I think:

Let $ x,y,z > 0,t \in N,f(t) \equal{} \frac {x^t}{y^t} \plus{} \frac {y^t}{z^t} \plus{} \frac {z^t}{x^t},$
then:
$ f(t) \le \frac {3}{3^r}\prod_{i \equal{} 1}^r{f(i)}$
inside,$ t \geq 5).$

BQ

and I think agin:

Let $ x,y,z > 0,t \in N,f(t) \equal{} \frac {x^t}{y^t} \plus{} \frac {y^t}{z^t} \plus{} \frac {z^t}{x^t},$
then:
$ f(n) \geq \frac{1}{3}f(m)f(n\minus{}m).$
inside,$ m \le [\frac{n}{2}]$

BQ.

[quote="kuing"][quote="xzlbq"]$ a,b,c > 0,t \in R$,Prove that
${ f(t) = \frac {{a}^{t}}{{b}^{t}}} + {\frac {{b}^{t}} {{c}^{t}}} + {\frac {{c}^{t}}{{a}^{t}}}$,

if
$ t_1\geq t_2$,
then
$ f(t_1) \le f(t_2).$

BQ[/quote]

not hold.

when $ t_1 = 1, t_2 = - 1$, ineq obvious not hold.

but I think if $ t_1\geq t_2 \ge 0$, will have $ f(t_1) \ge f(t_2).$[/quote]

I got a proof of this ineq for $ n$ variable (all variable are positive and cyclic, and $ t \ge 0$).  :lol:

[quote="kuing"][quote="xzlbq"]$ a,b,c > 0,t \in R$,Prove that
${ f(t) = \frac {{a}^{t}}{{b}^{t}}} + {\frac {{b}^{t}} {{c}^{t}}} + {\frac {{c}^{t}}{{a}^{t}}}$,

if
$ t_1\geq t_2$,
then
$ f(t_1) \le f(t_2).$

BQ[/quote]

not hold.

when $ t_1 = 1, t_2 = - 1$, ineq obvious not hold.

but I think if $ t_1\geq t_2 \ge 0$, will have $ f(t_1) \ge f(t_2).$[/quote]

$ t \in R^{+}$

if
$ a_1 \geq \frac {3}{3^t}a_3^t$
holds,then
$ t_{max} \equal{} 12.$
BQ

We have nice:
$ a_1 \minus{} 3 \geq \sum_{i \equal{} 2}^r{(a_i \minus{} 3)}$
r=10,
I think :$ r{\to}\infty$

BQ

[quote="xzlbq"]Now it seems,We should also think of some more wide.
Let $ x,y,z > 0,a_1 \equal{} f_1(x,y,z) \equal{} g(x,y,z)\geq0,g(x,y,z)$is a aymmetric
$ a_{i \plus{} 1} \equal{} f_{i \plus{} 1}(x,y,z) \equal{} f_i(y \plus{} z,z \plus{} x,x \plus{} y)(i \equal{} 1,2,...r,r\geq 2)$,

then:

$ a_1\geq \sum_{i \equal{} 2}^r{a_i}(r\geq2)$.
BQ

Now give some examples：

eg1:
Let$ r \equal{} 5,g(x,y,z) \equal{} \frac {x}{y \plus{} z} \plus{} \frac {y}{z \plus{} x} \plus{} \frac {z}{x \plus{} y} \minus{} \frac {3}{2}.$
then
$ a_1 \plus{} 9 \minus{} a_2 \minus{} a_3 \minus{} a_4 \minus{} a_5\geq0.$,
inside,
$ (a_1, a_2, a_3, a_4, a_5) \equal{} (\frac {x}{y \plus{} z} \plus{} \frac {y}{z \plus{} x} \plus{} \frac {z}{x \plus{} y}, \frac {y \plus{} z}{z \plus{} 2x \plus{} y} \plus{} \frac {z \plus{} x}{x \plus{} 2y \plus{} z} \plus{} \frac {x \plus{} y}{y \plus{} 2z \plus{} x}, \frac {z \plus{} 2x \plus{} y}{2x \plus{} 3y \plus{} 3z} \plus{} \frac {x \plus{} 2y \plus{} z}{2y \plus{} 3z \plus{} 3x} \plus{} \frac {y \plus{} 2z \plus{} x}{2z \plus{} 3x \plus{} 3y}, \frac {2x \plus{} 3y \plus{} 3z}{5y \plus{} 5z \plus{} 6x} \plus{} \frac {2y \plus{} 3z \plus{} 3x}{5z \plus{} 5x \plus{} 6y} \plus{} \frac {2z \plus{} 3x \plus{} 3y}{5x \plus{} 5y \plus{} 6z}, \frac {5y \plus{} 5z \plus{} 6x}{11z \plus{} 10x \plus{} 11y} \plus{} \frac {5z \plus{} 5x \plus{} 6y}{11x \plus{} 10y \plus{} 11z} \plus{} \frac {5x \plus{} 5y \plus{} 6z}{11y \plus{} 10z \plus{} 11x}).$

in factor,We have:

$ a_1\geq \sum_{i \equal{} 2}^{12}{a_i}$  holds.[/quote]

But  $ g(x,y,z)\equal{}x(x\minus{}y)(x\minus{}z)\plus{}y(y\minus{}z)(y\minus{}x)\plus{}z(z\minus{}x)*(z\minus{}y)$ does not holds.
ask:
 what is $ g(x,y,z)$ conditions?
BQ

Note this form:
$ \frac{x}{y\plus{}z}\plus{}\frac{y}{z\plus{}x}\plus{}\frac{z}{x\plus{}y}\minus{}\frac{3}{2}\geq (\frac{y\plus{}z}{2x\plus{}y\plus{}z}\plus{}\frac{z\plus{}x}{2y\plus{}z\plus{}x}\plus{}\frac{x\plus{}y}{2z\plus{}x\plus{}y}\minus{}\frac{3}{2})^t$,
$ t\equal{}20$ holds.

[quote="xzlbq"]Examples of holds:

(1)
$ g(x,y,z) \equal{} \frac {x^2 \plus{} y^2 \plus{} z^2}{xy \plus{} yz \plus{} zx}$;

Let r=3,
$ a_1 \minus{} 1 \geq \sum_{i \equal{} 2}^3{a_i}$<=>
$ \frac {x^2 \plus{} y^2 \plus{} z^2}{xy \plus{} zx \plus{} yz} \plus{} 1\geq \frac {(y \plus{} z)^2 \plus{} (z \plus{} x)^2 \plus{} (x \plus{} y)^2)}{(y \plus{} z)(z \plus{} x) \plus{} (x \plus{} y)(y \plus{} z) \plus{} (z \plus{} x)(x \plus{} y)} \plus{} \frac {(2x \plus{} z \plus{} y)^2 \plus{} (2y \plus{} x \plus{} z)^2 \plus{} (2z \plus{} y \plus{} x)^2}{(2x \plus{} z \plus{} y)(2y \plus{} x \plus{} z) \plus{} (2z \plus{} y \plus{} x)(2x \plus{} z \plus{} y) \plus{} (2y \plus{} x \plus{} z)*(2z \plus{} y \plus{} x)}$;

$ a_1\geq a_2a_3$.

in factor,r=20 holds.

very very very great great great interesting inequality is:

$ a_1\geq a_2^3a_3^3a_5a_6^3;$

[size=200]nice and nice![/size]

But Not finished.

You should pay attention to this problem!

You MUST pay attention to this problem!![/quote]

[size=84]we have:
$ a_1\geq a_2^3a_3^3a_4^0a_5a_6^3a_7^0a_8a_9a_{10}^2a_{11}^2a_{12}.$---------------------(*)

For each $ i,a_i$degree is best[/size].

$ a_1 \equal{} \frac {x^2 \plus{} y^2 \plus{} z^2}{yz \plus{} zx \plus{} xy},$

$ a_2 \equal{} \frac {(y \plus{} z)^2 \plus{} (x \plus{} z)^2 \plus{} (x \plus{} y)^2}{(x \plus{} z)(x \plus{} y) \plus{} (x \plus{} y)(y \plus{} z) \plus{} (y \plus{} z)(x \plus{} z)},$
$ a_3 \equal{} \frac {(2x \plus{} z \plus{} y)^2 \plus{} (2y \plus{} z \plus{} x)^2 \plus{} (y \plus{} 2z \plus{} x)^2}{(2y \plus{} z \plus{} x)(y \plus{} 2z \plus{} x) \plus{} (y \plus{} 2z \plus{} x)(2x \plus{} z \plus{} y) \plus{} (2x \plus{} z \plus{} y)(2y \plus{} z \plus{} x)},$
...

in program lhdc3_4(http://www.mathlinks.ro/Forum/viewtopic.php?t=321135,lhdc3_4.txt),
inequalyty (*) inti ($ \sigma_1,\sigma_2,\sigma_3$) is:

$ \minus{} 53748795701289807798991678762028404869081476808\sigma_1^22\sigma_2^7$

$ \plus{} 99812958214001304847011561168056091436220\sigma_1^{16}\sigma_2^{10}$

$ \minus{} 150761124721\sigma_1^2\sigma_2^{17} \minus{} 19550539832414211475950040585265375016931910\sigma_1^{18}\sigma_2^9$

$ \plus{} 65545873796437050430283487312209884396822265625\sigma_1^{36} \minus{} 10037760610802733178947186442364\sigma_1^{10}\sigma_2^{13}$

$ \plus{} 473447166275390578990550013634190682898860371536\sigma_1^{24}\sigma_2^6$

$ \plus{} 4543844468985097091810024676954102152158004891500\sigma_1^{28}\sigma_2^4 \minus{} 818856027771727049893934602339168509989099329809\sigma_1^{34}\sigma_2$

$ \plus{} 1615468505816562254819133418707829645631483302\sigma_1^{20}\sigma_2^8 \plus{} 131070\sigma_2^{18}$

$ \plus{} 87815719647089095203647161860135052\sigma_1^{12}\sigma_2^{12} \minus{} 6902307429709163393400\sigma_1^6\sigma_2^{15}$

$ \plus{} 3591345498808599831242421393947897607271156641526\sigma_1^{32}\sigma_2^2 \plus{} 403340464404520478832796800\sigma_1^8\sigma_2^{14}$

$ \plus{} 50408207480210633\sigma_1^4\sigma_2^{16} \minus{} 207643313838042935128724556482454056200\sigma_1^{14}\sigma_2^{11}$

$ \minus{} 6452725731480378450958025155803759135636124833336\sigma_1^{30}\sigma_2^3$

$ \minus{} 2275542599381596237447057751635054148427552963900\sigma_1^{26}\sigma_2^5\geq0.$

$ \sigma_1 \equal{} x \plus{} y \plus{} z,\sigma_2 \equal{} xy \plus{} yz \plus{} zx,\sigma_3 \equal{} xyz.$

BQ

\sigam_1

Let $ x,y,z > 0,a_1 \equal{} f_1(x,y,z) \equal{} \frac {x^2}{y} \plus{} \frac {y^2}{z} \plus{} \frac {z^2}{x} \minus{} x \minus{} y \minus{} z\geq0,$
$ a_{i \plus{} 1} \equal{} f_{i \plus{} 1}(x,y,z) \equal{} f_i(y \plus{} z,z \plus{} x,x \plus{} y)(i \equal{} 1,2,...r,r\geq 2)$,

then:

$ a_1 \equal{} \frac {x^2}{y} \plus{} \frac {y^2}{z} \plus{} \frac {z^2}{x} \minus{} x \minus{} y \minus{} z,$;
$ a_2 \equal{} \frac {(y \plus{} z)^2}{z \plus{} x} \plus{} \frac {(z \plus{} x)^2}{x \plus{} y} \plus{} \frac {(x \plus{} y)^2}{y \plus{} z} \minus{} 2y \minus{} 2z \minus{} 2x,$
$ a_3 \equal{} \frac {(z \plus{} 2x \plus{} y)^2}{x \plus{} 2y \plus{} z} \plus{} \frac {(x \plus{} 2y \plus{} z)^2}{y \plus{} 2z \plus{} x} \plus{} \frac {(y \plus{} 2z \plus{} x)^2}{z \plus{} 2x \plus{} y} \minus{} 4z \minus{} 4x \minus{} 4y.$
...
prove that:$ a_1\geq a_2$
<=>

$ \frac {x^2}{y} \plus{} \frac {y^2}{z} \plus{} \frac {z^2}{x} \plus{} x \plus{} y \plus{} z \minus{} \frac {(y \plus{} z)^2}{z \plus{} x} \minus{} \frac {(z \plus{} x)^2} {x \plus{} y} \minus{} \frac {(x \plus{} y)^2}{y \plus{} z} \geq0$

<=>
$ \sigma_1^3\sigma_2^2 \minus{} 2\sigma_2(3\sigma_3 \plus{} \sigma_{31})\sigma_1^2 \plus{} (6\sigma_3^2 \plus{} 2\sigma_3\sigma_{31} \plus{} \sigma_{31}^2)\sigma_1 \plus{} 2\sigma_2^2\sigma_3\geq0.$
$ \sigma_1 \equal{} x \plus{} y \plus{} z,\sigma_2 \equal{} xy \plus{} yz \plus{} zx,\sigma_3 \equal{} xyz,\sigma_{31} \equal{} x^2y \plus{} y^2z \plus{} z^2x$.

remark:
$ a_1\geq a_2 \plus{} a_3$ does not holds.

**********************************************************
But :
Let $ x,y,z > 0,a_1 \equal{} f_1(x,y,z) \equal{} \frac {x^2}{y(x \plus{} y \plus{} z)} \plus{} \frac {y^2}{z(x \plus{} y \plus{} z)} \plus{} \frac {z^2}{x(x \plus{} y \plus{} z)} \minus{} 1\geq0,$
$ a_{i \plus{} 1} \equal{} f_{i \plus{} 1}(x,y,z) \equal{} f_i(y \plus{} z,z \plus{} x,x \plus{} y)(i \equal{} 1,2,...r,r\geq 2)$,

then:

$ a_1 \equal{} \frac {x^2}{y(x \plus{} y \plus{} z)} \plus{} \frac {y^2}{z(x \plus{} y \plus{} z)} \plus{} \frac {z^2}{x(x \plus{} y \plus{} z)} \minus{} 1,$;
$ a_2 \equal{} \frac { (y \plus{} z)^2}{(z \plus{} x)(2y \plus{} 2z \plus{} 2x} \plus{} \frac {(z \plus{} x)^2}{(x \plus{} y)(2y \plus{} 2z \plus{} 2x)} \plus{} \frac {(x \plus{} y)^2}{(y \plus{} z)(2y \plus{} 2z \plus{} 2x)} \minus{} 1,$
...

prove that:
$ a_1\geq \sum_{i \equal{} 2}^{17}{a_i}$

BQ

BQ

Let $ x,y,z > 0,a_1 \equal{} f_1(x,y,z) \equal{} \frac{x^2}{y(x\plus{}y\plus{}z)}\plus{}\frac{y^2}{z(x\plus{}y\plus{}z)}\plus{}\frac{z^2}{x(x\plus{}y\plus{}z)}\geq0,$
$ a_{i \plus{} 1} \equal{} f_{i \plus{} 1}(x,y,z) \equal{} f_i(y \plus{} z,z \plus{} x,x \plus{} y)(i \equal{} 1,2,...r,r\geq 2)$,

then:

$ a_1\equal{}\frac{x^2}{y(x\plus{}y\plus{}z)}\plus{}\frac{y^2}{z(x\plus{}y\plus{}z)}\plus{}\frac{z^2}{x(x\plus{}y\plus{}z)},$

$ a_2\equal{}\frac{(y\plus{}z)^2}{(z\plus{}x)(2y\plus{}2z\plus{}2x)}\plus{}\frac{(z\plus{}x)^2}{(x\plus{}y)(2y\plus{}2z\plus{}2x)}\plus{}\frac{(x\plus{}y)^2}{(y\plus{}z)(2y\plus{}2z\plus{}2x)},$

prove that

$ a_1\geq a_2a_3^2a_4^0a_5^3.$
BQ

xzlbq, mathematics isn't only calculations...

yes,I agree with your think,
But calculation allows us to see more,
There is no need that assume that they do not exist...

What do we see？

We first saw it，
Then put in the hands 。
Happy new year！


For a fixed integer $n\geq 1$ compute the min value of:

    $x_1 + \frac{x_2^2}{2} + \frac{x_3^3}{3} + ... + \frac{x_n^n}{n} $,

given that  $x_1, x_2,..., x_n $  are positive numbers satistying the condition:

     $ \frac{1}{x_1} + \frac{1}{x_2} +...+ \frac{1}{x_n} = n.$

Let $a=1+\frac12+\frac13+\cdots+\frac1n.$ By weighted AM-GM, we have

\[
\sum_{k=1}^n \frac{1}{ka}x_k^k
\ge
\prod_{k=1}^n \left(x_k^k\right)^{\frac{1}{ka}}
=
\left(\prod_{k=1}^n x_k\right)^{\frac1a}
\]
i.e. $(\text{big given sum})\ge a\cdot\left(x_1\cdots x_n\right)^{\frac1a}$. Then, by the given condition and AM-GM,

\[
n=\sum_{k=1}^n \frac{1}{x_k}\ge n\cdot\sqrt[n]{\prod_{k=1}^n \frac1{x_k}}
\]
i.e. $x_1\cdots x_n\ge1$. Therefore,

\[
(\text{big given sum})\ge a\cdot\left(x_1\cdots x_n\right)^{\frac1a} \ge a
\]
with equality only when $x_1=\cdots=x_n=1$.


quadratics
inequalities
inequalities unsolved

[quote="sunchips"]Given $ x_1x_2x_3..x_n \equal{} 1$

Prove:

$ (x_1 \plus{} .. \plus{} x_n)^2 \leq n \plus{} (n \minus{} 1)(x_1^2 \plus{} ... \plus{} x_n^2)$[/quote]

Siranyi's inequality make this problem very easily.

[quote="me@home"]
\[ (x_i)^2\leq n \plus{} (n \minus{} 1)(x_i^2)\leq (x_i)^2/n \plus{} (n \minus{} 1)(x_i^2)\iff (x_i)^2\geq n(x_i^2)
\]
where $ i$'s imply summing over[/quote]

should be $ (x_i)^2\leq n(x_i^2)$

[quote="me@home"]this is trivial:
\[ (x_i)^2\leq n \plus{} (n \minus{} 1)(x_i^2)\leq (x_i)^2/n \plus{} (n \minus{} 1)(x_i^2)\iff (x_i)^2\geq n(x_i^2)
\]
where $ i$'s imply summing over[/quote]

When you sum of all $ i$, you get $ n^2$ as the constant. Your solution doesn't make sense anyway... $ 1\le n$ is false. So is $ n\le x_i^2/n$...

For two variables, this is obvious. Then $ n\ge 3$. Let $ f(x) \equal{} 1 \plus{} (n \minus{} 1)x^2$. $ f$ is obviously differentiable on $ (0,\infty)$. Suppose that $ x_1 \plus{} ... \plus{} x_n \equal{} d$ which is constant. Note $ x_1...x_n \equal{} 1$ is constant. Let $ g(x) \equal{} f'(x^{\frac {1}{0 \minus{} 1}})$.

Observe that $ g(x) \equal{} ( \minus{} 2)(n \minus{} 1)x^{ \minus{} 3}$. $ g''(x) \equal{} ( \minus{} 2)( \minus{} 3)( \minus{} 4)(n \minus{} 1)x^{ \minus{} 5} < 0$. So $ g(x)$ is strictly concave on $ (0,\infty)$.

Thus $ f(x_1) \plus{} f(x_2) \plus{} ... \plus{} f(x_n)$ is minimal for $ 0 < x_1 \equal{} x_2 \equal{} ... \equal{} x_{n \minus{} 1}\le x_n$ by the $ n \minus{} 1$ equal variable theorem.

In other words, with $ a\le b$, $ a^{n \minus{} 1}b \equal{} 1$. So $ 1 \equal{} a^{n \minus{} 1}b\ge a^n\implies a\le 1$, and we must prove
\[ n \plus{} (n \minus{} 1)( (n \minus{} 1)a^2 \plus{} b^2)\ge ( (n \minus{} 1)a \plus{} b)^2
\]

\[ n \plus{} (n \minus{} 1)b^2\ge 2(n \minus{} 1)ab \plus{} b^2
\]

\[ (n \minus{} 2)a^{ \minus{} 2(n \minus{} 1)} \plus{} n \minus{} 2(n \minus{} 1)a^{ \minus{} (n \minus{} 2)}\ge 0
\]

\[ na^{2(n \minus{} 1)} \minus{} 2(n \minus{} 1)a^{n} \plus{} (n \minus{} 2)\ge 0
\]
which can be proven easily enough (I only see how to do it with a bit of calculus). Equality occurs when $ a \equal{} b \equal{} 1$, i.e. $ x_1 \equal{} ... \equal{} x_n \equal{} 1$.


$ ABCD$ is a rectangle. $ F\in [BD]$ and  $ E\in [AB]$
$ |AE|\equal{}2$
$ |EF|\equal{}4$
$ |DF|\equal{}4$

$ m\widehat{(EFB)}\equal{}?$

[hide=" click to reveal hidden content"]

Not sure but I think it might be $ 30^{\circ}$.

Let $ A$ be the origin and $ E$ be $ (2, 0)$. Draw the circle, centre at $ (2, 0)$ and radius $ 4$, that is $ (x\minus{}2)^{2} \plus{} y^{2} \equal{} 16$, cutting the $ y\minus{}$axis at $ (0, \pm 2\sqrt{3})$. $ EF$ is a radius of this circle.

Without loss of generality (I think), let $ F$ be $ (2, 4)$.

Then $ D$ is where the circle, centre at $ F$ and radius $ 4$ cuts the $ y$-axis. This circle is $ (x\minus{}2)^{2} \plus{} (y\minus{}4)^{2} \equal{} 16$, so $ D$ is $ (0, 4 \plus{} \sqrt{3})$.

Draw the perpendicular from $ F$ to the $ y$-axis, meeting it at $ G$. Produce $ DF$ to meet the $ x\minus{}$axis at $ B$.

Triangles $ DGF$ and $ FEB$ are similar and so angle $ EFB \equal{}$ angle $ GDF \equal{} \sin^{\minus{}1}(\frac{2}{4}) \equal{} 30^{\circ}$.

[/hide]

[quote="AndrewTom]
Without loss of generality (I think), let $ F$ be $ (2, 4)$.

Then $ D$ is where the circle, centre at $ F$ and radius $ 4$ cuts the $ y$-axis. This circle is $ (x \minus{} 2)^{2} \plus{} (y \minus{} 4)^{2} \equal{} 16$, so $ D$ is $ (0, 4 \plus{} \sqrt {3})$.

[/hide][/quote]

Then $ DF^2\equal{} 2^2\plus{}3$ which means DF is not 4, a contradiction to the given.

This is an interesting problem though. I would love to see a solution...


geometry
inequalities
Euler
trigonometry
circumcircle
incenter
cyclic quadrilateral

Hi Everyone,

This is an attempt to make a nice [b][color=red]Geometry[/color][/b] marathon at the Pre-Olympiad level.

If you post a problem, please don't forget to indicate its [b][color=red]number[/color][/b] and if you write a solution please indicate to which problem it is and also quote the original problem.Please post a new problem after the pending problem is solved.

Please show a [b][color=red]complete solution[/color][/b] and try to include figures wherever needed.

Lastly, I was inspired by the ongoing inequalities marathon.I hope it will also be a success like its counterpart.

Btw , its advisable to install Java as it is needed to view and edit geogebra applets.Here's the link [url]http://java.sun.com/javase/downloads/index.jsp[/url]

Now to the first problem:

[b]Problem 1:[/b]
In a triangle $ \Delta ABC$, the sides $ AB$ and $ AC$ are tangent to a circle with diameter along $ BC$ at the points $ Q$ and $ P$ respectively.Let $ E$ and $ F$ be the extremities of the diameter along $ BC$. $ EP$ and $ FQ$ intersect at $ K$.Prove that $ K$ lies on the altitude from $ A$ to $ BC$ of $ \Delta ABC$.

[b]Problem 2:[/b]
$ ABCD$ is a cyclic quadrilateral and there exists a circle centered on the side $ AB$ that is tangent to $ BC,CD,DA$. Prove that $ AB \equal{} AD \plus{} BC$.

[hide="Solution to Problem 2"]
[b]Problem 2:[/b]
$ ABCD$ is a cyclic quadrilateral and there exists a circle centered on the side $AB$ that is tangent to $ BC,CD,DA$. Prove that $ AB \equal{} AD \plus{} BC$.

Let's euler the problem .

Let the center of the inscribed circle be $ O$ on $ AB$ and its radius be $ r$ .Let the points of tangency of $ BC, CD ,DA$ be $ P,Q,R$ respectively.
We have $ OA\equal{}\frac{r}{\sin A}$; $ OB\equal{}\frac{r}{\sin B}$
Again making use of the property that $ ABCD$ is cyclic we get ,
$ AD\equal{}r\left(\frac{1}{\tan A}\plus{}\tan \frac{B}{2}\right)$;

$ BC\equal{}r\left(\frac{1}{\tan B}\plus{}\tan \frac{A}{2}\right)$
We use the following identity: $ \frac{1}{\sin A}\equal{}\frac{1\plus{}\tan^2 \frac{A}{2}}{2\tan \frac{A}{2}}\equal{}\frac{1\minus{}\tan^2 \frac{A}{2}\plus{}2\tan^2 \frac{A}{2}}{2\tan \frac{A}{2}}\equal{}\frac{1}{\tan A}\plus{}\tan \frac{A}{2}$

Thus $ \frac{r}{\sin A}\plus{}\frac{r}{\sin B}\equal{}r\left(\frac{1}{\tan A}\plus{}\tan \frac{A}{2}\plus{}\frac{1}{\tan B}\plus{}\tan \frac{B}{2}\right)$

And, $ AB\equal{}AD\plus{}BC$[/hide]

[b]Problem 3:[/b]
A cyclic quadrilateral $ ABCD$ is given. The lines $ AD$ and $ BC$ intersect at $ E$, with $ C$ between $ B$ and $ E$; the diagonals $ AC$ and $ BD$ intersect at $ F$. Let $ M$ be the midpoint of the side $ CD$, and let $ N \ne M$ be a point on the circumcircle of the triangle $ \Delta ABM$ such that $ AN/BN \equal{} AM/BM$. Prove that the points $ E, F$ and $ N$ are collinear.

[b]Problem 4:[/b]
Let $ (O)$ be a circumference, $ P$ an internal point and $ AA'$  a chord passing through $ P$. Show that the circles that go through $ P$ tangent to $ (O)$ at $ A,A',$ respectively and the circle with diameter $ OP$ are coaxal (they concur at another point  P).

[b]Solution to Problem 4[/b]

Let circles $ (O_1)$ and $ (O_2)$ be tangent to the circle $ (O)$ at $ A$ and $ A'$ respectively.
Let $ (O_1)$and $ (O_2)$ meet at $ Q$ other than $ P$.
We then have, $ PO_1OO_2$ is a parallelogram.
So,  $ OP$ is bisected by $ O_1O_2$ at $ O'$(say).
Again $ O_1O_2$ bisects the radical axis $ PQ$ of $ (O_1)$and $ (O_2)$ at right angles at $ R$(say).So, $ O'R$ $ \parallel{}$ $ OQ$ .
And thus, $ \angle OQP \equal{} \frac {\pi}{2}$.
Therefore, the circle through $ OP$ as diameter passes through $ Q$.

[b]Problem 5:[/b]
Let $ I$ be the incenter of a given $ \Delta ABC$ and let $ D,E,F$ where the incircle touches the sides $ BC,CA,AB$ of $ \Delta ABC$ . Now let $ X,Y,Z$ be three points on the lines $ ID,IE,IF$ such that directed segments $ IX,IY,IZ$ are congruent. Prove that the lines $ AX,BY,CZ$ are concurrent.

PS. I waited for about 3 days, but no one seems interested in the Geometry Marathon :(


I remember reading somewhere something about the number of primes up to a certain n being bounded the powers of 2. It was something like there are <= n primes between 2^(n-1) and 2^n (but of course that isn't true). Can someone remind me what the theorem actually was?

There are so many results about prime numbers.

Some theorems related to powers of two :

. p(n) <= 2^(2^(n-1))         (p(n) is the nth prime number)

. (prod (p <= n, p prime) p) <= 4^n

. Bertrand's theorem :

for each n there is a prime number between n and 2*n

That's true, but I don't think the first one is very useful, since the bound is ridiculously large. However, Erdos did use the second one in his proof of Bertrand's Postulate.


[quote="Peter"]Show that there are infinitely many composite numbers $n$ such that $3^{n-1}-2^{n-1}$ is divisible by $n$.[/quote]

 We prove that for any positive integer $m$ the number $n=3^{2^{m}}-2^{2^{m}}$ satisfies the problems condition. Indeed, we have that if $a|b$ then $3^{a}-2^{a}|3^{b}-2^{b}$ so to prove that $n|3^{n-1}-2^{n-1}$ it is sufficent to prove that $2^{m}|n-1$,that is $2^{m}|(3^{2^{m}}-1)-2^{2^{m}}$. Since $2^{m}\geq m$, then $2^{m}|2^{2^{m}}$ and we only need to verify that $2^{m}|3^{2^{m}}-1$. It is easy to see that \[3^{2^{m}}-1=(3-1)(3+1)(3^{2}+1)...(3^{2^{m-1}}+1)\] which shows that $3^{2^{m}}-1$ is a product of $8=(3-1)(3+1)$ and $m-1$ even numbers and thus is divisible by $2^{m}$(actually we have proved that it is divisible by $2^{m+2}$).

I will post bellow a solution for a generalization of this problem.

Let $a$ and $b$ be positive integers such that $a>b$ and they are coprime.
Show that there are infinitely many composite numbers $n$ such that
 $a^{n-1}-b^{n-1}$ is divisible by $n$.

Solution:
We take an arbitrary prime number $p$ such that $a^2-b^2$ isn't divisible by $p$.
We will show that the number $n = \frac{a^{2p}-b^{2p}}{a^2-b^2}$ satisfies the problem's condition.

$n = \frac{a^{2p}-b^{2p}}{a^2-b^2} = \frac{a^{p}-b^{p}}{a-b}\cdot\frac{a^{p}+b^{p}}{a+b}\newline
= (a^{p-1}+a^{p-2}b+\cdots+b^{p-1})(a^{p-1}-a^{p-2}b+\cdots+b^{p-1})$.
Obviously,we have
$a^{p-1}+a^{p-2}b+\cdots+b^{p-1}>a^{p-1}-a^{p-2}b+\cdots+b^{p-1}>1$.
Therefore $n$ is a composite number.

$(a^2-b^2)n \equiv a^{2p}-b^{2p} \equiv a^2-b^2 \pmod{p}$.
Since $a^2-b^2$ and $p$ are coprime, we have $n \equiv 1 \pmod{p}$.
I claim that $n$ is an odd number. Otherwise, $n$ is an even number.
The equation $(a^2-b^2)n = a^{2p}-b^{2p}$ implies that $a^{2p}-b^{2p}$ is an even number.
So we have that $a$ and $b$ have same parity.
Since they are coprime, they are odd numbers.
But we have $n \equiv a^{2p-2}+a^{2p-4}b^2+\cdots+b^{2p} \equiv 1+1+\cdots+1 \equiv p \equiv 1 \pmod{2}$.
This is a contradiction.

From the above, $n-1$ is divisible by $2p$.
Or there is a positive integer $k$ such that $n-1=2pk$.
The equation $(a^2-b^2)n = a^{2p}-b^{2p}$ implies that $a^{2p} \equiv b^{2p} \pmod{n}$.
Therefore, we have $a^{n-1} \equiv a^{2pk} \equiv b^{2pk} \equiv b^{n-1} \pmod{n}$.

So,we have done.


Now that mathcounts is all over, when are you moving to the high school forum (if you don't regularly go there already)? It looks pretty boring and lifeless over there... no fun.

Just about as soon as I'm in high school for me.

[quote="leoxnlin"]It looks pretty boring and lifeless over there... no fun.[/quote]

You mean it isn't as spammy as this forum.  :wink:  (Oops was that outloud?!)

Haha, I wouldn't stick to the MATHCOUNTS forum for too long though, the curious nature of HSB is always acalling  :P .

For those of you who are now past MATHCOUNTS, the HS Intermediate and Pre-Olympiad forums are the next logical step.  High School Basics is suppose to basically be the high school equivalent of Classroom Math - much of which you probably already know.  Granted, like classroom math, there are many off-topic posts that are really more like high school contest problems (we kind of allow the basic ones but would prefer they were better classified), so you might find some interesting stuff there.

Those are contest forums, remember, so you might not get some of the material right away. HSB/Intermediate/Pre-Olympiad (according to levans) contains general topics you can ask.

There's only one AMC forum. It doesn't have many problems. Actually, none apart from the problems that show up on the real contests.

none of the forums are as fun as the mathcounts forum...except GFF...

and loosen up people, nothing better than spam thread with more spam thread criticizing the original spam threads and other spam threads talkign about how you should stop posting spam threads so the topic will die off to make this forum one of the best ever, if not the best.

and just don't go to round table, they can turn any topic into a 5 page long essay proving their point over and over again.

I don't really get about AMC stuff...can you guys tell me?

(Is my avatar irritating btw?)

Well if you're bored in the High School forums, you can always host mock contets for the newbies in the MATHCOUNTS forum.

[quote="nyanga-nyanga"]I don't really get about AMC stuff...can you guys tell me?[/quote]

Do you not get the AMC material or the contest itself? If you're having difficulties with the material, I suggest you familiarize yourself with regular high school mathematics, and then tackle AMC. Or pick up a copy of AoPS vol 2.

If you don't get the contest, here's a nice [url=http://www.artofproblemsolving.com/Wiki/index.php/AMC]link[/url].


A function is defined:
$ f(1)\equal{}1$
$ f(2n) \equal{} f(n)$
$ f(2n\plus{}1) \equal{} f(2n)\plus{}1$

Find the maximum value that the function can attain within $ 1 \leq n \leq 2009$.

Find for how many values of $ n$ this maximum is achieved.

[quote="Identity"]A function is defined:
$ f(1) \equal{} 1$
$ f(2n) \equal{} f(n)$
$ f(2n \plus{} 1) \equal{} f(2n) \plus{} 1$

Find the maximum value that the function can attain within $ 1 \leq n \leq 2009$.

Find for how many values of $ n$ this maximum is achieved.[/quote]

It's easy to see that $ f(n) \equal{}$ the sum of digits in the binary expression of $ n$.

Then $ 2009 \equal{} 11111011001_2$, so it's easy to see that the max. of $ f(n)$ in the range is $ 10$

Number of values of $ n \equal{}5$


linear algebra
matrix
quadratics
combinatorics unsolved
combinatorics

let E={1,2...,n} ,and let A_1,A_2....,A_n ,be a family of subsets of E such that:
for all i#j ,A_i#Aj and the cardinale of there intersection is 1,
show that the union of all Ai is equale to E

If one of the $ A_i$'s is a singleton, say $ A_1 \equal{} \{1 \}$, then $ 1$ belongs to each other $ A_i$.
Let $ B_i \equal{} A_i \minus{} \{1 \}$, for $ i\equal{}2, \cdots, n$.
Since $ A_i \neq A_1$, it follows that $ B_i \neq \emptyset$. Moreover the $ B_i$'s are pairwise disjoint.
Thus, for each each $ i$, there exists $ b_i \in B_i$, and the $ b_i$'s are pairwise distinct. This gives $ n\minus{}1$ distinct element from $ \{2, \cdots, n \}$, which forces $ \{ b_2, \cdots, b_n \} \equal{} \{2, \cdots, n \}$, and we are done in that case.

Now, assume that $ |A_i| \geq 2$ for each $ i$.
Let $ M \equal{} (m_{i,j})$ be the $ n \times n$ matrix, with $ m_{i,j} \equal{} 1$ if $ x_i \in A_j$ and $ m_{i,j} \equal{} 0$ otherwise.
Let $ U \equal{} tM \cdot M \equal{} (u_{i,j})$, where $ tM$ is the transpose of $ M$.
Then $ u_{i,j} \equal{} |A_i \cap A_j |1$ so that for $ u_{i,j} \equal{} 1$ for $ i \neq j$, and $ u_{i,i} \equal{} |A_i| \equal{} a_i \geq 2$.

It follows that $ U$ is the matrix of the bilinear form associated to the quadratic form $ f(x_1, \cdots , x_n) \equal{} (x_1 \plus{} \cdots \plus{} x_n )^2 \plus{} \sum_{i\equal{}1}^n (a_i \minus{}1)x_i^2$. Clearly, $ f$ is definite and positive so it is non-degenerated.
It follows that $ U$ is inversible, and so is $ M$.
In particular, there is no row of $ 0$ in $ M$, which ensures that each element of $ E$ belongs to at least one of the $ A_i$'s.

Pierre.


search function AMC AIME USA(J)MO USAMO AwesomeMath

I'm in tenth grade and AMC 10B was my first real AMC competition.  I was able to get 114 points, but this was  not enough to go on to the other rounds. This was the first year I really discovered my hidden passion for math ( came around October 2008) and I worked really hard to get to my score. I read both AoPS v1 and am working on v2. I don't want to stop though. I want to bring my game to the next level, and if anyone can help me, I know it would be you guys.  Any suggestions? :D

First of all, congratulations on your score. It takes a lot of effort to get such a score with such a short amount of time to work with. Second of all:
Step 1. Do problems
Step 2. Do more problems
Step 3. Do even more problems
But honestly, this is what you need to do. What is more important than that is that you must understand the solutions. Read and comprehend the solutions as if you were going to dictate it to a class tomorrow. If you want to raise your AMC score then I would suggest going over Volume 1 again, especially geometry. When you can consistently make your target score (I would suggest at least 130) on practice AMC 10's then move onto volume 2. Volume 2 will help you prepare for AMC 12 more as well as AIME. Additionally, if you list your location, we could give the names of a few local competitions. Getting into the competitive atmosphere many times helps with pressure and time management.

Be organized on the test. I'm normally fairly sloppy, but on this test I followed the recommendations on this site and made a numbered box for each question, and circled the answers. Because I was able to access all my previous work quickly and easily on questions I came back to, I was able to answer a few more questions. This was very important, because I got exactly 120.

Thank you for all the responses!
Well one problem I found when covering the AoPS books is that, although they have very nice potpourri sections, the problems are grouped together too well.
For example, if I read say, a chapter on number theory (my favorite), I know what the following problems will entail, in fact, EXACTLY what they will entail. I know what techniques are to be used before I read the problem.  I need AMCs. I went through most of the problems on this site ( contest section ) but I need like a general book that will focus on problem solving more than teaching.
Don't get me wrong, AoPS v1 and v2 are the best math texbooks I've ever had by a longshot, but if I really need to focus on more problems, I need something new. Can someone recommend a book to me? This would be greatly appreciated. I've saved enough money for a book less than or equal to 100USD ( drain my piggy bank lol ) but if its a little over my parents will probably help me there.

There's always the contests section, up on the top of your screen. Click it, then click on USA (at the bottom right), then click one of the several contests (AMC 10 for you I suppose), and you'll see several years you can practice on.
Also, you might be able to find some mock contests created by other AoPS users (Warning: Usually significantly harder than actual exam). Try the search function for that.

It sounds like you really enjoy math, which is what you need more than anything else.  I would continue to split your time between solving AMC problems and working through the AoPS books.  Always push yourself to solve harder and harder problems - maybe even spend some time trying to solve AIME problems!

You will become better at problem solving in general by solving problems yourself and by carefully reading solutions to problems you couldn't solve (this means figuring out how you might have found the solution, not just understanding the steps).  The only book I know of that addresses problem solving in general is Art and Craft of Problem Solving, but this book seems to be geared towards people making the transition from AIME to USAMO level, and I don't like it much anyway.

Yeah I do really enjoy math, it just took me some time to figure it out! Do you think that the book [i]First Steps for Math Olympians[/i] will do me good? I am about to order it. Or should I go with The Art and Craft of Problem  Solving anways, matt? I really want to make the USAMO, and I've got 2 years to do it. I really think I can if I keep working at the rate I've been going at.
Thanks in advance!
Oh yes, and crazypianist, I am from Louisiana.

I don't really know anything about First Steps for Math Olympians.  It may be good, but it seems like just a compilation of AMC problems (which you can find in the Contests section of this site) along with some instruction (which you can find in AoPS Volume 1 and 2, which you already have).  Unless you are not satisfied with AoPS Volume 1 and 2, I wouldn't rush to buy another book.

I didn't qualify for the AIME one year, and then I qualified for the USAMO the next year, and that was back when only 250 people qualified for the USAMO, so it's definitely possible.  The only resources I used during that time were AoPS Volume 1 and 2, problems people posted on AoPS, and lots of AIME problems.  If you get to the point where you are trying to transition into solving Olympiad (USAMO) problems, I would highly recommend Problem Solving Strategies instead of Art and Craft of Problem Solving, but other people might disagree.  I also may just be saying this because I think I got Art and Craft of Problem Solving too late for it to be useful to me.  If you want more opinions on this, search the forums - there have been many topics like this in the past.

The Art and Craft of Problem Solving (ACoPS) was the book that helped me transition from short answer to olympiad-style problems. I highly, highly recommend it; it's got pretty much everything you need to succeed at a basic level on USAMO, etc. It's well-organized and well-written, too, in my opinion. I really liked it.

I wouldn't recommend doing PSS immediately. It's mainly a collection of problems, with little instruction, which isn't too useful if you are new to the olympiad style.

Thanks guys, I'll very satisfied with AoPS v1 and v2, so I won't get another book just yet. I might move into AaCoPS next, but only later.

Do you think that going to AwesomeMath summer program would be a good idea for a student like me? I definitely like the idea of an intensive maths program geared towards problem solving. I know this question may seem pretty personal but any opinions are helpful. Once again, thanks. This forum is so great and its people like you who make it so!

I've also heard that AwesomeMath is helpful, and there is a lot more information on that in the [url=http://www.artofproblemsolving.com/Forum/index.php?f=442]AwesomeMath section[/url].

Well for me its a question of getting in or not. I was looking at the test A and I was able to solve quite a few problems. I am also very passionate about math. Do you guys think I could get in?
The thing is, at my school, I don't think there are 2 mathematics teachers to give me letters of recommendation.

I am also applying this year, and I'm pretty sure you don't need recommendations from two math teachers.


A is a subset of the real line and m(A)=0(lebesgue measure) then:

             m({x^2 : x in A})=0

suppose that $A$ is included in $]-n;+n[$.
Then, for any $e>0$, one can find intervals $]a_i;a_i+L_i[$ such that:
*$L_i<1$
*$|a_i|<n$
*$\sum L_i<e$
*$A$ included in $\cup_{\{\mathbb{N}\}}]a_i;a_i+L_i[$

But then $\{x^2|x \in A\}$ is included in $\cup_{\{\mathbb{N}\}}]a_i^2;a_i^2+L_i^2+2a_iL_i[=\cup_{\{\mathbb{N}\}}]a_i^2;a_i^2+D_i[$ where $|D_i|=L_i^2+2a_iL_i<L_i^2+2nL_i$.
Conclusion will follows.

Why can you assume that $A$ is bounded? :?  $\mathbb{N}$ has measure $0$ and is not bounded. I'm wrong?  :?  :?

I don't assume $A$ bounded. I just want to show that $m(]-n;+n[ \cap A)=0$ for any $n \in \mathbb{N}$  :)


x, y and z are real positive (one of them  is different from 0) such that xyz=a (a is a reel ).
 find all the values of a  checking the following inequality:
 1/(x+root (y+z)) + 1/(y+root (z+x)) + 1/(z+root (x+y)) =< 3/4

Um what do you mean by different from 0?


quadratics
inequalities
algebra
polynomial
number theory
number theory unsolved

Let p be an odd prime. Show that there are at most (p+1)/4 consecutive quadratic residues mod p. For which p is this bound attained ?

>3, or else 0 and 1 are 2>(3+1)/4 consecutive quadratic residues.

I don't see my problem :functionnal guest so I prove in this
      f(f(x)+y)=xf(xy+1) => f(2f(1))=f(1+f(1))
    =>f(1+y)=f(f(1)+y)=>f(1)=1
 f(f(1/k)+k)=1/kf(f(2))
 f(f(k)+1/k)=kf(2)
    we prove f(x)=1/x
      is this prove true ! :?  :D

Maybe we have to show that there are at most (p+1)/4 non-zero quadratic residues, because otherwise 7 doesn't work either because we have 0, 1, 2 as quadratic residues, and that's 3>2=(7+1)/4.

I think the case p=M4+1 is a bit easier, considering that we can't even have equality in this case. Assume the block of consecutive residues doesn't "cross" the pair (p-1)/2, (p+1)/2. If x is a quadratic residue then so is -x, so we have at least (p+1)/2>(p-1)/2 non-zero quadratic residues, which is obviously absurd. This means that the block contains both (p-1)/2 and (p+1)/2, so 2 is also a quadratic residue, so if x is a quadratic residue then so is 2x. Now take M to be a set of >(p+1)/4 quadratic residues centered at (p-1)/2, (p+1)/2. It's easy to see why M and 2M are disjoint, so we have at least |M|+|2M|=(p+1)/2>(p-1)/2 non-zero quadratic residues, which is again absurd, so we're done with the case p=M4+1.

maybe he is not including the zero residue, because then the inequality will hold.

I'm not quite sure I understand what you mean..

Anyway, any ideas for the case p=M4+3? Maybe the fact that the quadratic residues are the roots of the polynomial x<sup>(p-1)/2</sup>-1 helps, but I don't exactly know how..

I think that the original question may have been to show that there are at most (p+1)/4 consecutive quadratic residues in {1, 2, 3, ..., p-1} which can be proven by looking at pairs of consecutive quadratic residues.

Yesterday i was looking the book "A CLASSICAL INTRODUCTION TO MODERN NUMBER THEORY"  and i found the following exercises:

1. Let RR denote the number of consecutive quadratic residues modulo p, RN th number of pairs of consecutive numbers for which n is a quadratic residue and n+1 is not. and NR the number of pairs for which n is a non-quadratic residue, but n+1 is a quadratic residue. and NN the number of consecutive non-quadratic residues, everything modulo p (obviously, pairs in the set 1,2,3...,p-1). then the book asks to find a formula for:
RR+RN, NR+NN, RR+NR, RN+NN.

2.prove rthat RR+NN-RN-NR= \sum n(n+1)/p with n running from 1 to p-1, and prove that this sum is equal to -1.

3. Use the results to show that RR=1/4(p-4-a) where a=(-1)^[p-1]/2

I think the result is clear from this exercises

Pascual

Oh, this means that when p=M4+3 we have equality. I only thought about the first task and only for primes of the form 4k+3 (the interesting case; we use an identical line of reasoning for p=4k+1).

It's easy to see that RR+RN=(p-1)/2= the number of QR, because after any QR we either have a QR or a NQR and p-1 is a NQR.. The same type of reasoning shows us that RR+NR=NN+NR=(p-3)/2 and NN+RN=(p-1)/2. Remember that this is for the case p=M4+3.

As for the second task, RR+NN-RN-NR=\sum{i=1->p-2}(i/p)((i+1)/p)=\sum{i=1->p-2}(i(i+1))<sup>(p-1)/2</sup> (mod p). Maybe this helps. I didn't continue because I had no more obvious ideas.

I guess the second task wasn't that hard either:

The fact that S=RR+NN-RN-NR=\sum{i=1->p-1}(i(i+1)/p) is obvious and needs no proof (assume (p/p)=0; (a/p) is, as usual, the Legendre symbol). Then we use Euler's Criterion: (a/p)=a(p-1)/2 (p) (x (p) means x (mod p)).

We thus have S=\sum{k=0->(p-1)/2}([(p-1)/2]Ck*\sum{i=1->p-1}i(p-1)/2+k) (p). Since \sum{i=1->p-1}it=0 (p) for all t from 1 to p-2 and -1 (p) for t=p-1, we quickly get S=-1 (p) and we can see that it can't possibly be >=p-1 or <=-p-1 (we use some inequalities: if S>=p-1 then NN+RR>p-1 so one of them must be >(p-1)/2 but RR+RN=(p-1)/2 so RR<=(p-1)/2 and NN+RN=(p-1)/2 so NN<=(p-1)/2, so we have a contradiction; the same goes for S<=-p-1). This means that the only possibility is S=-1 and we're done.


[quote="vishalarul"][hide]Use your Newton's sums:
The sum of the roots is -4:
s+4=0
s=-4
The sum of the square of the roots is 0
s-16+8(2)=0
s=0[/hide][/quote]
correct

A slightly more elaborate method is:
Let $f(x) = x^{2}+4x+8$.
The polynomial whose roots are the squares of the roots of $f(x)$ is
$g(x) = f(\sqrt x)*f(-\sqrt x) = (x+4 \sqrt x+8)(x-4 \sqrt x+8) = (x+8)^{2}-16x = x^{2}+8x+64-8x = x^{2}+64$
Now, by Vieta's Sums, the sum of the roots of $g(x)$ is 0. Thus, the sum of the squares of the roots of $f(x)$ is 0.

[quote="bpms"]Find the sum of the squares of the roots of $x^{2}+4x+8$.[/quote]

[hide]Let the roots be a and b.

Note that a^2+b^2=(a+b)^2-2ab.

So the sum of the squares since a+b=-4 and ab=8 is 0.

$0$[/hide]


Well ,guys This problem is trivial  but one of my friends needs different solutions for this,so please post your solutions .Thanks

[b]Problem[/b]:two circles are externally tangent at $A$. common external tangent touches the circles at $B$ and $C$.$BA$ is extended to intersect the other circle at $D$.prove that $CD$ is a diameter.

Let the circle with C on it have center O, and the circle with B on it have center P. Draw OP and extend CA through to intercent the other circle at E.

$\angle CAO = \angle PAE = \beta$
$\angle DAO = \angle BAP = \alpha$

Since BPA, APE, CAO, and DOA are all isosceles,

$\angle ABP = \angle ADO = \alpha$
$\angle PEA = \angle ACO = \beta$

Also, BAE is a triangle, so

$\angle BAE + \angle EBA + \angle AEB = 2\alpha + 2\beta = 180^{\circ} \Rightarrow \alpha + \beta = 90^{\circ}$

Since $\angle CAD = \alpha + \beta$, it equals $90^{\circ}$ and CD is a diameter.

Prove that $\triangle{ABC}$ is a right triangle.
Construct the common tangent line throught $A$ and result follows.


Why does USNCO restrict participation for the national exam to only CITIZENS??!?!?!

Math doesn't do that. Physics doesn't do that. I think Biology doesn't do that. WHY CHEMISTRY??? I mean even Canadians can take the USAMO, no? Ack, I feel left out...

At least all that work wasn't all for nothing, because I don't have to study as much for the AP and SAT II tests.

I think the camp is on an US airforce base so NON-US citizens are not allowed.

But why ban noncitizens from the National Exam, which is taken on campuses across the nation?

Chinaboy, I agree with your sentiment.  I have heard that the reason for this is about the Airforce Academy, and I have also heard that it is because the ACS wishes to have American citizens represent the US in the IChO.  Still, I feel that the should at least let non-citizens TAKE the national exam.  One of the purposes of the competition is to foster interest in chemistry for everyone, not just people who happen to have an American passport.

On the other hand, it is possible that money is an issue--I don't know the details about who pays for the national exam (I know I've never had to pay for it)--but it is likely harder to justify spending money on setting up and grading a test for a person who will not be able to attend IChO.  That being said, I still think that you have a very good point.


Show that if $A=(a_{ij})$ is a matric such that $\sum_{j}a_{ij}=1$ for all $j$, then $1$ is an eigenvalue of $A$.


A) Let a, b, c be 3 angles such that a + b + c = pi. If we are given that tan(a)tan(b) = csc($ \pi$/3), determine
$ \frac{\cos(a)\cos(b)}{\cos(c)}$

[hide="Solution"]
\[ \minus{}\dfrac{\sin(c)}{\cos(c)}\equal{}\minus{}\tan(c)\equal{}\tan(\pi\minus{}c)\equal{}\tan(a\plus{}b)\equal{}\dfrac{\tan(a)\plus{}\tan(b)}{1\minus{}\tan(a)\tan(b)}\equal{}\]
\[ \dfrac{1}{\cos(a)\cos(b)}\dfrac{\sin(a)\cos(b)\plus{}\sin(b)\cos(a)}{1\minus{}\csc(\frac{\pi}{3})}\equal{}\dfrac{\sqrt{3}}{\cos(a)\cos(b)}\dfrac{\sin(a\plus{}b)}{\sqrt{3}\minus{}2}\equal{}\]
\[ \dfrac{\sin(c)}{\cos(a)\cos(b)}\dfrac{\sqrt{3}}{\sqrt{3}\minus{}2}\]

Then

\[ \dfrac{\cos(a)\cos(b)}{\cos(c)}\equal{}\dfrac{\sqrt{3}}{2\minus{}\sqrt{3}}\]

[/hide]

:)

[quote="ElChapin"][hide="Solution"]
\[ \minus{} \dfrac{\sin(c)}{\cos(c)} \equal{} \minus{} \tan(c) \equal{} \tan(\pi \minus{} c) \equal{} \tan(a \plus{} b) \equal{} \dfrac{\tan(a) \plus{} \tan(b)}{1 \minus{} \tan(a)\tan(b)} \equal{}
\]

\[ \dfrac{1}{\cos(a)\cos(b)}\dfrac{\sin(a)\cos(b) \plus{} \sin(b)\cos(a)}{1 \minus{} \csc(\frac {\pi}{3})} \equal{} \dfrac{\sqrt {3}}{\cos(a)\cos(b)}\dfrac{\sin(a \plus{} b)}{\sqrt {3} \minus{} 2} \equal{}
\]

\[ \dfrac{\sin(c)}{\cos(a)\cos(b)}\dfrac{\sqrt {3}}{\sqrt {3} \minus{} 2}
\]
Then
\[ \dfrac{\cos(a)\cos(b)}{\cos(c)} \equal{} \dfrac{\sqrt {3}}{2 \minus{} \sqrt {3}}
\]
[/hide]

:)[/quote]

Could you explain this in further detail please? I am really lost right now.

Thank you very much for any help!

a+b+c = pi
this problem is really easy if u realized that
cos(c) = -cos(a+b)
(you can see this on a unit circle easly)

cos(a)cos(b)/cos(c) = cos(a)cos(b)/-cos(a+b)

then use the sum of the angle forumal on -cos(a+b)

cos(a)cos(b)/(sin(a)sin(b)-cos(a)cos(b)) = x

(sin(a)sin(b) - cos(a)cos(b)) / cos(a)cos(b) = 1/x

break the parts down

sin(a)sin(b)/(cos(a)cos(b)) - cos(a)cos(b)/(cos(a)cos(b)) = 1/x

tan(a)tan(b) - 1 = 1/x

now tan(a)tan(b) = = csc(pi/3) = 2/3^(1/2)   given

so (2-3^(1/2))/3^(1/2) = 1/x

x = 3^(1/2) / (2-3^(1/2)

sorry it  is really messy, i dont know how to use latex...

[quote="lgd0612"]$ a+b+c = \pi$
this problem is really easy if u realized that
$ \cos{c} = -cos(a+b)$
(you can see this on a unit circle easily)

$ \frac{\cos{a}\cos{b}}{\cos{c}} = -\frac{\cos{a}\cos{b}}{\cos(a+b)}$

then use the sum of the angle formula on $ -\cos(a+b)$

$ \frac{\cos{a}\cos{b}}{\sin{a}\sin{b}-\cos{a}\cos{b}} = x$

$ \frac{\sin{a}\sin{b}-\cos{a}\cos{b}}{\cos{a}\cos{b}}=\frac1x$

break the parts down

$ \frac{\sin{a}\sin{b}}{\cos{a}\cos{b}}-\frac{\cos{a}\cos{b}}{\cos{a}\cos{b}}=\frac1x$

$ \tan{a}\tan{b} - 1 = \frac1x$

now $ \tan{a}\tan{b} =\csc\left(\frac{\pi}3\right) = \frac2{\sqrt3}$   given

so $ \frac{2-\sqrt3}{\sqrt3}=\frac1x$

$ x = 3+2\sqrt3$

sorry it  is really messy, i dont know how to use latex...[/quote]

$ \text{\LaTeX}$ed and slightly edited...you can learn a lot of it just from this.

thank you so much, math154!

that really helped!

Don't thank me--it was lgd0612's solution.


group theory
abstract algebra
superior algebra
superior algebra solved

$L$ is a nilpotent left ideal of $R$, show that there exist a nilpotent ideal $\mathfrak I\supseteq L$ in $R$.

Let $\mathfrak I=L+LR$.
$LR$ is the subgroup generated by $\{lr|l\in L,\ r\in R\}$.
It is easy to check that $\mathfrak I$ is an ideal of $R$.

I guess we also have to prove that $\mathfrak I\neq R$.
Suppose $L+LR=R$.
Then $L+LR=L+L(L+LR)\subseteq L+L^2+L^2R=L+L^2(L+LR)\subseteq\ldots\subseteq L+L^nR\subseteq\ldots$.
Because $L$ is nilpotent and because of the previous line there exists $n\in\mathbb{N}$ such that $R=L+LR\subseteq L+L^nR=L$. This is a contradiction to $L\neq R$ (which should be in the definition of $L$).

oops...I'm sorry...I've modified the problem.


In a triangle $ ABC$ , if $ O\equal{}H\equal{}G$ (specific notations) it is true that $ ABC$ is equilateral ?? If yes ,can I use that in problems without proof ?

It is true. I would not use it without proof though... memorizing it and writing it down is always safer.

[hide="Idea of the Proof"]The medians, perpendiculars and perpendicular bisectors will all coincide, and if you use their distinctive properties it's fairly obvious that the triangle has to be equilateral.[/hide]


Hi everyone!!

My name is Makis and i am an international student. I attend the international baccalaureate(IB) in a greek school and this year i am finishing the first os the two(2) years. I am seriously thinking of applying in American universities for engineering courses, and i would like your opinion of whether i can get there. Here are the universities i am thinking of applying to:
-Columbia(1st choice)
-Cornell(2nd choice)
-Georgetown(3rd choice)

[u]Biographie in short:[/u]

In Greece the mathematical competitions are done in three(3) rounds(in every round a number of students is eliminated), and the third is the national mathematical olympiad. I have participated four(4) times and in the two(2) first i made it to the second round, and in the two(2) last i made it on the mathematical olympiad(3rd round, top 50 of my grade participated whereas on the first round 2000 from my grade participate). Also i have participated in the  mathematical competition kangaroo and got extra prizes due to my performance, and i have once been given a 800$ scholarship from my school due to the fact that i finished 1st in a mathematical competition that my school organized.

I have a Lower, an Advanced and a Profiecency in English and a B2 in German. I also have computer certificates from the university of cambridge exams.

I had excellent marks in maths, physics and chemistry at high school and i also have now in IB good grades. In IB students are grades in scale of 7, 1 being the worse grade and 7 being the best. Here are my grades in the subjects i attend:
-Maths, high level:6
-Physics, high level:7
-English as a foreign language, high level:6
-Chemistry, standard level:6
-Business and Management, standard level:6
-Modern Greek, standard level:5
-Total 36/42

I am also planning of taking the sat reasong test, and the sat subject tests in maths and physics(where i think i am capable of getting a 800 at both).

As an activity I had been playing tennis for 7 years but now i have stopped.

So what do you think can i make it to these universities?

Competition at top tier universities is extremely tough for American students, and even tougher for international students.  Unfortunately, I don't see anything that would cause you to stand out.  Nothing is spectacular.  Sorry.

International students always think that highlighting their grades and academic results is the most important part of their application. But there are plenty of domestic students with excellent grades. Colleges like international students because they add diversity to the campus. So you have to seem like an interesting person. They also are very careful to pick such international students that will "mesh" with their college environment. It may help to know what they're looking for:

Columbia looks for something "more" than an engineer. Someone with a perspective of what's going on in the world, a potential leader, perhaps a businessman, who wishes to be enabled/empowered with an engineering education. It is important to show well-roundedness, ambition, and initiative. They also like research experience. Finally, show some interest in New York City. Columbia considers it's location in New York City, and its relationship with the city, to be one of its defining characteristics.

Cornell will look to see that you have a balanced, amicable personality who can survive the demanding course-load at Cornell, and will not be held back by such things as the size of the school (and the resulting red-tape), the weather (don't seem very moody), and the fact that Ithaca is somewhat isolated (show interest in some of the clubs at Cornell). So, it helps to have recommendations that highlight your personality, your ambition (without seeming overly competitive), and your initiative. Research experience is helpful.

It is also helpful to show a wide range of interests, and to seem interested in the diversity of the students at Cornell. Cornell's motto is a statement by Ezra Cornell; "I would found an institution where any person can receive instruction in any study."

One thing that American colleges in general value that many international students often don't highlight is work experience. If you've ever been employed, put it down, and highlight the skills you've gained from it. Also put down any internships, any labwork, and any *significant* personal projects.

If you have any particular questions about the program at Cornell, PM me.

Thanks to both of you for your replies.

@JRav: OK, lets take it the other way around. In what univerities in New York City do you think i could be accepted?

@TZF: Hmm, what do you exactly mean by research experience? For example, during the summer i will do an extended essay(4000 words) on waves and motion of springs which have weight(in school we only examine weightless springs). Is this considered some kind of research experience.

Could you also name some other prestigious universities in New York?


vector
linear algebra
linear algebra unsolved

Let T: V -- V be a linear transformation, where V is a finite dimensional vector space. If dim(range(T))=dim(range(T2)) show that the range and null space of T only have the zero vector in common.

T2 is T square couldnot get the 2 as superscript!
T:V has an arrow pointing to V

Suppose $ v \in Im(T) \cap Ker(T)$, and let $ T(w)\equal{}v$, then $ T^2(w) \equal{} 0$, so $ w \in Ker (T^2) \equal{} Ker (T)$, so $ v \equal{} 0$

$ Ker(T^2) \equal{} Ker(T)$ because their dimensions are the same.


Hôm nay em nhận được 1 ps từ 1 người bạn nước ngoài có nói là số bài của bạn ấy giảm. Em nói không có gì lạ cả nhưng nhìn lại mình bị mất hơn 70 posts , các bài gần đây koo bị mất .Vô lý quá  :mad:

[quote="N.T.TUAN"]Anh cũng thế, mất gần 500 bài, để anh hỏi xem. :D[/quote]

Có thật là mất 500 koo  :D

Gần 300 post , vào đây này http://www.mathlinks.ro/Forum/viewtopic.php?t=172668 . Locked!


inequalities
calculus
AMC
three variable inequality

okay here is a solution:

sum((2ab+2ac)/(a^2+b^2+c^2+2bc))<=24/5 (Note that the sum is sym).
lets denote s=a^2+b^2+c^2 for easy calculus! :D
5s^2(sum(ab))+10s(sum(a^2bc))+20(sum(a^3b^2c))<=6s^3+6s^2(sum(ab))+12s(sum(a^2bc))+48a^2b^2c^2 (all sums are sym)
which after some simplifacations and calculus we get:
6s^3+s^2sum(ab)+2s(sum(a^2bc)+8sum(a^2b^2c^2)>=10s(sum(a^2bc))+20(sum(a^3b^2c). (sums are sym)
we now multiply out the powers of s=a^2+b^2+c^2 and get:
sym sum 3a^6+2a^5b-2ab^2+3abc+2a^3b^3-12a^3b^2c+4a^2b^2c^2>=0
now using Schur's inequality after multiplication with 4abc:
sym sum 4abc-8a^3b^2c+4a^2b^2c^2>=0  and this reduces our claim to

sym sum 3a^6+2a^5b-2ab^2-abc+2a^3b^3-4a^3b^2c>=0.

now we have a sum of four expressions which are positive by weighted AM-GM:

2sym sum(2a^6+b^6)/3-ab^2>=0
sym sum(4a^6+b^6+c^6)/6-abc>=0
2 sym sum (2a^3b^3+c^3a^3)/3-a^3b^2c>=0
2 sys sum (2a^5b+a^5c+ab^5+ac^5)/6-a^3b^2c>=0

equality is when a=b=c ( we see this from Shur's ineq and AM-GM)

cheers hope this is good! :D

I think I have an interesting solution for your problem . Here goes :
(all sums are cyclic , not symmetric)

sum (b+c-a)^2/[(b+c)^2 + a^2] >= 3/5  <=>
S = sum a(b+c)/[(b+c)^2 + a^2] <= 6/5  , after a few trivial computations

Now , (b+c)^2 + a^2 >= (a+2b+2c)^2/5 , by cauchy applied to the numbers (b+c)/2 , (b+c)/2 , (b+c)/2 , (b+c)/2 , a - we apply it in this manner in order to have equality .

This means that S <= sum 5a(b+c)/(a+2b+2c)^2 = S'

We prove that S' <= 6/5 . Assume that a+b+c=1 since the inequality is homogenous , and you end up with :

sum a(1-a)/(2-a)^2 <= 6/25

Consider the function f(x) = (x-x^2)/(2-x)^2 . If it is concave on (0,1) we are done . But this is true by a trivial computation (show that f''(x)<0 on (0,1)) . So we can apply Jensen's inequality and we are done .

Voila , no horrendous computations ! :)

well my solution is the so called "official solution" that was given at the contest! cheers!

Hey, see my soln for poland 1996 inequality :D

[Moderator edit: Also compare with the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=53672]USAMO 2003[/url] inequality.]

I also have a short solution. Put x=(b+c)/a, y=(c+a)/b and z=(a+b)/c. Apply Cauchy to find that the LHS is sum (x-1)^2/(x^2+1)>=(x+y+z-3)^2/(sum x^2+3). This becomes (x+y+z)^2-15(x+y+z)+18+3(xy+yz+zx)>=0.But xy+yz+zx>=2(x+y+z) (since this one reduces to (a+b-c)(b+c-a)(c+a-b)<=abc), so it is enough to prove that (sum x)^2-9sumx+18>=0 and it is true since sumx>=6.

[quote="alekk"]hey, how did you find 23/25 and -54/25??
(1-2a)2/(a2 + (1-a)2)>= (23-54a)/25[/quote]

I think my solution basically employs the same idea. First, say $a+b+c=3$ so that equality will hold at $a=b=c=1$. Essentially, one conjectures that

$\frac{(b+c-a)^2}{(b+c)^2+a^2} = \frac{(3-2a)^2}{(3-a)^2+a^2} \ge \frac{1}{5} + ka - k$

for some $k$. As the equation is parametrically contrived to yield equality where $a=1$, we need the $k$ for which there is a double root. That can be solved for by equating derivatives, etc. and you get $k = \frac{-18}{25}$. Factoring for this $k$, we see that the inequality is always true. Then add the three expressions to finish.

If you want prove  it, you should use [b]SOS inequality[/b]

we may assume that $ a+b+c=2$ , the inequality becomes
$ \sum\frac{(a-1)^{2}}{1+(a-1)^{2}}\geq \frac{3}{10}$
$ \leftrightarrow \sum \frac{1}{1+(a-1)^{2}}\leq \frac{27}{10}$
now we denote $ x=a-1;y=b-1;z=c-1$
the inequality becomes $ \sum \frac{1}{1+x^{2}}\leq \frac{27}{10}$ with $ x+y+z=1$ and $ x,y,z <1$
by the fact $ 27(1-3t)^{2}(4-3t) \geq 0$ we obtain $ \frac{1}{1+x^{2}}\leq \frac{27}{50}(2-x)$
now we easily get the desire result

[quote="Mildorf"][quote="alekk"]hey, how did you find 23/25 and -54/25??
(1-2a)2/(a2 + (1-a)2)>= (23-54a)/25[/quote]

I think my solution basically employs the same idea. First, say $a+b+c=3$ so that equality will hold at $a=b=c=1$. Essentially, one conjectures that

$\frac{(b+c-a)^{2}}{(b+c)^{2}+a^{2}}= \frac{(3-2a)^{2}}{(3-a)^{2}+a^{2}}\ge \frac{1}{5}+ka-k$

for some $k$. As the equation is parametrically contrived to yield equality where $a=1$, we need the $k$ for which there is a double root. That can be solved for by equating derivatives, etc. and you get $k = \frac{-18}{25}$. Factoring for this $k$, we see that the inequality is always true. Then add the three expressions to finish.[/quote]
I tried this one with isolated fudging of the form $\frac{a^{p}}{a^{p}+b^{p}+c^{p}}$ and the exponent is the same as your k... coincidence?

But isolated fudging doesn't work so that's why I requested a link to here :P

There is actually a short solution:
$ \frac{2a(b+c)}{(b+c)^{2}+a^{2}}=\frac{2a(b+c)}{\frac{3}{4}(b+c)^{2}+\frac{1}{4}(b+c)^{2}+a^{2}}\le \frac{2a(b+c)}{\frac{3}{4}(b+c)^{2}+(b+c)a}=\frac{2a}{\frac{3}{4}(b+c)+a}$
Now if we assume $ a \le b \le c$ then $ \frac{1}{\frac{3}{4}(b+c)+a}\ge \frac{1}{\frac{3}{4}(a+c)+b}\ge \frac{1}{\frac{3}{4}(b+a)+c}$
So by rearrangement, we deduce that this cyclic sum is $ \le 12/5$
The rest follows easily.

[quote=arqady][quote=hxtung]Prove that
$ \frac{\left(b+c-a\right)^{2}}{\left(b+c\right)^{2}+a^{2}}+\frac{\left(c+a-b\right)^{2}}{\left(c+a\right)^{2}+b^{2}}+\frac{\left(a+b-c\right)^{2}}{\left(a+b\right)^{2}+c^{2}}\geq\frac35$
for any positive real numbers $ a$, $ b$, $ c$.[/quote]
It's true for any reals $a$, $b$ and $c$ such that $\prod_{cyc}((a+b)^2+c^2)\neq0.$[/quote]
See my [url="https://artofproblemsolving.com/community/c6h28678p14450762"]solution[/url]  :D

[quote=SBM]Better: https://artofproblemsolving.com/community/u493456h1939876p14491686  :P[/quote]
Your proof is #20

Since the inequality is homogeneous, we can assume the condition $a+b+c=1$ and scale up later. The inequality then becomes

\begin{align*}
\sum \frac{(1-2a)^2}{a^2 + (1-a)^2} \ge \frac{3}{5}
\end{align*}

which rearranges to

\begin{align*}
\frac{27}{5} \ge \sum \frac{1}{2a^2 - 2a + 1}.
\end{align*}

By the tangent line trick, we derive the inequality

\[\frac{1}{2a^2-2a+1} \le \frac{9}{5} + \frac{54}{25}\left(a - \frac{1}{3}\right) \Longleftrightarrow 2(3a-1)^2(6a+1) \ge 0.\]

Summing this cyclicly yields the desired inequality. $\square$

Since the inequality is homogeneous therefore WLOG we assume $a+b+c=3$. Now the inequality becomes
\begin{align*}
\sum_{cyc}\frac{(3-2a)^2}{a^2+(3-a)^2}\geq \frac{3}{5}
\end{align*}
Consider the function $f(x)=\frac{(3-2x)^2}{x^2+(3-x)^2}$. So we have to prove that $$f(a)+f(b)+f(c)\geq \frac{3}{5}$$ Notice that (Tangent line trick) $$f(x)\geq \frac{23-18x}{25}\Longleftrightarrow \frac{(x-1)^2(2x+1)}{2x^2-6x+9}\geq 0$$ is true (the denominator $2x^2-6x+9>0$ because the coefficient of $x^2$ is positive and least value is $\left(\frac{3}{2}, \frac{9}{2}\right)$.) for all value of positive $x$. Now just summing up them we get our desired result.

Same as everyone else. Nice application of TLT.
-----
Since both sides of the inequality are of same degree, the inequality is homogeneous. Since the domain is positive reals as well, along with the previous fact, we can assume $a+b+c=3.$ Our inequality we want to prove rewrites to  \[\sum_{\text{cyc}} \frac{(3-2a)^2}{a^2+(3-a)^2} \geq \frac{3}{5}.\] Recall the Tangent Line Trick, which is if we fix $m=\tfrac{a_1+\cdots+a_n}{n},$ and if $f$ is nonconvex, then if $f(x) \geq f(m)+f'(m)(x-m)$ holds for all $x,$ summing cyclically produces the desired result. In our case, consider the function $f(x)=\tfrac{(3-2x)^2}{x^2+(3-x)^2},$ then we want to prove $f(a)+f(b)+f(c) \geq \tfrac{3}{5}.$ Since the second derivative of $f$ is negative, the function is nonconvex. By applying the Tangent Line Trick, \[\frac{(3-2a)^2}{(3-a)^2+a^2} \geq \frac{1}{5}-\frac{18}{25}(a-1) \implies\]\[ \frac{18}{25}(a-1)-\frac{1}{5}+\frac{(3-2a)^2}{(3-a)^2+a^2} = \frac{18(a-1)^2(2a+1)}{25(2a^2-6a+9)} \geq 0,\] of which the last inequality is clearly true because $a$ is a positive real. Thus by Tangent Line Trick, summing cyclically implies the result.

Prove that
$$ \frac{a\left(a+b-c\right)^{2}}{\left(a+b\right)^{3}+c^{3}}+\frac{b\left(b+c-a\right)^{2}}{\left(b+c\right)^{3}+a^{3}}+\frac{c\left(c+a-b\right)^{2}}{\left(c+a\right)^{3}+b^{3}}\geq\frac1{3}$$
$$ \frac{\left(b+c-a\right)^{3}}{\left(b+c\right)^{3}+a^{3}}+\frac{\left(c+a-b\right)^{3}}{\left(c+a\right)^{3}+b^{3}}+\frac{\left(a+b-c\right)^{3}}{\left(a+b\right)^{3}+c^{3}}\geq\frac1{3}$$
$$ \frac{a\left(a+b-c\right)^{2}}{\left(a+b\right)^{3}+2c^{3}}+\frac{b\left(b+c-a\right)^{2}}{\left(b+c\right)^{3}+2a^{3}}+\frac{c\left(c+a-b\right)^{2}}{\left(c+a\right)^{3}+2b^{3}}\geq\frac3{10}$$
$$ \frac{\left(b+c-a\right)^{3}}{\left(b+c\right)^{3}+2a^{3}}+\frac{\left(c+a-b\right)^{3}}{\left(c+a\right)^{3}+2b^{3}}+\frac{\left(a+b-c\right)^{3}}{\left(a+b\right)^{3}+2c^{3}}\geq\frac3{10}$$
for any positive real numbers $ a$, $ b$, $ c$.
[hide=Thanks.][quote=arqady][quote=sqing]Prove that
$$ \frac{(b+c-5a)^2}{(b+c)^2+5a^2}+\frac{(c+a-5b)^2}{(c+a)^2+5b^2}+\frac{(a+b-5c)^2}{(a+b)^2+5c^2}\geq 3$$
for any positive real numbers $ a$, $ b$, $ c$.[/quote]
It's true for any reals $a$, $b$ and $c$ such that $\prod\limits_{cyc}((a+b)^2+5c^2)\neq0.$[/quote]
[/hide]

De-homogenize with $a+b+c = 1$. Now, notice that $$\frac{(1-2a)^2}{a^2+(1-a)^2} \geq \frac{23-54a}{25} \iff \frac{25(3a-1)^2(6a+1)}{\geq 0} \geq 0,$$ so sum to get the result.

 Set $f(x):= \frac{3-2a}{a^2+(3-a)^2}$. Due to the homogeneity of the inequality, we can assume $a+b+c=3$. The inequality can now be rewritten as follows:

 $$\sum_{cyc} f(a) \geq \frac{3}{5}$$

This is true by tangent trick lemma since:
$$f(x)\geq f'(1)(x-1)+f(1)$$
$$\iff \frac{3-2x}{x^2+(3-x)^2} \geq \frac{1}{5}-\frac{18}{25}(x-1) $$

$$\iff \frac{18(x-1)^2(2x+1)}{25(2x^2-6x+9)} \geq 0$$

 Which holds for all $0<x<3$

  $$\mathbb{Q.E.D.}$$

[quote=shaaam]I think, cauchy-Swarchz inequality can be used to prove this inequality...

[color=blue]Problem.[/color]

[color=blue]$ \frac {\left(b \plus{} c \minus{} a\right)^{2}}{\left(b \plus{} c\right)^{2} \plus{} a^{2}} \plus{} \frac {\left(c \plus{} a \minus{} b\right)^{2}}{\left(c \plus{} a\right)^{2} \plus{} b^{2}} \plus{} \frac {\left(a \plus{} b \minus{} c\right)^{2}}{\left(a \plus{} b\right)^{2} \plus{} c^{2}}\geq\frac {3}{5}$[/color]

Solution: Let :

$ P \equal{} \frac {\left(b \plus{} c \minus{} a\right)^{2}}{\left(b \plus{} c\right)^{2} \plus{} a^{2}} \plus{} \frac {\left(c \plus{} a \minus{} b\right)^{2}}{\left(c \plus{} a\right)^{2} \plus{} b^{2}} \plus{} \frac {\left(a \plus{} b \minus{} c\right)^{2}}{\left(a \plus{} b\right)^{2} \plus{} c^{2}}$
 By Cauchy Swarchz inequality, we have

$ (a^2 \plus{} b^2 \plus{} c^2 \plus{} (a \plus{} b)^2 \plus{} (b \plus{} c)^2 \plus{} (c \plus{} a)^2)(P) \geq (a \plus{} b \plus{} c)^2$

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

$ (3(a \plus{} b \plus{} c)^2 \minus{} 4(ab \plus{} bc \plus{} ca))(P) \geq (a \plus{} b \plus{} c)^2$

$ P \geq \frac {(a \plus{} b \plus{} c)^2}{3(a \plus{} b \plus{} c)^2 \minus{} 4(ab \plus{} bc \plus{} ca)^2}$

but we know that $ (a \plus{} b \plus{} c)^2 \geq 3(ab \plus{} bc \plus{} ca)$

$ P\geq \frac {3(ab \plus{} bc \plus{} ca)}{9(ab \plus{} bc \plus{} ca) \minus{} 4(ab \plus{} bc \plus{} ca)} \equal{} \frac{3}{5}$

If my solution contains error, can you please correct it? Thanks .[/quote]

however, you can not turn into $ P\geq \frac {3(ab \plus{} bc \plus{} ca)}{9(ab \plus{} bc \plus{} ca) \minus{} 4(ab \plus{} bc \plus{} ca)} \equal{} \frac{3}{5}$,cause it’s on the denominator

Homogenous so we let $a+b+c = 3$. We get $\frac{(b+c-a)^2}{a^2+(b+c)^2} = \frac{(3-2a)^2}{2a^2 - 6a + 9} = 2 - \frac{9}{2a^2-6a+9}$. Thus we need to prove that \[\sum_{\text{cyc}}  \frac{1}{2a^2-6a+9} \leq \frac{3}{5}\] We claim that \[\frac{1}{2a^2-6a+9} \leq \frac{2a+3}{25}\] Obviously this proves the problem. Multiplying both sides by $50a^2-150a+225$ we get \[4a^3-6a^2+2 \geq 0 \Rightarrow 2(x-1)^2(2x+1) \geq 0\] This is obviously true over $[0, 3]$ and we have the result. $\blacksquare$


integration
function
calculus
real analysis
real analysis solved

Let $f:[0,1]\to\mathbb R$ be an increasing function. Prove that if $\int_0^1f(x)dx=\int_0^1\left(\int_0^xf(t)dt\right)dx=0$ then $f(x)=0,\forall x\in(0,1)$.

[i]Mihai Piticari[/i]

$F(x)=\int^x_0 f(t)dt$ is a continuous function which is also convex. This is because it's continuous and semiconvex, i.e. $F(\frac {x+y}2)\le\frac{F(x)+F(y)}2$. To show the semiconvexity, translate it to $\int^{\frac {x+y}2}_0 f(t)dt\le\frac{\int^x_0 f(t)dt+\int^y_0 f(t)dt}2$. Assuming $x\le y$, what we have to show is $2\int^{\frac{x+y}2}_x f(t)dt\le\int^y_x f(t)dt$, which follows immediately from the fact that $f$ is increasing.

Now, since a convex function on an interval reaches its maximum in one of the interval's ends, and our function $F$ is $0$ in both ends, it means that it's $\le 0$. Since its integral over $[0,1]$ is also $0$, we get $F\equiv 0$, and from here the fact that $f$ is identically null on $(0,1)$ follows.


How should one go about factoring multivariable polynomials?

For example, how should I factor $ f(x,y)\equal{}x^2y^3\minus{}y^4\minus{}x^3\plus{}x^2\plus{}xy\minus{}y$?

Arrange $ f(x,\ y)$ with respect to $ x$, because the degree of $ x$ is lower than that of $ y$, so regard $ y$ as if it were constant,

$ f(x,\ y) \equal{}\minus{}x^3\plus{}(y^3\plus{}1)x^2\plus{}yx\minus{}y^4\minus{}y$

$ \equal{}\minus{}x^3\plus{}(y^3\plus{}1)x^2\plus{}yx\minus{}y(y^3\plus{}1)$, Now you can proceed from here.

I did the folowing but I like kunny's idea:

$ x^{2}y^{3} \minus{} y^{4} \minus{} x^{3} \plus{} x^{2} \plus{} xy \minus{} y$

$ \equal{} x^{2}y^{3} \minus{} x^{3} \plus{} x^{2} \plus{} xy \minus{} y^{4} \minus{} y$

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

$ \equal{} (x^{2} \minus{} y)(y^{3} \minus{} x \plus{} 1)$.

If you need help finishing it (Kunny's way):

[hide]Group them by common terms:

$ \minus{} x^3 \plus{} xy \plus{} (y^3 \plus{} 1)x^2 \minus{} (y^3 \plus{} 1)y$

$ \minus{} x(x^2 \minus{} y) \plus{} (y^3 \plus{} 1)(x^2 \minus{} y)$

$ \boxed{f(x,y) \equal{} (y^3 \plus{} 1 \minus{} x)(x^2 \minus{} y)}$[/hide]

Since $ f(x,x^2)\equal{}x^8\minus{}x^8\minus{}x^3\plus{}x^2\plus{}x^3\minus{}x^2\equal{}0$, $ x^2\minus{}y$ is a factor of $ f(x,y)$.

Now factor out $ x^2\minus{}y$ from $ f(x,y)$, and we have $ f(x,y)\equal{}(x^2\minus{}y)(y^3\minus{}x\plus{}1)$.

well you could try grouping

$ y(x^2 y^2$ $ \minus{}y^3$ $ \plus{} x\minus{}1)$+$ x(\minus{}x^2$ $ \plus{}x)$
 well since you have it broken down into groups you basically factor the groups and then you combine them


function
integration
calculus
probability and stats

let $ B_{t}$ be Brownian motion on R, $ B_{0}\equal{}0$ . Put $ E\equal{}E^{0}$
 a) Prove
$ E\left [ exp(iuB_{t}) \right ]\equal{}exp(\minus{}\frac{1}{2}u^{2}t)$ for all $ u\epsilon R$

 b) Use power series expansion of the exponential function on both sides, compare the terms with the same power of u and deduse that
$ E\left [ B_{t}^{4} \right ]\equal{}3t^{2}$

c) Prove that
$ E\left [ f(B_{t}) \right ]\equal{}\frac{1}{\sqrt{2\pi t}}\int f\left ( x \right )e^{\minus{}\frac{x^{2}}{2t}}dx$
for all functions f such that the integral of the right converges.

no need of brownian motion here: all you need is to compute stuffs like $ E[\exp(iuN)]$ where $ N$ is a Gaussian variable. Have you done that before ?

No, I haven't done that so I have no idea how to solve this example.
I would be grateful if you help me with this.

do you know that if $ X$ is a random variable with density $ \Phi(x) \, dx$ then for any (well behaved) function $ \phi$:
\[ E[\phi(X)] \equal{} \int \phi(x) \Phi(x) \, dx\]

From that, you can see that:
\[ E[\exp(iuB_t)] \equal{} \frac{1}{\sqrt{2 \pi t}} \int_{\minus{}\infty}^{\plus{}\infty} \exp(iux) \exp(\minus{}\frac{x^2}{2t}) \, dx\]
Can you compute that ?

I tried to compute this integral using integration by parts, but I got more complicated integral  :blush:
Here what I've done:
$ I=\int e^{iux}e^{-\frac{x^{2}}{2t}}dx$

$ U=e^{-\frac{x^{2}}{2t}}$ and $ dV=e^{iux}dx$
and I got
$ I=UV-\int VdU=\frac{1}{iu}e^{iux}e^{-\frac{x^{2}}{2t}}\left |_{-\infty }^{\infty }+\frac{1}{iut}\int xe^{iux}e^{-\frac{x^{2}}{2t}}dx$

Maybe I should use some substitution?


find all cotinuous fuction $ f: R \to R$ such that :$ f(\sqrt{x^2\plus{}y^2}) \equal{}f(x) \plus{} f(y)$  ,$ \forall x \in R$

note that $ f(x)\equal{}f(\minus{}x)$ for all $ x$.

consider $ g: \mathbb{R}^{\plus{}} \rightarrow \mathbb{R}^{\plus{}}.$  such that $ g(x^{2}) \equal{} f(x)$ for all $ x \ge 0$.

note that $ g(x^{2} \plus{} y^{2}) \equal{} g(x^{2}) \plus{} g(y^{2}).$

this implies that $ g(x) \equal{} kx$ for some $ k$, this implies that $ f(x) \equal{} kx^2$ for some $ k$. :ddr:


Anyone from IL going to NSF nationals, held at Maryland?

Wish I did, but I'm not. :(

But I know suma_milli is going.

(she'd have to thank me for being her spokesperson :rotfl: )

:rotfl:  :rotfl:

Wait asethi you live in IL?  That's great!  Obviously because IL pwns :D lol

lol.

Also, there are a whole bunch of people who attended our national Mathcounts practices as guests/spectators, some of whom I know are going...

IL does seem to have a lot of people going for the Math Bee, which is utterly awesome. :D

uhh, pretty sure that michigan will have a top three sweep composed of athunder, wiz, and dafjbomb (somehow if i can jump in there...). but yes, i'm going, for math and vocab. and i'm determined to get top three in both.

(ps: michigan is gonna pwn you guys.)

I hope to do well at the bee. The thing is that its easier than MATHCOUNTS, about state-easy national level. I got to state at Mathcounts last year, and have one year left.
Anyways good luck to all.

      last year's champion at Math Bee level 3 was Anirudh Sailesh(IL), who got a perfect score. :D

      Oh yeah, nikabella don't guarentee on Michigan's top three sweep, even though they got to Mathcounts Nationals. There are other people out there who can surprise all.

         Go Kennedy jr.high for Mathcounts 2010!

A test easier than Mathcounts can often times show results that seem pretty strange, because a huge place difference can be made if a person known to be beastly makes a stupid mistake and plunges down a lot.

BTW, asethi01, athunder wasn't on this year's MI National Mathcounts team, just so you know.

dafjbomb get second last year, athunder either won or got second in level two, two years ago, and wiz&i both beat both of them in nearly every competition. i'm not guaranteeing a sweep, but i'm pretty sure that michigan will have a good number (4 or 5?) of the top 10 spots.

Lets hope for a great bee, full of competition and fun! Good luck to all!
Hope to see you there!

Go Kennedy Jr.High for Mathcounts 2010

Does anyone know what kind of questions the nats test will have?

Based on what others have told me, stuff that is perhaps more trivial than State Mathcounts :D  :D  :D yeah.

sumamilli, i've taken nationals last year, its state-easy national MATHCOUNTS LEVEL!

its comming up...

Why? Do you need help preparing?
I have last year's NSF nats questions.

Yeah I was just wondering so I could know what to practice.

If it's okay can you send me a copy of the questions please?

hmm the practice questions on the website are nats level, which creeps me out.
i really can't fail this.

asethi, can you send them to me too?
 (or suma, after you get them, can you send them to me?)

Woah what?  Those practice questions were nats level?  I thought they said regional level?

Sure I can send them to you after I get it.

ok, i'm confused. Can someone tell me whats going on. What board problems. Whats the link? There are other qs I have. If they're the same, no point of giving you them.

These math questions: http://www.northsouth.org/uscontests/2009reg/mathSyllabus.asp

And they do say 2009 Math Regionals =/

[quote="AIME15"]Er, have you guys noticed that NSF [b]copies MATHCOUNTS problems word for word[/b]?  Is that even legal?[/quote]

Hmm I don't know after all these are indian people we're talking about :P

Also thanks for the problems asethi!  I glanced over them and they don't seem to be that hard.  [Though I mean I haven't really looked at them since I'm going to try seeing what I'll get on it] Have you already received them already nike?

@AIME15, of course not.
If you just mem mathcounts you are a winner. The winner last year I don't even think got top 25 in state!

lol probably not exactly memming but if you understand how to do every typical mathcounts question I think you'll have a good shot at winning.

Hmm NSF should get some creative problem writers otherwise it's just mathcounts for indians :P

What do you mean by memming only if you come in the top ten?

Abhi, the NSF problems are Mathcounts problems, but the competition *used* to be fun. Unfortunately, its become very political lately.

P.S:When does our school get SAT results?


Let's $a,$ $b$ and $c$ are positive numbers such that $abc=1.$ Prove that:
$a^{4}+b^{4}+c^{4}+33\geq12(ab+ac+bc).$
$12$ is a maximal natural number such that this inequality is true.

Let $f(a,b,c)=a^{4}+b^{4}+c^{4}+33-12(ab+bc+ca)$
Without loss of generation, assume that $a\leb\le c$ so $a\le 1$, $bc\ge 1$
By computer, it's easy to check that
$f(a,b,c)\ge f(a,\sqrt{bc},\sqrt{bc})$ and $f(a^{2},\frac{1}{a},\frac{1}{a})\ge 0$. Q-E-D


AMC
USA(J)MO
USAMO
geometry
3D geometry
inequalities
analytic geometry

Let k be the real cube root of 2. Find integers A, B, C, a, b, c such that

|(Ax^2 + Bx + C)/(ax^2 + bx + c) - k| < |x - k|

for all non-negative rational x. (USAMO 1972/4)

Ideas?

That is one nasty looking problem...[hide]I'm guessing that we can find integers such that (Ax^3 + Bx^2 + Cx)/(ax^2 + bx + c) < 1 for x > 1 and that (Ax^3 + Bx^2 + Cx)/(ax^2 + bx + c) > 1 for x < 1. [/hide]

Hmm, where do we go from there?

[quote]That is one nasty looking problem[/quote]

Exactly what I thought when I saw it, so I have little idea about how to approach it as well.

I've seen this before, but I don't really remember it, so I'm just going to toss out some ideas.  First of all, if (ax^2 + bx + c) has any real roots, there will exists a sequence of rational numbers approaching those roots, which will make the left side of our inequality blow up unless (Ax^2 + Bx + C) has the same root.  This means that (ax^2 + bx + c) probably shouldn't have any real roots, or at least not anywhere near the cube root of 2.  I'm going to keep thinking about this -- it looks like it uses a lot of pseudo-calculus.

By the way, Masamune, I don't see how your suggestion will help at all.

Two hints (in spoiler):

[hide]

1. When x = k, the left-hand side must be 0.  That determines A, B, and C in terms of a, b, and c.

2.  Both sides now have a factor of x - k.  So cancel those factors.

[/hide]

Ravi -- [hide]did you note that it was for rational x, not real?  Because there would definitely be other steps involved in showing that it generalizes, if only a continuity argument of some sort.[/hide]

Joel, [hide]yes, I noticed it was for rational only.  My hint was a bit cryptic.  But, as you say, by continuity, the inequality must be true with < replaced by :le:.  In the end, we are going to prove the inequality for all nonnegative real x <> k anyway.

Actually, we don't even need to prove continuity here, as long as we believe that it is true.  That's because we only need to find a particular choice that works (and prove that it does).  We don't need to prove that we have found the only possible choices.

[/hide]

[hide="Solution"] Taking Ravi B's hint, we have

$ Ak^2 + Bk + C = ak^3 + bk^2 + ck = bk^2 + ck + 2a$

hence $ A = b, B = c, C = 2a$.  We now have

$ | \frac {bx^2 + cx + 2a}{ax^2 + bx + c} - k | \le |x - k| \Leftrightarrow$
$ | \frac {(b - ka)x^2 + (c - kb)x + (2a - kc)}{ax^2 + bx + c} | \le |x - k|$.

Dividing out and multiplying by $ ax^2 + bx + c$, we obtain

$ | (b - ka) x + (c - k^2 a) | \le |ax^2 + bx + c|$

Now, if $ ax^2 + bx + c$ has real roots, then they must be equal (or else $ b - ka = 0$) and rational, but a rational value of $ x$ produces an irrational value of the LHS unless $ a = 0$ (which makes the LHS trivial).  We conclude that $ ax^2 + bx + c$ has only complex roots.

A line is below a convex curve iff the line is below the curve at the point where their slopes are equal. The slope of the tangent line to the quadratic at $ x$ is $ 2ax + b = b - ka$, hence $ x = - \frac {k}{2}$.  Plugging in and multiplying by $ 4$, we obtain

$ | - 2bk + 2k^2 a + 4c - 4k^2 a | \le |ak^2 - 2bk + 4c| \Leftrightarrow$
$ | - 2k^2 a - 2bk + 4c | \le | ak^2 - 2bk + 4c |$

We can easily find $ a, b, c$ that satisfy this inequality:  let's take $ a = 1, b = 0, c = 1$, for example.  Then our final answer is (unless I have slipped up somewhere)

$ \boxed{ A = 0, B = 1, C = 2, a = 1, b = 0, c = 1 }$

so that the inequality becomes

$ | \frac {x + 2}{x^2 + 1} - k | < |x - k|$ [/hide]
Edit:  A follow-up.  Does a choice of $ A, B, C, a, b, c$ exist such that rational convergents of $ k$ are taken to rational convergents of $ k$?


geometry
circumcircle
incenter
search
angle bisector
geometry proposed

Let $ T$ be the midpoint of arc $ BC$ in circumcircle $ \omega$ of triangle $ \triangle ABC$,that doesn't contain point $ A$.Also let $ M$ be the midpoint of arc $ ABT$ in the same circle.
Suppose that $ I$,$ I_A$ and $ A'$ are incenter,$ A$-excenter and feet of angle bisector of $ \angle BAC$,respectively.Let $ P,Q$ be intersection points of $ MI,MI_A$ with $ \omega$.
Prove that $ P,A',Q$ are collinear.


inequalities
function
calculus
derivative
inequalities proposed

I give another genelization for the Iran 96 inequality.
For all nonnegative real numbers $ a,b,c$ such that $ ab\plus{}bc\plus{}ca\equal{}1$ and $ k \ge \minus{}2$ is a real number, find the minimum value of the following expression
\[ F_k(a,b,c)\equal{}\frac{1}{a^2\plus{}kab\plus{}b^2}\plus{}\frac{1}{b^2\plus{}kbc\plus{}c^2}\plus{}\frac{1}{c^2\plus{}kca\plus{}a^2}
\]

[hide]Results.
+) For all $ \minus{}2 \le k \le \minus{}1 \ \min \equal{} 2\minus{}k$
+) For all $ \minus{}1 \le k \le 2 \ \min \equal{} 2\plus{}\frac{1}{k\plus{}2}$
+) For all $ 2 \le k \ \min \equal{} \frac{9}{k\plus{}2}$[/hide]

[hide]A proof for $ k\equal{}\minus{}2$:
Assume $ c\equal{}\min(a,b,c)$ and by AM-GM:
$ F_{\minus{}2}(a,b,c)\equal{}\frac{1}{ (a\minus{}b)^2}\plus{}\frac{(a\minus{}b)^2}{(b\minus{}c)^2(a\minus{}c)^2} \plus{}\frac{2}{(a\minus{}c)(b\minus{}c)} \ge \frac{2}{(a\minus{}c)(b\minus{}c)} \plus{}\frac{2}{(a\minus{}c)(b\minus{}c)} \ge \frac{4}{ab\plus{}bc\plus{}ca}$[/hide]

:(
The forum Inequality is very boring these days.

The hardest is the case $ \minus{}1\le k\le2$. We can prove it using the inequality

$ \frac {3k\plus{}6}{a^2\plus{}kab\plus{}b^2}\ge \frac{2\minus{}k}{a^2\minus{}ab\plus{}b^2}\plus{}\frac{4k\plus{}4}{(a\plus{}b)^2}$.

[quote="Vasc"]The hardest is the case $ \minus{} 1\le k\le2$. We can prove it using the inequality

$ \frac {3k \plus{} 6}{a^2 \plus{} kab \plus{} b^2}\ge \frac {2 \minus{} k}{a^2 \minus{} ab \plus{} b^2} \plus{} \frac {4k \plus{} 4}{(a \plus{} b)^2}$.[/quote]
Cool, Vasc !!!  :coolspeak:
Thank you very much!

[quote="behemont"][quote="Vasc"]The hardest is the case $ \minus{} 1\le k\le2$. We can prove it using the inequality

$ \frac {3k \plus{} 6}{a^2 \plus{} kab \plus{} b^2}\ge \frac {2 \minus{} k}{a^2 \minus{} ab \plus{} b^2} \plus{} \frac {4k \plus{} 4}{(a \plus{} b)^2}$.[/quote]

how do you come up with sth like this vasc? i think it's extraordinary :)[/quote]
Very nice, vasc  :wink:
\[ \frac {3k \plus{} 6}{a^2 \plus{} kab \plus{} b^2}\ge \frac {2 \minus{} k}{a^2 \minus{} ab \plus{} b^2} \plus{} \frac {4k \plus{} 4}{(a \plus{} b)^2}\]
\[ \iff \frac{\minus{}3ab(a\minus{}b)^2(k\plus{}1)}{(a^2 \plus{} kab \plus{} b^2)(a^2 \minus{} ab \plus{} b^2)(a \plus{} b)^2}\ge 0\]

[quote="zaizai-hoang"][quote="behemont"][quote="Vasc"]The hardest is the case $ \minus{} 1\le k\le2$. We can prove it using the inequality

$ \frac {3k \plus{} 6}{a^2 \plus{} kab \plus{} b^2}\ge \frac {2 \minus{} k}{a^2 \minus{} ab \plus{} b^2} \plus{} \frac {4k \plus{} 4}{(a \plus{} b)^2}$.[/quote]

how do you come up with sth like this vasc? i think it's extraordinary :)[/quote]
Very nice, vasc  :wink:
\[ \frac {3k \plus{} 6}{a^2 \plus{} kab \plus{} b^2}\ge \frac {2 \minus{} k}{a^2 \minus{} ab \plus{} b^2} \plus{} \frac {4k \plus{} 4}{(a \plus{} b)^2}
\]

\[ \iff \frac { \minus{} 3ab(a \minus{} b)^2(k \plus{} 1)}{(a^2 \plus{} kab \plus{} b^2)(a^2 \minus{} ab \plus{} b^2)(a \plus{} b)^2}\ge 0
\]
[/quote]

i know the equivalence :wink:  but how does he think of something like that :?:

[quote="ehku"]I give another genelization for the Iran 96 inequality.
For all nonnegative real numbers $ a,b,c$ such that $ ab \plus{} bc \plus{} ca \equal{} 1$ and $ k \ge \minus{} 2$ is a real number, find the minimum value of the following expression
\[ F_k(a,b,c) \equal{} \frac {1}{a^2 \plus{} kab \plus{} b^2} \plus{} \frac {1}{b^2 \plus{} kbc \plus{} c^2} \plus{} \frac {1}{c^2 \plus{} kca \plus{} a^2}
\]
Results.
+) For all $ \minus{} 2 \le k \le \minus{} 1 \ \min \equal{} 2 \minus{} k$........

[/quote]

I think this is wrong, ehku. And for $ k\equal{}\minus{}1$ we know that $ min \equal{} 9$

Sorry for my previous msg, it was wrong because I confused it. :blush:

I would like only to know if it exists an analog formula for the case $ \minus{}2 \leq k \leq \minus{}1$ similar to

[quote="Vasc"]The hardest is the case $ \minus{} 1\le k\le2$. We can prove it using the inequality

$ \frac {3k \plus{} 6}{a^2 \plus{} kab \plus{} b^2}\ge \frac {2 \minus{} k}{a^2 \minus{} ab \plus{} b^2} \plus{} \frac {4k \plus{} 4}{(a \plus{} b)^2}$.[/quote]

Thank you very much.

Let $ f(k)\equal{}\frac 1{a^2\plus{}kab\plus{}b^2}$. Since $ f$ is concave, by Jensen's inequality we have for $ \minus{}2\le k\le\minus{}1$:

$ f(k) \ge (k\plus{}2)f(\minus{}1)\plus{}(\minus{}k\minus{}1)f(\minus{}2)$.

[quote="Vasc"]In Algebraic Inequalities (page 315), there is the following statement:

If $ E(a,b,c) \equal{} \sum \frac {pab \plus{} bc \plus{} ca}{a^2 \plus{} kab \plus{} b ^2}$ and $ k > \minus{} 2$, then

(a) $ E(a,b,c) \ge \frac {3(p \plus{} 2)}{k \plus{} 2}$, for $ p\le k \minus{} 1$;

(b) $ E(a,b,c) \ge \frac {p}{k \plus{} 2} \plus{} 2$, for $ k \minus{} 1\le p \le (k \plus{} 2)^2$;

(c) $ E(a,b,c) \ge 2\sqrt {p} \minus{} k$, for $ p\ge (k \plus{} 2)^2$.[/quote]
Ok, your result is better. I haven't any complete solution now, but it suffits to prove in three cases $ p \equal{} k \minus{} 1$, $ p\equal{}(k\plus{}2)^2$ and $ k\equal{}\minus{}2$ (for (a) and (b) it is sure, but I am not sure about (c)).

Please tell me if the solution in your book is nice. Thank you!

[quote="ehku"] I am not sure about (c). Please tell me if the solution in your book is nice. Thank you![/quote]
Yes, it is nice. :wink:
The idea is to show that $ E(a,b,c)\ge E(0,b,c)\ge 2\sqrt p\minus{}k$ for the case $ pb^2\ge (k\minus{}1)bc\plus{}ca$, with $ a\le b\le c$.

[quote="Vasc"]Let $ f(k) \equal{} \frac 1{a^2 \plus{} kab \plus{} b^2}$. Since $ f$ is concave, by Jensen's inequality we have for $ \minus{} 2\le k\le \minus{} 1$:

$ f(k) \ge (k \plus{} 2)f( \minus{} 1) \plus{} ( \minus{} k \minus{} 1)f( \minus{} 2)$.[/quote]

Dear Vasc, I have checked it and it is false as the function $ f(k)$ has second
derivative negative only for $ k \leq \frac{\minus{}a^2\minus{}b^2}{ab}$ and $ \minus{}2 \geq \frac{\minus{}a^2\minus{}b^2}{ab}$ so $ f(k) \equal{} \frac 1{a^2 \plus{} kab \plus{} b^2}$ doesn't satisfy $ f(k) \ge (k \plus{} 2)f( \minus{} 1) \plus{} ( \minus{} k \minus{} 1)f( \minus{} 2)$ for $ \minus{} 2\le k\le \minus{} 1$ . I have also checked $ f(k) \ge (k \plus{} 2)f( \minus{} 1) \plus{} ( \minus{} k \minus{} 1)f( \minus{} 2)$ by direct calculation and for $ \minus{} 2\le k\le \minus{} 1$ it is false. Am I right, Vasc ?

You are right, Manlio: $ f(k)$ is convex, not concave.  Therefore, we can't prove the inequality for $ \minus{}2\le k\le \minus{}1$ using only the result for $ k\equal{}\minus{}2$ and $ k\equal{}\minus{}1$.

[quote="ehku"]For all nonnegative real numbers $ a,b,c$ such that $ bc \plus{} ca \plus{} ab > 0$ and $ k \ge \minus{} 2$ is a real number, we have

$ \frac {1}{b^2 \plus{} kbc \plus{} c^2} \plus{} \frac {1}{c^2 \plus{} kca \plus{} a^2} \plus{} \frac {1}{a^2 \plus{} kab \plus{} b^2} \geq \frac {2 \minus{} k}{bc \plus{} ca \plus{} ab},$

with equality if $ a \equal{} 0,b \equal{} 2,c \equal{} 1 \minus{} k \minus{} \sqrt {(k \plus{} 1)(k \minus{} 3)}$ when $ k\in [ \minus{} 2, \minus{} 1].$[/quote]$ \sum{\frac {1}{b^2 \plus{} kbc \plus{} c^2}} \minus{} \frac {2 \minus{} k}{bc \plus{} ca \plus{} ab}\equiv\frac {F(a,b,c,k)}{(bc \plus{} ca \plus{} ab)\prod{\left(b^2 \plus{} kbc \plus{} c^2\right)}}.$

$ F(a,b,c,k) \equal{} F(a,a \plus{} s,a \plus{} t,r \minus{} 2)$

$ \equal{} r^2\left(r^2 \minus{} 4r \plus{} 9\right)a^6 \plus{} 2r^2\left(r^2 \minus{} 4r \plus{} 9\right)(s \plus{} t)a^5$

$ \plus{} r\left[\left(r^3 \minus{} 2r^2 \plus{} 3r \plus{} 12\right)(s \minus{} t)^2 \plus{} \left(6r^3 \minus{} 22r^2 \plus{} 48r \plus{} 12\right)st\right]a^4$

$ \plus{} r(s \plus{} t)\left[2\left(r^2 \minus{} 3r \plus{} 7\right)(s \minus{} t)^2 \plus{} \left(2r^3 \minus{} 5r^2 \plus{} 9r \plus{} 17\right)st\right]a^3$

$ \plus{} \left \{(s \minus{} t)^2\left[\left(r^2 \plus{} 3\right)s^2 \plus{} 3r\left(r^2 \minus{} 3r \plus{} 7\right)st \plus{} \left(r^2 \plus{} 3\right)t^2\right] \plus{} \left(r^4 \plus{} 2r^3 \minus{} 8r^2 \plus{} 30r \plus{} 3\right)s^2t^2\right\}a^2$

$ \plus{} (s \plus{} t)\left[(s \minus{} t)^2\left(2s^2 \plus{} r^2st \plus{} 2t^2\right) \plus{} \left(r^4 \plus{} 2r^3 \minus{} 8r^2 \plus{} 30r \plus{} 3\right)s^2t^2\right]a$

$ \plus{} st\left[s^2 \minus{} (3 \minus{} r)st \plus{} t^2\right]^2\geq0,$

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


Can you figure this out?

I was walking down the street when I saw a person who never lies. She was holding 4 oranges, but said there were only 3. How could this have happened?

just could see 3 (one of them hidden), so she thought she just had 3

No, she is aware of how many oranges she has. Good try, though!

Maybe she said there were only 3.......boxes or hats or something other than oranges?

Her words exactly: "Hello, Isabella, I am randomly telling you that I have 3 oranges."

Nice try, but she can count. Her abilities are those of the average person.

Orange is the name of her cat that she is holding with her and she says she has 3 oranges the fruit?

is she talking about some one else's or her oranges?

She is speaking in some language where 4 maps to 3...

In the time between when the person saw her and when she talked to him/her, she ate an orange?

Those are all good guesses, but none of them are right.

she thought that one of the oranges was a tangerine but it really wasnt?

lame, I know :P

[quote="isabella2296"]but said there were only 3.[/quote]
[quote="isabella2296"]"Hello, Isabella, I am randomly telling you that I have 3 oranges." [/quote]

These two statements are not quite the same.
In the first case, she could have been holding 4 oranges and 3 apples and saying that she only had 3 (apples).
In the second case, she could have been holding 4 apples, and she says she has 3, and 4>3.

I apologize for not being clear. She is referring to the oranges.

Should I give the answer, or let people keep guessing?

HINT
[hide]It's the lamest answer you can think of.[/hide]

She saw a person who never lies but was acctually talking to someone who does lie?

[quote="isabella2296"]

HINT
[hide]It's the lamest answer you can think of.[/hide][/quote]

And yet we haven't guessed it. :rotfl:
This riddle will definitely be an "ohhh!"er.

I'm fresh out of ideas...

[quote="t0rajir0u"]You're the one that's lying.[/quote]

Woah. You're right!!! Nice!!

wait, it's contradictory.

You are lying. so, you didn't meet someone that says truth. She didn't say that she had 3 oranges, and she didn't have four. So, what is the meaning of this?

well he might only be lying about one part.

You meet someone who never lies. You say the person has 4 oranges. You are lying about that. The person says that he/she has 3 oranges and he/she is telling the truth.

You go, t0rajir0u. Simple, and no one could get it  :wink:

[quote="dynamo729"]Ohhhhhh.......... cough*[b]thats stupid[/b]*]cough[/quote]

Thank you for your amazingly polite comment. That makes me feel so good inside.

NICE RIDDLE I was thinking about it and.....
OMG IZZY IS A LIAR??????????? :?:
jkjkjkjkjkjkjkjkjkjkjkjkjkjkjkjkjkjkjkjk   :)

[quote="D3m0n Shad0w"]Hey, dynamo, if you couldn't get it, then why is it stupid? What does it make you? [b]That[/b] is my riddle.[/quote]

The riddle is totally lame (as said by bella) and I was not thinking at the right level. And for your information, I was technically correct in my first guess (before bella modified the riddle), so...

riddles can have various answers. most of them are pretty vague

it was clever because it required looking at it at a different angle.  It is true that any riddle could have this answer... but the fact that the riddle involved a person who never lies made it appropriate I suppose.


ratio
projective geometry
geometry unsolved
geometry

One possible approach is to notice that the intersection point $ A$ of the common tangents to 2 circles lies on the line joining their centers $ O_1$ and $ O_2$ and $ \frac{|O_1A|}{|O_2A|}$ equals to the ratio of the radii. Now just use the Menelaus theorem.

[quote="Finesse"]Circles seem to be the bane of my mathematical existence!..

If anyone could please help me, this would be very much appreciated..

Consider the following..

Consider three circles. For every couple of circles, consider their two common external tangent lines and take its point of intersection.
Need to show that these 3 points of intersection belong to one line.

Thanks![/quote]
[hide="Solution"]
Let the circles have centers $ A$, $ B$, and $ C$ and let the common tangents of the circles with centers $ A$ and $ B$ intersect at $ X$, the common tangents of circles with centers $ A$ and $ C$ intersect at $ Y$, and the common tangents of circles with centers $ B$ and $ C$ intersect at $ Z$.  Furthermore, allow the circle with center $ A$ to have radius $ r_1$, and those with centers $ B$ and $ C$ to have radii of $ r_2$ and $ r_3$ respectively.  Consider the circles with centers $ A$ and $ B$ with common tangents $ MN$ and $ M'N'$ such that $ M$ and $ M'$ lie on the circle with center $ A$ and $ N$ and $ N'$ lie on the circle with center $ N$.  Notice that $ MN$, $ M'N'$, and $ AB$ are all concurrent at $ X$.  Hence, we have that $ \triangle XAM\sim \triangle XBN\implies \frac {XA}{XB} \equal{} \frac {AM}{BN} \equal{} \frac {r_1}{r_2}$.  Similarly, we can see that $ \frac {BZ}{CZ} \equal{} \frac {r_2}{r_3}$ and $ \frac {CY}{AY} \equal{} \frac {r_3}{r_1}$.  Hence, $ \frac {XA}{XB}\cdot \frac {BZ}{CZ}\cdot \frac {CY}{AY} \equal{} \frac {r_1}{r_2}\cdot \frac {r_2}{r_3}\cdot \frac {r_3}{r_1} \equal{} 1$, and by the converse of Menalaus's Theorem, $ X$, $ Y$, and $ Z$ are collinear, as desired.  [/hide]

Known as Monge's theorem and proven at http://www.mathlinks.ro/viewtopic.php?t=39186 as well. As you see by most of its proofs, it has not too much to do with circles...

  darij


The amount of wine in the water is 0%, and the amount of wine in the water is also 0%. :D

[hide="real life solution"]
Nowhere in the problem did it say the water and wine were evenly distributed. In this case it would be no answer.
[/hide]
[hide="if it is evenly distributed"]
, then you can say:
After first pour
wine:$\frac{3}{13}$ water

After second pour,
There are $13 \frac{1}{13}$ oz. wine in the water.
There are $23 \frac{1}{13}$ oz. water in the wine.
[/hide]

Actually the problem says

Compare the amount of wine in A cup of water and amount of water in A cup of wine after I do this.

I doesn't say to use the same cups he poured into, just to use a cup of pure water and a cup of pure wine.

[quote="ZzZzZzZzZzZz"][quote="hunter34"][quote="kohjhsd"]Consider a cup filled with 100 oz. of water and another cup filled with 100 oz. of wine. I pour 30 oz. of water into the wine. Then I pour 30 oz. of this mixture back to the water, again making it 100 oz. each. Compare the amount of wine in a cup of water and amount of water in a cup of wine after I do this.[/quote]

So on the first first pour, you have $70%
$ ounces of water in the water glass and $\frac{3}{13}$ of water in the wine glass.

On the second pour there is $13\frac{1}{13}$of wine in the water glass
There is $23\frac{1}{13}$ water in the wine glass.[/quote]

There's nothing in the problem about using the same cups that you poured into.  It says to compare with a cup of water and a cup of wine, so the cup of water is 100% water, and the cup of wine is 100% wine.  It's a logic question, so it's supposed to be a trick, I guess.[/quote]

Actually, I meant to use the same cups with mixed liquids in it. There's no trick there(I even had to specify that?).

[quote="mchoi815"]arent they the same

think...
they each have 100 oz.  also, if wine is missing k, then that k is in the water.  this k amount of wine displaces k in the water cup, which will end up in the wine cup.  then both have k of the other.

p.s. since when did cups have 100 oz. make it a container

Also, those who said 23+1/3 in one of the glasses, it isnt a glass.  its a cup so make it proportional.[/quote]

That's the intended logical conclusion (no calculation needed).


Further information about $a_1,a_2,...,a_6$ is needed, otherwise there is no max value for $s= a_{1}a_{2}a_{3}+a_{2}a_{3}a_{4}+a_{3}a_{4}a_{5}+a_{4}a_{5}a_{6}+a_{5}a_{6}a_{1}+a_{6}a_{1}a_{2}$. Indeed, if we take $a_1=...=a_5=-x$ and $a_6=5x+6$ we obtain
$s=12x^3+18x^2$ and this expression can be as large as we wish for x sufficiently large.

I suppose $a_1,a_2,...,a_6$ should have been assumed positive.

I think that it is safe to assume \[ a_1,a_2,...a_6 \ge 0  \]

I strongly suspect that the maximum value (when all variables are non-negative) is 8.

(three variables are 2, and the other three are 0)

We just had this at [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=399698#p399698]http://www.mathlinks.ro/Forum/viewtopic.php?t=1832 post #5[/url]: For any six nonnegative reals $a_1$, $a_2$, $a_3$, $a_4$, $a_5$, $a_6$, we have

$a_1a_2a_3+a_2a_3a_4+a_3a_4a_5+a_4a_5a_6+a_5a_6a_1+a_6a_1a_2\leq\frac{1}{27}\left(a_1+a_2+a_3+a_4+a_5+a_6\right)^3$,

and equality can be assumed (take $a_1=a_2=0$, $a_3=a_6=1$, $a_4=a_5=2$).

  Darij


linear algebra
matrix
geometry
geometric transformation
rotation
linear algebra unsolved

Hello:

I need help with this isometry problem.
a) Let $ D$ be a triangle defined by $ A(1,1), \,B(4,1),\, C(4,3)$.

Determine an isometry $ T: R^2 \rightarrow R^2$. Find $ T(D)$ and explain what geometric relation exists between $ D$ and $ T(D)$.

b) Generalise part a) to a any region in the plane.

Thanks for helping.
I have problems with appllying the definition of isometry.

Something is missing from the statement, perhaps some description of $ T$.

That's all. I have to define the transformation $ T$ such that $ T$ is an isometry.

Hi thrasher.
You don't understand the definition of an isometry. If you don't choose explicitly an isometry (as you want), then you can only say what follows:
$ T(ABC)$ is a triangle $ A'B'C'$ s.t. $ AB\equal{}A'B',AC\equal{}A'C',BC\equal{}B'C'$.
If $ det(T)\equal{}1$ then $ ABC,A'B'C'$ have the same orientation (you can slide $ ABC$ over $ A'B'C'$).If $ det(T)\equal{}\minus{}1$ then $ ABC,A'B'C'$ have opposite orientations (you can't slide $ ABC$ over $ A'B'C'$).

thanks , now I have another question:

Will I always obtain +1 and -1 in the determinant?

Let $ T$ be an affine isometry; we choose $ Oxy$ an orthonormal basis fo $ \mathbb{R}^2$. There exists $ f$ a vectorial isometry s.t. $ T(M)\equal{}T(O)\plus{}f(\vec{OM})$; there exist $ a,b$ with $ a^2\plus{}b^2\equal{}1$ s.t. the $ f$-matrix is in one of the following 2 forms: $ \begin{pmatrix}a&\minus{}b\\b&a\end{pmatrix}$ (a rotation), $ \begin{pmatrix}a&b\\b&\minus{}a\end{pmatrix}$ (a symmetry).


Is there a pattern to the minimum amount of move required to complete the Tower of Hanoi with x number of rings?
The original Tower of Hanoi is inspired by a legend...
[quote="The Regents of the University of California"]legend that tells of a Hindu temple where the pyramid puzzle might have been used for the mental discipline of young priests. Legend says that at the beginning of time the priests in the temple were given a stack of 64 gold disks, each one a little smaller than the one beneath it. Their assignment was to transfer the 64 disks from one of the three poles to another, with one important proviso large disk could never be placed on top of a smaller one. The priests worked very efficiently, day and night. When they finished their work, the myth said, the temple would crumble into dust and the world would vanish.[/quote]
How many moves will it take for the priests to complete this task? If the priest made each move in one second, how long will it take for them to be finished? (Please round to the nearest billion years)

Neal and I did this in class last year...

[hide]Ok... for

1, it was 1 move

2,          3

3,          7

4...........15

and so forth... it took the moves needed to move n-1 pieces times two + 1(moving the largest or nth disk)

It was 2^x - 1

so 2^64 - 1 = 18446744073709551615 moves... and if it took 1 second for each move... (Ans)/3600/24/365.25  :approx: 585 billion years???[/hide]

Correct!

...so you did this in class. hmm...

yah, our teacher didn't follow the textbook(thank goodness) since it's like first 5 chapters-review...


Find all square numbers with the property that all their digits ( in base 10) are odd.

The only such squares are $ 1, 9$.

Suppose such a square of at least two digits existed.  Such a square is $ \equiv 1 \bmod 8$ and $ \equiv 0, 1, 4 \bmod 5$, so the last three digits are a three-digit number $ abc$ also congruent to $ 1 \bmod 8$ and $ c \equal{} 1, 9$.  But then $ ab$ is an odd number divisible by $ 4$ - contradiction.


Prove that for every positive integer n there exists an n -digit number divisible by 5^n  all of whose digits are odd.

We simply use the mathematical induction.

Suppose that for $ n$, we have $ B$ $ \vdots$ $ 5^n$

$ \Rightarrow B\in\{ k5^{n+1},k5^{n+1}+5^n,k5^{n+1}+2 \cdot 5^n,k5^{n+1}+3 \cdot 5^n,k5^{n+1}+4 \cdot 5^n\}$

Now, $ 1B, 3B, 5B, 7B, 9B$ $ \vdots$ $ 5^n$. Suppose that none of these numbers is divisible by $ 5^{n+1}$

$ \Rightarrow \{1B, 3B, 5B, 7B, 9B\}=\{k5^{n+1}+5^n,k5^{n+1}+2 \cdot 5^n,k5^{n+1}+3 \cdot 5^n,k5^{n+1}+4 \cdot 5^n\}$

$ \Rightarrow \exists i,j \in \{ 1,3,5,7,9 \}, i\not= j$ such that $ iB-jB$ $ \vdots$ $ 5^{n+1}$ $ \Rightarrow (i-j)$ $ \vdots$ $ 5$, impossible

$ \Rightarrow A \in \{1B, 3B, 5B, 7B, 9B\}$

See the related problem [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=317971887&t=228211]here[/url].