii)  G has no element of order 2.
Prove that G is abelian."

Απο το ι) δεν συνεπαγεται οτι η G ειναι αβελιανη?

Στοιχεια ταξης 2 τι ειναι ακριβως?

Νομίζω ότι πρέπει να αποδείξεις ότι xψ = ψx για να καθιστά τη G αβελιανή.....Επίσης mailo η "τάξη" (στα αγγλικά order) που αναφέρω αφορά στο πλήθος των στοιχείων μιας ομάδας, δλδ πόσα στοιχεία έχει.....

Για τη συγκεκριμένη έχω δύο λύσεις αν και εσάς μπορεί να σας φανεί ιδιαίτερα εύκολη σαν άσκηση....

Επειδή διαβάζω Abstract Algebra στα αγγλικά δεν ξέρω να τα μεταφράζω σωστά και μπορεί να λέγεται αλλιώς οπότε χίλια σόρι...............

Αφού η άσκηση δν άγγιξε κανέναν θα βάλω μία λύση τώρα.......

Έστω $ e$ το ουδέτερο στοιχείο. Για οποιαδήποτε δύο στοιχεία της ομάδας $ x ,y$ έχουμε:
Από i) : $ (xy)^2 = (yx)^2$ Έτσι $ (xy)^ - ^1(xy)^2(yx)^ - ^1 = (xy)^ - ^1(yx)^2(yx)^ - ^1$. Δλδ $ (xy)(yx)^ - ^1 = (xy)^ - ^1(yx)$.Τετραγωνίζουμε και από i) πάλι:
$ ((xy)(yx)^ - {}^1){}^2 = ((yx)(xy)^ - {}^1){}^2$. Αφού
$ ((yx)(xy)^ - ^1)^ - ^1 = (xy)(yx)^ - ^1$ , συνεπάγεται ότι $ ((xy)(yx)^ - {}^1){}^4 = e$
Γράφοντας το αυτό ως: $ [((xy)(yx)^ - {}^1){}^2]^2 = e$ και από ii) έχουμε:
$ ((xy)(yx)^ - {}^1){}^2 = e$. Από ii) πάλι προκύπτει: $ (xy)(yx)^ - ^1 = e$.
Άρα $ xy = yx$ που καθιστά τη G αβελιανή...

Για τυχόν typo ή λάθος σόρι προκαταβολικά...Να μου το επισημάνετε....


calculus
integration
calculus computations

Hi ;) !!!

$ Let$ $ \sqrt [4]{x} \equal{} t \Rightarrow t^4 \equal{} x$

than: $ H \equal{} \int \frac {\sqrt [3]{1 \plus{} \sqrt [4]{x}}}{\sqrt {x}}dx \equal{} \int 4t \sqrt [3]{1 \plus{} t}dt$

$ Let$ $ y \equal{} \sqrt [3]{1 \plus{} t} \Rightarrow t \equal{} y^3 \minus{} 1$

than $ H \equal{} \int 12y^3(y^3 \minus{} 1)dy \equal{} \int (12y^6 \minus{} 12y^3) dy \equal{} \frac {12}{7}y^7 \minus{} 3y^4 \plus{} Cte$ .

finaly:

$ \int \frac {\sqrt [3]{1 \plus{} \sqrt [4]{x}}}{\sqrt {x}}dx \equal{} \frac {12}{7} \left( \sqrt [3]{1 \plus{} \sqrt [4]{x}}\right)^7 \minus{} 3 \left( \sqrt [3]{1 \plus{} \sqrt [4]{x}}\right)^4 \plus{} Cte$

________________________
lahoucine

$ \sqrt [3]{1 \plus{} \sqrt [4]{x}} \equal{} t\Longrightarrow \sqrt {x} \equal{} (t^3 \minus{} 1)^2\Longrightarrow \frac {1}{2\sqrt {x}}dx \equal{} 6t^2(t^3 \minus{} 1) dt$

$ \therefore \int \frac {\sqrt [3]{1 \plus{} \sqrt [4]{x}}}{\sqrt {x}} \equal{} 12\int t\cdot (t^6 \minus{} t^3)dt \equal{} \frac {12}{7}t^7 \minus{} 3t^4 \plus{} Const.$

$ \equal{} \boxed{\frac {12}{7}(1 \plus{} \sqrt [4]{x})^{\frac {7}{3}} \minus{} 3(1 \plus{} \sqrt [4]{x})^{\frac {4}{3}} \plus{} Const.}$

hello, simplifying your result i got $ \frac{3}{7}\left(\sqrt[3]{1\plus{}\sqrt[4]{x}}\right)^4(\minus{}3\plus{}4\sqrt[4]{x})\plus{}C$
Sonnhard.


$ABCD$ is circumscribed qadrilateral. The angel bisectors of $\angle ADC$ and $\angle BCD$ intersect in $E \in AB$. If $\frac{AB}{BC}=k$ find $\frac{S_{ADE}}{S_{BCE}}$

I think $\frac{AD}{BC}=k$ replace  $\frac{AB}{BC}=k$
Then  result is $k$

If AD/BC were what was given, the solution wouldn't require the use of cyclicity.

[hide]Take $F\in AB$such that $AD = AF$. We can prove $AFEC$ be cyclic. Hence $\angle BFC = \angle EDC=\frac{180-\angle B}{2}$. From that $\angle BFC = \angle BCF$ or $BF = BC$. We have $AD +BC = AB$. Implying $\frac{S_{ADE}}{S_{BCE}}=k-1$[/hide]


How would you show that f is constant if f(z) is entire and Im(f(z))>0 for all z?

Apply a linear fractional transformation: $ \frac{f(z)\minus{}i}{f(z)\plus{}i}$ is entire and bounded.

Generalize that, and we get that the image of any nonconstant entire function is dense in the plane.

Check out the [url=http://en.wikipedia.org/wiki/Picard_theorem]Little Picard theorem[/url].Its little too strong for this question  but would otherwise help you.


[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14012[/img]

 :(

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14041[/img]

 :clap:


Let $ a;b;c$ be positive real numbers satisfying $ abc \equal{} 1$
Prove that:
$ 1)\sum\frac{1}{\sqrt{13a\plus{}5}}\leq\frac{1}{\sqrt{2}}$
$ 2)\sum\frac{1}{\sqrt{7\plus{}2a}}\leq 1$

Let's generize the problems.

Let $ p,\ q$ be  positive real constants. For $ a > 0,\ b > 0,\ c > 0$ with $ abc \equal{} 1$,

prove that $ \frac{1}{\sqrt{pa\plus{}q}}\plus{}\frac{1}{\sqrt{pb\plus{}q}}\plus{}\frac{1}{\sqrt{pc\plus{}q}}\leq\frac{3}{\sqrt{p\plus{}q}}$.


Pumping stations deliver raw waste to a dump site at the rate modeled by the function D, given by
$ D(t)\equal{}\frac{10t}{(1\plus{}2t)}$
  Cleaning pumps remove the waste at a rate modeled by the function R, given by
$ R(t)\equal{}3\plus{}4cos(\frac{2{\pi}t}{15})$
Both D(t) and R(t) have units of tons per hour and t is measured in hours for [0,8] . At time t = 0, the dump site contains 3000 tons of waste.
1.Write an expression for W (t), the total number of tons of waste in the site at time t.
2. Then find a critical value for W(t) in the interval 0 < t < 8.  Give 3 decimal places and explain what this critical value means to the amount of waste in the dump at this time.
3.  For [0,8] , at which time t is the amount of waste in the site a minimum? What is this minimum value? Justify your answer.
Do those question could be done without the calculator? If it has have to use calculator, how I am use the calculator to get critical value?

Note that $ W'(t) \equal{} D(t) \minus{} R(t).$ That means that a critical point for $ W$ would be a time $ t$ such that $ D(t) \minus{} R(t) \equal{} 0.$

Now if you examine the equation $ D(t) \minus{} R(t) \equal{} 0,$ you'll realize that you don't know any tools that would enable you to find an algebraic solution in closed form. That's what will require numerical methods - solving this equation. Once you realize that it's an issue of equation-solving, there are many different techniques that can be brought to bear. A graphing calculator, or its computer equivalent, can be used to sketch the graphs and to estimate in general terms where an intersection might like; many graphing calculators then have routines for finding intersections.

I have enough mistrust of "black boxes" that I tend not to use such things. For one thing, what do the graphs of the functions look like?

$ D(t)$ increases monotonically from $ 10$ when $ t \equal{} 0$ to $ \frac {80}{17},$ which is just barely smaller than $ 5,$ at $ t \equal{} 8.$ The increase is rapid at first but slows down; near the end of the interval, the curve is quite flat.

$ R(t)$ decreases from a value of $ 7$ at $ t \equal{} 0$ to $ \minus{} 1$ when $ t \equal{} 7.5$ and then increases very slightly after that.

Very clearly, these curve intersect. Because $ D(t)$ is increasing, and $ R(t)$ is very nearly decreasing, and the segment on which it is not decreasing has no chance of containing an intersection, these graphs intersect exactly once. There will be exactly one critical point.

When I'm sitting at a computer screen, I tend to use a spreadsheet rather than a calculator; depending on my mood, I might try to find the root to such an equation by successive bisection or some modification, or I might take the trouble to set up Newton's method. Let's just say that the place to look is fairly near $ t \equal{} 3.$


Maynard has one penny, one nickel and one dime. How many different amounts of money can he make by using one or more of his three coins?

If he uses 1 coin, then he could either have 1, 5, or 10 cents
...so 3 possibilities.
If he uses 2 coins, then he could either have 6, 11, or 15 cents
...also 3 possibilities.
If he uses all 3 coins, then he can only have 16 cents
... so 1 possibility.
so altogether, he can make $ 3\plus{}3\plus{}1 different amounts of money \equal{} \boxed{7} amounts$

(note that this is also 3C1 + 3C2 + 3C3)

Or you could just do 2^3 subsets and subtract one since you can't pick 0 coins


Let $a,b,c,x,y,z,p$ be real numbers such that $ a^2-b^2-c^2>0, ax+by+cz=p,ap<0,x>0$.
Determin the sign of $x^2-y^2-z^2$.

hello, is the problem so correct?
Let $a,b,c,x,y,z,p$ be real numbers such that
$a^2-b^2-c^2>0,ax+by+cz=p,ap>0,x>0$.
Determin the sign of $x^2-y^2-z^2$.
Sonnhard.

Hello! Sonnhard
Yes,This problem is no doubt in sentence.

kunny

If a>0,
ax < by+cz
a^2x^2 < (by+cz)^2 <= (b^2+c^2)(y^2+z^2) < a^2(y^2+z^2)
therefore x^2-y^2-z^2<0

If a<0
ax+by+cz > 0
We also have
by+cz > |a|x
By the same way we get x^2-y^2-z^2<0


Show that all monic polynomials of degree $2n$ can be written as the sum $q^2 + r$ for polynomials $q$, $r$, deg $r$ < $n$

Let $p(x) = x^{2n} + ... + a_1 x + a_0$. Then we set $q(x) = x^n + ... + b_1 x +b_0$ with:
We can calculate the coefficients of $q(x)^2=x^{2n}+...+c_1x+c_0$ and see that starting from $b_{n-1}$ we can choose $b_{n-i}$ such that $c_{2n-i}=a_{2n-i}$ for $i=1,2,...,n$ (note that $c_{2n-i} = 2 b_{n-1} + \text{terms}$ with the terms only containing $b_{n-k}$ for $k>i$).
Choosing them that way gives $q$ such that $p-q^2$ has degree $<n$.


Let $ABC$ be a triangle and $O$ be its circumcircle. Draw the median $AA'$ intersects $O$ in $A\"$ . Let $AP$ be diameter of $O$ .Draw the perpendicular from $A'$ to $AP$ and name this line $l$ . Draw the tangent from $A\"$ to the $O$ and name this line $l\"$ . let $l'$ intersects $l\"$ at $X_a$ . $X_b , X_c$ are maken such as $X_a$. Prove that $X_a,X_b,X_c$ are collinear .

See http://www.mathlinks.ro/Forum/viewtopic.php?t=16566 .

  darij

Bad problem!!! :D
here is my solution:
Draw the circles $(X_a,X_aA''),..$ , easily we can prove that triangles $X_aA'A'',...$ are isocles and now prove that the power of $G$(centroid) than three circles are equal,and we are done.


superior algebra
superior algebra solved

Let $A$ be a ring and $x^6=x$ for any $x\in A$.
Prove that $A$ is a commutative ring
[ps... it's a particular case of the Jacobson theorem which I posted here few days ago...  but seems that no one have said anything about it ]

From the fact that $x^{6} = x,$ $( \forall ) x \in $ A it follows that $x^{2}=x.$  But this is a well-known theorem.. if $x^{2}=x$ $( \forall ) x \in $ A then $A$ is commutative ring.

$x^{6} = x,$ $( \forall ) x \in $ A follows that $x^{2}=x$

from $x= -1 $ follows that $1+1=0.$ Then $(x+1)^{6} = x+1.$ Also $x^{6}+x^{4}+x^{2}+1 = x+1.$ So we found that $x^{4}=x^{2}.$ Now we multyply by $x^{2}$ and get that $x^{6}=x^{4}=x^{2}.$ But $x^{6}=x,$ thus we have $x^{2}=x.$

the rest is simple!

cheers! :D :D

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=24885]Here[/url]'s a more interesting related question posted by vess. And $x^6=x,\ \forall x\Rightarrow x^2=x\ \forall x$ is also proven there, in pretty much the same way Lagrangia did it here :).


algebra
polynomial
group theory
number theory unsolved
number theory

$ p$ is a prime number.

Prove $ p | (k\minus{}1)C_p^k (k\equal{}2,3,\cdots,p\minus{}1)$ .

Thanks

$ f(x)\equal{}\sum_{k\equal{}1}^{p\minus{}1} \binom{p}{k}x^k\equal{}(x\plus{}1)^p\minus{}(x^p\plus{}1)$
We have $ f(x)\equiv 0(mod p) \Rightarrow f(x)$ have $ p$ roots modulo $ p$
Moreover, $ deg f\equal{}p\minus{}1$, it means all coefficients of $ f$ is divisible by $ p$
$ \Rightarrow p| \binom{p}{k}$  $ \forall 1\leq k \leq p\minus{}1$

The polynomial argument is just a decategorification of the following combinatorial proof:  the action of the cyclic group $ C_p$ on necklaces with $ p$ beads, each of which can be colored one of $ 2$ colors, has two fixed points.  The necklaces with at least one bead of one color and one bead of the other color are in orbits of size $ p$.

[quote="water2011"][quote="forevercontinue"]We have a well known result :
If $ p$ is a prime number , then
$ \binom{p}{k} \vdots p$ $ \forall k \equal{} 1,2,...,p \minus{} 1$
Hence ,the problem is obvious true following this result[/quote]

How to prove it ? Or where can I find detailed instructions about that ? Or  in which book?

thanks a lot !![/quote]

Just write $ \binom{p}{k} \equal{}\frac{p!}{k!(p\minus{}k)!}$

$ ord_p(p!)\equal{}1$ while $ ord_p(k!)\equal{}ord_p(p\minus{}k)!\equal{}0$

Pco, I think it should be $ v_p(p!)\equal{}1; v_p(k!)\equal{}v_p((p\minus{}k)!)\equal{}0$. :(

[quote="pco"][quote="water2011"][quote="forevercontinue"]We have a well known result :
If $ p$ is a prime number , then
$ \binom{p}{k} \vdots p$ $ \forall k \equal{} 1,2,...,p \minus{} 1$
Hence ,the problem is obvious true following this result[/quote]

How to prove it ? Or where can I find detailed instructions about that ? Or  in which book?

thanks a lot !![/quote]

Just write $ \binom{p}{k} \equal{} \frac {p!}{k!(p \minus{} k)!}$

$ ord_p(p!) \equal{} 1$ while $ ord_p(k!) \equal{} ord_p(p \minus{} k)! \equal{} 0$[/quote]

what is $ ord_p(p!)$ ?

[quote="forevercontinue"]$ f(x) \equal{} \sum_{k \equal{} 1}^{p \minus{} 1} \binom{p}{k}x^k \equal{} (x \plus{} 1)^p \minus{} (x^p \plus{} 1)$
We have $ f(x)\equiv 0(mod p) \Rightarrow f(x)$ have $ p$ roots modulo $ p$
Moreover, $ deg f \equal{} p \minus{} 1$, it means all coefficients of $ f$ is divisible by $ p$
$ \Rightarrow p| \binom{p}{k}$  $ \forall 1\leq k \leq p \minus{} 1$[/quote]

why $ f(x)\equiv 0(mod p)$ ?

what is deg ?

thanks

[quote="tronghieu"]Pco, I think it should be $ v_p(p!) \equal{} 1; v_p(k!) \equal{} v_p((p \minus{} k)!) \equal{} 0$. :([/quote]

what is $ v_p(p!)$ ? thanks

Stop multiposting!  :mad:

$ v_p$ is the $ p$-adic valuation of an integer and $ \mathrm{deg}$ is the degree of a polynomial.

[quote="water2011"]
what is $ ord_p(p!)$ ?
what is $ v_p(p!)$ ?
what is $ deg(f)$ ?
thanks[/quote]

$ deg(f(x))$ is the degree of the polynomial $ f(x)$
$ v_p(x)$, where $ x$ is a positive integer and $ p$ is a prime, is the greatest power of $ p$ which divides $ x$. Example : $ v_2(28) \equal{} 2$ since $ 28 \equal{} 2^27$
$ ord_p(x)$ is used by myself with the same meaning than $ v_p$ but I think the usual notation is $ v_p$, as tronghieu said.


hello, let $ f(x) \equal{} x^{\frac {\ln(30)}{\ln(x)}} \minus{} x^3 \minus{} (x^3 \plus{} x)^{\frac {\ln(3)}{\ln(30)}}$ then we have
$ \lim_{x \to 0 \plus{} 0}f(x) \equal{} 30$
$ \lim_{x \to \infty}f(x) \equal{} \minus{} \infty$ and
$ f^{'}(x) \equal{} \minus{} 3x^2 \minus{} \frac {(x^3 \plus{} x)^{\frac {\ln(3)}{\ln(30)}}\ln(3)\ln(3x^2 \plus{} 1)}{\ln(30)\ln(x^3 \plus{} x)} < 0$ for $ x > 0$, hence the only solution is $ x \equal{} 3$ as kunny said.
Sonnhard.

hello kunny, ok, that's clear. What want you to tell me posting this?
Sonnhard.


algebra
polynomial
limit
number theory unsolved
number theory

Find a monic polynomial  $p\in{\mathbb{Z}[x]}$, such that $\deg p =2k$, $k\in{\mathbb{N}}$, and for infinitely many  $n\in\mathbb{N}$ we have $\sqrt{p(n)}\in \mathbb{N}$ and $\sqrt{p}\not \in \mathbb{Z}[x]$.

Anything of the form $ax^{2b+1}$ for $b \in \mathbb{N}$ does it...

oh, excuse me $deg(p)$ shoud be even and positive.

all the answer above are wrong
and i am afraid there is no such answer

I don't know what is the meanning of "monic".
But I think $p(x)=2x^{2}-1$ is a solution.
And I have proved that there is no solution $p(x)$ with the form $p(x)=x^{2k}+a_{2k-1}x^{2k-1}+\dots+a_{0}$

Monic means the leading coefficient is 1.
I can't find a solution with proof...

very nice Rust!
lemma. if deg(p)=kd then $p(x)=Q(x)^{d}+R(x),deg(R)<k(d-1)$
 is this lemma true for d>2?
if yes then:
if $p\in\mathbb{Z}[x]$ such that $deg(p)=kd$and for infinite n
$\sqrt[d]{p(n)}\in\mathbb{N}$ then $\sqrt[d]{p}\in\mathbb{Z}[x]$

Rust's proof is eaxcatly same as mine  :)

Yes $Q(x)\in Q[x]$. If $P(x)=y^{2}, y\in Z$, then $\{Q(x)\}\in S$ (S finite set), $y=[Q(x)+\frac{1}{2}]$ if x is big. Because $R(x)=0$ had only finite roots, for big x $y=Q(x)+Q(x)(\sqrt{1+\frac{R(x)}{Q(x)^{2}}}-1)=Q(x)+\epsilon$ give contradition with $y\in Z$ and $S$ - finite.


If theres an inequality problem on the USAMO, will we have to state when equality holds?

If it asks you to. Sometimes it will and sometimes it won't.


i dont know y hav u posted it?
is there anything special about f(x)=1

hahahhahahahahahahhahahahahahah!
now thats another spam i believe!

check out the spam in bio section
titled"just read it" :rotfl:  :rotfl:  :rotfl:


geometry circumcircle inradius trigonometry

In a right angled-triangle $ABC$, $\angle{ACB} = 90^o$. Its incircle $O$ meets $BC$, $AC$, $AB$ at $D$,$E$,$F$ respectively. $AD$ cuts $O$ at $P$. If $\angle{BPC} = 90^o$, prove $AE + AP = PD$.

Let $p-a=x$, $p-b=y$, $p-c=z$, $AP=n$, $PD=m$ then $x^2=n(n+m)$.
Since the circumcircle  of $\triangle{CPB}$ is tangent to $AC$  we
can find that: $\displaystyle{\frac{n}{m}=\frac{2x+z}{y}}$.(1)
Using this we conclude that
$\displaystyle{(m+n)=\frac{x(2x+y+z)}{\sqrt{(2x+y+z)(2x+z)}}}$.
But from Steward's Theorem:
$\displaystyle{\frac{4xyz}{y+z}=m(n+m)}$. Therefore
$\displaystyle{\frac{4xyz}{y+z}+x^2=(m+n)^2=\frac{{x^2}{(2x+y+z)}}{2x+z}}$
or $4z^2+7xz=xy$(2). Since $\angle{ACB}=90^\cdot$ it follows that
$xy=z^2+zx+zy$ and (2) reduces to $3z+6x=y$. Using this and (1)
we get that $\displaystyle{\frac{n}{m}=\frac{1}{3}}$.
This implies $x=2n$, consequently $x+n=m$ and we are done.

[u]Thank you[/u], [b]Shobber[/b] (for the your [u]proposed problem[/u]) and [b]Sailor[/b] (for the your [u]solution[/u]) ! [u]See [b]PP12[/b][/u] from [url=http://www.artofproblemsolving.com/blog/101045][color=red][b]here[/b][/color][/url] a clearing up of the nice [b]Sailor's solution[/b].

CMO 2006= Chinese Mathmatical Olympiad 2006.

The claim is true regardless the fact that trangle ABC be rectangle. I will post later a sketch of a synthetic  prove of that.

I presume some facts which can be proved using some well known projective results.

Fact 1.  If X is a generic point on AD and XB and XC cut (O) at Q and R one can prove that QR pass through S where S= FE.BC

Fact 2. If RB cuts (O) in T one can prove that RQ, PT and DF concurr at one point.

Fact 3. If U is the pole of RT, as B lies on RT the polar of B which is FD pass through U.

Dim.

From fact 1, as RQ pass trhough O it results that RQ is orthogonal to PD.  Then from fact 2 we can deduce that D and T are the simmetrical points of P and F wrt diameter RQ.  If UR intersect AB at V, as UV// AD we have that FR bisects VU, and from Fact 3 as F,D and U are colinears, FR bisects AD too at, let's say, M. Furthermore, as VR=VF and VRF~AMF, AM=AF. From Those it follows that AD = 2AF = 2 AP which is equivalent to the thesis.

[quote="sprmnt21"]I presume some facts which can be proved using some well known projective results.

Fact 1.  If X is a generic point on AD and XB and XC cut (O) at Q and R one can prove that QR pass through S where S= FE.BC

Fact 2. If RB cuts (O) in T one can prove that RQ, PT and DF concurr at one point.

Fact 3. If U is the pole of RT, as B lies on RT the polar of B which is FD pass through U.

[/quote]

Can someone show the proofs of these results using projective geometry?

I thought that Sangaku is more of a Japaneese specialty, rather than Chineese.

  I quess that, like most other things, it was smuggled into closed at that period

  Japan from the friendly neighbour China. We should look carefully into the true

  origins of Sangaku.

  Anyway, if one looks at the configuration, it is rather difficult not to notice the

 squares  rotated around a common vertex and inscribed and circumscribed circles.

  Reminds of the famous Sangaku of folded square with inscribed circle in the

  triangle with the inradius equal to the overhang of the folded part.

  Billzhao is right, it would be interesting to get a projective solution for  a

  Sangaku that traditionally was solved by algebraic methods.

  Dear Sprmnt21, please share with us the details of your projective solution.

  T.Y.

  M.T.

Sorry. Lately I didn't watched the forum.

Here some more details about the "facts":

Fact 1. If X is a generic point on AD and XB and XC cut (O) at Q and R one can prove that QR pass through S where S= FE.BC

Proof 1.

As S is harmonically conjugated with  D wrt (B,C),  then, if Y = AD/\QS and R’ = XC/\QS, {(Q,R’);(Y,S)} is a harmonic quadruple. If R” = (O)/\QS then, is well known, {(Q,R”);(Y,S)} )} is a harmonic quadruple too, i.e. R’ = R” = R, i.e. Q,R and S are collinear.

Fact 2. If RB cuts (O) in T one can prove that RQ, PT and DF concur at one point.

Proof 2.
It is a well known fact that, if PQTR is an inscribed quadrilateral, the intersection of the diagonals PT, QR is the pole of the line through the points of intersection of the opposite sides. From this and the De la Hire theorem it follows what claimed.

For this fact and other  similar let have a look at
http://www.maths.gla.ac.uk/~wws/cabripages/quad.html

Fact 3. If U is the pole of RT, as B lies on RT the polar of B which is FD pass through U.

Proof 3.

This is simply  the  De La Hire theorem.

========

I use this post to correct a typo:
The phrase:" From Those it follows that AD = 2AF = 2 AP which is equivalent to the thesis."

should be:

From Those it follows that AD = 2AF  and AF = 2 AP which is equivalent to the thesis.

[hide="Bashy solution"]Let $\angle{CAD}=\theta$.It is not hard to see that $EC=CD=r$ where $r$ is the inradius.Thus we have

$cos\theta=\frac{b}{\sqrt{b^2+r^2}}$
$cos(A-\theta)=cosAcos\theta+sinAsin\theta=\frac{ar+b^2}{c\sqrt{b^2+r^2}}$.

We also have $AP \times AD=AE^2 \Rightarrow AP=\frac{(b-r)^2}{\sqrt{b^2+r^2}}$.

By cosine rule in $\triangle{CAP}$ we get

$CP^2=b^2+AP^2-2bAPcos\theta=b^2+\frac{(b-r)^4}{b^2+r^2}-\frac{2b^2(b-r)^2}{b^2+r^2}$.

Similarly by cosine rule in $\triangle{BAP}$ we get

$BP^2=\frac{(b-r)^4}{b^2+r^2}+a^2+b^2-\frac{2(b-r)^2(ar+b^2)}{b^2+r^2}$.

Thus $\angle{BPC}=90^{\circ}$
$\Rightarrow BP^2+CP^2=a^2$
$\Rightarrow \frac{(b-r)^4}{b^2+r^2}+b^2-\frac{(b-r)^2(ar+2b^2)}{b^2+r^2}=0$
$\Rightarrow b^2+r^2=\frac{(b-r)^2(ar+b^2+2br-r^2)}{b^2}$.

Note that the condition $AE+AP=PD$ yeilds $b^2+r^2=\frac{4br-b^2-r^2}{b-r}$.

and so it suffices to show that the RHS of the two equations are same.Putting $r=\frac{a+b-c}{2}$ we get ${a^2+b^2=c^2}$ so done!![/hide]

anyone can post a solution with pure geometry


Two different prime numbers between 4 and 18 are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?
(A) 21
(B) 60
(C) 119
(D) 180
(E) 231

Old AMC problems?  http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1404835349&t=71774 and http://www.artofproblemsolving.com/Forum/viewtopic.php?p=312621#312621

(I searched for "prime sum product subtracted" to find those threads.)

Well, the prime numbers between 4 and 18 must be odd number, odd*odd=odd, and odd-odd=even, and odd-even=odd, so it can't be B or D.
The minimum of the difference:7*5-(7+5)=23, which is greater than A
the maximum of the difference:17*13-(17+13)=191which is less than E
so it must be C
13*11-(13+11)=119


USAMTS
calculus
derivative
function
algorithm
number theory
greatest common divisor

What did you get?  I got 5/5/5/5/5, same as last time.

5/5/5/5/4. :(
Apparently, my solution wasn't exact for the last problem. an approximation haha.

Yea I used a decimal approximation on the last one, 4/5 on that :(

5/5/5/5/0=20. I assumed that $ P$ was between $ C$ and the center of the circle for #5, which lead to the maximum being 2.5, so I got a 0.

4/5/5/5/5 : 24.

Hmm. I should not omit details.

So for #5, if you used mathematica and got an exact answer to the derivative and an exact value of the function, then you got full credit?  Cause that's what one of my friends did and he got a 5...  And everyone who doesn't have mathematica and used a calculator approximation got a point off?

I think I'm gonna go pirate this mathematica thing. brb

I used Mathematica and got an exact value, but I only got 3/5 for #5.

The grading rubric says "A student who did the calculus via computer but did not prove that it worked gets at most 3/5." How would I "prove that it worked"? Do I have to disassemble Mathematica and verify that its algorithms are accurate? =_=

I got a 3 on #5, and I was confused as to why at first: my comment was, "Alright, but see the official solutions for an elementary bound that does not use calculus." I did the calculus by hand, but used Wolfram Alpha to find exact values for the derivative's zeroes. I didn't want to go through the process of solving for a cubic's roots by hand >_> I guess that's why I lost two points; same general idea of doing something by computer and not showing that it works.

[quote="math154"]Currently 5/5/5/5/4, but I'm protesting #5 because it's definitely rigorous. Grader said that it was a good approximation but see the solution for an exact answer. :|

http://www.usamts.org/Papers/Year21/Round2/10511-1258937571-solutions.pdf[/quote]

Exactly the words in mine...

Anyway, 4 (?!)/5/4(what the heck i put almost half page of explanation)/1(yay, barely any work)/4(approximation?? Was I [i]supposed[/i] to prove my cubic formula solution?)

=18.

I guess we should have done the cubic formula by hand...

What is funny that I didn't use calculus at all and I found two solutions, one which easily gives a stronger bound of $ 175\sqrt2 \minus{} 100\sqrt6\approx2.538$ (which is the only $ \approx$ sign I put in the whole thing! It has the exact value anyway) while the maximum is $ \approx2.534$, a lot of people proved $ 125/49\approx2.551$, and $ 25/(4\sqrt6)\approx2.552$. Hopefully protest goes through... :P

[quote="BOGTRO"][quote="math154"]Currently 5/5/5/5/4, but I'm protesting #5 because it's definitely rigorous. Grader said that it was a good approximation but see the solution for an exact answer. :|

http://www.usamts.org/Papers/Year21/Round2/10511-1258937571-solutions.pdf[/quote]

Exactly the words in mine...
[/quote]

Exactly the words in mine....

My grader said Nice; see official solution for a way to bound PQ without decimal approximation.  So he apparently thought my solution was okay without the decimal part.

5/1/3/3/3 = 15

Ack, not the greatest. Didn't show calculus for #5 and they didn't think that my proofs were stringent enough for #3 and #4B. Completely botched my way through #2, so not too sad about that. Ack.

I think the main lesson (at least for me) is to show all steps, even if they were done through Mathematica / CAS, etc.

[quote="iYOA"]for #1 i got a 4

and the examiner's comments were: good!

     :mad:
so idk what i did wrong but i doubt theres any point in protesting only for 1 point.[/quote]

Exactly what I got... "good!" usually means 5...

Should I protest? Nothing in the comments or grading criteria seems to suggest that I would have lost a point, as (as far as I know), I did (a), proved that there was a smallest solution to $ a\plus{}n \leq bn$, and there was no flaw in my proof.

The graders seem to be biased against 5,1,1,1 as opposed to 1,1,1,5.

[quote="NuncChaos"]They prolly aren' biased that way...

I did 5 111, and got 5....[/quote]

From the grading criteria: "Many students did it the opposite way as the solution above: they first multiplied and then repeatedly added 1. This is valid but slightly more difficult to make rigorous."

I received 5/5/3/5/4, I protested both #3 and #5, both were accepted, and my final score was 5/5/5/5/5.

[quote="buzzer11"]@ajd: why did they take points off #5 originally?[/quote]
Comment read "Good approximation. See solution for exact proof." I protested, something to the effect of it wasn't an approximation; it was exact integer arithmetic. I got an email back saying that my score had been raised to 5 with a comment that "Computer use was minor, appropriate, and non-fatal." So I imagine the grader thought that my silly trick with the discriminant constituted an approximation when in reality it [i]technically[/i] wasn't (although it was certainly extremely inelegant).

[quote="236factorial"][quote="BOGTRO"][quote="iYOA"]for #1 i got a 4

and the examiner's comments were: good!

     :mad:
so idk what i did wrong but i doubt theres any point in protesting only for 1 point.[/quote]

Exactly what I got... "good!" usually means 5...

Should I protest? Nothing in the comments or grading criteria seems to suggest that I would have lost a point, as (as far as I know), I did (a), proved that there was a smallest solution to $ a \plus{} n \leq bn$, and there was no flaw in my proof.

The graders seem to be biased against 5,1,1,1 as opposed to 1,1,1,5.[/quote]

Remember that a point may be deducted for writing style. While I highly doubt that was the case, you may want to share your solution (if you haven't already).[/quote]

I didn't have time to make copies, but he basic idea was:

Note that if a>b, we simply choose 1 b-a times, which is clearly less than b.

Thus, the statement must be proved for a<b.

Consider the smallest integer x such that ax>b+x. We wish to show that $ ax\minus{}(b\plus{}x) \leq b\minus{}2$.

If $ ax\minus{}(b\plus{}x)\equal{}b\minus{}1$, then x is not the smallest integer satisfying the equation, x-1 is, since we have
$ ax\minus{}a>b\plus{}x\minus{}1$
$ ax\minus{}(b\plus{}x)>a\minus{}1$
$ b\minus{}1>a\minus{}1$
$ b>a$
working backwards gives the result.

Thus it takes at most b-1 turns to achieve a from b, and we are done.

Oh, is this is poorly written, but it looked much better in my actual paper.

hm, that looks exactly the same as what I did (i got a 5). Assuming that you showed that there exists a smallest integer x such that ax>b+x, you should definitely protest.

[quote="buzzer11"]I used calculus, and used an approximation. But what if our approximation really doesn't affect our final answer? I mean, if there is no loss of accuracy because of the approximation: I only used an approximation to show that the maximum length was indeed less than 25/4rt(6).[/quote]

I think that is a problem with the grading criteria. Unfortunately, it is not protestable.

[quote="buzzer11"]I used calculus, and used an approximation. But what if our approximation really doesn't affect our final answer? I mean, if there is no loss of accuracy because of the approximation: I only used an approximation to show that the maximum length was indeed less than 25/4rt(6).[/quote]
I think you'd need to put bounds on the approximation - show that the error involved in approximating it could not be greater than a certain value and then show that even with that amount of error, you're still below 25/4sqrt6.

[quote="buzzer11"]Its not that kind of approximation.
The FIRST decimal in my proof appears when I find the root of the cubic. I just used a computer for that.[/quote]
Wouldn't that still be an approximation, given that you didn't put an infinite number of digits for the decimal in the proof?


This update is already sent to all Math Zoomers.

The Math Zoom Summer Program 2009 Opening Ceremony will start at 7PM sharp.  It will be held at Bronco Student Center (Building 35) directly.  The room is Ursa Major AB.  Check-in for the Opening Ceremony starts at 6:30PM at the room.  The ceremony features a Math Magic Show, performed by the famous Math Magician Arthur Benjamin.  Arthur Benjamin is a Math Professor at Harvey Mudd College.  He brought the fun of Mathematics to formal Magic stages like the Magic Castle in Hollywood.  If you can't wait to see how he does it, please follow this link [url]http://www.ted.com/talks/arthur_benjamin_does_mathemagic.html[/url] to see his video.  If you are a Math Zoom camper, don't rush into buying his DVD, as Math Zoom has purchased his DVD for each camper as a gift.  Meanwhile, prepare to beat him at the show! :)

Campus map is available at the link:
[url]http://www.csupomona.edu/map/campus/central_campus_small.pdf[/url]

The open ceremony is open to general public.  Everyone is welcome!

It was pretty cool, since I already had that person's book.

hmm...it seemed like that was just a repeat of his other shows :(


It says here:
[url]http://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_Advanced_Configuration#Using_Asymptote_in_LaTeX[/url]
that it is possible to create a batch file to do these three things in order:

1. Pass it once through LaTeX without viewing the output (click Build in TeXnicCenter; this extracts your asy code into a .asy file.)
2. Run Asymptote on the new file example.asy (pressing Ctrl+A as normal in TeXnicCenter will work, since it only looks at the title example.)
3. Finally, run it again through LaTeX (click Build and View Current File in TeXnicCenter).

However, on the AoPSWiki, it did not tell me how to make such a batch file. It did give me a [url=http://www.computerhope.com/batch.htm]site[/url], but I still don't understand how to make the batch file.

So, could you tell me what to put in the batch file, and what to put in the commands, arguments, and directory fields in TeXnicCenter?

[quote="stevem"]You can see an example of a batch file (which is just a text file with an extension of .bat) just below that paragraph for asymptote.bat

You need to put in the lines you have in Build Profile for compilation, followed by the asymptote command then repeat the first line again. You pass the name of the file using placeholders like %pm (see TexnicCenter's help) and then each one is put in the batch file using %1, %2 etc.

The batch file asymptote.bat should make this clear. Start with that and make changes to it as appropriate.[/quote]

I'm not sure what you mean by "pass the name of the file". I looked up placeholders in TexnicCenter's help, and I have a rough idea of what placeholders are.

Will this batch file work?

[code]"C:\Program Files\MiKTeX 2.6\miktex\bin\pdflatex.exe"-interaction=nonstopmode "%pm" "C:\Program Files\MiKTeX 2.6\miktex\bin\bibtex.exe" "%bm" "C:\Program Files\MiKTeX 2.6\miktex\bin\makeindex.exe" "bm"
@echo off
echo Asymptote is working ... please wait
"C:\Program Files\Asymptote\redir.exe" -e %1 "C:\progra~1\Asymptote\asy.exe" -batchView -tex "pdflatex" -align C -f "pdf" %2

find /i ".asy" %3 >nul
If errorlevel 1 goto end
notepad %3
:end
"C:\Program Files\MiKTeX 2.6\miktex\bin\pdflatex.exe"-interaction=nonstopmode "%pm" "C:\Program Files\MiKTeX 2.6\miktex\bin\bibtex.exe" "%bm" "C:\Program Files\MiKTeX 2.6\miktex\bin\makeindex.exe" "bm"[/code]

Or, would it just be easier to call the asymptote.bat file?

I think I should put:
Command: (the place where I saved my batch file)
Initial Directory: %dc

What should I put for the Arguments?

[quote="stevem"]You can only use %pm etc in TexnicCenter as they are not DOS commands. Look at the asymptote example where the arguments are
%tc_asy.log %tc "%dc\%tc_asy.log"
These 3 arguments are expanded by TexnicCenter to give the file and directory names and then they are included in the DOS file with %1, %2, %3
So wherever you want %tc_asy.log in the batch file you put %1, wherever you want %tc you put %2 etc. You can have up to 9 of these. So if you want %pm and %bm as well you put
%tc_asy.log %tc "%dc\%tc_asy.log" %pm %bm
as arguments and in the batch file put %4 instead of %pm and %5 instead of %bm.

Just play with it, you can't do any harm. That will be a good intro to batch files.[/quote]

I'm starting to understand how this works.

I started playing with the batch file for a while, and this is the best I could do:

[code]@echo off
"C:\Program Files\MiKTeX 2.6\miktex\bin\pdflatex.exe" -interaction=nonstopmode "%4" %2
"C:\Program Files\MiKTeX 2.6\miktex\bin\bibtex.exe" "%5" %2
"C:\Program Files\MiKTeX 2.6\miktex\bin\makeindex.exe" "%5" %2
echo Asymptote is working ... please wait
"C:\Program Files\Asymptote\redir.exe" -e %1 "C:\progra~1\Asymptote\asy.exe" -batchView -tex "pdflatex" -align C -f "pdf" %2

find /i ".asy" %3 >nul
If errorlevel 1 goto end
notepad %3
:end
"C:\Program Files\MiKTeX 2.6\miktex\bin\pdflatex.exe" -interaction=nonstopmode "%4" %2
"C:\Program Files\MiKTeX 2.6\miktex\bin\bibtex.exe" "%5" %2
"C:\Program Files\MiKTeX 2.6\miktex\bin\makeindex.exe" "%5" %2[/code]

I stumbled across this error (sorry, I couldn't get the whole thing since it passes so quickly):
[code]
This is pdfTeX, Version 3.141592-1.40.4 (MiKTeX 2.6)
entering extended mode
! I can't find file '" solutions '".
<*> " solutions

Please type in another input file name
! Emergency stop.
<*> " solutions

! ==> Fatal error occurred, no output PDF file produced!
Transcript written on texput.log
Need exactly one file argument
Couldn't find input index file nor .idx.
Usage: C:\Program Files\MiKTeX 2.6\miktex\bin\makeindex.exe [-ilqrcgLT] [-s sty]
 [-o ind] [-t log] [-p num] [idx0 idx1 ...]
Asymptote is working... Please wait[/code]

I think it has something to do with the fact that %2 refers it to solutions. I'm not sure what kind of placeholder to make so that it refers it to the correct thing, since there are so many of them.

Thanks.

EDIT: Well, I fixed one thing, I forgot to put the quotes at the end!

[quote="fedja"]It is a bit off topic but why do you need to do things in such a complicated way? :? Why not just to define the tool "run asymptote" or whatever in TeXnic center that does nothing but runs the batch file that allows you to see the errors (see earlier stevem's posts) with argument "%tc.asy". Then all you'll need to do will be to click "Build", "run asymptote", "Build and view" in TeXnic center to get the file with all the pictures (provided that your asy code is correct; if it isn't, it may be a bit hard to interpret the error message line numbers just looking at the TeX-file, but you can define another tool that views the asy file and you should be able to do everything easily that way. At least, I find this "modus operandi" rather convenient. Maybe I'm missing something? :?[/quote]

That's how I normally do it. I was reading the instructions on AoPSWiki when I saw this, and I became curious on how to implement batch files so that I could do that.


I'm in college right now so i've been doing other stuff than math. I'm suppose to take over the stuff I learned in highschool because i've forgotten it. It's sad that i've fallen behind to broaden my mind into the arts. I'm gifted to mathematics.

The topics I would need help with would be Intermediate Algebra and Geometry.

I understand I need to review these things once I noticed I didn't know how to multiply fractions until I went to purplemath.com

There are some books at the college I go to that are college level textbooks with the odd number answers in the back. I don't want to bother all of the people in the forum with intermediate algebra questions asking for answers, unless it would be alright with people in the forum. If it would be alright with someone of you, I would be glad to hear a response. Otherwise, i'm looking for some online tutors or people who would be able to answer my questions. I would make sure most definitely that i try to finish the problem to the best of my ability, however, not knowing the answer is the problem.

I would probably have over 30 answers to questions that i would like to be reviewed by my peers so that i don't have an incorrect answer to a question within the textbook. Would anyone know where I could get a [u][b]free[/b][/u] online tutor to help me with my math skills?

As far of questions, you should ask any questions you have. It may take a little while for the response, but just specify that you'd like a full explanation.

I don't know much about online tutors, but I suggest going back and reviewing what you've learned in the past to try to refresh. I love this forum because it can answer pretty much any question I could ever have on math, plus all the people are really nice and helpful.

If you have any questions about anything, just ask. I been on the forums for about a month, and they're hella great.


geometry
rectangle
logarithms
combinatorics proposed
combinatorics

$f(n)$ is the least number that there exist a $f(n)-$mino that contains every $n-$mino.
Prove that $10000\leq f(1384)\leq960000$.
Find some bound for $f(n)$

A little typo which i will correct :
[quote="Omid Hatami"]$10000 \le f(n) \le 960000$[/quote]
Should be :
$10000 \le f(1384) \le 960000$
and a little defination for $n-$mino:
We call a shape, a $n-$mino If and only if, it contains only $n$ equal squares  such that from each vertic ,by using the sides ,you will be able to reach anyother vertices.

Is 1384 from the Persian calendar equivalent of 2005?

upper bound: $n^2/2$

A rectangle of sides $n/2$ and $n$

yes its a bound.
I think the coordinators said the best bound is $nlog(n)=f(n)$ ,but i dont get the shape due :!:  :!: (how on the earth did they get $nlog(n)$  :( ?

Of course we said that the best bound we got is $n\log\ n$


Find $m$ such that
$f(a)=\frac{3k-1}{k-a}-\frac{3k+1}{k+a}+m(a-\frac{1}{3})$
and $f'(\frac{1}{3})=0$

I am sure it works but I am lasy to make the computations  :D

we can simply use chebyshev (or chebyshov as mentioned by arqady :P ), like what zhaobin did in the other thread.

so the inequality is true for any $k\geq 1$ :wink:


geometry
circumcircle
search
power of a point
radical axis
geometry unsolved

problem is very very hard i think.

prove that in a triangle, circumcirle and nine point circle and a circle with diameter $HG$ are coaxal circles.

Mrs britney, i dont have time to post complete solution.sorry,u can do a little search to find them.

theorem:suppose $(C)$ and $(C')$ are two circles and  $O$ and $O'$ are their inner and outher homotecy centers.

a circle with diameter$OO'$is coaxal with them.

ok.

u should also know another fact.

$H$ and $G$ are the outher and inner homotecy center of circumcircle and nine point circle.

then a circle with diameter $HG$, as theorem said is coaxal with circumcircle and nine point circle.

and we name their radical axis ,ortho pole.

why does ure teacher say these hard things to u?

as i know ure under $13$.

any way.u can ask me to paraphraise each part u want. ;)


arithmetic sequence
algebra unsolved
algebra

Let ${x_1, x_2, \cdot\cdot\cdot , x_n, \cdot\cdot\cdot}$ be a sequence of nonzero real numbers. Show that the sequence is an arithmetic
progression if and only if, for each integer $n\ge 2$:
$\frac{1}{x_1x_2}+\frac{1}{x_2x_3}+ \cdot\cdot\cdot +\frac{1}{x_{n-1}x_n}=\frac{n-1}{x_1x_n}$.

Also you can help me solve this problem:
http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=x_ky_k&t=70728

I think the "if" part can be done inductively:

a) Let $n=3$. Then $\frac{1}{x_1x_2}+\frac{1}{x_2x_3}=\frac{2}{x_1x_3}\implies x_1+x_3=2x_2$

b) Assume that first $n$ terms belong to an arithmetic progression. Then $x_n=x_1+(n-1)d$ for some real $d$ and

\begin{eqnarray*}\frac{1}{x_1x_2}+\dots+\frac{1}{x_{n-1}x_n}+\frac{1}{x_nx_{n+1}} &=& \frac{n-1}{x_1x_n}+\frac{1}{x_nx_{n+1}}\\ &=& \frac{(n-1)x_{n+1}+x_1}{x_1x_nx_{n+1}}\end{eqnarray*}

From $\frac{(n-1)x_{n+1}+x_1}{x_1x_nx_{n+1}}=\frac{n}{x_1x_{n+1}}$ it follows $x_{n+1}=\frac{nx_n-x_1}{n-1}=\frac{(n-1)x_1+n(n-1)d}{n-1}=x_1+nd$

The "only if" part can be shown thus:

Let $x_i-x_{i-1}=d$. Then the desired sum is

\begin{eqnarray*}\frac{1}{d}\left(\frac{1}{x_1}-\frac{1}{x_2}+\frac{1}{x_2}-\frac{1}{x_3}+\dots+\frac{1}{x_{n-1}}-\frac{1}{x_n}\right)=\frac{x_n-x_1}{dx_1x_n}=\frac{x_1+(n-1)d-x_1}{dx_1x_n}=\frac{n-1}{x_1x_n}\end{eqnarray*}


Is there any team willing to take me in? My school team got disbanded and I need a team to join.

I'm fairly sure that my school team will not be reformed.

The competition is already over.
Are you talking about next year's? :maybe:

My team should have room (I think we only have 13 people going so far, maybe even less, and I was planning to bring two teams.)  Contact me for information.

My school is highly considering to refund our team to go to HMMT. I'll come back with final decision.

I too would like to go for the 2007 competition, so if your school is short ppl, I'm here, lol. And how many people are on a team?


Ring Theory
superior algebra
superior algebra unsolved

If K is a number field with class number 2. Prove that if p is irreducible in O (the ring of integers) and <p> (the principal ideal generated by p) is not a prime ideal, then

<p>=ab
for some prime ideals a and b in O, not necessarily

Otherwise $ p\mathcal{O} \equal{} ABC$ with some proper ideals $ A,B,C$ of $ R$.
(if $ p\mathcal{O}$ is the product of $ \ge 3$ prime ideals.)
Then neither $ A,B$ or $ C$ are principal ideals, for if for example $ A \equal{} \alpha \mathcal{O}$,
then $ p\in ABC \equal{} \alpha BC$ and $ p \equal{} \alpha \beta$ for some $ \beta\in BC\subseteq B$,
and since $ p$ is irreducible we get $ \alpha\in\mathcal{O}^{\times}$ or $ \beta\in\mathcal{O}^{\times}\Rightarrow A \equal{} R$ or $ B \equal{} R$. Cotradiction!

But the class number of $ \mathcal{O}$ is $ 2$, so $ A,B,C$ must all represent the involution in the ideal class group,
and so does $ ABC$, which contradicts $ ABC \equal{} p\mathcal{O}$ principal.

sry i've never seen the symbol at the end, what does  [img]http://alt2.mathlinks.ro/Forum/latexrender/pictures/9/d/0/9d04ff985f5185f57c0fa8de9a744de48a3229a1.gif[/img] mean?

what does it mean for A,B, and C to represent the involution in the ideal class group?

$ \mathcal{O}^{\times}$ is the set of units of $ \mathcal{O}$. (divisors of $ 1$)
A group of order $ 2$ has only one element $ \not \equal{}1$, which has order $ 2$.
Elements of order $ 2$ are simply called involutions.
By assumption the ideal class group has order $ 2$.


1. In teh following sequence, there are 10 numbers. Each number has one more 2 than the preceding number: 2, 22, 222, 2222....., 2222222222. What is the hundreds digit of the sum of these ten numbers?

2.  this is one of my favorites: Mark has the first three Harry POtter books. Each cover is 1/8 inch thick. Each of the first two volumes is 2 inches thick between teh covers. The third volume is 2 and 1/2 inches thick between the covers. The three volumes are arranged in order on a shelf. How far in inches is it from the first page of volume I to the last page of volume III?

3. What is the value of the following expression in simplest terms? (do NOT multiply it out, try to find an easier way): (20*24*28*32)/(10*12*14*16)

4. A bus can hold 24 adults or 30 children. 25 children are already on the bus. What is the largest number of adults that can stilll get on the bus?

5. The average of a group of 20 numbers is 20. The average of a different group of 60 numbers is 60. THe two groups of numbers are combined into a single group. What is the average of the combined group?

if you want, I can post more later.

after looking at these, i realized i had mostly all the answers to these.......

I gave you every single sheet of MOEM.... but you said you lost half of them

hellooo? i'm in math Olmpiad right now even if i lost ur sheets.....

1. 100 * 2 =200

2. I don't get what you're talking about..

3. 2*2*2*2 = 16

4. 1A = 30 / 24 = 1.25C
4adults = 5 children

25 children = 16 adults

4 more adults are able to get on the bus

5. 400 + 3600 = 4000
   4000 / 80 = 50
   50

this problem confuse people because the average number and the group of numbers are the same. people might think since the numbers and the average are the same, the two groups of numbers would equal the average...

What does MOEM stand for?

[quote="shinwoo"]1. 100 * 2 =200

[/quote]

I don't think you are getting the point of the problem, it's asking what's the hundreds digit of the SUM of the ten numbers, so it wouldn't be 200

I think MOEM stands for Math Olmypaid Elementry something.... don't know what the M stand for, but it's Math Olmypaid for 5th and 6th grade...  you have 5 contests with 4 or 5 questions each. you get like a certificate for participating and if you get a lot right and is the top of your school, you get like a trophy.... or if you are in the top 2% nationally

[quote="shinwoo"]

2. I don't get what you're talking about..

What does MOEM stand for?[/quote]

first:
number 2 is saying.... there's three Harry Potter books on the shelf....placed next to each other in the order from the 1st harry potter book to the 3rd...... and it's asking how thick is it from the cover of volume one to the back cover of volume 3....

MOEM stands for Mathematical Olympiads for Elementary and Middle school students.

[quote="LuCy4EvA"]

2.  this is one of my favorites: Mark has the first three Harry POtter books. Each cover is 1/8 inch thick. Each of the first two volumes is 2 inches thick between teh covers. The third volume is 2 and 1/2 inches thick between the covers. The three volumes are arranged in order on a shelf. How far in inches is it from the first page of volume I to the last page of volume III?[/quote]

check the wording there.... it's supposed to be from the FRONT COVER of the first volume...not the first PAGE...
to the Back cover of the 3rd volume....not the LAST PAGE....

eh... Kristy, I have the question right in front of me and it says first page and last page, NOT first cover and last cover.... if it was from cover to cover, it would have been way easy... on this one, you actually have to think about how it's placed on the shelf...

think about how books are placed on a shelf, it's not what you would think it would be, actually get out three books and see where the first and last page is.

well i have the question in front of me too.. and mine is from this year's contest!! If according to ur question....then wouldn't u have to know how thick the covers are :? ??? think....

[quote]1. In teh following sequence, there are 10 numbers. Each number has one more 2 than the preceding number: 2, 22, 222, 2222....., 2222222222. What is the hundreds digit of the sum of these ten numbers? [/quote]

oooo, I missunderstood the question..I was actually chatting with my friend through AIM hee hee..

now I get it...

it's 8...

obviously....

#2, I didn't feel like solving it.. I couldn't read...- I actually didn't want to read.. well...

it's
2 + 1/8 + 1/8 + 2 + 1/8 + 2and1/2

then it's..( ahh!! I hate arithmatics )
6 + 7/8= 6and7/8 inches

[quote="KBabe"][quote="LuCy4EvA"]um... kristy, just because our questions are different doesn't mean that mine's wrong... and besides, my questions needs to know the cover too. it's a do-able questions therefore it's not wrong....  maybe they change the question this year because it's too hard... u'll never know.[/quote]

yea....well according to ur question...it doesn't mention how thick each cover is.....and maybe i'm just stupid :blush:  but can u show me how u find the thickness of the covers?? :P[/quote]

[quote="LuCy4EvA"]2. this is one of my favorites: Mark has the first three Harry POtter books. [u][b]Each cover is 1/8 inch thick.[/b][/u] Each of the first two volumes is 2 inches thick between teh covers. The third volume is 2 and 1/2 inches thick between the covers. The three volumes are arranged in order on a shelf. How far in inches is it from the first page of volume I to the last page of volume III? [/quote]

I think this is how LuCy4EvA found the thickness. :P  :D

that wasn't there the first time she posted the question......was it? :?

She's not a mod, so she couldn't have edited w/o us knowing after someone posted, unless either A+Math or Treething edited it... Treething might do that...

it was there when I first posted, and I'm gonna look over Shinwoo solution, having doubts, but don't remember the answer.....

here's some hints about number 2:

don't look at the spoilers if you want to try it out on your own:

[hide]hint number 1: the best hint I can give you is to actually get three books and put them on a shelf and figure out where the first and last pages are.... that might sound dumb right now, but trust me, it'll suprise you

[/hide]

[hide]hint number 2: well, if you don't believe me about the actually doing it thing, I'm gonna give you a clue on what you might find.... the FIRST page of the FIRST book is right next to theLAST page of the SECOND book, and the LAST page of the THIRD book is next to the FIRST page of the SECOND book. when I say next to it, I mean that there's a cover(s) between it... and you must read what I jsut typed REALLY carefully to get it [/hide]

[hide]hint number 3: okay, final clue: the answer is the sum of four covers and the thickness of the second book. if you still don't get it, look back at the first hint, or tell me and I'll further explain[/hide]

[quote="LuCy4EvA"]wait, I remember putting this in the Classroom Math section... did someone move it? is it too hard for Classroom Math?[/quote]

In the classroom forum there isn't any thing that says [Moved] MOEM, so probably not. But I don't think that contest problems are supposed to be in Classroom Math.

Classroom math is supossed to be math that your teacher would give you for homework or something, so even 8th grade math could be in it.  The Basics is for math that would be in contests.

[quote="LuCy4EvA"]well, this is MOEM, so it's like 5th and 6th grade. and KBabe told me that Classroom math is like suppose to be sixth grade...[/quote]

i said it was supposedly the easiest.......
 :?  :P

1. In teh following sequence, there are 10 numbers. Each number has one more 2 than the preceding number: 2, 22, 222, 2222....., 2222222222. What is the hundreds digit of the sum of these ten numbers?

[hide]2+22+222*8=1800
so the answer is 8[/hide]

wow, that's a fast way of doing it, I actually add them all up, it took a while

3. What is the value of the following expression in simplest terms? (do NOT multiply it out, try to find an easier way): (20*24*28*32)/(10*12*14*16)

[hide]2*2*2*2=[hide]16[/hide][/hide]

4. A bus can hold 24 adults or 30 children. 25 children are already on the bus. What is the largest number of adults that can stilll get on the bus?

[hide]24A=30C
5C=4A

the answer is 4 [/hide]

5. The average of a group of 20 numbers is 20. The average of a different group of 60 numbers is 60. THe two groups of numbers are combined into a single group. What is the average of the combined group?

[hide]the first group adds to be 400
the second group adds to be 3600
3600+400=4000
4000/(20+60)=4000/80= 50 [/hide]

[quote="jli"]4. A bus can hold 24 adults or 30 children. 25 children are already on the bus. What is the largest number of adults that can stilll get on the bus?

[hide]24A=30C
5C=4A

[hide]the answer is 4[/hide][/hide][/quote]
You have got the correct answer, but I don't think you have explained what A and C are.

[quote]but I don't think you have explained what A and C are.[/quote]

I think  they are: (Click on them) :

[size=150] [hide="A"]dult [img]http://www.magicmakers.com/retail/adult%20animal%20masks/dog.jpg[/img] [/hide]  [hide="C"]hild  [img]http://home.att.net/~vlea/Image14.gif[/img]  [/hide][/size]

Yes, but what about the adult and child?  Is it the number of them, their age, their size?

I think A was meant to equal the amount of space that an adult takes up.  And C was meant to equal the amount of space that a child takes up.


algebra
polynomial
function
calculus
calculus computations

Let $n\in \mathbb{N},n\geq3$ and $a_{1},a_{2},\dots ,a_{n}\in \mathbb{R},a_{1}\neq a_{2}\neq \dots \neq a_{n}$ Find
$\sum_{k=1}^{n}{a_{k}\prod_{j\in \{1,2,\dots ,n\}\backslash \{k\}}{\frac{1}{a_{k}-a_{j}}}}$

Please help  :lol:

Also,find

$f(n)=\sum_{n=1}^{\infty}\arctan{\frac{1}{n^{2}+n+1}}$

I have tried $\frac{d}{dn}f(n)$  to work something out ,but i had no luck with solving t.

[quote="Mr.Doe"]Also,find

$f(n)=\sum_{n=1}^{\infty}\arctan{\frac{1}{n^{2}+n+1}}$
[/quote]
$\arctan\frac{1}{n^{2}+n+1}=\arctan{(n+1)}-\arctan{n}$

For the sum, try looking into [url=http://mathworld.wolfram.com/LagrangeInterpolatingPolynomial.html]Lagrange interpolating polynomials[/url] (not sure of it, but seems relevant).

More precise, use Lagrange interpolating polynomial for function $f(x)=x$


MATHCOUNTS
calculus
logarithms
number theory
relatively prime
geometric sequence
calculus computations

$ 1\minus{}\frac{1}{2}\plus{}\frac{1}{3}\minus{}\frac{1}{4}\plus{}\cdots\plus{}\frac{1}{39}\minus{}\frac{1}{40}\equal{}\frac{q}{p}$ ; $ p, q$ are relatively prime, find the value of $ q$. :P

Is this a joke? $ \frac qp \equal{} \frac {\boxed{3208915}}{4713424}$, at least as far as my calculator goes... (probably wrong) :P

Umm, this is DEFINITELY NOT a Mathcounts level problem. Also, this would be extremely hard to do without a calculator. I think that if this was an infinite series, it could probably be solved with calculus, because I remember that $ \pi \equal{} 1 \minus{} \frac {1}{3} \plus{} \frac {1}{5} \minus{} \frac {1}{7} \plus{} \frac {1}{9} \minus{} ...$ , so I think that the solution to the infinite sequence has to do with $ \pi$, therefore making it impossible to find integers $ q$ and $ p$ (Because $ \pi$ is irrational.)

I'm going to go ask my dad  :maybe:

EDIT: w00t!!! Ok, my dad helped me with this. Anyway, IF it was an infinite sequence, then:

Because $ 1 \plus{} x \plus{} x^2 \plus{} x^3 \plus{} x^4 \plus{} ... \equal{} \frac {1}{1 \minus{} x}$ (infinite geometric sequence),

Then $ \frac {1}{1 \plus{} x} \equal{} 1 \minus{} x \plus{} x^2 \minus{} x^3 \plus{} x^4 \minus{} x^5 \plus{} ...$

Also, note that $ \frac {1}{1 \plus{} x} \equal{} (\ln(x \plus{} 1))'$

Find the antiderivative to get $ x \minus{} \frac {1}{2}x^2 \plus{} \frac {1}{3}x^3 \minus{} \frac {1}{4}x^4 \plus{} \frac {1}{5}x^5 \minus{} ...$

When $ x \equal{} 1$, then you get:

$ 1 \minus{} \frac {1}{2} \plus{} \frac {1}{3} \minus{} \frac {1}{4} \plus{} \frac {1}{5} \minus{} ... \equal{} \ln(1 \plus{} 1) \equal{} \boxed{\ln 2}$.

For reference, $ \ln 2$ is about $ 0.693147181$.

The problem given is only a finite series, but it's basically calculator bash

A mod really needs to move this to another forum :P


If $ 9^{x \plus{} 2} \equal{} 240 \plus{} 9^x$, then the value of $ x$ is:

$ \textbf{(A)}\ 0.1 \qquad
\textbf{(B)}\ 0.2\qquad
\textbf{(C)}\ 0.3\qquad
\textbf{(D)}\ 0.4\qquad
\textbf{(E)}\ 0.5$

minus 9^x from both sides and then factor to get 80(9^x)=240. You then get 9^x=3 or x=.5. Alternatively, we know that 9 to a,b,c, or d would be extremely hard to calculate so we could do e by elimination.


[quote="silouan"]Sorry the fast[/quote]

What do you mean by that??

I was in hurry and I make a big mistake. Sorry

For $\frac{1}{2}<a<1$, we can prove that $\frac{a}{a^3+bc}>1$

so, what's the use? By the way, nttu has posted his beautiful solution in other thread.

Here is solution of my student

+ If all terms in LHS > 1 then of course we have done.

+ If, for example $a  \leq a^3 + bc $, then we act as follow

$\frac{b}{b^3+ac}+\frac{c}{c^3+ab} = \frac{1}{b^2+ac/b}+\frac{1}{c^2+ab/c} \geq \frac{4}{b^2+c^2+ac/b+ab/c} > \frac{4}{1+ac/b+ab/c}$

+ Put t = ac/b + ab/c then it is easy to show that $ \frac{a}{a^3+bc} \geq t $
(Just substitute t = a(1-a)/bc and tranform we come to  $a - a^3 - bc \leq 0$)

+ Finally $ LHS >  t + \frac{4}{1+t}\geq 3$ (AM - GM)

Have fun


Let $ x,y,z\in R$ such that:
$ \frac {2}{5}\leq z\leq Min\{x,y\}$
$ yz\geq 1/5$
$ xy\geq 4/5$
Find the Maximum of $ P \equal{} \frac {1}{x} \plus{} \frac {2}{y} \plus{} \frac {3}{z}$


Here is the question from AOPS II:

How mnay positive integers less than 101 are multiples of 5 or 7, but not both?

I just did [101/5]+[101/7]-[101/35] = 32, but the solution manual actually says that we "have counted the multiples of 35 twice", so we need to subtract another [101/35] or 2 from 32, giving 30. Can someone explain this? Thanks.

How many multiples of 35 are included in [101/5]+[101/7]  :)

[quote="keta"]How mnay positive integers less than 101 are multiples of 5 or 7, [b]but not both[/b]?[/quote]

[101/5]+[101/7]-2[101/35] = 30

you counted 35 and 70 in computing [101/5]
and again in computing [101/7]
we want neither
so we subtract 2 times


[hide]
$a-7b+8c=4 8a+4b-c=7$
there is some symmetry between the coefficients of a and c and the coefficent of b and the constant

multiply 3 to the second equation
$24a+12b-3c=21$
add it to the first equation, get:
$25a+5b+5c=25\\ 5a+b+c=5 b+c=-5(a-1)$
multiply -3 to the first equation
$-3a+21b-24c=-12$
add it to the second equation, get:
$5a+25b-25c=-5\\ a+5b-5c=-1\\ b-c=\frac{-1-a}{5}\\ \\ (b+c)(b-c) = -5(a-1)(\frac{-1-a}{5}) = a^2-1 = b^2-c^2\\ c^2-b^2=1-a^2\\ a^2-b^2+c^2 = 1\\ B$
[/hide]

[hide]
Rewrite the first equation as $a+8c=4+7b$, and square both sides: $a^2+16ac+64c^2=16+56b+49b^2$

Rewrite the second equation as $8a-c=7-4b$, and square both sides: $64a^2-16ac+c^2=49-56b+16b^2$

Adding these two equations, we obtain $65a^2+65c^2=65+65b^2\Longleftrightarrow a^2-b^2+c^2=1$

$\boxed{\text{B}}$
[/hide]


LaTeX
inequalities unsolved
inequalities

1.Let a,b,c>0 be the real numbers such abc=1. Prove:

(1/sqrt(a+ 1/b + 1/2)) + (1/sqrt(b + 1/c + 1/2)) + (1/sqrt(c + 1/a + 1/2)) is greater than sqrt(2) (or equal ?)

2. Let n be positive integer and a(1), a(2), ... , a(n) be the positive real numbers satisfying:

sum( a(1) + a(2) + ...+ a(n) ) = sum( 1/a(1) + 1/a(2) + ... + 1/a(n) )
Prove:

sum( (1/( n - 1 + a(1))) + (1/(n - 1 + a(2))) + ... + (1/(n - 1 + a(n))) ) =< 1

Sorry for my bad english and not using Latex  :blush:

\[ \sqrt{\frac{1}{\frac{2ab+2+b}{2b}}}+\sqrt{\frac{1}{\frac{2bc+2+c}{2c}}}+\sqrt{\frac{1}{\frac{2ac+2+a}{2a}}}> \sqrt{2}\]
\[ \sqrt{\frac{a}{a+2ac+2}}+\sqrt{\frac{b}{b+2ba+2}}+\sqrt{\frac{c}{c+2cb+2}}> 1\]
$ a=\frac{x}{y}$, $ b=\frac{y}{z}$, $ c=\frac{z}{x}$; $ x,y,z \in \mathbb{R}^{+}$
\[ \sqrt{\frac{x}{x+2y+2z}}+\sqrt{\frac{y}{y+2z+2x}}+\sqrt{\frac{z}{z+2x+2y}}> 1\]
WLOG  $ x+y+z=1$
We must prove:
\[ \sqrt{\frac{x}{2-x}}+\sqrt{\frac{y}{2-y}}+\sqrt{\frac{z}{2-z}}> 1\ \ \ \ (*)\]
We have:
\[ \sqrt{\frac{x}{2-x}}> x\ \ \ \ (**)\]
so $ (*)$ holds and we are done.
[hide="Proof for $ (**)$"] First notice that $ x \in \left\langle 0,1 \right\rangle$.
\[ \sqrt{\frac{x}{2-x}}> x \Leftrightarrow \frac{x}{2-x}> x^{2}\Leftrightarrow \frac{1}{2-x}-x > 0 \Leftrightarrow \frac{(1-x)^{2}}{2-x}> 0\] [/hide]
\[ Q.E.D.\]

[quote="JANKRI"]
\[ \sqrt{\frac{1}{\frac{2ab+2+b}{2b}}}+\sqrt{\frac{1}{\frac{2bc+2+c}{2c}}}+\sqrt{\frac{1}{\frac{2ac+2+a}{2a}}}> \sqrt{2}\]

\[ \sqrt{\frac{a}{a+2ac+2}}+\sqrt{\frac{b}{b+2ba+2}}+\sqrt{\frac{c}{c+2cb+2}}> 1 \]
$ a=\frac{x}{y}$, $ b=\frac{y}{z}$, $ c=\frac{z}{x}$; $ x,y,z \in \mathbb{R}^{+}$
\[ \sqrt{\frac{x}{x+2y+2z}}+\sqrt{\frac{y}{y+2z+2x}}+\sqrt{\frac{z}{z+2x+2y}}> 1 \]
WLOG  $ x+y+z=1$
We must prove:
\[ \sqrt{\frac{x}{2-x}}+\sqrt{\frac{y}{2-y}}+\sqrt{\frac{z}{2-z}}> 1\ \ \ \ (*) \]
We have:
\[ \sqrt{\frac{x}{2-x}}> x\ \ \ \ (**) \]
so $ (*)$ holds and we are done.
[hide="Proof for $ (**)$"] First notice that $ x \in \left\langle 0,1 \right\rangle$.
\[ \sqrt{\frac{x}{2-x}}> x \Leftrightarrow \frac{x}{2-x}> x^{2}\Leftrightarrow \frac{1}{2-x}-x > 0 \Leftrightarrow \frac{(1-x)^{2}}{2-x}> 0 \]
[/hide]
\[ Q.E.D. \]
[/quote]

but , how  x+y+z = 1 :wink:

[quote="Kasparov"] but , how  x+y+z = 1 :wink: [/quote]

$ \sum_{cyc}{x}=S \mid : S \Rightarrow \sum_{cyc}{x_{1}}=1$, and now we have: $ x=S\cdot x_{1}$ and so one, and now we work with $ x_{1}$, $ y_{1}$, $ z_{1}$ which satisfies $ \sum_{cyc}{x_{1}}=1$, $ \sqrt{\frac{x}{2S-x}}=\sqrt{\frac{Sx_{1}}{2S-Sx_{1}}}=\sqrt{\frac{x_{1}}{2-x_{1}}}$, and that's because we may assume that $ \sum_{cyc}{x}=1$. :wink:


In Mathematics and Youth Magazine, issue 4/2007, I propose the following inequality

[i]Let $ a,b,c$ be positive real numbers. Prove that[/i]
\[ \left(1\plus{}\frac{2a}{b}\right)^{2}\plus{}\left(1\plus{}\frac{2b}{c}\right)^{2}\plus{}\left(1\plus{}\frac{2c}{a}\right)^{2}\ge\frac{9(a\plus{}b\plus{}c)^{2}}{ab\plus{}bc\plus{}ca}.\]
This problem is slight easy, but the following problem in Secrets in Inequalities is harder

[i]Let $ a,b,c$ be positive real numbers. Prove that[/i]
\[ \left(2\plus{}\frac{a}{b}\right)^{2}\plus{}\left(2\plus{}\frac{b}{c}\right)^{2}\plus{}\left(2\plus{}\frac{c}{a}\right)^{2}\ge\frac{9(a\plus{}b\plus{}c)^{2}}{ab\plus{}bc\plus{}ca}.\]
Please try to prove it. Many my problems in mathlinks are unsolved now  :(

[quote="Kath"]No.1
Apply Cauchy-Schwarz 's Inequality:
$ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\geq\frac{(a\plus{}b\plus{}c)^{2}}{ab\plus{}bc\plus{}ca}$

Hence,it suffices to prove that:
$ \sum_{cyc}(1\plus{}\frac{2a}{b})^{2}\geq 9\sum_{cyc}\frac{a}{b}$
Moreover:
$ \sum_{cyc}(1\plus{}\frac{2a}{b})^{2}\geq\frac{(2x\plus{}3)^{2}}{3}\geq 9x$ with $ \ x \equal{}\sum_{cyc}\frac{a}{b}\geq 3$
So,we have done!
The equality occurs when $ \ a \equal{} b \equal{} c$[/quote]

The solution of Kath is the same as mine. However, the problem in Secrets is really harder. Previous way is inapplicable.

Maybe it is useful this inequality, with it you can transform the second inequality into a symmetric one, and you may apply Muirhead.

If $ abc\equal{}1$,

$ 3\plus{}\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\geq a\plus{}b\plus{}c\plus{}\frac{1}a\plus{}\frac{1}b\plus{}\frac{1}c$.

(it appeared in Crux, problem 3237)

Good observation, xavi. This way is very useful. But my solution is not as that. I use the 10maths- method.

[quote="xavi"]What is the 10maths- method?¿[/quote]

Hi, it is just my habit of calling. 10math-method is the method that you will rewrite inequalities to
\[ M(a\minus{}b)^{2}\plus{}N(c\minus{}a)(c\minus{}b)\ge 0.\]

Does anyone catch the idea to solve this one?

Hmmm, i'm quite annoyed about this "new" methods in inequalities. To be honest, I don't like them. These methods kill the beuaty of inequalities.
 I think harazi was right here about many of the inequalities that have appeared on the forum lately. Oki, let's get back to our stuff here. I will solve the following inequality very gently. Here it goes:

  [b] Problem. [/b] Let $ a, b, c$ three positive reals such that $ abc\equal{}1$. Prove that

                             \[ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\plus{}3\geq a\plus{}b\plus{}c\plus{}\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}.\]

  [i] Solution. [/i]

     By the substitutions $ a\equal{}\frac{y}{x}, b\equal{}\frac{z}{y}$ and $ c\equal{}\frac{x}{z}$, the given inequality transforms into

                \[ \frac{yz}{x^{2}}\plus{}\frac{zx}{y^{2}}\plus{}\frac{xy}{z^{2}}\plus{}3\geq\frac{x\plus{}y}{z}\plus{}\frac{z\plus{}x}{y}\plus{}\frac{y\plus{}z}{x}.\]
     By elementary calculations, the inequality above is equivalent to
               \[ x^{3}y^{3}\plus{}y^{3}z^{3}\plus{}z^{3}x^{3}\plus{}3x^{2}y^{2}z^{2}\geq y^{2}z^{2}(xy\plus{}yz)\plus{}z^{2}x^{2}(xy\plus{}xz)\plus{}x^{2}y^{2}(yz\plus{}zx).\]
     Now, we denote $ m\equal{}xy, n\equal{}yz$ and $ p\equal{}zx$ the inequality finally reduces to

              \[ m^{3}\plus{}n^{3}\plus{}p^{3}\plus{}3mnp\geq mn(m\plus{}n)\plus{}np(n\plus{}p)\plus{}pm(p\plus{}m),\]
    which is Schur.

Yes actually, I dont like brute force method, but sometime we must use sthing. For example, when you meet a very very nice and beautiful problem, and "nice" method can not kill them? what happens?

For the recitation of Cezar, well, your problem is nice, but I think it is pretty easy, since I know that it is exactly Schur Inequality at the first look.

[quote="hungkhtn"]Yes actually, I dont like brute force method, but sometime we must use sthing. For example, when you meet a very very nice and beautiful problem, and "nice" method can not kill them? what happens?
[/quote]

Yes,i do agree with you hungkhtn.Like you said in the past :some problems are really hard to solve by romantic ways.  :P
Btw,is the 10maths method,the new methods of yours??
I found a difficulty to write the inequality in that form,so could you please tell me your solution and also your method to write
inequality in that form???  :maybe:

Not a "method" of mine actually. My friend (with his nick "10math_0609") discovers it. I think Arqady knows this method very well.

Anyway, I hope solutions of other people, especially a "romantic" solution. I will post my solution after that...

Is the following inequality true?\[ \frac{a^{3}}{b}\plus{}\frac{b^{3}}{c}\plus{}\frac{c^{3}}{a}\minus{}(a^{2}\plus{}b^{2}\plus{}c^{2})\ge (a\minus{}b)^{2}\plus{}(b\minus{}c)^{2}\plus{}(c\minus{}a)^{2}\]
If it is true I might have a solution.

[quote="hungkhtn"]

[i]Let $ a,b,c$ be positive real numbers. Prove that[/i]
\[ \left(2\plus{}\frac{a}{b}\right)^{2}\plus{}\left(2\plus{}\frac{b}{c}\right)^{2}\plus{}\left(2\plus{}\frac{c}{a}\right)^{2}\ge\frac{9(a\plus{}b\plus{}c)^{2}}{ab\plus{}bc\plus{}ca}.\]
Please try to prove it. Many my problems in mathlinks are unsolved now  :([/quote]
We can rewrite the inequality in the following form :
$ (a\minus{}b)^{2}M\plus{}(c\minus{}a)(a\minus{}b)N\geq 0$
where $ M\equal{}\frac{4}{ab}\plus{}\frac{(a\plus{}b)^{2}}{a^{2}b^{2}}\minus{}\frac{9}{ab\plus{}bc\plus{}ca}$
$ N\equal{}\frac{4}{ac}\plus{}\frac{(c\plus{}a)(c\plus{}b)}{a^{2}c^{2}}\minus{}\frac{9}{ab\plus{}bc\plus{}ca}$
 :blush:

[quote="hungkhtn"]In Mathematics and Youth Magazine, issue 4/2007, I propose the following inequality

[i]Let $ a,b,c$ be positive real numbers. Prove that[/i]
\[ \left(2\plus{}\frac{a}{b}\right)^{2}\plus{}\left(2\plus{}\frac{b}{c}\right)^{2}\plus{}\left(2\plus{}\frac{c}{a}\right)^{2}\ge\frac{9(a\plus{}b\plus{}c)^{2}}{ab\plus{}bc\plus{}ca}.\]
Please try to prove it. Many my problems in mathlinks are unsolved now  :([/quote]

Assume that $ c \equal{} max(a,b,c)$, then the inequality can be rewritten as the following form:
$ LHS\minus{}RHS \equal{} M(a\minus{}b)^{2}\plus{}N(a\minus{}c)(b\minus{}c)$
In which:
$ M \equal{}\frac{a^{3}bc\plus{}b^{2}a^{2}c\plus{}5b^{3}ac\plus{}7c^{2}ba^{2}\plus{}3c^{2}b^{2}a\plus{}b^{3}c^{2}\plus{}c^{2}a^{3}\minus{}4b^{3}a^{2}}{a^{2}b^{2}c(ab\plus{}bc\plus{}ca)}\ge 0$
$ N \equal{}\frac{b^{2}a^{2}\minus{}2bca^{2}\plus{}6b^{2}ca\plus{}7bc^{2}a\plus{}c^{2}b^{2}\plus{}bc^{3}\plus{}ac(2a\minus{}c)^{2}}{a^{2}bc^{2}(ab\plus{}bc\plus{}ca)}\ge 0$

[quote="nayel"]Is the following inequality true?
\[ \frac{a^{3}}{b}\plus{}\frac{b^{3}}{c}\plus{}\frac{c^{3}}{a}\minus{}(a^{2}\plus{}b^{2}\plus{}c^{2})\ge (a\minus{}b)^{2}\plus{}(b\minus{}c)^{2}\plus{}(c\minus{}a)^{2}\]
If it is true I might have a solution.[/quote]

Rewritte the inequality as the following form ( assume that $ c \equal{} min(a,b,c)$):
$ LHS\minus{}RHS \equal{} M(a\minus{}b)^{2}\plus{}N(a\minus{}c)(b\minus{}c)$
For:
$ M \equal{}\frac{(a\minus{}b)^{2}\plus{}ac\plus{}bc\minus{}c^{2}}{ab}\ge 0$
$ N \equal{}\frac{c(a\minus{}2b)^{2}\plus{}b(b\minus{}c)^{2}}{abc}\ge 0$

[quote="hungkhtn"]Such a great solution, thank you my brother!

@ 10maths. How are you? A long time I can not contact![/quote]

Yes, long time no see. Nice to meet you again here, I can not contact too. I'm fine now, I don't have much time to do inequalities anymore, but I still do it when I have time because it is really my favorite. You have gone abroad, I wish you success.  Contact me when you come back, it's really to meet you again.

[quote="hungkhtn"][i]Let $ a,b,c$ be positive real numbers. Prove that[/i]
\[ \left(2\plus{}\frac{a}{b}\right)^{2}\plus{}\left(2\plus{}\frac{b}{c}\right)^{2}\plus{}\left(2\plus{}\frac{c}{a}\right)^{2}\ge\frac{9(a\plus{}b\plus{}c)^{2}}{ab\plus{}bc\plus{}ca}.\][/quote]
[quote="10maths_tp0609"][quote="nayel"]Is the following inequality true?
\[ \frac{a^{3}}{b}\plus{}\frac{b^{3}}{c}\plus{}\frac{c^{3}}{a}\minus{}(a^{2}\plus{}b^{2}\plus{}c^{2})\ge (a\minus{}b)^{2}\plus{}(b\minus{}c)^{2}\plus{}(c\minus{}a)^{2}\]
If it is true I might have a solution.[/quote]

Rewritte the inequality as the following form ( assume that $ c \equal{} min(a,b,c)$):
$ LHS\minus{}RHS \equal{} M(a\minus{}b)^{2}\plus{}N(a\minus{}c)(b\minus{}c)$
For:
$ M \equal{}\frac{(a\minus{}b)^{2}\plus{}ac\plus{}bc\minus{}c^{2}}{ab}\ge 0$
$ N \equal{}\frac{c(a\minus{}2b)^{2}\plus{}b(b\minus{}c)^{2}}{abc}\ge 0$[/quote]

Thank you very much! :) This is my solution: (all sums are cyclic)
By Cauchy-Schwarz,
\[ (ab\plus{}bc\plus{}ca)\left(\sum\left(2\plus{}\frac{a}{b}\right)^{2}\right)\ge\left(\sum\sqrt{ab}\left(2\plus{}\frac{a}{b}\right)\right)^{2}\equal{}\left(\sum 2\sqrt{ab}\plus{}\sum\sqrt{\frac{a^{3}}{b}}\right)^{2}\]
Thus it suffices to show that \[ 2\sum\sqrt{ab}\plus{}\sum\sqrt{\frac{a^{3}}{b}}\ge 3(a\plus{}b\plus{}c)\]
which was just shown by 10maths.

I really like your solution, Nayel. It is very natural and nice  :)


If n and m are whole numbers, $ 99 < n < 401$, and $ 19 < m < 81$, what is the greatest possible value of $ \frac{n}{m}$ ?

to maximize n/m minimize m and maximize n. Thus 400/20 is the maximum or 20.


geometry
geometric transformation
reflection

In a closed economy, if the prices of all goods and services increases, then the expenditure and consumption will rise in exactly the same volume, so inflation has no net effect on spending decisions in the economy

Can someone provide an explanation as to why inflation is relevant?

Inflation decreases the value of saved money.

Inflation matters due to price-setting rigidities.
We do not observe prices being changed constantly. They are
usually set for given period. Also, wage contracts define how
much you expect to earn in the next year or mabye longer.
Inflation causes uncertainty about future prices and the real wage (wages / price level), which causes problems.
It has been shown that countries whose central bank is more independent have lower inflation. The best example of this is
the E.U. , the ECB being completely independent of any government and its sole task being to keep inflation down (It's not supposed to be concerned with unemployment).
A stable-price environment will foster economic activity and employment as employers and workers have more confidence in
what the real wage will be during the life of the wage agreement. Increasing CB independence should lead to a left shift in the Phillips Curve [url]http://en.wikipedia.org/wiki/Phillips_curve[/url]

Those were both excellent points:
1. Money in my wallet and in my checking and low-interest savings accounts becomes less valuable over time and makes money a bad place to store wealth.  This may not be much for one person, but think of 300 million people combined.

2. Unexpected inflation leads to sticky/rigid wages and prices.

3. Expected inflation can also cause a couple problems.  First of all, if inflation is expected, then the nominal interest rate should reflect it.  So I'd rather leave my money in a bank than carry it in my wallet.  This means more trips to the bank to make withdrawals if I want to carry cash.  Of course, this isn't a big problem now since we can do everything on line.
There are also what's called Menu Costs.  For example, a restaurant has to reprint it's menu every few months.  This costs money.  Think of how much it costs a store like Walmart to constantly be changing all their price tags.  Aggregate this accross a whole economy and it's very costly.  So basically, expected inflation is inefficient.
On the other hand, since companies usually have long contracts with employees, it gives them an edge.  They will have relatively lower wage costs as the year progresses; this saves them money.

4. Unexpected inflation also redistributes wealth.  To whom it gives and from whom it takes depends if it's going up or down.  E.G., if you have a long term fixed interest loan and inflation unexpectedly goes up, your broker gets hurt.  If it goes the down, you get hurt.  On that note, all the people with adjustable rate mortgages get hurt when inflation unexpectedly goes up and they didn't plan on it.

EconMan
[url]http://economics.fundamentalfinance.com[/url]


induction
combinatorics proposed
combinatorics

Consider $n$ points $A_1,A_2,A_3,\ldots, A_n$ ($n\geq 4$) in the plane, such that any three are not collinear. Some pairs of distinct points among $A_1,A_2,A_3,\ldots, A_n$ are connected by segments, such that every point is connected with at least three different points. Prove that there exists $k>1$ and the distinct points $X_1,X_2,\ldots, X_{2k}$ in the set $\{A_1,A_2,A_3,\ldots, A_n\}$, such that for every $i\in \overline{1,2k-1}$ the point $X_i$ is connected with $X_{i+1}$, and $X_{2k}$ is connected with $X_1$.

We can prove it by induction on the number $n$ of vertices.

The assertion is easy to prove for $n=4$. Our graph $G$ must have a cycle (otherwise it would be a forest, and a terminal vertex of one of the trees which are its components would have degree $1$), so let $C=(y_1,\ldots,y_{2t+1})$ be an odd cycle. No $y_i,y_j,\ i\ne j$ can be connected through a path different that the one in $C$, because that path, together with one of the two different paths which connect them in $C$, would form an even cycle. This means that a cycle in $G/C$ (the graph we get by collapsing the vertices in $C$ to a single vertex) is a cycle in $G$ as well. Since $G/C$ still has the property that every vertex has degree $\ge 3$ (and it has $\ge 4$ vertices), we can use induction.

grobber, but you are proving that the graph contains an odd cycle.. the problem asks to prove that it contains an even cycle

I don't think so :). I'm assuming the graph has no even cycles. Since it does have a cycle, it must be odd, and I'm using that cycle in the proof. I think you misunderstood something in there, but it's hard to tell what.

oops yea i was trying to delete my post right now

Shorter solution without induction:

consider the longest path $X_1X_2\dots X_m$ (with different vertices) in our graph. A vertex $X_1$ may be joined only withother $X_i$ (else the path may be increased). Say, $X_1$ is joined with $X_k$ and $X_l$, $2<k<l$. Two of three numbers $2,k,l$ are of the same parity, redenote them as  $p,\,q,\,p<q$. There is an even cycle $X_1X_pX_{p+1}\dots X_qX_1$.


geometry circumcircle incenter inradius rotation trapezoid parallelogram

In triangle $ABC$ a line through $A$ perpendicular to $IO$ meet $BC$ at $K$.prove that:

$|(\text{circumradius of }AKB)-(\text{cirumradius of }AKC)|=IO$

Let $O, I$ be the circumcenter and incenter, $R, r$ the circumradius and inradius of the triangle $\triangle ABC$ and $d = IO$. Let's "rotate" the triangle $\triangle ABC$ around the fixed circumcircle (O) and incircle (I) according to Poncelet's porism. Let $O_1, O_2$ be the circumcenters and $R_1, R_2$ the circumradii of the triangles $\triangle AKB,\ \triangle AKC$. It is trivial that $|R_2 - R_1| = d$, when the triangle $\triangle ABC$ becomes isosceles with AB = BC, because then $K \equiv C,$ $(O_1) \equiv (O),$ $O_2$ is the midpoint of the arc CA opposite to B, hence $O_2 \in OI \equiv OO_1$ and $I \in (O_2).$ Likewise, it is trivial that $|R_2 - R_1| = d$, when the triangle $\triangle ABC$ becomes isosceles with BC = CA, because then $K \equiv B,$ $(O_2) \equiv (O),$ $O_1$ is the midpoint of the arc AB opposite to C, hence $O_1 \in OI \equiv OO_2$ and $I \in (O_1).$ When the triangle $\triangle ABC$ becomes isosceles with AB = CA, then $K \equiv A$ and the circumcircles $(O_1), (O_2)$ degenerate into the lines AB, CA, respectively, and anything is possible. That's where the change of sign of $R_2 - R_1$ occurs.

Let $(O_1), (O_2)$ be 2 arbitrary fixed circles with radii $R_1 < R_2$ intersecting at points A, K. Let an arbitrary line through K meet the circles  $(O_1), (O_2)$ again at B, C, respectively, and let B move on the circle $(O_1).$ The triangle $\triangle ABC$ remains similar for all positions of $B \in (O_1)$, because $\angle ABC = 180^\circ - \angle ABK = 180^\circ - \text{const} = \text{const}$ and $\angle ACB \equiv \angle ACK = \text{const}.$ Since A is fixed and B, C move on circles $(O_1), (O_2)$, respectively, the loci of the circumcenter O and incenter I of the triangle $\triangle ABC$ are also some circles (P), (J). (That's a theorem from Yaglom's Geometric Transformations II, pp. 68-69.) When B coincides with A, the triangle $\triangle ABC$ degenerates into the point A, hence both the circles (P), (J) pass through the point A. When B coincides with K, $(O)$ coincides with $(O_2)$, hence $O_2 \in (P)$. Likewise, when C coincides with K, $(O)$ coincides with $(O_1)$, hence $O_1 \in (P)$. Thus the locus of circumcenters O of the triangles $\triangle ABC$ is the circumcircle (P) of the triangle $\triangle AO_1O_2.$ Let the circle $(O_1)$ meet the center line at $U_1, V_1$ and the circle $(O_2)$ at $U_2, V_2,$ the points $U_1, U_2$ lying inside the circles $(O_2), (O_1)$ and the points $V_1, V_2$ outside the circles $(O_2), (O_1),$ respectively. When B coincides with $V_1$, the center line bisects the angle $\angle ABC \equiv \angle AV_1K$ and the line $CU_2$ bisects the angle $\angle BCA \equiv \angle KCA$, because $U_2$ is the midpoint of the arc $AK$ of the circumcircle $(O_2)$ of the triangle $\triangle AKC$ opposite to the vertex C. Hence $U_2$ is the incenter of this triangle $\triangle ABC$ and $U_2 \in (J)$. Likewise, when C coincides with $V_2$, the center line bisects the angle $\angle ACB \equiv \angle AV_2K$ and the line $BU_1$ bisects the angle $\angle CBA \equiv \angle KBA$, because $U_1$ is the midpoint of the arc $AK$ of the circumcircle $(O_1)$ of the triangle $\triangle AKB$ opposite to the vertex B. Hence $U_1$ is the incenter of this triangle $\triangle ABC$ and $U_1 \in (J)$. Thus the locus of incenters I of the triangles $\triangle ABC$ is the circumcircle (J) of the triangle $\triangle AU_1U_2.$

The locus circles (P), (J) intersect at the point A and at another point Z. The line BC of the similar triangles $\triangle ABC$ passes through the fixed point K different from their common vertex A. Consequently, the diacentral line OI of these triangles also passes through some fixed point. (That's another theorem from Yaglom's Geometric Transformations II, p. 69.) Obviously, this fixed point is the other intersection Z of the locus circles (P), (J) - we can move the triangle $\triangle ABC$ into a position $\triangle AB'C',$ when $O' \equiv Z$ or into a position $\triangle AB''C'',$ when $I'' \equiv Z$. The lines $O'I' \equiv ZI'$ and $O''I'' \equiv O''Z$ must both pass through the same fixed point and they have only the point Z in common.

Since $AK \perp O_1O_2$ for the fixed circles $(O_1), (O_2)$, then for a triangle $\triangle ABC$ satisfying the problem condition $AK \perp OI$, it must be $OI \parallel O_1O_2$. Thus the locus circles (P), (J) are intersected by 2 parallel lines $OI \equiv OZ \equiv IZ$ and $O_1O_2 \equiv U_1U_2$ forming 2 cyclic (isosceles) trapezoids $OZO_1O_2$ and $IZU_2U_1$ with the common vertex Z and identical base lines. It is clear that

$O_2U_1 - O_1U_2 = O_2U_2 - U_1U_2 - O_1U_1 + U_1U_2 = O_2U_2 - O_1U_1 = R_2 - R_1$

On the other hand, if a parallel to $OO_2$ through I meet the center line $O_1O_2$ at a point W, then $OIWO_2$ is a parallelogram, $OI = O_2W$ and the triangles $\triangle IU_1W \cong \triangle ZU_2O_1$ are (oppositely) congruent, so that $WU_1 = O_1U_2.$ Consequently,

$OI = O_2W = O_2U_1 - WU_1 = O_2U_1 - O_1U_2 = R_2 - R_1$

which is what we were supposed to prove.

[hide="Here is a short metrical proof for this very nice problem."]
[color=darkblue][b][u]A preliminary study.[/u][/b] Suppose $K\in (BC)\ ,\ AK\perp OI$ and $b\ne c\ .$ Denote

$\alpha =m(\widehat {BAK})$ and $\beta =m(\widehat {CAK})\ .$ Using an elementary metrics we obtain :[/color]

$1.\blacktriangleright \sum a(a-b)(a-c)=4pr(R-2r)\ ;$

$2.\blacktriangleright\frac{KB}{c(b-a)}=\frac{KC}{b(a-c)}=\frac{1}{b-c}\ ;$

$3.\blacktriangleright \boxed {\ AK^2=\frac{bc}{a(b-c)^2}\cdot \sum a(a-b)(a-c)\ }\ ;$

$4.\blacktriangleright\tan \alpha=\frac{\sin A-\sin B}{\cos A+\cos B-1}\ ,$ $\tan \beta =\frac{\sin A-\sin C}{\cos A+\cos C-1}\ .$

[color=darkred][b][u]Proof of the proposed problem.[/u][/b] Denote the circumradius $R_1$ of the triangle $BAK$ and

the circumradius $R_2$ of the triangle $CAK\ .$ Using the relations $(1)$ and $(3)$ we obtain :[/color]

$AK=2R_1\sin B=2R_2\sin C\Longrightarrow (R_1-R_2)^2=AK^2\cdot \frac{\sin B-\sin C)^2}{4\sin^2B\sin^2C}=$
$\frac{bc}{a(b-c)^2}\sum a(a-b)(a-c)\cdot \frac{R^2(b-c)^2}{b^2c^2}=\frac{R^2}{abc}\sum a(a-b)(a-c)=$
$\frac{R^2}{4Rpr}\cdot 4pr(R-2r)=R(R-2r)=OI^2\Longrightarrow \left|R_1-R_2\right|=OI\ .$

[color=red][b][u]Remark.[/u][/b] If you wish to prove the above properties $1\div 4\ ,$ with pleasure ![/color][/hide]

Let $R_1, R_2$ be circumradii of the triangles $\triangle ABK,\ \triangle ACK,$ respectively, a = BC, b = AC, c = AB, d = OI and assume b < c.

$\frac{c}{2R_1} = \sin \widehat{AKB} = \sin \widehat{AKC} = \frac{b}{2R_2}\ \Longrightarrow\ \sin \widehat{(AK, BC)} = \frac{c - b}{2(R_1 - R_2)}$

Let T be the tangency point of the incircle (I) with BC and M the midpoint of BC, $MT = \frac a 2 - (s - c) = \frac{c - b}{2}$ and $\sin \widehat{(OI, IT)} = \frac{MT}{OI} = \frac{c - b}{2d}.$

$AK \perp OI\ \ \&\ \ IT \perp BC\ \Longrightarrow \widehat{(AK, BC)} = \widehat{(OI, IT)}$

$\frac{c - b}{2(R_1 - R_2)} = \frac{c - b}{2d}\ \Longrightarrow\ R_1 - R_2 = d$


[hide="Solution"]
We can use Jensen's inequality for both a) and b) or:
a) $  \tan^{2}\frac{A}{2}=\frac{(s-b)(s-c)}{s(s-a)}, \ \tan^{2}\frac{B}{2}=\frac{(s-a)(s-c)}{s(s-b)}, \ \tan^{2}\frac{C}{2}=\frac{(s-a)(s-b)}{s(s-c)}$, where $ s$ is semiperimeter.
and the inequality becomes
$ (s-b)^{2}(s-c)^{2}+(s-a)^{2}(s-c)^{2}+(s-a)^{2}(s-b)^{2}\geq s(s-a)(s-b)(s-c)$
Let $ x=s-a,y=s-b,z=s-c\Rightarrow x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2}\geq xyz(x+y+z)$ which is true.

b)$ s[(s-a)^{2}+(s-b)^{2}+(s-c)^{2}]\geq 9(s-a)(s-b)(s-c)\Rightarrow$
$ \Rightarrow (x+y+z)(x^{2}+y^{2}+z^{2})\geq 9xyz$, which is true by AM-GM.
[/hide]

[quote="red_dog"][hide="Solution"]
We can use Jensen's inequality for both a) and b) or...

[/hide][/quote]

nice solution...

how could we use Jensen?

3 different solutions by now...


Show that if $ S \equal{} \{\mathbf v_1,\mathbf v_2,\cdots,\mathbf v_r\}$ is a linearly independent set of vectors, then so is every nonempty subset of $ S$.

My way is to prove the following: If there exists a linearly dependent subset of $ S \equal{} \{\mathbf v_1,\mathbf v_2,\cdots, \mathbf v_r\}$, then $ S$ is linearly dependent.

That's right, use a proof by contrapositive.

Yes, and quite efficiently.

Independence of $ S:$ Any linear combination of the vectors in $ S$ which equals zero is the trivial linear combination. (That is, all coefficients equal zero.)

Independence of $ P\subset S:$ Any linear combination of the vectors in $ S,$ in which the only nonzero coefficients are for vectors in $ P,$ is the trivial linear combination.

Prof Kent Merryfield, I don't understand the second.

It's really that a linear combination of the vectors in $ P$ could be thought of as a linear combination of the vectors in $ S.$ All you have to do is add a bunch of zeros.

I still don't know what you mean. But the following is my proof by contrapositive. And I want if there is a direct proof.

The original proposition is equivalent to the following: [i]If there exists a linearly dependent subset of $ S \equal{} \{\mathbf v_1,\mathbf v_2,\cdots, \mathbf v_r\}$, then $ S$ is linearly dependent.[/i]

Now we prove the equivalent proposition. Without loss of generality, we let $ \{\mathbf v_1,\mathbf v_2,\cdots,\mathbf v_n\}$ be a linearly dependent subset of $ S$, where $ n\leq r$. Then the vector equation

$ \alpha_1\mathbf v_1 \plus{} \alpha_2\mathbf v_2 \plus{} \cdots \plus{} \alpha_n\mathbf v_n \equal{} \mathbf 0$

has a non-trivial solution, denoted by

$ \alpha_i \equal{} k_i,\ i \equal{} 1,2,\cdots,n,$

where some $ k_i$'s are nonzero. Consider the fact that

$ 0\mathbf v_1 \plus{} 0\mathbf v_2 \plus{} \cdots \plus{} 0\mathbf v_n \plus{} 0\mathbf v_{n \plus{} 1} \plus{} \cdots \plus{} 0\mathbf v_r \equal{} \mathbf0.$

Since

$ 0\mathbf v_1 \plus{} 0\mathbf v_2 \plus{} \cdots \plus{} 0\mathbf v_n \equal{} \mathbf0 \equal{} k_1\mathbf v_1 \plus{} k_2\mathbf v_2 \plus{} \cdots \plus{} k_n\mathbf v_n,$

the above equation can be written as

$ k_1\mathbf v_1 \plus{} k_2\mathbf v_2 \plus{} \cdots \plus{} k_n\mathbf v_n \plus{} 0\mathbf v_{n \plus{} 1} \plus{} \cdots \plus{} 0\mathbf v_r \equal{} \mathbf0.$

That is, the vector equation

$ \beta_1\mathbf v_1 \plus{} \beta_2\mathbf v_2 \plus{} \cdots \plus{} \beta_r\mathbf v_r \equal{} \mathbf0$

has a non-trivial solution. Hence $ S$ is linearly dependent. As a result, the original proposition is also ture.

Perhaps you understand independence differently. For me, linear independence of a set S means that every finite subset of S spans 0 only trivially. The answer to your question is then trivial. But perhaps you understand it differently.

So. Given a set T in your linear space (not known whether is finite) what is the definition you have for independence?

Assume to the contrary that there exists a subset of a linearly independent that is linearly dependent, that is, suppose that $ \{w_1,\hdots, w_s\}\equal{}R\subset S$, $ s\leq r$, is linearly dependent.  Then we have a nonzero vector $ (\lambda_1\hdots, \lambda_s,0,..,0)$ such that $ 0\equal{}\sum_{i\equal{}1}^s\lambda_iw_i\plus{}\sum_{v_j\in S\setminus R}^r0\cdot v_j$.  Contradiction!


advanced fields
advanced fields unsolved

Show that if $ \mu$ is a finite measure, there cannot be uncountably many disjoint sets $ A$ such that $ \mu(A) > 0$

Let’s suppose {Aa} is a family of disjoint sets A such that  m(A) >0
A0 = {A |A belongs to {Aa} and    1<m(A)  }
Ai = {A |A belongs to {Aa} and    1/(i+1)<m(A)<=1/i  }
{Aa} = UAi(i=0  to infinity )
Each set Ai is finite unless measure of union of its members must be infinite so {Aa} is a countable union of finite sets so it is countable .


Let n > 2, and let f(x) be the polynomial x^2006 + x + n. Prove that all the roots a of
f(x) have absolute value |a| > 1. Prove that if n = p is prime then f(x) is irreducible.

If $ a$ is a root of $ f(x)$ with $ \minus{}1\le a\le 1$, then $ a^{2006}\plus{}a\plus{}n\equal{}0$ and $ a^{2006}\plus{}a\plus{}n\ge (\minus{}1)\plus{}(\minus{}1)\plus{}n\equal{}n\minus{}2 >0$ which is absurd.  So, such a root $ a$ cannot exist.  I.e. all the roots of $ f(x)$ must satisfy $ |a|>1$.  Now, if $ n\equal{}p$ where $ p$ is prime, use the rational root theorem.  The only possible rational roots are $ \pm 1,\pm p$, all of which fail to be roots.  Hence, $ f(x)$ is irreducible over the rationals.

[quote="zgorkster"]The second part needs more work.  Just because a polynomial doesn't have a rational root doesn't mean that it is irreducible in the field $ \mathbb{Q}[x]$.[/quote]
This is true.  One needs to show that there is no polynomial of degree $ \ge 2$ dividing $ f(x)$.  But, why don't we just use that any root $ a$ satisfies $ |a|>1$ forcing $ a^{2006}\plus{}a\plus{}n\ge\min_{\{|a|>1\}}(a^{2006}\plus{}a)\plus{}n >0\plus{}n>0$ i.e. there are no real roots.

[quote="baz"][quote="zgorkster"]The second part needs more work.  Just because a polynomial doesn't have a rational root doesn't mean that it is irreducible in the field $ \mathbb{Q}[x]$.[/quote]
This is true.  One needs to show that there is no polynomial of degree $ \ge 2$ dividing $ f(x)$.  But, why don't we just use that any root $ a$ satisfies $ |a| > 1$ forcing $ a^{2006}\plus{}a\plus{}n\ge\min_{\{|a| > 1\}}(a^{2006}\plus{}a)\plus{}n > 0\plus{}n > 0$ i.e. there are no real roots.[/quote]
$ (1\plus{}x\plus{}x^{2})(1\plus{}x\plus{}x^{2}\plus{}x^{3}\plus{}x^{4})$
doesn't have any real roots

[quote="junggi"][quote="baz"][quote="zgorkster"]The second part needs more work.  Just because a polynomial doesn't have a rational root doesn't mean that it is irreducible in the field $ \mathbb{Q}[x]$.[/quote]
This is true.  One needs to show that there is no polynomial of degree $ \ge 2$ dividing $ f(x)$.  Could we just use that any root $ a$ satisfies $ |a| > 1$ forcing $ a^{2006}\plus{}a\plus{}n\ge\min_{\{|a| > 1\}}(a^{2006}\plus{}a)\plus{}n > 0\plus{}n > 0$.[/quote]
$ (1\plus{}x\plus{}x^{2})(1\plus{}x\plus{}x^{2}\plus{}x^{3}\plus{}x^{4})$
doesn't have any real roots[/quote]

Excuse me, I meant could we use...  I am not sure what has been going through my head  :blush:  Sorry.  Maybe there is a better approach.

Hopefully, I can redeem my foolishness  :) :

Suppose that $ f(x)\equal{}g(x)h(x)$ where $ g,h\in\mathbb{Z}[x]$ with degrees of $ g,h\ge 1$.  Then, $ p\equal{}n\equal{}f(0)\equal{}g(0)h(0)$ implies that WLOG $ |g(0)|\equal{}1$.  Then, we can write $ g(x)\equal{}c\prod_{k\equal{}1}^{m}(x\minus{}x_{k})$ where $ m\ge 1$, $ 0\neq c\in\mathbb{Z}$ and $ x_{k}$ are the zeros of $ g(x)$.  So,  \[ 1\equal{}|g(0)|\equal{}|c|\prod_{k\equal{}1}^{m}|x_{k}|\] which forces some $ x_{k}$ to be $ \le 1$.

Note: we showed that the roots of $ f(x)$ are all $ >1$ in absolute value.  Thus, $ f$ must be irreducible.


Sorry if this is above level I will post it in another forum.
Moderators please delete this. Thanks

Generally, proofs don't go in High School Basics.   :huh:


geometry
3D geometry
sphere
symmetry
calculus
calculus computations

Is it possible to calculate the volume of solids of revolution by calculating the volume of infinitesimally small spheres bounded by the surface of the solid? Or, since the spheres would be acting as points essentially, would the ray of spheres together just act as the radius of a cylinder, and then it'd be just a plain old solid of revolution type equation?

Do you mean something like [url=http://en.wikipedia.org/wiki/Sphere_packing]sphere packing[/url]?

Not quite sure. Basically, I mean instead of using circular discs or cylindrical shells to calculate volume, could you use spheres approaching zero volume? So it wouldn't be an approximation, it'd tell the actual volume.

Perhaps what you want could be called "the method of spherical shells". But, just like the method of cylindrical shells applies only to solids with an axis of symmetry, the method of spherical shells would apply only to solids with rotational symmetry about a point. And there aren't that many of those.


calculus
integration
function
geometry
trigonometry
calculus computations

Hi, what's the integral of [u][i][b](cos x)/x[/u][/b][/i]?  can you also please show workings on how its done. thanks:)

Another classic function whose primitive cannot be written as a finite sum of elementary functions. However, one can compute several [i]definite[/i] integrals of that function.

In my reference book it is given that these integrals cannot be computed in the indefinite way....

Can they be computed as a definite integral......
If so how...

Oh. the original question is find the volume of the solid generated by revolving the area between the curve y=(cos x)/x and the x-axis for pi/6 <= x <= pi/2 about the y-axis. How do you do this? thanks!:)

That volume does not require you to integrate $ \frac{\cos x}{x}.$ The integral is of a different function, one which can be computed in closed form.

Set up the integral properly for this problem.

oh okay. well, can you just help me answer this problem?:D thanks

Take an infinitesimally thin vertical strip of the region under the graph. This strip is located $ x$ units from the vertical axis, has thickness $ dx$ and has height $ \frac {\cos x}{x}.$ Revolve that strip around the vertical axis. It sweeps through a region which is an infinitesimally thin cylindrical shell. What is the (infinitesimal) volume of that shell? That will be the thing that we integrate.

(Don't worry about any lack of rigor in the notion of "infinitesimal" quantities; we can retreat in the direction of Riemann sums and defend ourselves if we have to. I use this language because it appeals to the intuition.)

sorry Kent merryfield, but i really dont understand what you're saying. can you show the workings? thanks?:D

You probably have formulas that look something like this
$ \displaystyle V=2\pi\int {x f(x)} dx$

$ \displaystyle V=\pi\int {(f(x))^2 dx}$

in your calc book. You have to use one of these to solve the problem.