trigonometry
ratio
Math Kangaroo
geometry
angle bisector
trig identities
Law of Sines

Number 30 on this year's Grade 7-8 Math Kangaroo test was:

30. In $ \triangle ABC$ the measure of angle $ B$ is $ 20^o$ and the measure of angle $ C$ is $ 40^o$. Moreover, the length of the bisector of angle $ A$ is equal to $ 2$. What is the value of $ |BC|\minus{}|AB|$?

I got that wrong. Looking back, the only way I can solve it is using trigonometry, but calculators weren't allowed (the calculations using trigonometry couldn't be done by hand). This was on a middle school test, so I think there should be an easier way. Can I have some help?

Ok the answer is $ \boxed{2}$ the solution i used is that angle b is 20, angle c is 60 so angle a is 120. When you bisect it it becomes 60. So if you draw a little triangle where you bisected it you have a 60,20,and 100 angle triangle. The side opposite the 20 is 2. So you can make a ratio and find out that AB is 10. Half of BC would then be 6. So 12-10=2[/hide]

Sorry for reviving such an old thread, but after coming back to this, I realized that the answer SN gave was correct, but the solution was wrong.

The solution assumed that the angle bisector was also a median and that the ratio of the sides of a triangle is equal to the ratio of its angles.

In [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=283526]this[/url] thread, there was a discussion about the relationship between the angles of a triangle and its sides. Also, the angle bisector is not the median unless the triangle is isoscoles, but it isn't. I don't think the solution works.

Does anybody have another solution?

hmm... Possibly stewart's theorem. Although that might complicate things.

Yes, well I started with the law of sines and that didn't go to far because sin of 100 is not a nice thing to evaluate...

Also, you can use stewart's theorem if you can use trig to find side lengths. However, I guess it's not a very good idea... Hmmm... I don't think you're supposed to use trig on math kangaroo though...

As I said in the original post, calculators aren't allowed, and the trig that has to be used on this problem can't be evaluated by hand.

After the test, I was able to solve it with trig and a calc. I am looking for a non-trig solution though.

Put E on BC such that AB=BE.So triangle ADE is isosceles,so AE=2.Triangle AEC is isosceles too.So AE=EC=2.
therefore BC-AB=2.Point D is the foot of angle A bissector

^ Nice!  Clever problem, thanks for pointing it out PoP.


algebra polynomial ARML quadratics articles

Given that $ a_0 \equal{} 1, a_1 \equal{} 3, a_2 \equal{} 7$, and $ a_{n \plus{} 3} \equal{} 3a_{n \plus{} 2} \minus{} 3a_{n \plus{} 1} \plus{} a_n$ for $ n \ge 0$, find a general polynomial expression for $ f(n) \equal{} a_n$.

I don't know why this is in a local forum, but...

The given recurrence is equivalent to saying that the third difference is zero, so $ a_n$ is quadratic.  It's straightforward from there.

I think it's here in the local forum because we're trying to get more South Carolinians, and in particular ARML team members, active right now. So I think its better to leave the problems posted here for South Carolinians only -- unless, of course, someone has something particularly novel or insightful to contribute.

Yeah, we have its characteristic polynomial is $ (x^3\minus{}1)\equal{}0$, so it just has a root of one in the closed-form, which we find to be have $ a_n\equal{}n^2c_2\plus{}nc_1\plus{}1$, so we set $ a_1\equal{}3\equal{}c_2\plus{}c_2\plus{}1$ , or $ a_2\equal{}4c_2\plus{}2c_1\plus{}1$, so $ c_1\equal{}1$, $ c_2\equal{}1$. Thus we have $ n^2\plus{}n\plus{}1$.

Or the easy way is to note that all quadratics satisfy the recurrence, and we simply need to solve for the constants :).

A discussion about how to solve recurrence relations can be found here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=59858

In particular, for a nice technique for solving them look at the article that I linked to in that thread.


trigonometry
inequalities
trig identities
Law of Cosines

probably not the best way... i don't do much trig :(

$ sin^2(\frac{A}{2})\plus{}sin^2(\frac{B}{2})\plus{}sin^2(\frac{C}{2})<1$
$ 3\minus{}(cos(A)\plus{}cos(B)\plus{}cos(C))<2$
$ cos(A)\plus{}cos(B)\plus{}cos(C)>1$
$ cos(A)\plus{}cos(B)\minus{}cos(A\plus{}B)>1$
$ 2cos(\frac{A\plus{}B}{2})cos(\frac{A\minus{}B}{2})\minus{}cos(A\plus{}B)>1$
$ 2cos(\frac{A\plus{}B}{2})cos(\frac{A\minus{}B}{2})\minus{}2cos^2(\frac{A\plus{}B}{2})\plus{}1>1$
$ 2cos(\frac{A\plus{}B}{2})(cos(\frac{A\minus{}B}{2})\minus{}cos(\frac{A\plus{}B}{2}))>0$
$ cos(\frac{A\plus{}B}{2})sin(\frac{A}{2})sin(\frac{B}{2})>0$

which is true as each of the terms is positive.

[quote="nayel"]For any triangle $ ABC$ prove that
\[ \sin^2\frac A2 \plus{} \sin^2\frac B2 \plus{} \sin^2\frac C2 < 1
\]
[/quote]
[hide="Alternate Solution"]
Call $ \frac {A}{2} \equal{} x$, $ \frac {B}{2} \equal{} y$, $ \frac {C}{2} \equal{} z \equal{} 90 \minus{} x \minus{} y$.  So, we want to prove that $ \sin^2 x \plus{} \sin^2 y \plus{} \sin^2 (90 \minus{} x \minus{} y) < 1\rightarrow \sin^2 x \plus{} \sin^2 y \plus{} \cos^2 (x \plus{} y) < 1$.  So, we want to show that
\[ \sin^2 x \plus{} \sin^2 y \plus{} (\cos x\cos y \minus{} \sin x\sin y)^2 < 1
\]

\[ (\sin^2 x \plus{} \sin^2 y) \plus{} (\cos^2 x\cos^2 y \plus{} \sin^2 x\sin^2 y \minus{} 2\sin x\cos x\sin y\cos y) < 1
\]

\[ (\cos^2 x\cos^2 \plus{} 1 \minus{} \cos^2 x \plus{} 1 \minus{} \cos^2 y) \plus{} \sin^2 x\sin^2 y \minus{} 2\sin x\cos x\sin y\cos y < 1
\]

\[ (1 \minus{} \cos^2 x)(1 \minus{} \cos^2 y) \plus{} \sin^2 x\sin^2 y \minus{} 2\sin x\cos x\sin y\cos y < 0
\]

\[ 2\sin^2 x\sin^2 y \minus{} 2\sin x\sin y\cos x\cos y < 0
\]

\[ \sin x\sin y \minus{} \cos x\cos y < 0
\]

\[ \cos (x \plus{} y) > 0
\]
This is true since $ x \plus{} y< 90$ and thus we are done.   [/hide]

[quote="nayel"][color=darkred]For any triangle $ ABC$ prove that $ \frac 34\le \sin^2\frac A2 \plus{} \sin^2\frac B2 \plus{} \sin^2\frac C2 < 1\ .$[/color][/quote]
 [color=darkblue] [b][u]Proof.[/u][/b] The relations $ \sum\cos A\equal{}1\plus{}\frac rR$ and $ 2r\le R$ are well-known or prove easily it.
$ S\equiv\sum\sin^2\frac A2\equal{}\frac 12\cdot\sum\left(1\minus{}\cos A\right)\equal{}1\minus{}\frac {r}{2R}\ .$ Therefore, $ \frac 34\le S < 1\ .$[/color]

Thanks everyone for your nice solutions. :)

My solution:

Using $ \cos 2x\equal{}1\minus{}2\sin^2x$ we can rewrite the inequality as \[ \cos A\plus{}\cos B\plus{}\cos C>1\]
By the law of cosines this is equivalent to \[ \sum\frac{b^2\plus{}c^2\minus{}a^2}{2bc}>1\]
Or \[ \sum_{sym}a^2b>a^3\plus{}b^3\plus{}c^3\plus{}2abc\]
Which is just $ (b\plus{}c\minus{}a)(c\plus{}a\minus{}b)(a\plus{}b\minus{}c)>0$.


Find the minimum value of $ f(x, y, z)$ = $ \sqrt{x+y+z}$ subject to the constrains
$ x, y, z > 0,$, $ 1/x+1/y+1/z=1$
.

[hide]
$ (x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\geq 9\Rightarrow x+y+z\geq 9\Rightarrow \sqrt{x+y+z}\geq 3$
[/hide]

[hide="my attempt"]
suppose x is 3
$ z=\frac{3y}{2y-3}$
then $ f(x,y)=\sqrt{x+y+\frac{3y}{2y-3}}$
$ \frac{df}{dy}=(1-\frac{-9}{4y^{2}-12y+9})(x+y+\frac{3y}{2y-3})^{-.5}$
which equals 0 when y=3
therefore, the maximum occurs at 3,3,3
[/hide]


inequalities
calculus
derivative
logarithms
inequalities proposed

Let $ a,b,c,d$ be real numbers satisfying the condition $ abcd\equal{}1.$ Prove that
$ \frac{1}{(1\plus{}a)^2}\plus{}\frac{1}{(1\plus{}b)^2}\plus{}\frac{1}{(1\plus{}c)^2}\plus{}\frac{1}{(1\plus{}d)^2} \ge 1.$

I am waiting for nice solution using classical inequalities as Cauchy Schwarz or something like that.

let $ a\equal{}\frac{wxy}{z^3}, b\equal{}\frac{xyz}{w^3}, c\equal{}\frac{yzw}{x^3}, d\equal{}\frac{zwx}{y^3}.$

The inequality becomes $ \sum \frac{w^6}{(xyz\plus{}w^{3})^{2}} \ge 1.$

by Cauchy-Schwarz, $ LHS\ge \frac{(\sum w^3)^2}{\sum (w^{2}x^{2}y^{2} \plus{} 2w^{3}xyz \plus{} w^6)}.$

it suffices to show $ \frac{1}{6}[6,0,0,0] \plus{} \frac{1}{2}[3,3,0,0] \ge \frac{1}{6}[6,0,0,0] \plus{} \frac{1}{3}[3,1,1,1] \plus{} \frac{1}{6}[2,2,2,0]$, which is obvious by AM-GM. :pilot:

It is known the following lemma :

$ \frac{1}{(1\plus{}x)^2}\plus{}\frac{1}{(1\plus{}y)^2}\geq \frac{1}{1\plus{}xy}\ ,\ (\forall)x,y\in \mathbb{R}_\plus{}$

[hide="Proof"]$ \frac{1}{(1\plus{}x)^2}\plus{}\frac{1}{(1\plus{}y)^2}\geq \frac{1}{1\plus{}xy}\Leftrightarrow xy(x\minus{}y)^2\plus{}(1\minus{}xy)^2\geq 0$[/hide]

Therefore :

$ \left(\frac{1}{(1\plus{}a)^2}\plus{}\frac{1}{(1\plus{}b)^2}\right)\plus{}\left(\frac{1}{(1\plus{}c)^2}\plus{}\frac{1}{(1\plus{}d)^2}\right)\geq \frac{1}{1\plus{}ab}\plus{}\frac{1}{1\plus{}cd}$ $ \equal{}1.$

[quote="can_hang2007"]Let $ a,b,c,d$ be real numbers satisfying the condition $ abcd \equal{} 1.$ Prove that
$ \frac {1}{(1 \plus{} a)^2} \plus{} \frac {1}{(1 \plus{} b)^2} \plus{} \frac {1}{(1 \plus{} c)^2} \plus{} \frac {1}{(1 \plus{} d)^2} \ge 1.$

I am waiting for nice solution using classical inequalities as Cauchy Schwarz or something like that.[/quote]
We have $ LHS\equal{}\sum {\frac{{bcd}}{{\left( {1 \plus{} a} \right)\left( {1 \plus{} bcd} \right)}}}  \ge \sum {\frac{{bcd}}{{{{\left( {a \plus{} bcd} \right)}^2}}}}  \equal{} \sum {\frac{a}{{{a^2} \plus{} 1}} \minus{} \sum {\frac{{{a^3}}}{{{{\left( {{a^2} \plus{} 1} \right)}^2}}}} }$
Now AM-GM works here.
[hide="@can_hang2007"]Lời giải của anh thế nào ạ? [/hide]

\[ \frac {1}{(1 \plus{} x)^2} \plus{} \frac {1}{(1 \plus{} y)^2}\geq\frac {1}{1 \plus{} xy}\]
holds if I'm not mistaken and the inequality follows.
EDIT: 2 new replies while I was posting?  :lol:

Thanks again for replying the topic, friends. For the proof using the inequality
$ \frac{1}{(1\plus{}x)^2}\plus{}\frac{1}{(1\plus{}y)^2} \ge \frac{1}{1\plus{}xy},$ we can use Cauchy Schwarz Inequality as follows
First, $ (1\plus{}x)^2 \le (1\plus{}xy)\left(1\plus{}\frac{x}{y}\right),$ implying $ \frac{1}{(1\plus{}x)^2} \ge \frac{y}{(1\plus{}xy)(x\plus{}y)}.$
Similarly, we find that $ \frac{1}{(1\plus{}y)^2} \ge \frac{x}{(1\plus{}xy)(x\plus{}y)},$ and by adding to the above inequality, we get the claim.

Now, I will present my proof:

Because $ abcd\equal{}1,$ one may put $ a\equal{}\frac{y}{x},b\equal{}\frac{z}{y},c\equal{}\frac{t}{z},d\equal{}\frac{x}{t},$ where $ x,y,z,t$ are some positive numbers. Then, our inequality becomes
$ \frac{x^2}{(x\plus{}y)^2}\plus{}\frac{y^2}{(y\plus{}z)^2}\plus{}\frac{z^2}{(z\plus{}t)^2}\plus{}\frac{t^2}{(t\plus{}x)^2} \ge 1.$
From the Cauchy Schwarz Inequality, we get
$ \left[\sum \frac{x^2}{(x\plus{}y)^2}\right] \left[ \sum (x\plus{}y)^2(x\plus{}t)^2\right] \ge \left[ \sum x(x\plus{}t)\right]^2.$
So, it suffices to prove that
$ (x^2\plus{}y^2\plus{}z^2\plus{}t^2\plus{}xy\plus{}yz\plus{}zt\plus{}tx)^2 \ge \sum (x\plus{}y)^2(x\plus{}t)^2.$
Because
$ x^2\plus{}y^2\plus{}z^2\plus{}t^2\plus{}xy\plus{}yz\plus{}zt\plus{}tx \equal{}\frac{1}{2} \sum (x\plus{}y)^2,$
this inequality is equivalent to
$ (m\plus{}n\plus{}p\plus{}w)^2 \ge 4(mn\plus{}np\plus{}pw\plus{}wm),$
where $ m\equal{}(x\plus{}y)^2,n\equal{}(y\plus{}z)^2,p\equal{}(z\plus{}t)^2,w\equal{}(t\plus{}x)^2.$
From the AM-GM Inequality, we find that
$ 4(mn\plus{}np\plus{}pw\plus{}wm) \equal{}4(m\plus{}p)(n\plus{}w) \le (m\plus{}p\plus{}n\plus{}w)^2,$
as desired. This ends our proof.

@leedt26: I dont really understand your solution, could you please explain more? Thanks.

[quote="Albanian Eagle"]
\[ \frac {1}{(1 \plus{} x)^2} \plus{} \frac {1}{(1 \plus{} y)^2}\geq\frac {1}{1 \plus{} xy}\]
holds if I'm not mistaken and the inequality follows.
[/quote]

Another proof of this lemma is using Cauchy Schwartz :

$ (x \plus{} y)(1 \plus{} xy)\ge (\sqrt {x} \plus{} \sqrt {x}y)^2 \equal{} x(1 \plus{} y)^2\Rightarrow \frac {1}{(1 \plus{} y)^2} \ge \frac {x}{x \plus{} y}\cdot \frac {1}{1 \plus{} xy}$

In the same way $ \frac {1}{(1 \plus{} x)^2}\ge \frac {y}{x \plus{} y}\cdot \frac {1}{1 \plus{} xy}$ . Suming we are done .

[quote="Albanian Eagle"]EDIT: 2 new replies while I was posting?  :lol:[/quote]

While i was posting it happened the same thing and earldbest was faster  :lol:

Edit : this time can_hang was faster  :o

Thank you for quoting me twice!  :lol:

I wanted to mention that I think I remember an inequality of Vasc that said
If $ x_1x_2\cdots x_n\equal{}1$ , $ 0<k\le n\minus{}1$ and $ c\equal{}n^{\frac{1}{k}}\minus{}1$ then
\[ \sum_{i\equal{}1}^n\frac{1}{(1\plus{}cx_i)^k}\geq 1\]
holds.

[quote="Albanian Eagle"]Thank you for quoting me twice!  :lol:

I wanted to mention that I think I remember an inequality of Vasc that said
If $ x_1x_2\cdots x_n \equal{} 1$ , $ 0 < k\le n \minus{} 1$ and $ c \equal{} n^{\frac {1}{k}} \minus{} 1$ then
\[ \sum_{i \equal{} 1}^n\frac {1}{(1 \plus{} cx_i)^k}\geq 1\]
holds.[/quote]
I think we should first discuss the following inequality. This is much simpler than the above generalization but it is also stronger than the China Inequality:

(Vasc) If $ a,b,c,d > 0,abcd \equal{} 1,$ then
$ \frac {1}{(1 \plus{} a)(1 \plus{} a^2)} \plus{} \frac {1}{(1 \plus{} b)(1 \plus{} b^2)} \plus{} \frac {1}{(1 \plus{} c)(1 \plus{} c^2)} \plus{} \frac {1}{(1 \plus{} d)(1 \plus{} d^2)} \ge 1.$

Are there any solutions using classical inequalities? I am waiting.

[quote="can_hang2007"]Thanks again for replying the topic, friends. For the proof using the inequality
$ \frac {1}{(1 \plus{} x)^2} \plus{} \frac {1}{(1 \plus{} y)^2} \ge \frac {1}{1 \plus{} xy},$ we can use Cauchy Schwarz Inequality as follows
First, $ (1 \plus{} x)^2 \le (1 \plus{} xy)\left(1 \plus{} \frac {x}{y}\right),$ implying $ \frac {1}{(1 \plus{} x)^2} \ge \frac {y}{(1 \plus{} xy)(x \plus{} y)}.$
Similarly, we find that $ \frac {1}{(1 \plus{} y)^2} \ge \frac {x}{(1 \plus{} xy)(x \plus{} y)},$ and by adding to the above inequality, we get the claim.

Now, I will present my proof:

Because $ abcd \equal{} 1,$ one may put $ a \equal{} \frac {y}{x},b \equal{} \frac {z}{y},c \equal{} \frac {t}{z},d \equal{} \frac {x}{t},$ where $ x,y,z,t$ are some positive numbers. Then, our inequality becomes
$ \frac {x^2}{(x \plus{} y)^2} \plus{} \frac {y^2}{(y \plus{} z)^2} \plus{} \frac {z^2}{(z \plus{} t)^2} \plus{} \frac {t^2}{(t \plus{} x)^2} \ge 1.$
From the Cauchy Schwarz Inequality, we get
$ \left[\sum \frac {x^2}{(x \plus{} y)^2}\right] \left[ \sum (x \plus{} y)^2(x \plus{} t)^2\right] \ge \left[ \sum x(x \plus{} t)\right]^2.$
So, it suffices to prove that
$ (x^2 \plus{} y^2 \plus{} z^2 \plus{} t^2 \plus{} xy \plus{} yz \plus{} zt \plus{} tx)^2 \ge \sum (x \plus{} y)^2(x \plus{} t)^2.$
Because
$ x^2 \plus{} y^2 \plus{} z^2 \plus{} t^2 \plus{} xy \plus{} yz \plus{} zt \plus{} tx \equal{} \frac {1}{2} \sum (x \plus{} y)^2,$
this inequality is equivalent to
$ (m \plus{} n \plus{} p \plus{} w)^2 \ge 4(mn \plus{} np \plus{} pw \plus{} wm),$
where $ m \equal{} (x \plus{} y)^2,n \equal{} (y \plus{} z)^2,p \equal{} (z \plus{} t)^2,w \equal{} (t \plus{} x)^2.$
From the AM-GM Inequality, we find that
$ 4(mn \plus{} np \plus{} pw \plus{} wm) \equal{} 4(m \plus{} p)(n \plus{} w) \le (m \plus{} p \plus{} n \plus{} w)^2,$
as desired. This ends our proof.

@leedt26: I dont really understand your solution, could you please explain more? Thanks.[/quote]
Very sorry. My solution was wrong.
This is my solution for this one:
Because $ abcd\equal{}1$ we can put $ a\equal{}\frac{{yz}}{{{x^2}}}$ , $ b \equal{} \frac{{zt}}{{{y^2}}}$,$ c \equal{} \frac{{tx}}{{{z^2}}}$,$ d \equal{} \frac{{xy}}{{{t^2}}}$
Then $ LHS\equal{}\frac{{{x^4}}}{{{{\left( {{x^2} \plus{} yz} \right)}^2}}} \plus{} \frac{{{y^4}}}{{{{\left( {{y^2} \plus{} zt} \right)}^2}}} \plus{} \frac{{{z^4}}}{{{{\left( {{z^2} \plus{} tx} \right)}^2}}} \plus{} \frac{{{t^4}}}{{{{\left( {{t^2} \plus{} xy} \right)}^2}}}$
From the Cauchy Schwarz Inequality, we get
$ \frac{{{x^4}}}{{{{\left( {{x^2} \plus{} yz} \right)}^2}}} \plus{} \frac{{{y^4}}}{{{{\left( {{y^2} \plus{} zt} \right)}^2}}} \plus{} \frac{{{z^4}}}{{{{\left( {{z^2} \plus{} tx} \right)}^2}}} \plus{} \frac{{{t^4}}}{{{{\left( {{t^2} \plus{} xy} \right)}^2}}}\ge\frac{{{{\left( {{x^2} \plus{} {y^2} \plus{} {z^2} \plus{} {t^2}} \right)}^2}}}{{{{\left( {{x^2} \plus{} yz} \right)}^2} \plus{} {{\left( {{y^2} \plus{} zt} \right)}^2} \plus{} {{\left( {{z^2} \plus{} tx} \right)}^2} \plus{} {{\left( {{t^2} \plus{} xy} \right)}^2}}}$
$ \ge\frac{{{{\left( {{x^2} \plus{} {y^2} \plus{} {z^2} \plus{} {t^2}} \right)}^2}}}{{\left( {{x^2} \plus{} {y^2}} \right)\left( {{x^2} \plus{} {z^2}} \right) \plus{} \left( {{y^2} \plus{} {z^2}} \right)\left( {{y^2} \plus{} {t^2}} \right) \plus{} \left( {{z^2} \plus{} {t^2}} \right)\left( {{z^2} \plus{} {x^2}} \right) \plus{} \left( {{t^2} \plus{} {x^2}} \right)\left( {{t^2} \plus{} {y^2}} \right)}} \equal{} 1$

I have an ugly solution and I think it works for the generalized one  :wink:  ,

lemma :
for all $ 0\leq a <1$ we have :
$ 4a^2 \geq   ln(\frac{a}{1\minus{}a})\plus{}1$
Proof : easy with derivative

now go back to the original problem :

let $ a\equal{} \frac{1}{1\plus{}x}$ etc....hence : $ (1\minus{}a)(1\minus{}b)(1\minus{}c)(1\minus{}d)\equal{}abcd$   and the poblem is equivalent to prove that :
$ \sum{a^2} \geq 1$ using the above lemma we are done  :)  :wink:

[quote="can_hang2007"]
Because $ abcd \equal{} 1,$ one may put $ a \equal{} \frac {y}{x},b \equal{} \frac {z}{y},c \equal{} \frac {t}{z},d \equal{} \frac {x}{t},$ where $ x,y,z,t$ are some positive numbers. [/quote]

Why $ x,y,z,t$ are positive numbers ? :wink:

You choose them to be  :)
For example take $ x\equal{}1,y\equal{}a$ etc..

[quote="leedt26"][quote="can_hang2007"]Thanks again for replying the topic, friends. For the proof using the inequality
$ \frac {1}{(1 \plus{} x)^2} \plus{} \frac {1}{(1 \plus{} y)^2} \ge \frac {1}{1 \plus{} xy},$ we can use Cauchy Schwarz Inequality as follows
First, $ (1 \plus{} x)^2 \le (1 \plus{} xy)\left(1 \plus{} \frac {x}{y}\right),$ implying $ \frac {1}{(1 \plus{} x)^2} \ge \frac {y}{(1 \plus{} xy)(x \plus{} y)}.$
Similarly, we find that $ \frac {1}{(1 \plus{} y)^2} \ge \frac {x}{(1 \plus{} xy)(x \plus{} y)},$ and by adding to the above inequality, we get the claim.

Now, I will present my proof:

Because $ abcd \equal{} 1,$ one may put $ a \equal{} \frac {y}{x},b \equal{} \frac {z}{y},c \equal{} \frac {t}{z},d \equal{} \frac {x}{t},$ where $ x,y,z,t$ are some positive numbers. Then, our inequality becomes
$ \frac {x^2}{(x \plus{} y)^2} \plus{} \frac {y^2}{(y \plus{} z)^2} \plus{} \frac {z^2}{(z \plus{} t)^2} \plus{} \frac {t^2}{(t \plus{} x)^2} \ge 1.$
From the Cauchy Schwarz Inequality, we get
$ \left[\sum \frac {x^2}{(x \plus{} y)^2}\right] \left[ \sum (x \plus{} y)^2(x \plus{} t)^2\right] \ge \left[ \sum x(x \plus{} t)\right]^2.$
So, it suffices to prove that
$ (x^2 \plus{} y^2 \plus{} z^2 \plus{} t^2 \plus{} xy \plus{} yz \plus{} zt \plus{} tx)^2 \ge \sum (x \plus{} y)^2(x \plus{} t)^2.$
Because
$ x^2 \plus{} y^2 \plus{} z^2 \plus{} t^2 \plus{} xy \plus{} yz \plus{} zt \plus{} tx \equal{} \frac {1}{2} \sum (x \plus{} y)^2,$
this inequality is equivalent to
$ (m \plus{} n \plus{} p \plus{} w)^2 \ge 4(mn \plus{} np \plus{} pw \plus{} wm),$
where $ m \equal{} (x \plus{} y)^2,n \equal{} (y \plus{} z)^2,p \equal{} (z \plus{} t)^2,w \equal{} (t \plus{} x)^2.$
From the AM-GM Inequality, we find that
$ 4(mn \plus{} np \plus{} pw \plus{} wm) \equal{} 4(m \plus{} p)(n \plus{} w) \le (m \plus{} p \plus{} n \plus{} w)^2,$
as desired. This ends our proof.

@leedt26: I dont really understand your solution, could you please explain more? Thanks.[/quote]
Very sorry. My solution was wrong.
This is my solution for this one:
Because $ abcd \equal{} 1$ we can put $ a \equal{} \frac {{yz}}{{{x^2}}}$ , $ b \equal{} \frac {{zt}}{{{y^2}}}$,$ c \equal{} \frac {{tx}}{{{z^2}}}$,$ d \equal{} \frac {{xy}}{{{t^2}}}$
Then $ LHS \equal{} \frac {{{x^4}}}{{{{\left( {{x^2} \plus{} yz} \right)}^2}}} \plus{} \frac {{{y^4}}}{{{{\left( {{y^2} \plus{} zt} \right)}^2}}} \plus{} \frac {{{z^4}}}{{{{\left( {{z^2} \plus{} tx} \right)}^2}}} \plus{} \frac {{{t^4}}}{{{{\left( {{t^2} \plus{} xy} \right)}^2}}}$
From the Cauchy Schwarz Inequality, we get
$ \frac {{{x^4}}}{{{{\left( {{x^2} \plus{} yz} \right)}^2}}} \plus{} \frac {{{y^4}}}{{{{\left( {{y^2} \plus{} zt} \right)}^2}}} \plus{} \frac {{{z^4}}}{{{{\left( {{z^2} \plus{} tx} \right)}^2}}} \plus{} \frac {{{t^4}}}{{{{\left( {{t^2} \plus{} xy} \right)}^2}}}\ge\frac {{{{\left( {{x^2} \plus{} {y^2} \plus{} {z^2} \plus{} {t^2}} \right)}^2}}}{{{{\left( {{x^2} \plus{} yz} \right)}^2} \plus{} {{\left( {{y^2} \plus{} zt} \right)}^2} \plus{} {{\left( {{z^2} \plus{} tx} \right)}^2} \plus{} {{\left( {{t^2} \plus{} xy} \right)}^2}}}$
$ \ge\frac {{{{\left( {{x^2} \plus{} {y^2} \plus{} {z^2} \plus{} {t^2}} \right)}^2}}}{{\left( {{x^2} \plus{} {y^2}} \right)\left( {{x^2} \plus{} {z^2}} \right) \plus{} \left( {{y^2} \plus{} {z^2}} \right)\left( {{y^2} \plus{} {t^2}} \right) \plus{} \left( {{z^2} \plus{} {t^2}} \right)\left( {{z^2} \plus{} {x^2}} \right) \plus{} \left( {{t^2} \plus{} {x^2}} \right)\left( {{t^2} \plus{} {y^2}} \right)}} \equal{} 1$[/quote]
Very nice your proof, leedt26. Thanks. Now, please try to prove the inequality I mentioned above
$ \frac{1}{(1\plus{}a)(1\plus{}a^2)}\plus{}\frac{1}{(1\plus{}b)(1\plus{}b^2)}\plus{}\frac{1}{(1\plus{}c)(1\plus{}c^2)}\plus{}\frac{1}{(1\plus{}d)(1\plus{}d^2)} \ge 1,$
where $ a,b,c,d$ are positive reals and $ abcd\equal{}1.$

I also used Cauchy Schwarz to prove it but this time, my proof is not elegant. I am waiting for another nice proof (because Albanian Eagle has notice a nice solution above).

[quote="Vasc"][quote="can_hang2007"]Let $ a,b,c,d$ be real numbers satisfying the condition $ abcd \equal{} 1.$ Prove that
$ \frac {1}{(1 \plus{} a)^2} \plus{} \frac {1}{(1 \plus{} b)^2} \plus{} \frac {1}{(1 \plus{} c)^2} \plus{} \frac {1}{(1 \plus{} d)^2} \ge 1.$

I am waiting for nice solution using classical inequalities as Cauchy Schwarz or something like that.[/quote]
Actually, I published this nice inequality in the romanian journal Gazeta Matematica, for the first time in No. 6/1996, and for the second time in No. 11/1999. The second publication is problem O:922 for Senior Level, which has the following statement:

[b]O:922.[/b] Let $ a,b,c,d$ be positive numbers such that $ abcd \equal{} 1$. Prove that:

$ (1)\ \ \ \ \sum \frac {1}{1 \plus{} a \plus{} a^2 \plus{} a^3} \ge 1$;

$ (2)\ \ \ \ \sum \frac {1}{(1 \plus{} a)^2} \ge 1$;

$ (3)\ \ \ \ \sum \frac {1}{1 \plus{} 3a} \ge 1$.

Of course, the 2004 China Contest has made the inequality (2) to be well-known in Mathematics Community.   :) :lol:[/quote]
Thanks Vasc for your information. :) For (3), as I know, you have proposed its generalization for the Romania TST 1999. Am I right?

[quote="can_hang2007"] For (3), as I know, you have proposed its generalization for the Romania TST 1999. Am I right?[/quote]
Actually, its generalization was proposed for Romania TST 1999 by Gh. Eckstein  :oops:, after what I publised it in Gazeta Matematica No. 10/1991 in the following form:

If $ a_1,a_2,...,a_n$ are positive numbers such that $ \sqrt[n]{a_1a_2...a_n}\equal{}r\ge n\minus{}1$, then

$ \frac 1{1\plus{}a_1}\plus{}\frac 1{1\plus{}a_2}\plus{}...\plus{}\frac 1{1\plus{}a_n} \ge \frac {n}{1\plus{}r}$.

[quote="can_hang2007"](Vasc) If $ a,b,c,d > 0,abcd \equal{} 1,$ then
$ \frac {1}{(1 \plus{} a)(1 \plus{} a^2)} \plus{} \frac {1}{(1 \plus{} b)(1 \plus{} b^2)} \plus{} \frac {1}{(1 \plus{} c)(1 \plus{} c^2)} \plus{} \frac {1}{(1 \plus{} d)(1 \plus{} d^2)} \ge 1.$[/quote]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=17246&start=20&search_id=2094400416]here[/url]

[quote="anas"]I have an ugly solution and I think it works for the generalized one  :wink:  ,

lemma :
for all $ 0\leq a < 1$ we have :
$ 4a^2 \geq ln(\frac {a}{1 \minus{} a}) \plus{} 1$
Proof : easy with derivative

now go back to the original problem :

let $ a \equal{} \frac {1}{1 \plus{} x}$ etc....hence : $ (1 \minus{} a)(1 \minus{} b)(1 \minus{} c)(1 \minus{} d) \equal{} abcd$   and the poblem is equivalent to prove that :
$ \sum{a^2} \geq 1$ using the above lemma we are done  :)  :wink:[/quote]
Dear anas, I checked your solution and I think it is incorrect, because the main lemma:
$ 4a^2 \ge \ln \frac{a}{1\minus{}a}\plus{}1,$
does not hold. For example, you can take $ a\equal{}0.99.$

you are right I made a silly mistake of computation , sorry :blush:


email
geometry
geometric transformation
reflection
HMMT
ARML

From an email message I got today:

[quote="SLAdirian1"]Team Names

This is a New Policy for ARML 2009

To make the scoring process friendlier to participants, coaches, parents and
ARML officials for the 2009 ARML competition, we are requiring all coaches
to apply the following conventions when naming their teams:

*    Division A teams will use the code A1, A2, A3, etc to indicate the
strength of their teams with A1 being the strongest team and subsequent
teams decreasing in strength.
*    Division B teams will use the code B1, B2, B3, etc to indicate the
strength of their teams with B1 being the strongest team and subsequent
teams decreasing in strength.
*    Example: Let us say that Wyoming has 7 teams, four are in Division A
and three are in Division B.

*    The four Division A teams will be named as follows:

*    Wyoming A1: Red Sox,
*    Wyoming A2: White Sox
*    Wyoming A3: Mariners
*    Wyoming A4: Yankees

*    The three Division B teams will be named as follows:

*    Wyoming B1: Braves
*    Wyoming B2: Phillies
*    Wyoming B3: Mets

This will enable everyone, including team members, coaches, scorers and
parents to easily determine the strength of a team and the division that
they will be participating in. All other rules are not affected by this
convention. [/quote]
Uh - is this [i]really[/i] necessary?

Here's our organization's issue with this: Of the teams we had last year, and expect to have this year, two competed in Division A and two competed in Division B. We called the two Division A teams the A team and the B team. This proposal would rename them A1 and A2. OK, that's clunky, but not that big an issue. But then there are the other two teams - the two Division B teams. Last year we called them the C team and the Gamma team. We did that explicitly to [i]not[/i] imply a relative strength ranking between the two teams. The assignment of individuals between these two teams was, well, non-monotonic. Decisions between these two teams were made to reflect other factors, and with an eye to making the teams near-equal in strength. (And they finished within a point of each other.) I [i]don't[/i] want to give them names that imply a strength ranking. I don't want them to be B1 and B2.

Does anyone else have anything similar to say?

I completely agree. Perhaps it could be a suggestion to the coaches instead of a requirement, for them to follow as they wish.

Especially as I'll be sad when I'm placed in the SoCal B2 team. :whistling:

That still does allow for internal names; Colorado can call their teams "Granite" and "Quartz", for example. This just means that there's an official sheet correlating codes and names, so we can tell that Granite is the A team and Quartz is the B team quickly.

Non-ordered teams should still be allowed; they just have an official code with a fictitious order for ease in sorting.

[quote="Kent Merryfield"]Does anyone else have anything similar to say?[/quote]  New York City always has an N, Y and C team*, and the ranking has definitely not been consistently linear.  (In fact, at NYSML this year, I think they're dividing the teams based on borough and not at all by strength.)  To have teams have two names (one official, one unofficial) seems really confusing.  On the other hand, anything that makes the naming scheme less like that of PUMaC and HMMT is probably a good thing.

* I recall fondly the look on a fellow-competitor's face when he heard we had a Y team and assumed it was our 25th.

I will not complain about this change, if and only if it doesn't stifle creativity in team naming.  Naming your teams A, B, C,... is boring; naming them A, B, N, Y, C is a little better; naming them Fire, Ice, Lightning, and Thunder is great.  I think it will be confusing to see entries on the list like:
New York City A1: A
New York City A2: N

But maybe for that reason, teams that usually have less interesting team names will have an incentive to come up with something better.

I always assumed that the assignment of students to NYC N, Y, and C was, as I put it, "non-monotonic."

[quote="PaulDreyer"]Kent,

The statement about placing teams in decreasing strength was really only intended to apply in the correct separation between division A and B teams.  If your B2 is better than your B1, no big deal.  If your B2 is better than your A2, there's a problem.

No creativity is meant to be stifled.  It does, however, make it easier to differentiate, for example, Colorado Quartz and Granite, which are two teams whose divisions we've gotten wrong in the scoring room in the past.

Paul[/quote]

Then you can simply make the "A" and "B" tags mandatory.

[quote="Differ"]Then you can simply make the "A" and "B" tags mandatory.[/quote]

Not every team puts clever monikers after their region names.  Then, you end up forcing regions to do so if they have multiple teams in a division.

Having a unique tag of <region> <A/B> <#> makes things much easier for the scorers and for figuring out whether teams are in the correct division.  Despite our best efforts, we've had issues with this in the past, and if a small change in naming avoids larger inconveniences down the road, I think it is worth it.

Again, there's no reason for team B1 to be better or worse than team B2.  I seriously don't care if Kent does a intra-division shuffle of names and doesn't tell anyone which is the "worst" team, not that there really is one.

[quote="PaulDreyer"]Not every team puts clever monikers after their region names.  Then, you end up forcing regions to do so if they have multiple teams in a division.

Having a unique tag of <region> <A/B> <#> makes things much easier for the scorers and for figuring out whether teams are in the correct division.  Despite our best efforts, we've had issues with this in the past, and if a small change in naming avoids larger inconveniences down the road, I think it is worth it.

Again, there's no reason for team B1 to be better or worse than team B2.  I seriously don't care if Kent does a intra-division shuffle of names and doesn't tell anyone which is the "worst" team, not that there really is one.[/quote]

If just "A" and "B" are mandatory, they can put the number if they wish to instead of any custom name. The division sorting problems will still be resolved.

Also, it's true that there is no reason for B1 to have any strength relationship with B2; however, as ARML is telling coaches to list $ B1 > B2 > \cdots$, people are going to assume there's a strength comparison when there isn't one. Hence it's better to stick with the most minimal solution that completely solves the problem, to avoid having extra complications (like that of NYC and SoCal).

I can see how the system would work and I don't have many problems with it.

However, I'm hoping that the subsequent decreasing in strength isn't always the best policy.

In a local competition, our school sent two and a half teams.  Our first team was supposed to be way better than the second or third team, but the second team came out ahead (even though only the first team's score counts)

subsequent strength isn't exactly the best policy, to put it that way

for all we know, a "lower" team may finish higher...it could happen (although very unlikely at ARML)

I agree with Differ (as odd as that sounds) and Kent.

I don't understand exactly the nature of why errors are occurring in the scoring room.  If you're willing to elaborate, I'm sure you can get a better solution from the collective effort of members on this forum.  The main problem is that your solution to this problem is not user friendly at all and has failed to consider the emotional value of the students.  I'm also skeptical as to how well it'll work.  If possible, it's preferable to do as much work behind the scenes as efficiently as possible to reach an optimal solution.

[quote]It does, however, make it easier to differentiate, for example, Colorado Quartz and Granite, which are two teams whose divisions we've gotten wrong in the scoring room in the past. [/quote]

That was entirely my fault.  Last year I mistakenly put the two teams in the wrong competition rooms for Power and Team Rounds, which resulted in some swapping of scoresheets in the scoring room the next day.  The proposed naming change won't prevent this problem; teams are assigned numbers for Power Round grading.


[hide]$1=t(2k\pm7t)$, so $t|1$, from which $t=\pm1$ follows.  Then $1\pm7=\pm2k$, so $k=\pm3$ or $k=\pm4$.[/hide]


On the hypotenuse BC of the right triangle ABC consider an interior point D. Let o1 be the circumcenter of ABD, and o2 of ACD. If Bo2 and Co1 intersect at E prove that <BAD=<CAE.

Kvant is not a valid source, please also give the year and the problem number of the source. Thanks.  :)

Using Ceva theorem in trigonometric form BO2, CO1, AE are concurrent iff

 sin CEB/sin EBA * sin BAE/sin EAC * sin ACE/sin ECB =1

We denote angle BAD = x, angle CAD = 90 -x, so it is enough to prove

sin BAE/sin EAC = ctg x, because sin BAD/sin DAC = tgx.

Using facts that:    CO2/sin CEB = BO2/sin BCO2 = BO2 /sin x

AO2/sin EBA = BO2/sin BAO2 = BO2/sin (b+x) or

sin CEB/sin EBA = sinx/sin(b+x)

Analogously sin ACE/sin ECB = sin (j+90-x)/sin (90-x)

sin(b+x) = sin(j+90-x) so from firdst equation we get

sin BAE/sin EAC = ctg x. Hence <BAD = <CAE.

Firstly, thanks prowler for solution. Unfortunately, dear orl, I don't know the details.
 A friend of mine told me the problem with the source.

By the way prowler, I suppose you meant <CBE in first relation.

This is Kvant 2004/3 as far as I remember.

[quote="DusT"]Since it is a a thread with posts only from oe country and even school, I thought to put a post here too. I like it, someone takes at school a problem from Kvant, posts it on Mathlinks, another guy posts the solution, and the other 2 guys comment all this :) :D[/quote]

I see that Orlando is from Moldova too ;) :D .

[quote="Valiowk"]

I see that Orlando is from Moldova too ;) :D .[/quote]

I would rather say I am cosmopolitan.  ;)


geometry
circumcircle
ratio
geometry proposed

Let $ ABC$ be an acute triangle.Let $ A_1,B_1,C_1$ be the projections of $ A,B,C$ onto $ BC,CA,AB$, respectively. $ B_1C_1$ intersects $ AA_1$ by the point $ H_a$. Denote $ K$ the midpoint of $ AH$, where $ H$ is the orthocenter of triangle $ ABC$. Prove that: $ H_a$ is the orthocenter of triangle $ KBC$. :lol:

It's OK if $ K$ or $ H_a$ is the orthocenter of triangle $ H_aBC$, or $ KBC$, respectively!!. I even can state that $ B$ is the orthocenter of triangle $ KH_aC$, they're all OK, my dear friend. :lol: . Anyway, very nice solution, after lunch, I will post my solution using harmonic bundle and Pascal+Desgaurter theorem. :blush:  :D

$\mathcal{K}$ is the circumcircle of the cyclic quadrangle $AC_1HB_1.$ Then $ BH_a$ is the polar of $C$ WRT  $\mathcal{K}$ and $ CH_a$ is the polar of $B$ WRT  $\mathcal{K}.$ Since the polar is perpendicular to the line connecting the center of the circle and the pole, it follows that the center $K$ of $\mathcal{K}$ is the orthocenter of $\triangle BCH_a$ $\Longrightarrow$ $H_a$ is the orthocenter of $\triangle KBC.$

From $ K$, let $ xKy\bot AC, Kz\bot CC_1$. Let $ X$ be the intersection of $ Kz$ with $ BB_1$, then it is easy to notice that $ X$ is the midpoint of $ HB$. Therefore, $ (Kx,KX,KH,KB) \equal{} \minus{} 1$. It is a very well- known result that $ (A,H,H_a,A_1) \equal{} \minus{} 1$. Therefore, $ (CA,CH,CH_a,CA_1) \equal{} \minus{} 1$. Hence, $ (Kx,KX,KH,KB) \equal{} (CA,CH,CH_a,CA_1) \equal{} \minus{} 1$, but $ Kx\bot CA, KX\bot CH, CA_1\bot KH$. Therefore, $ CH_a\bot KB$,   which leads to the fact that   $ H_a$  is the orthocenter of triangle  $ KBC$.
Our proof is completed. :D

There is a shorter solution with cross ratio:

Let $ H_a$ be the orthocenter of $ \triangle KBC$ then $ H_a$ is a unique point lies on $ AA_1$. We are to prove that $ H_a,B_1,C_1$ are collinear. Indeed:
$ (BC_1,BH_a,BB_1,BC) \equal{} (AH_aHA_1) \equal{} (CC_1,CH_a,CB_1,CB)$, implies that $ H_a,B_1,C_1$ are collinear and we are done.

I want to introduce you another solution of this problem (A little longer) :lol: :
[b]Solution[/b]:
Let $ (O)$ be the circumcircle of triangle $ ABC$. Denote $ A_2$ the intersection of $ AA_1$ with $ (O)$. It is obviously that $ \overline {A_1A}. \overline {A_1A_2}\equal{} \overline {A_1B}. \overline {A_1C}\equal{} \overline {A_1H}. \overline {A_1A}$. But, $ (A,H,H_a,A_1)\equal{}\minus{}1$, hence, we get, $ \overline {A_1H}. \overline {A_1A}\equal{} \overline {A_1H_a}. \overline {A_1K}$. Therefore, $ A_1B.A_1C\equal{}A_1H_a.A_1K$.
Hence, $ \frac {A_1B}{A_1K}\equal{} \frac {A_1K}{A_1C}$, which implies, triangle $ A_1BK$ is similar to triangle $ A_1H_aC$, which implies the fact that $ CH_a \bot KB$.
Our proof is completed :lol:


Okay, so I have to do a presentation over mass point geometry. While I was working through the examples shown at the website I was learning it from (http://mathcircle.berkeley.edu/mpgeo/mpgeo.html) I came into trouble with these two questions. Could someone please help me understand them?

1. In triangle ABC, D, E, and F are the trisection points of AB, BC, and CA nearer A, B, and C respectivly. Let the intesection of BF and AE be J. Show that BJ:JF = 3:4 and AJ:JE = 6:1.

2. For an altitude, say AD in triangle ABC, note that CDcotB = DBcotC. Therefore, assign weights proportional to cot B and cot C to C and B, respectivly. Let angle B = 45 degrees, angle C = 60 degrees, and let the angle bisector of angle B intersect AD in E and AC in F. Find AE:ED and BE:EF

I someone could explain the answers it would be much appreciated.

Here's a sketch of how to solve Question 1.  By trisection, we have $3E = 2B + C$ and $3F = 2C + A$.  Eliminating $C$, we get $4B + 3F = A + 6E$.  The left side (when divided by 7) is on $\overline{BF}$, and the right side (when divided by 7) is on $\overline{AE}$, so both sides are equal to $7J$.  In other words, $7J = 4B + 3F$ and $7J = A + 6E$.  The first equation gives the first ratio, and the second equation gives the second ratio.

What extent of mass points do you understand?  The first one is pretty straightforward.  If they are trisection points, you can weight them in 2:1 ratios.

[hide]Placing 1 on A and 2 on C will thus balance out, with a total of 3 on F.  Then consider segment CB.  If 2 is on C then you need 4 on B to balance, making a total of 6 on E.  7 on J balances both AE and BF so you're good to go, and your 6:1 and 3:4's are right there.[/hide]

If you can't follow this, PM me before reading the next solution.

[hide]Doing what it says in the problem a put $\cot{45}= 1$ on C and $cot60 = \frac{\sqrt{3}}{3}$ on B.  Adding these we get a total mass of $\frac{3+\sqrt{3}}{3}$ on point D.   Note that ABD is a 45-45-90 triangle so knowing that AB:BD = $\sqrt{2} :1$ and using the angle bisector theorem, we know that the masses on the points D and A must be in the ratio $\sqrt{2}:1$ but since the current mass on point D is not a multiple of $\sqrt{2}$, I multiplied the masses on B and C by $\sqrt{2}$ to make it so.

So now we have $\frac{\sqrt{6}}{3}$ on B, $\sqrt{2}$ on C, $\frac{3\sqrt{2}+\sqrt{6}}{3}$ on D, and $\frac{3+\sqrt{3}}{3}$ on A, then we add to find the mass on F to be $\frac{3+3\sqrt{2}+\sqrt{3}}{3}$.  These masses make the same pressure on point E, so ratio BE:EF can now be calculated to I believe $\frac{\sqrt{2}+2\sqrt{3}+\sqrt{6}}{2}$

(AE:ED was already determined to be $\sqrt{2}:1$)
((Also, I did this kinda quick without checking and I'm prone to arith mistakes, so work through the numbers to make sure everything checks))[/hide]


f(x)= x^2ln5x,  find f'(3)

so i called  x2= a(X) and ln5x= b(x)

so f'(X)= a(X)b'(x)-a'(x)b(X)

so ........x^2(5/x)- 2x (ln5+lnx) ?????
so is this correct

f'(3) = 3^2(5/3)- 2(3)(ln5+ln3)

i dont know why but i have a feeling i am doing something wrong

The product rule is +, not -.
Also, there's no reason to mix $x$ and $X$; there's only one variable here.


Oh man, triple summation?

For more information, [url=http://mathworld.wolfram.com/Series.html]this[/url] might help.

This should be in HSB.

Thew sigma is a summation.

Take for instance $ \sum{x}$.

Using your example of 1 at the bottom and 10 at the top...

$ \displaystyle\sum_{x \equal{} 1}^{10} x$.

Would mean, starting with 1 and counting up to 10, plugging each of those values in for x, and add them all together.

1+2+3+4+5+6+7+8+9+10 = 55 or $ \displaystyle\sum_{x \equal{} 1}^{10} x \equal{} 55$.  Assuming I made no typing or calculation errors :)

Sigmas are really used more at the intermediate level, such as the evaluated part of the expression you posted. so if you desire more info, post in the High School Intermediate Topics forum.

I think this sounds better in the Intermediate forum  :) .

[quote="shaun ong"]There is a sigma ok.  Above the sigma, there is a 10, below the sigma there is a i=1 . The question goes like this => If x,y are real numbers, try the minimum of: first sigma above thers a 10 ,below theres a i=1 , 2nd sigma above thers a 10, below  j=1, 3rd sigma above  is 10 ,below k=1 |k(x+y-10i)(3x-6y-36j(19x+95y=95k| what is this and how do i read the sigma.[/quote]

$ \sum_{i\equal{}1}^{10} \sum_{j\equal{}1}^{10}\sum_{k\equal{}1}^{10} |k(x\plus{}y\minus{}10i)(3x\minus{}6y\minus{}36j)(19x\plus{}95y\plus{}95k)|$

Is that what you wanted?

Hey wait, didn't JRav post something before?  :o

whats that? I dun really understand....... And that was wat i wanted

Is that even computable? You don't know x or y, do you?

Err in some instances you'll leave your final answer in terms of $ x$ and $ y$.

But there is an absolute value symbol, and I don't want to have to write out 1000 long terms.

May i ask a question? x! is x absolute right? Does absolute means its sure to be at least or more then x? or wat? and |  y | whats this? I remember is something like the absolute sign thing

x!=x factorial

x factorial =1*2*3*...*x

|x|=absolute value of x

|x|=x if x is non-negative
|x|=-x if x is negative.

[quote="1=2"]||x|=x if x is non-negative
|x|=-x if x is negative.[/quote]

Now I may not be the sanest puppet in this show, but I'm pretty sure that
$ |1|\equal{}|\minus{}1|\equal{}1\Leftrightarrow |x|\equal{}|\minus{}x|\equal{}x$

Right!?!?!? :huh:

[quote="igiul"][quote="1=2"]||x|=x if x is non-negative
|x|=-x if x is negative.[/quote]

Now I may not be the sanest puppet in this show, but I'm pretty sure that
$ |1| \equal{} | \minus{} 1| \equal{} 1\Leftrightarrow |x| \equal{} | \minus{} x| \equal{} x$

Right!?!?!? :huh:[/quote]

What's your point?

[quote="igiul"][quote="1=2"]||x|=x if x is non-negative
|x|=-x if x is negative.[/quote]

Now I may not be the sanest puppet in this show, but I'm pretty sure that
$ |1| \equal{} | \minus{} 1| \equal{} 1\Leftrightarrow |x| \equal{} | \minus{} x| \equal{} x$

Right!?!?!? :huh:[/quote]
Technically, $ |x|\equal{}|\minus{}x|\equal{}x$, but if you have a negative number and an absolute value, you might as well put -x.

er........Im still waiting for the answer.

You basically sum up that term in the middle with all 1000 possible combinations of i,j, and k, each ranging from 1 to 10.

[quote="igiul"][quote="1=2"]||x|=x if x is non-negative
|x|=-x if x is negative.[/quote]

Now I may not be the sanest puppet in this show, but I'm pretty sure that
$ |1| \equal{} | \minus{} 1| \equal{} 1\Leftrightarrow |x| \equal{} | \minus{} x| \equal{} x$

Right!?!?!? :huh:[/quote]

Therefore, $ | \minus{} 1| \equal{} |1| \equal{} \minus{} 1$, and $ 1 \equal{} \minus{} 1$. *universe collapses*

@shaun ong: Could you possibly scan the problem, because I can't find a nice and simple way to do this problem, so I think I misinterpreted it...

[quote="igiul"][quote="1=2"]||x|=x if x is non-negative
|x|=-x if x is negative.[/quote]

Now I may not be the sanest puppet in this show, but I'm pretty sure that
$ |1| \equal{} | \minus{} 1| \equal{} 1\Leftrightarrow |x| \equal{} | \minus{} x| \equal{} x$

Right!?!?!? :huh:[/quote]

You have implicitly assumed that $ a |b| \equal{} |ab|$, which is patently false if $ a$ is negative.

[b]shaun:[/b]  What's the source of the problem?  It doesn't look very easy at all.


Let $ C(n) \equal{} \begin{cases} 3n \plus{} 1 & \text{if } n \equiv 1 \mod{2} \\
n/2 & \text{if } n \equiv 0 \mod{2}\end{cases}$

Prove that $ \forall n \in \mathbb{Z}^\plus{}$, there exists a finite integer $ I$ such that $ C^I(n) \equal{} 1$.

Sorry, $ \forall n \in \mathbb{Z}^\plus{}$.

OK well what is the opposite of positive? Negative. As 0 is not negative, it must be the opposite, i.e. positive.

nice, good job, mewto.

and @Zhero that article is a bunch of BS, as we just proved on this thread.


I gave Rmo in 2007
I am not selected but i Want to Know my RMO Marks
So please give a contact no.
or
any website

it depends in which state u r .Please contact ur state coordinator for the purpose.u will get the list [url=http://www.isid.ac.in/~rbb/rmoinmo07.pdf]here[/url]
hence on u can post these kind of india related questions in the [url=http://www.mathlinks.ro/index.php?f=362]India forum[/url]


In the never-ending story, I start out with 1-5 sentences.  The next person writes another phrase, and so on.  I hope to create a LOOOOOOOOOONG story.  Have fun!

[b]START:[/b]

Once upon a time, there lived a ferret whose life was very uneventful.  One day, he was taking a short stroll when suddenly, he saw a wolf who had glinting eyes and a watering mouth.

Once upon a time, there lived a ferret whose life was very uneventful. One day, he was taking a short stroll when suddenly, he saw a wolf who had glinting eyes and a watering mouth.
[b]He looked at professordad at started to lick his lips.  MMM...Tasty![/b]

Once upon a time, there lived a ferret whose life was very uneventful. One day, he was taking a short stroll when suddenly, he saw a wolf who had glinting eyes and a watering mouth.
He looked at professordad at started to lick his lips. MMM...Tasty!
[b]Suddenly, a snarling wolverine named Jongao landed out of nowhere, on the wolf's back!  The wolf turned to Jongao, whereupon the frightened ferret, professordad, ran for his life.[/b]

Once upon a time, there lived a ferret whose life was very uneventful. One day, he was taking a short stroll when suddenly, he saw a wolf who had glinting eyes and a watering mouth.
He looked at professordad at started to lick his lips. MMM...Tasty!
Suddenly, a snarling wolverine named Jongao landed out of nowhere, on the wolf's back! The wolf turned to Jongao, whereupon the frightened ferret, professordad, ran for his life.
[b]"ROAR!" came a scream from behind. Out came the king of all kings KING1 the lion. Gaping at the king of all kings, the wolverine and the wolf starting whimpering and ran for their life. The ferocious lion climbed its crag aloft in the sky, and a sense of satisfaction overwhelmed him as it protected the savanna. He was the gaurdian of the Savanna. Mostly know as the "Dark King".[/b]


hey all national mathletes,
i REALLY want to make nationals and so i was wondering how you guys trained for it and everything. im doin mathcounts over the summer so can you help me?
thanks

I'm sure that the Art of Problem Solving books are very helpful for mathcounts nationals and beyond, but I had never even heard of them either time when I went to nationals.  I practiced with old mathcounts national tests, and got ahead in AlgII/Trig (this really helped, for some reason - maybe made me look at things in a different way), and when I came to a problem I didn't know how to do, I'd look for patterns and eventually derive a formula from them.

Anyway, the main thing is to put lots of time into doing problems, problems, problems, and the distance you put in will pay off in the end (a similar analogy to running, swimming, etc.)

[quote="ComplexZeta"]``You aren't allowed to do a problem you already know how to do.''[/quote]

Naturally, you wouldn't want to use this logic on an actual exam, since in the ideal situation you would then have to leave the test blank.  Of course, you can't know how to do every problem in the world, so you have to improvise.

[quote="ComplexZeta"]Yes, and to quote Professor Ravi Vakil of Stanford, who runs a Putnam seminar, ``You aren't allowed to do a problem you already know how to do.'' I think I already said that, but it's really important.[/quote]

Why?

Because, why do that when you could do a problem that you don't know... and maybe *gasp* learn something!

Only us Indians can really understand what we're sayin'

All your base, your base are belong to us.

IW.OAV

There's not only one possible way to solve a problem. That's one of math's many beauties.

[quote="Idiot_without_a_village"]All your base, your base are belong to us.
IW.OAV[/quote]

you have no chance to survive make your time.

AoPS should help you. also, if you can, find past MC questions, get past handbooks, stuff like that.

Solving a problem you already know how to solve is an escape. You gain so much more from solving a new problem than from reviewing an old one.

Solve a problem in another way than you when you solved it first. Also, I consider generalisations of problems as "another solution".

While I generally agree with Ravi's sentiment, you gotta be careful about telling yourself you know how to solve a problem you really don't know how to solve (even if you solved it once before, doesn't mean you can do it again).


topology
geometry
3D geometry
sphere
function
real analysis
real analysis unsolved

I've two topology questions to ask:

1) Show that for n>=2, any map from RP^n to S1 is null homotopic

2) Let X belongs to R3 be the union of S2 with a chold joining the north and south poles. Describe the universal covering space of X and compute pi1(x)

The first question:
You will need the monodromy theorem or an equivalent. This is a hard problem, and I had it on a final.
[hide="hint"]Do it for the sphere first[/hide]

[hide="Part 1"]We start by proving that every continuous map from $S$ to $\mathbb{S}^{1}$ is a composition $\exp\circ g$, where $S$ is compact, path-connected, and simply connected, $g$ is a continuous function from $S$ to $\mathbb{R}$, and $\exp$ is the standard covering map from $\mathbb{R}$ to the circle.

This lift $g$ is constructed from $f$ by choosing a value at one point and then lifting along curves in $S$; by the monodromy theorem, $g$ is well-defined. This $g$ will also be continuous since $\exp$ is a local homeomorphism, and therefore the image of $g$ will be bounded.
Then $f_{t}(x)=\exp((1-t)g(x))$ is a homotopy between $f$ and a constant map.[/hide]

[hide="Part 2"]For this problem, note that the sphere $\mathbb{S}^{n}$ is compact and simply connected. $\mathbb{P}^{n}$ is $\mathbb{S}^{n}/\{1,-1\}$, so every continuous $f$ from $\mathbb{P}^{n}$ to $\mathbb{S}_{1}$ gives $f'$ from $\mathbb{S}^{n}$ to $\mathbb{S}^{1}$. Lifting this $f'$ to $g$ as before, we can show that $g(x)=g(-x)$ for all $x$, and the homotopy we construct will descend to projective space.[/hide]


Please send a solution, because i really cant solve that one :blush: .

[quote="primoz2"]A trangle ABC is given. The circle which pases through A and B intersects AC, BC at D and E. AB and DE intersect at F; FC and BD intersect at M. Prove that MC^2=MD*MB if and only if MC=FC.
I havent solved it yet, because i just can draw a good picture of it.[/quote]
Look here: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=336205#p336205


[b]Correct Me if I am wrong , I will be greatful :)[/b]

using Pigeon whole principle we have 50 hole , each hole corresponding to person who shake hand with him , Hence the problem is solved . I still very poor in explaining hope anyone can provide better solution

the number of sent cards is $ 50\cdot 25$.on the other hand if there are no such students,then the number of cards is less or equal to
$ \binom{50}{2}$,contradiction...

This is easy to prove with contradiction.

Assume that it is possible that no one would greeted each other.

Let's look at one of them. He sends out $ 25$ cards. So if such configuration is possible he must not recive any cards from those $ 24$ whom he didn't send mail. The same argument is true for any of those $ 50$ sendrs. But that means that every sneder can recive at most $ 24$ cards. So

there were $ 25*50 \equal{} 1250$ cards send, however only $ 24*50 \equal{} 1200$ cards could be recived. Contradiction.

This can be easily generalized to $ n$ people sending $ \frac {n}{2}$ cards to each other.

P.s. I wonder why tha LaTeX isn't working here is my message:[code]This is easy to prove with contradiction.

Assume that it is possible that no one would greeted each other.

Let's look at one of them. He sends out $25$ cards. So if such configuration is possible he must not recive any cards from those $24$ whom he didn't send mail. The same argument is true for any of those $50$ sendrs. But that means that every sneder can recive at most $24$ cards. So

there were $25*50 = 1250$ cards send, however only $24*50 = 1200$ cards could be recived. Contradiction.

This can be easily generalized to $n$ people sending $\frac {n}{2}$ cards to each other.[/code]

[quote="maestronam"]we can use contradiction to solve this problem
suppose that the given proposition is wrong.
then, there must be 25*50=1250cards, which is contradiction that there is less or the same number of 50C2 cards...(cause there is only one card between two people)

this is my first answer, so I'm glad to be corrected by others[/quote]
not the correction, though it would be great if you firstly had a look at previous posts.(especially at mine)

[hide="solution"]
Let $ P$ be the set of possible (undirected) greetings. Let $ \mathcal{F}$ be the family of sets $ G_i$ of "actual" (undirected) greetings sent by each student $ s_i$. It is sufficient to show that $ \bigcap_i \mathcal{F}\neq \emptyset$.
[list]
$ a)\ |P| \equal{} 50\times 49$

$ b)\ \sum_{i \equal{} 1}{|G_i|}^50 \equal{} 50\times 25 > 50 \times 49$.
[/list]
$ \blacksquare$
[/hide]

Yes, I realize this is pretty much exactly the same (albeit more formal) as what everyone else has posted. But come on! It's the first time I've actually managed to solve one of the problems of the day on my own. ... at least I hope. :)

[quote=April]Each of the $50$ students in a class sent greeting cards to $25$ of the others. Prove that there exist two students who greeted each other.[/quote]

Easy!
Total sent cards =$25$×$50$
Thus there exists a person $X$ who receives atleast $25$ cards.
Now $X$ sends $25$ cards and leaving the $25$ people who sent him a card there are at most $24$ people.
Thus he has to send one card to one of the person  who sent him a card.
Thus we get a pair of people who greeted each other.

Hence proved.


function
calculus
derivative
complex analysis

Show that a harmonic function satisfies the formal differential equation:
$ \frac { \partial ^2 u }{ \partial z \partial {\overline z} } \equal{} 0$.

What did you try? It's a direct calculation

I've thought about it a bit more and I think I have a rough idea, but I'm not sure if it is right. I'm not very familiar with partial derivatives.
If $ z \equal{} x \plus{} iy$, then
$ \frac {\partial u}{\partial z } \equal{} \frac {1}{2} ( \frac {\partial u}{\partial x} \minus{} i \frac {\partial u}{\partial y} )$
and
$ \frac {\partial u}{\partial \overline z } \equal{} \frac {1}{2} ( \frac {\partial u}{\partial x} \plus{} i \frac {\partial u}{\partial y} )$.
Then multiplying these two gives:
$ \frac {\partial ^ 2 u} {\partial z \partial \overline z } \equal{} \frac {1}{4} ( \frac {\partial^2 u }{\partial x^2 } \plus{} \frac {\partial^2 u}{\partial y^2 } ) \equal{} 0$.

But I'm not sure if I can just multiply the partial derivatives, i.e. is the following always true: $ \frac {\partial u}{\partial z } \frac {\partial u}{\partial \overline z } \equal{} \frac {\partial ^ 2 u} {\partial z \partial \overline z }$?
The first equation I found in my book without much justification; why is it true? Does it follow from $ z \equal{} x\plus{}iy$; where does the $ \frac {1}{2}$ come from?

Ah, no you cannot multiply the partial derivatives in general. The notation means to first "differentiate with respect to z" and then "differentiate with respect to $ \bar{z}$". The other order should produce the same result (and this is checked by calculation, because this whole thing is just formal).

So once you have the first expression, you now must differentiate it w.r.t to $ \bar{z}$ and see that it gives zero. Remember that the partial differentiation operators are linear both w.r.t to addition and constant multiplication.

The first equation is true because it's the definition. This is not really a derivative in the classical sense, it's just a linear operator (just like the true partial derivates in $ x$ and $ y$). In mathematics one likes to describe things in an abstract framework, and by saying that the partial derivative is a linear operator, one realizes that there is perhaps much to gain by considering other linear operators.


search
algebra
difference of squares
special factorizations

A positive integer is the product of n distinct primes. In how many ways can it be represented as the difference of two squares.

[hide="If 2 is one of them"] $ 0$ ways.  A difference of squares $ (a+b)(a-b)$ consists of two factors of the same parity, which is not possible if only one power of $ 2$ is present. [/hide]
[hide="If 2 is not one of them"] $ 2^{n-1}$ ways.  This is the number of subsets of primes that can be chosen for one factor out of $ (a+b), (a-b)$, but we divide by two because we get a negative value of $ b$ half of the time. [/hide]

you can take a look at [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=585855585&t=134105]this[/url] too.


Express as a common fraction the product of
\[ (1 \minus{} \frac{1}{3})(1 \minus{} \frac{1}{4})(1 \minus{} \frac{1}{5})\cdot \ldots \cdot (1 \minus{} \frac{1}{100})\]

[quote="GameBot"]Express as a common fraction the product of
\[ (1 \minus{} \frac {1}{3})(1 \minus{} \frac {1}{4})(1 \minus{} \frac {1}{5})\cdot \ldots \cdot (1 \minus{} \frac {1}{100})
\]
[/quote]

This is

\[ \frac{2}{3} \cdot \frac{3}{4} \cdot \frac{4}{5} \cdot \ldots \cdot \frac{99}{100} \]

As you can see, the terms cancel out, leaving us with $ \frac{2}{100} \implies \frac{1}{50}$.


[quote="k00lperson"]How many positive divisors does the number 2 times 3 to the power of 2 times 5 to the power of 3 have? :?:[/quote]

OK, so basically you wanna know how many factors $ 2^1*3^2*5^3$ has?

You have 2 choices for the power of 2 you want: 0 or 1
You have 3 choices for 3 and 4 for 5,
So the answer is $ 2*3*4\equal{}24$, i believe

There's a formula. Add 1 to all the exponents in the prime factorization and then multiply them. Your number is
\[ 2\cdot3^2\cdot5^3
\]
We now do
\[ (1 \plus{} 1)(2 \plus{} 1)(3 \plus{} 1) \equal{} 2\cdot3\cdot4 \equal{} \boxed{24}
\]
Hope that helped!


Let a,b,c are positive real numbers,and    {a + b}{b + c}{c + a} = 1
proof:$ {ba} \plus{} {ac} \plus{} {bc}\leq\frac {3}{4}$
thank you

[quote="Frank-Jun"]Let a,b,c are positive real numbers,and    (a + b)(b + c)(c + a) = 1
proof:$ ba \plus{} ac \plus{} bc\leq\frac {3}{4}$
thank you[/quote]
First,we prove that: $ 9(a\plus{}b)(b\plus{}c)(c\plus{}a) \ge 8(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)$
It is equivalent to:$ a(b\minus{}a)^2\plus{}b(c\minus{}a)^2\plus{}c(a\minus{}b)^2 \ge 0$
Which is obvious true.
But $ a\plus{}b\plus{}c \ge \sqrt{3(ab\plus{}bc\plus{}ca}$,then$ 9 \ge 8\sqrt{3(ab\plus{}bc\plus{}ca)^3}$
So $ ab\plus{}bc\plus{}ca \le \frac{3}{4}$ :)


I am looking for answer sheet for chapter competition for year 1997. Is there anyone who has that currently. I seem to have lost mine.

Thanks,

i believe you can find it on the internet


Again , I use my technique and found another inequality. It seems very nice to me and I hope I can receive a nice solution too.  Here it is:
If $a,b,c$ are the side of a triangle then:
$2[a(b^3+c^3)+b(a^3+c^3)+c(a^3+b^3)] \geq 2(a^2c^2+b^2c^2+b^2a^2)+(a^4+b^4+c^4)+abc(a+b+c)]$
  Cheers ! :D

If it's true , it can be solved similarly Gibb 1 ,2,3,4 .  ;)

Rewrite it as $\sum (ab+bc+ca-a^2-b^2)(a-b)^2 \ge 0$
WLOG assume $a \ge b \ge c$, if all the coefficients of the squares are above 0 then it's done. Obviously, (ab+bc+ca-b^2-c^2) and (ab+bc+ca-a^2-c^2) are $\ge 0$. If the remaining one is not larger than 0, then we use this
$(ab+bc+ca-b^2-c^2)(b-c)^2+(ab+bc+ca-a^2-c^2)(a-c)^2 +(ab+bc+ca-a^2-b^2)(a-b)^2$
$\ge (ab+bc+ca-b^2-c^2)(b-c)^2+(ab+bc+ca-a^2-c^2+ab+bc+ca-a^2-b^2)(a-b)^2$
$= (ab+bc+ca-b^2-c^2)(b-c)^2+(2ab+2bc+2ca-2a^2-c^2-b^2)(a-b)^2$
$=(ab+bc+ca-b^2-c^2)(b-c)^2+(2a(b+c-a)-(b-c)^2)(a-b)^2$
$=(ab+bc+ca-b^2-c^2-(a-b)^2)(b-c)^2+2a(b+c-a)(a-b)^2$
$=(a(b+c-a)+2b(a-b)+c(b-c))(b-c)^2+2a(b+c-a)(a-b)^2 \ge 0$

Soarer!
  Nice solutions. It is just perfectly!. I admire you. Well, In fact there is also one-line solution too but the way to find it is very hard and it would take a lot of time and skill. Yours is nice already!

Man, why don't you just post it if it's that short.

I think the above solution is better (because it's natural to me), but maybe this is what you mean.
$\sum (c^2-(a-b)^2)(a-b)^2 \ge 0$

Well, I think mine is exactly like Soar . Anyway, I wonder why Soar can have a lot of sols for that like this. Finding 1 of them is very hard already. :D
   Moreover, I am sorry you guys for my temper from the previous post! :blush:  I just don't like other who told me that my inequality is not made by mine and that my answear is copied from others . I do understand that there are billion of genius in the world can generate it and answear it easily.
   My technique is very simple.Dealing with triangle ineq, you have to focus on the fact that $a-b+c>0$... and when inequa is hold in other to put $(a-b)^2$ or $(a-2b)$ or $(a+b-2c)$... and also the way of homogenous and so on... Moreover, the skill of divide is also need to manipulate with them.
   But what happen if s.o do vice versa. You know what I mean ? If they just expand the obviously fact like $\sum(a+b-c)(a-b)^4 \geq 0$ or put more condition in oder to make that is true such as $a+b>3c$.... If you do like this, there will be ton of inequalities....
And I doubted that the author of IMO 83 using the expand  to generate his iequality.
   But anyway, my major is chemistry so I don't need to do much with inequa.   :lol:


ding, didnt you say youu werent going to any
camps?
if you're going to apply to AMSP thats great
because i am too....

[quote="korammath"]ding, didnt u say u werent going to ne camps?
if ur going 2 apply 2 AMSP thats gr8
cuz i am 2....[/quote]

It would be appreciative if you used proper english. Specifically, I've never seen "ur," "2", "gr8", "ne," or "cuz" in any credible dictionary.

whats wrong with the shortcuts?

and korammath, i might go, i don't know yet.

It's just not professional. You don't see that kind of language in published papers, and with good reason too. It's hard enough to follow someone's thought process at times; don't make someone work to follow your abbreviations as well.

Oh, I thought the discussion was about writing up solutions; I use abbreviations all the time in my conversations.

[quote="dingzhou"]basically, what the title says..

and i don't have much experience in writing proofs..

can someone give me a sample from admission test A or previous tests?[/quote]
why would you need old tests... just do this one. its not that hard and your better than me.... i didnt do the proofs but i did other ones. just show how you did the problems.

another question, how neat does the solution have to be? do they expect pretty much perfect spelling and grammar? and i wrote my solution, so theres some places that i had to scratch out cause i use pen, is that fine?

It doesn't have to be perfectly neat, but I think as long as they can read your handwriting, and can understand what you're trying to say, it should be okay. I hope scratches in pen is okay, since I did that quite a bit, and I already sent mine...

The whole point of a proof is to convince the reader that you have solved the problem. Hence it doesn't matter so much how well it is written as long as it is understandable. In general, it is in your best interest to write well because it makes it more likely that the grader will understand what you are writing. Scratch-outs are generally not a problem.

Just as a note, here is my admissions test from last year. Looking back at it, I think that it was okay, but I was pretty lazy and didn't make pictures for the geometry problems and such. Also I started the test the day before it was due so I sorta was rushing to finish haha.


calculus function limit trigonometry integration probability logarithms

I'm teaching a Calculus II course right now. We're studying the representation of functions by power series. My theme for the day was why I make very little routine use of L'Hôpital's rule. In the problems that follow (my classroom examples for the day), I don't forbid the use of L'Hôpital - I merely suggest that using it would often be the hard way. Why give yourself extra headaches?

$1.\ \lim_{n\to\infty}\left(\sqrt[3]{n^{3}+n^{2}}-\sqrt[3]{n^{3}-n^{2}}\right)$

$2.\ \lim_{x\to0}\left(\frac{\sin x}x\right)^{\frac1{x^{2}}}$

$3.\ \lim_{x\to 0}\frac{1-\cos(x^{3})}{\sin(x^{2})\sin^{4}(x)}$

$4.$ Let $m(t)=\frac13\int_{2}^{5}e^{tx}dx$ (It's from my other class, the probability class: the moment generating function of a particular uniform distribution.) Verify that $m(0)=1$ and compute $m'(0)$ and $m''(0).$

For the first limit we use the following trick:

$a-b=\frac{a^{3}-b^{3}}{a^{2}+ab+b^{2}}$, then factor out both numerator and the denominator by $n^{2}$.

2). $\left(\frac{\sin x}{x}\right)^{1/x^{2}}\rightarrow e^{\frac{\sin x-x}{x^{3}}}=e^{-1/6}$, since $\sin x \approx x-\frac{x^{3}}{6}$ near $x=0$.

3. $\frac{1-\cos x^{3}}{x^{6}}\frac{x^{2}}{\sin x^{2}}\frac{x^{4}}{\sin^{4}x}\rightarrow \frac{1}{2}$.

On the first one, given my emphasis for the day on series, I didn't want to use algebra, so I didn't use anything like that identity. Instead, I expanded the two roots using the binomial series.

Yes, that's the right answer for #3. It's supposed to be something you could do in your head.

That's the right answer for #2, but I'm not 100% sure I trust your intermediate step. My work involved taking the logarithm of the whole expression.

The steps are correct.
Calculate the two limits from the left and from the right and since the limits are equal the result follows.
Didi

The actual methods from my lecture earlier today:

1. Write $\sqrt[3]{n^{3}+n^{2}}=n\left(1+\frac1n\right)^{\frac13}= n\left(1+\frac13\cdot\frac1n+O(n^{-2})\right)$

$=n+\frac13+O(n^{-1})$

Similarly, $\sqrt[3]{n^{3}-n^{2}}=n-\frac13+O(n^{-1})$

Subtract to get $\frac23+O(n^{-1})$ and the limit is $\frac23.$

2. Both didilica and Virgil Nicula have assumed without further comment that if $f(x)\to 0$ and $g(x)\to\infty$ then $\lim\left(1+f(x)\right)^{g(x)}=e^{\lim f(x)g(x)}.$ This is true, but it's not something I would presume that my students know. Hence:

$L=\lim_{x\to0}\left(\frac{\sin x}x\right)^{1/x^{2}}$ so

$\ln L=\lim_{x\to0}\frac1{x^{2}}\ln\left(\frac{\sin x}x\right)$

$=\lim_{x\to0}\frac1{x^{2}}\ln\left(1-\frac{x^{2}}6+O(x^{4})\right)$ (that used the power series of the sine)

Now note that $\ln(1+u)=u+O(u^{2})$ as $u\to0$ to get

$\ln L=\lim_{x\to 0}\frac1{x^{2}}\left(-\frac{x^{2}}6+O(x^{4})\right)$

$=\lim_{x\to 0}-\frac16+O(x^{2})=-\frac16$ so $L=e^{-\frac16}.$

#3 was done by similar arguments to what didilica and Virgil Nicula used, with the additional comment that these arguments take no cleverness and can be employed quickly. We know from the power series that $1-\cos u$ looks like $\frac{u^{2}}2.$ We just have to see $1-\cos(x^{3})$ as essentially $\frac{x^{6}}2$ and that mess in the denominator as essentially $x^{6}.$

The example in #3 was chosen so that any attempt to use L'Hôpital blindly (with no algebraic cleverness) would be "rewarded" with the sort of notational nightmare that few beginning students could wade through - while all along, the experienced among us see that the answer is $\frac12$ without writing anything down.


Let ABC is acute-angled triangle and $ X \in \Delta ABC$
 $ AB\cdot BC\cdot CA \ \equal{} \ XA\cdot AB\cdot BX \plus{} XB\cdot BC\cdot CX \plus{} XC\cdot CA \cdot AX$
Find X

Well-knowm:
 $ XA\cdot AB\cdot BX \plus{} XB\cdot BC\cdot CX \plus{} XC\cdot CA \cdot AX \ge AB\cdot BC\cdot CA$
The quality holds if and only if H is orthocentre of ABC.


[quote="155488"]x,y,z>0; x+y+z=$ x^{3} + y^{3} + z^{3}$ Prove
${{{ \frac {z^{2}x}{y^{2}(x^{2} + 1)}} + \frac {x^{2}y}{z^{2}(y^{2} + 1)}} + \frac {y^{2}z}{x^{2}(z^{2} + 1)}}\ge\frac {x + y + z}{2}$[/quote]
Using the Cauchy-Schwarz Inequality, we have
$ [ \sum \frac{z^2x}{y^2(x^2+1)}] [ \sum x(x^2+1)] \ge ( \sum \frac{zx}{y})^2.$
Therefore, it suffices to prove that
$ (\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y})^2 \ge \frac{(x+y+z)(x^3+y^3+z^3+x+y+z)}{2}.$
Since $ x^3+y^3+z^3+x+y+z=2(x+y+z),$ this inequality can be written as
$ (\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y})^2 \ge (x+y+z)^2,$
or
$ \frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y} \ge x+y+z,$
which is a known result.

[quote="fjwxcsl"][quote="hedeng123"][quote="155488"]$ x,y,z > 0; x + y + z = x^{3} + y^{3} + z^{3}$ Prove
${{{ \frac {z^{2}x}{y^{2}(x^{2} + 1)}} + \frac {x^{2}y}{z^{2}(y^{2} + 1)}} + \frac {y^{2}z}{x^{2}(z^{2} + 1)}}\ge\frac {x + y + z}{2}$[/quote]

${{{ \frac {z^{2}x}{y^{2}(x^{2} + 1)}} + \frac {x^{2}y}{z^{2}(y^{2} + 1)}} + \frac {y^{2}z}{x^{2}(z^{2} + 1)}}\ge \frac {(\frac {z}{y} + \frac {x}{z} + \frac {y}{x})^2}{\sum {\frac {x^2 + 1}{x}}} = \frac {(\frac {z}{y} + \frac {x}{z} + \frac {y}{x})^2\frac {\sum {x^3}}{\sum x}}{\sum {\frac {x^2 + \frac {\sum x^3}{\sum x}}{x}}}$

we just need to prove

$ \frac {(\frac {z}{y} + \frac {x}{z} + \frac {y}{x})^2\frac {x^3 + y^3 + z^3}{x + y + z}}{\sum {\frac {x^2 + \frac {x^3 + y^3 + z^3}{x + y + z}}{x}}} \geq \frac {x + y + z}{2}$

the inequality holds,I have no way to prove that. :lol:[/quote]

>f:=csum(z/y)^2*csum(x^3)/csum(x)-csum(x)/2*csum((x^2+csum(x^3)/csum(x))/(x)):

>csymprove(f);

                         ff=subs({y=z,z=y},f)

                         f+ff  is symmetric for {z, x, y}

                              DEGU = 3
                               DEG = 0

                  The original form is psd!(SYM3(1))

                           它是平凡非负的!

                             TIME = 0.172

                         [f][ff] is symmetric for {z, x, y}

                               DEGU = 6
                               DEG = 0

                  The original form is psd!(SYM3(1))

                           它是平凡非负的!

                             TIME = 1.047

                   The original form is psd!(SYM32)[/quote]

I think I can understand your idea of proof.

 :lol:


On Fe 21, 2006 Ductrung proposed this nice ineq

$ \sum_{cycl}\sqrt{\frac{b\plus{}c}{a}}\ge\sqrt{6}\frac{a\plus{}b\plus{}c}{\sqrt{ab\plus{}bc\plus{}ca}}$

I tried to prove it using only Cauchy and AM-GM but I failed.

Dear Ductrung, can you please give me a hint to solve it?

Thank you very much.

P.S. I tried to solve it using the squaring form

$ \sum\frac{b\plus{}c}{a}\plus{}\sum\sqrt{\frac{b\plus{}c}{a}}\sqrt{\frac{c\plus{}a}{b}}\geq 6\frac{(\sum a)^{2}}{\sum ab}$

then I used Cauchy

 $ (\frac{b\plus{}c}{a})(\frac{c\plus{}a}{b})\equal{} (1\plus{}\frac{c}{a})(1\plus{}\frac{c}{b})\geq (1\plus{}\frac{c}{\sqrt{ab}})^{2}$

but I failed. :(

[quote="ductrung"][quote="manlio"]
$ \sum_{cycl}\sqrt{\frac{b\plus{}c}{a}}\ge\sqrt{\frac{6(a\plus{}b\plus{}c)^{2}}{ab\plus{}bc\plus{}ca}}$
[/quote]

Hint to Manlio:
[b]1st solution[/b]: Chebyshev and Cauchy
[b]2nd solution[/b]: Holder and AM-GM[/quote]
     your asteroid doesn't send out demob profit ?
    can you post you solution?

Thank you very much, Ductrung and Silouan. :)

[quote="ductrung"][quote="manlio"]
$ \sum_{cycl}\sqrt{\frac{b\plus{}c}{a}}\ge\sqrt{\frac{6(a\plus{}b\plus{}c)^{2}}{ab\plus{}bc\plus{}ca}}$
[/quote]

Hint to Manlio:
Chebyshev and Cauchy
[/quote]

WLOG $ a\geq b\geq c$ then $ a\plus{}b\geq a\plus{}c\geq b\plus{}c$ and
$ \frac{1}{c}\geq\frac{1}{b}\geq\frac{1}{a}$ so by Chebyshev

$ \sum\sqrt{\frac{b\plus{}c}{a}}\geq\frac{1}{3}\sum (\frac{1}{\sqrt{a}})\sum (\sqrt{b\plus{}c})$

This way do you use Chebychev?

Thank you very much.

[quote="silouan"]Very nice inequality
I will follow the second hint  :P  (I like Holder)
By holder we have
$ (\sum_{cycl}\sqrt{\frac{b\plus{}c}{a}})^{2}(\sum (a(b\plus{}c)^{2}))\geq 8(a\plus{}b\plus{}c)^{3}$

After some trivial munipulation we have to prove that $ 4(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)\geq 3\sum (a(b\plus{}c)^{2})$ which is equivalent to
$ \sum (a^{2}b\plus{}b^{2}a)\geq 6abc$  which is a plain AM-GM QED  :wink:[/quote]
Yes, silouan  :lol:

[quote="manlio"]
This way do you use Chebychev?
[/quote]
\[ \sum_{cycl}\sqrt{\frac{b\plus{}c}{a}}\equal{}\sum_{cycl}\frac{b\plus{}c}{\sqrt{a(b\plus{}c)}}\cdots\]

Thank you very much, Ductrung. Now it works.  :)

Very very nice problem!
My solution:
By Cauchy-Schwarz we have:
$ \sum{\sqrt{\frac{b+c}{a}}}=\sum{\frac{(b+c)^2}{(b+c)\sqrt{a(b+c}}\ge }$
$ \ge\frac{(b+c+c+a+a+b)^2}{\sum{(b+c)\sqrt{a(b+c)}}} $
By Chebyshev we have:
$ \sum{(b+c)\sqrt{a(b+c)}}\le\frac{1}{3}(\sum{(b+c)})(\sum{\sqrt{a(b+c)})}\le $
$ \le\frac{2(a+b+c)}{3}\sqrt{6(ab+bc+ca)} $
We used again Cauchy-Schwarz:
$ \sum{\sqrt{a(b+c)}\le\sqrt{6(ab+bc+ca)} }$
After some simple calculations we obtain:
$ \sum_{cyc}{\sqrt{\frac{b+c}{a}}}\ge (a+b+c)\sqrt{\frac{6}{ab+bc+ca}}\Leftrightarrow QED $

There are two inequalities stronger than ductrung's inequality:
(1)  $ \sum{\sqrt{\frac{b+c}{a}}}\ge 2(a+b+c)\sqrt{\frac{2(a+b+c)}{4abc+(b+c)(c+a)(a+b)}} $
(2)  $ \sum{\sqrt{\frac{b+c}{a}}}\ge\frac{6(a+b+c)}{\sum{\sqrt{a(b+c)}}} $
Amounts are cyclical!
The proofs for these are elementary!I will post them later !!
_____
Marin Sandu

[quote="sandumarin"]There are two inequalities stronger than ductrung's inequality:
(1)  $ \sum{\sqrt{\frac{b+c}{a}}}\ge 2(a+b+c)\sqrt{\frac{2(a+b+c)}{4abc+(b+c)(c+a)(a+b)}} $
[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3679349#p3679349

[quote="sandumarin"]
(2)  $ \sum{\sqrt{\frac{b+c}{a}}}\ge\frac{6(a+b+c)}{\sum{\sqrt{a(b+c)}}} $
[/quote]
We can prove this inequality by C-S and AM-GM.


limit
function
calculus
calculus computations

[quote="PatC"]Does anyone know how to do: Lim as X :arrow: -Infinity of X(e^X)

Test tomorrow help is greatly appreciated thanks in advance![/quote]

\[ \mathop {\lim }\limits_{x \to  \minus{} \infty } xe^x
\]

^^Right, that. Do you know how to do it?

you want to express that term as a quotient of two functions so you can actually apply l'hopitale's rule. I hope you know how L'Hs works, if not then google is your friend.

I understand the rule and have worked through a few problems but I am having trouble on this one. Would you mind working through it and telling me what you get?

Or, you could become familiar with the observation that the exponential function $ e^x$ decreases to $ 0$ faster than any power of $ x$ decreases to $ \minus{}\infty$ as $ x \to \minus{}\infty$.  Thus,

$ \lim_{x \to \infty} [x \ e^x] \equal{} 0$.

There may be other methods of evaluating this limit, but they all lead to the same result.  What I have described is a intuitive understanding that would be helpful to become very familiar with (among other "common" limits).

^^That's exactly what I was wondering because it says to apply L'Hospital's Rule but I didn't even think I needed it. Thanks!

I believe you could express the problem as limit bla bla of $ \frac{x}{e^{\minus{}x}}$ and use l'hopitale's from there.

And I arrive at 0 once again, thank you for confirming that. I was looking at it in the wrong way and confusing myself for no reason.


Let system of equation: $ \begin {cases} x^2y \plus{} a \equal{} y^2\\y^2x \plus{} a \equal{} x^2 \end {cases}$
Find all $ a < 0$ for this system have only one root $ (x;y)$ with $ x,y \in R$.

If $ (x,y)$ is a solution then $ (y,x)$ is also a solution $ \Rightarrow$  we must have $ x=y$
We have \[ x^3+a=x^2\]
\[ y: =\Leftrightarrow x^3-x^2=a\]
\[ y'=x^2-2x\]
\[ y'=0\Leftrightarrow \[
\begin{array}{ll}
x=0&\\
x=2&.
\end{array}
\]
\[ y(0)=0\]
\[ y(2)= 4\]
thus $ a\leq 0$ or $ a\ge 4$ then $ y=0$ have one only  solution
$ \Leftrightarrow$ This system have 1 solution.

If I am not mistaken,
\[ y' \equal{} 3x^{2} \minus{} 2x\]

Also, it doesn't follow that just because $ (x,y)$ is a solution, $ (y,x)$ is a solution that $ x\equal{}y$. Consider $ x\plus{}y\equal{}k$, which has that property but has infinitely many solutions that are not $ x\equal{}y$.


Hi it is really urgent to me: Can you please give me a short summary of the book Great Expectations, that is, to rewrite the book in 200--400 words.

Thanks in advance.

Sounds like homework...or someone forgot to read.  Anyway, sparknotes.com is the best place for slacking off (it's also good even if you're not slacking off, however).

Yes it is homework...... :P
Well, books by Dickens are really difficult for me to understand. I've got 1/3 of it and gave up... The homework is to read the symplified Great Expectations, which is about 50 pages long,  :o   but I don't like reading uncompleted novels, that's why I am having this problem...

Thanks for your help!  :D

This is my $2000's$ post!?  :huh:  :roll:  :D

[quote="rgep"]Time for a shortest summary competition.  Here's mine

Pip is scared by convict and batty old lady, inherits piles of cash, not from lady so gives it up, gets old friends into trouble, lives happily/unhappily ever after [depending on version].[/quote]You should compete in Monty Python's All-England Summarizing Proust Competition. Contestants are given 15 seconds to summarize Proust's A La Recherché du Temps Perdue, once in a swimsuit and once in evening dress. :rotfl: One of my favorite Python sketches.


vector
induction
linear algebra
linear algebra unsolved

Let $ A_1,A_2,\ldots\,A_d$ be real symmetric matrices and let them form a basis of for the linear vector space $ \mathcal{A}$.  Show that $ \mathcal{A}$ has a basis $ \{E_1,E_2,\ldots,E_d\}$ such that

\[ E_iE_j\equal{}
\begin{cases}
 0 & \text{if\  \(i\ne j\),}\\
E_i &\text{if\  \(i\equal{}j\).}
\end{cases}\]

If this implication holds, $ \mathcal A$ is necessarily a commutative algebra, i.e. $ A_iA_j\equal{}A_jA_i$ for every pair $ i,j$.

If the last condition is additionally assumed, then apply induction on $ d$ using in the induction step that $ \mathbb R^n\equal{}\bigoplus_{m\equal{}1}^{d\minus{}1}V_m$  and $ A_d$ maps each $ V_m$ to itself where  $ V_m\equal{}E_m(\mathbb R^n)$.

Sorry but i do not understand your proof.

Assuming that you have constructed $ E_1, E_2, \ldots, E_{d\minus{}1}$, how do you construct $ E_d$?

[quote="t0rajir0u"]I'm not sure I understand the result.  How can this be generically true for $ d \equal{} 1$?[/quote]Apologies for the late reply. Reading your lines I am not even sure that I understand my post. ;) I was driven by the fact that a commutative algebra of symmetric real matrices yields a decomposition of $ \mathbb R^n$ such that the algebra acts by scalar multiplication on each of the components. I tried to amend the statement by assuming the matrices commute with each other, apparently this is not enough. ... Certainly the statement as it is seems to fail for n=1 as you have pointed out. Thanks again.

My apologies for the late reply.  There are quite a few assumptions that must be made on $ A_i$. Please refer to a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=235437]new topic[/url] i just posted.

Thanks!


I am now going to start up my own contest on this forum.  And I will post the top scores later, answers need to be PMed to me.  The deadline is by Christmas Day.  I will score with 5 points for a correct answer, if the problem is left blank, then 2 points, and if it is wrong, then 0 points.

[b]Round 1[/b]

[b]1.[/b] If M is 40% of Q, Q is 15% of P, and N is 50% of P, then M/N = ...

--------------------------------------------------------------------------------------

[b]2.[/b] How many base 10 four-digit numbers, N = [u]a[/u] [u]b[/u] [u]c[/u] [u]d[/u], satisfy all three of the following conditions?
([i]i[/i]) 4,000 :le: N < 7,000; ([i]ii[/i]) N is a multiple of 5;
([i]iii[/i]) 3 :le: b < c :le: 7.

--------------------------------------------------------------------------------------

[b]3.[/b] The area of the triangle bounded by the lines y = x, y = -x and y = 18 is ...

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[b]4.[/b] Seven points on a circle are numbered 1, 2, 3, 4, 5, 6, and 7 in clockwise order.  A bug jumps in a clockwise direction from one point to another around the circle; if it is on an odd-numbered point, it moves one point, and if it is on an even-numbered point, it moves two points.  If the bug begins on point 6, after 1995 jumps it will be on the point numbered ...

--------------------------------------------------------------------------------------

[b]5.[/b]Two rays with common endpoint O form a 30 :^o: angle.  Point A lies on one ray, point B on the other ray, and AB = 2.  The maximum possible length of OB is ...

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[b]6.[/b] If log[size=75]24[/size]p = log[size=75]9[/size]q = log[size=75]64[/size](p + 2q), what is p/q?

--------------------------------------------------------------------------------------

[b]7.[/b] Two parallel chords in a circle have lengths 40 and 56, and the distance between them is 24.  The chord parallel to these chords and midway between them is of length :sqrt:a where a is ...

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[b]8.[/b] How many positive integer solutions (x, y) are there that satisfy 1/x + 1/y = 1/36?

--------------------------------------------------------------------------------------

[b]9.[/b] For how many three-element sets of positive integers {a, b, c} is it true that a * b * c = 2730?

--------------------------------------------------------------------------------------

[b]10.[/b] If :sqrt: (4 + :sqrt: x) =5, then x = ...

*If you have any questions or suggestions regarding this, contact me at WarpedKlown51435 on AIM or email me at soccerboyyp35@yahoo.com or yanman12@hotmail.com*

How are you going to score it?

-interesting_move

5 points for corrects, 2 points for blanks, and 0 points for wrong answers.

Seeing how the deadline has passed...

[hide]You could consider 2 cases (both chords are on the same side of the centre / the chords are on opposite sides of the centre), but it's really not necessary.

Construct a line through the centre, perpendicular to both chords. Let the distance between the chord of length 40 and the centre of the circle be x; let the distance between the chord of length 56 and the centre of the circle be y. Then,

x + y = 24

x2 + 400 = r2

y2 + 784 = r2

where r is the radius of the circle. Solving yields x = 20, y = 4, r2 = 800. So the desired chord will be 12 units from each given chord; that is 8 units from the centre. Let the length of the desired chord be z. Then,

64 + (z / 2)2 = 800

Solving yields z = :sqrt: 2944.[/hide]

two questions, what if both cords are on the same side of the circle?
and how did you get
x2 + 400 = r2
y2 + 784 = r2

he got those from the pythagorean theorem. I think if you assume both chords are on the same side of the circle, you get that some distance is negative or something. (while that distance would be positive if you had the chords on opposite sides)

Do you want me to have a separate topic about the AMC problems like what mathfanatic did before or do you like it how it is right now?  I want to give people some time to go over the problems before I have new problems, which in fact I hope will be done around sometime next week.


for all reals $ x,y,z > 0$ prove that:

$ \frac{x^{2}}{y^{2}} \plus{} \frac{y^{2}}{z^{2}} \plus{} \frac{z^{2}}{x^{2}} >\equal{} \frac{x}{y} \plus{} \frac{y}{z} \plus{}\frac{z}{x}$

it is preferably supposed to be solved by Hardy-Polya...(but other ideas are appreciated as well)

[quote="farbod_mp2002"]for all reals $ x,y,z > 0$ prove that:

$ \frac {x^{2}}{y^{2}} \plus{} \frac {y^{2}}{z^{2}} \plus{} \frac {z^{2}}{x^{2}} > \equal{} \frac {x}{y} \plus{} \frac {y}{z} \plus{} \frac {z}{x}$

it is preferably supposed to be solved by Hardy-Polya...(but other ideas are appreciated as well)[/quote]

$ \sum_{cyc}\left(\frac {x^2}{y^2} \minus{} \frac {x}{y}\right)\geq0\Leftrightarrow\sum_{cyc}\left(\frac {4x^2}{y^2} \plus{} \frac{y^2}{z^2}\plus{}\frac {z^2}{x^2} \minus{} \frac {6x}{y}\right)\geq0,$ which true by AM-GM.

Put $ a\equal{}x/y$, $ b\equal{}y/z$ , $ c\equal{}z/x$.
The inequality to prove now is:
Let $ a,b,c>0$ such that $ abc\equal{}1$, then
$ a2\plus{}b2\plus{}c2>\equal{}a\plus{}b\plus{}c$

Now, by Cauchy Schwarz we have
$ a2\plus{}b2\plus{}c2>\equal{}((a\plus{}b\plus{}c)2)/3>\equal{}a\plus{}b\plus{}c$

The last inequality because $ abc\equal{}1$ and $ (a\plus{}b\plus{}c)/3>\equal{}1$

Q.E.D.

 :lol:

Note: $ x2$ = x square

Multiply the inequality by $ x^2y^2z^2$ and get
$ x^4z^2 \plus{} y^4x^2 \plus{} z^4y^2 \ge x^3yz^2 \plus{} x^2y^3z \plus{} xy^2z^3$.

But it follows from
$ 4x^4z^2 \plus{} y^4x^2 \plus{} z^4y^2 \ge 6\sqrt {x^{18}y^6z^{12}} \equal{} 6x^3yz^2$
and other two similar inequalities. Add up these three inequalities and divide by 6. That gives our inequality
$ x^4z^2 \plus{} y^4x^2 \plus{} z^4y^2 \ge x^3yz^2 \plus{} x^2y^3z \plus{} xy^2z^3$.

More way:
Let $ z\equal{}\min\{x,y,z\}.$
Hence, $ \frac{x^{2}}{y^{2}} \plus{} \frac{y^{2}}{z^{2}} \plus{} \frac{z^{2}}{x^{2}}\minus{}\frac{x}{y} \minus{} \frac{y}{z} \minus{}\frac{z}{x}\equal{}$
$ \equal{}\frac{(x\minus{}y)^2(x^2\plus{}xy\plus{}y^2)}{x^2y^2}\plus{}\frac{(z\minus{}x)(z\minus{}y)(z^2\plus{}xy\plus{}yz)}{x^2z^2}\geq0.$

It is obvious Cauchy-schwarz:    for vektors:   x/y  y/z  z/x  and   y/z   z/x   x/y

[quote="greentreeroad"]arqady how do you try to factor the inequality into that complicate thing.
It looks a very tricky methods for me[/quote]
You can download file and see here: http://www.mathlinks.ro/viewtopic.php?t=168016

thanks everyone for the nice ideas... :)
arqady can you explain your first solution a little bit more??

Here is generalisation : http://www.mathlinks.ro/viewtopic.php?p=934040&sid=d62391e5f1cda8e2683742b3c77ddb9b#934040

another way of solving it would be $ \frac {x^2}{y^2} \plus{} 1 \geq 2\frac {x}{y}$ by AM-GM which solves the inequality since $ \frac {x}{y} \plus{} \frac {y}{z} \plus{} \frac {z}{x} \geq 3$ also by AM-GM

 [size=150][b]even more general[/b][/size]

[b]Inequality[/b]
let $ a_1,a_2,a_3...a_n$ be positive reals and $ r,A,B$ non-negative integers. than the folowing inequality holds  if $ A > B$
$ \sum_{cyc}^{n} \frac {a_i^{r \plus{} A}}{a_{i \plus{} 1}^A} \geq \sum_{cyc}^{n} \frac {a_i^{r \plus{} B}}{a_{i \plus{} 1}^B}$

[b]Proof[/b]
- we prove the first step with rearangement inequality
$ \sum_{cyc}^{n} \frac {a_i^{r \plus{} 1}}{a_{i \plus{} 1}} \geq \sum_{cyc}^{n} \frac {a_i^{r \plus{} 1}}{a_{i}} \equal{} \sum^{n} a_i^r$

-than assume the inequality holds for $ A$ up to some natural $ m \minus{} 1$. meaning $ f(m \minus{} 1) \geq f(m \minus{} 2)... \geq f(1) \geq f(0))$ where $ f(k) \equal{} \sum_{cyc}^{n} \frac {a_i^{r \plus{} k}}{a_{i \plus{} 1}^k}$

-than the following inequallity obviously holds by AM-GM
$ (m \minus{} 1) \sum_{cyc}^{n} \frac {a_i^{r \plus{} m}}{a_{i \plus{} 1}^m} \plus{} \sum^{n} a_i^r \geq m \sum_{cyc}^{n} \frac {a_i^{r \plus{} m \minus{} 1}}{a_{i \plus{} 1}^{m \minus{} 1}}$
because $ (m \minus{} 1) \frac {a_i^{r \plus{} m}}{a_{i \plus{} 1}^m} \plus{} a_i^r \geq m(\frac {a_i^{(r \plus{} m)(m \minus{} 1) \plus{} r}}{a_{i \plus{} 1}^{m(m \minus{} 1)}})^{\frac {1}{m}}\equal{}m \frac{a_i^{r\plus{}m\minus{}1}}{a_{i\plus{}1}^{m\minus{}1}}$

- therefore since by assumption $ \sum_{cyc}^{n} \frac {a_i^{r \plus{} m \minus{} 1}}{a_{i \plus{} 1}^{m \minus{} 1}} \geq \sum^{n} a_i^r$ this means that
$ \sum_{cyc}^{n} \frac {a_i^{r \plus{} m}}{a_{i \plus{} 1}^m} \geq \sum_{cyc}^{n} \frac {a_i^{r \plus{} m \minus{} 1}}{a_{i \plus{} 1}^{m \minus{} 1}}$

-therefore $ f(m) \geq f(m \minus{} 1)$
Q.E.D


for positive reals $a,b,c$
find the best constant $\lambda$ for the following inequality to hold
\[\frac{a^{2}-\lambda bc}{b+c}+\frac{b^{2}-\lambda ca}{c+a}+\frac{c^{2}-\lambda ab}{a+b}\ge \frac{1-\lambda}{2}(a+b+c). \]

If $\lambda >0$, the inequality is obvious. If $\lambda =-k0$, we let $c=0$ and $a=b$ then
\[\sum \frac{a^{2}+kbc}{b+c}=\frac{a^{2}}{b}+\frac{b^{2}}{a}+\frac{kab}{a+b}=(2+\frac{k}{2})a \]
so $k$ must satisfy $2+\frac{k}{2}\ge 1+k$ or $k\le 2$. If $k=2$, the inequality
\[\frac{a^{2}+2bc}{b+c}+\frac{b^{2}+2ca}{c+a}+\frac{c^{2}+2ab}{a+b}\ge \frac{3}{2}(a+b+c) \]
is so familiar. I suggest one solution by SOS and one solution by Cauchy+Schur. Easy.

[quote="anuj kumar"]for positive reals $a,b,c$
find the best constant $\lambda$ for the following inequality to hold

[/quote]
since it's given positive reals i suppose gackiem u can't put $c=0$

I say again and again that in proving inequalities, there is no difference between "positive" and "non-negative". It is just a simple property of limit.


Apply Cauchy_Schwarz inequality and find that the LHS is at most  \sqrt bc+(1-b)(1-c) which is smaller than 1.

In fact i was going to post this here too.
i didn't understand your solution harazi.
Any help please? ;)

Nickolas, LaTeX is really easy to learn, at least the few basics needed to formulate a problem like yours:

[i]If a, b, c are real numbers in the open interval ]0; 1[ (some people also denote this interval by (0; 1)), then prove the inequality

  $\sqrt{abc}+\sqrt{\left( 1-a\right) \left( 1-b\right) \left( 1-c\right) }<1$.[/i]

Harazi, I admit I don't understand your solution, but here is mine:

I will use the following inequality:

[i]Let k and n be natural numbers, and $a_{i;j}$ some positive reals for all natural i and j with $1\leq i\leq k$ and $1\leq j\leq n$. Moreover, let $p_i$ be positive reals for all natural i with $1\leq i\leq k$. We assume that $p_{1}+p_{2}+...+p_{k}=1$. Then,

$a_{1;1}^{p_{1}}a_{2;1}^{p_{2}}...a_{k;1}^{p_{k}}+a_{1;2}^{p_{1}}a_{2;2}^{p_{2}}...a_{k;2}^{p_{k}}+...+a_{1;n}^{p_{1}}a_{2;n}^{p_{2}}...a_{k;n}^{p_{k}}$
$\leq \left( a_{1;1}+a_{1;2}+...+a_{1;n}\right) ^{p_{1}}\cdot \left( a_{2;1}+a_{2;2}+...+a_{2;n}\right) ^{p_{2}}\cdot ...\cdot \left( a_{k;1}+a_{k;2}+...+a_{k;n}\right) ^{p_{k}}$.[/i]

This was proven in Problem 304 of D. O. Shkljarskij, N. N. Chenzov, I. M. Yaglom, [i]Izbrannye Zadachi i Teoremy Elementarnoj Matematiki, Chastj 1 (Arifmetika i Algebra)[/i], Moscow 1954 (I believe this book was translated into English and published as the "USSR olympiad problem book" or something like this). It is actually one form of (one possible?) generalization of the Hölder inequality. Now apply it to k = 3, n = 2, $a_{1;1}=a$, $a_{2;1}=b$, $a_{3;1}=c$, $a_{1;2}=1-a$, $a_{2;2}=1-b$, $a_{3;2}=1-c$ (note that all of these variables are positive), and $p_{1}=p_{2}=p_{3}=\frac{1}{3}$. Then you get

  $\root{3}\of{abc}+\root{3}\of{\left( 1-a\right) \left( 1-b\right) \left( 1-c\right) }$
  $\leq \root{3}\of{a+\left( 1-a\right) }\cdot \root{3}\of{b+\left( 1-b\right) }+\root{3}\of{c+\left( 1-c\right) }=1$.

Now the inequality

  $\sqrt{abc}+\sqrt{\left( 1-a\right) \left( 1-b\right) \left( 1-c\right) }<1$

trivially follows, since abc < 1 and (1 - a) (1 - b) (1 - c) < 1 imply

  $\sqrt{abc}+\sqrt{\left( 1-a\right) \left( 1-b\right) \left( 1-c\right) }<\root{3}\of{abc}+\root{3}\of{\left( 1-a\right) \left( 1-b\right) \left(1-c\right) }$.

  Darij

[quote="billzhao"]I think harazi meant

$
\sqrt {a} \sqrt{bc} + \sqrt {1-a} \sqrt{(1-b)(1-c)} \leq \sqrt{a+1-a}\sqrt{bc + (1-b)(1-c)} = \sqrt{1-b-c} < 1
$[/quote]

Thanks. But you have a little mistake in your last calculation: Actually,

  bc + (1 - b) (1 - c) = 1 - (b + c) + 2bc.

Therefore, the corrected proof should be

  $\sqrt{abc}+\sqrt{\left( 1-a\right) \left( 1-b\right) \left( 1-c\right) }=\sqrt{a}\sqrt{bc}+\sqrt{1-a}\sqrt{\left( 1-b\right) \left( 1-c\right) }$
  $\leq \sqrt{a+\left( 1-a\right) }\sqrt{bc+\left( 1-b\right) \left( 1-c\right) }=\sqrt{bc+\left( 1-b\right) \left( 1-c\right) }=\sqrt{1-\left( b+c\right) +2bc}<1$.

Hereby, $\sqrt{1-\left( b+c\right) +2bc}<1$ is because $1-\left( b+c\right) +2bc<1$, what itself follows from $2bc\leq b^{2}+c^{2}<b+c$ (in fact $b^2 < b$ and $c^2 < c$ since b < 1 and c < 1).

Thanks again, I understand now.

  Darij

[quote="darij grinberg"]  $\sqrt{abc}+\sqrt{\left( 1-a\right) \left( 1-b\right) \left( 1-c\right) }<\root{3}\of{abc}+\root{3}\of{\left( 1-a\right) \left( 1-b\right) \left(1-c\right) }$.

  Darij[/quote]

After noting this, we may use AM-GM here to finish the proof.

$\displaystyle \root{3}\of{abc}+\root{3}\of{\left( 1-a\right) \left( 1-b\right) \left(1-c\right) } \leq \frac{a+b+c}{3} + \frac{(1-a)+(1-b)+(1-c)}{3} = 1$.

[quote=parmenides51]If $a, b, c$ are real numbers in the open interval  $ (0,1)$ then prove the inequality
  $$\sqrt{abc}+\sqrt{\left( 1-a\right) \left( 1-b\right) \left( 1-c\right) }<1$$
[hide=PS]posted for the [url=https://artofproblemsolving.com/community/c1184234_2002_junior_balkan_team_selection_tests__romania]post collection [/url][/hide][/quote]
[b]Junior Balkan Team Selection Tests - Romania 2002[/b]
See also [url=https://artofproblemsolving.com/community/c6h459496p2578782]here[/url]
$$\sqrt{abc}+\sqrt{\left( 1-a\right) \left( 1-b\right) \left( 1-c\right) }< \sqrt{ab}+\sqrt{(1-a)(1-b)}\leq \frac{a+b}{2}+\frac{1-a+1-b}{2}=1$$


inequalities
integration
real analysis
real analysis unsolved

Let $f\colon[0,1]\to\mathbb R_{+}: =[0,+\infty)$ be a concave function. Prove that
\[\int_{0}^{1}xf(x)dx\le\frac23\int_{0}^{1}f(x)dx.\]

A stronger inequality holds:

We integrate by parts:

$\int_{0}^{1}xf(x)dx=x\int_{0}^{x}f(t)dt|_{0}^{1}-\int_{0}^{1}\int_{0}^{x}f(t)dtdx$.
Now $f$ is concave therefore we get that the following inequality is valid

$f(t)\geq \frac{f(x)-f(0)}{x}t+f(0)$ on $0\leq t\leq x\leq 1$.
Thus we get that

$\int_{0}^{1}xf(x)dx\leq \int_{0}^{1}f(t)dt-\int_{0}^{1}(\frac{f(x)-f(0)}{2}x+f(0)x)dx$,
and after simple calculations we get that

$\int_{0}^{1}xf(x)dx\leq \frac{2}{3}\int_{0}^{1}f(t)dt-\frac{f(0)}{6}\leq \frac{2}{3}\int_{0}^{1}f(t)dt$, since $f(0)\geq 0$.
Didi


A kite has diagonals of 14 inches and 9 inches. What is the area of the kite, in square inches?


Give [tex] g:R \to R [/tex] that saitsfying:
[tex] i) g(x)=-g(-x) \forall x \in R [/tex].
[tex] ii) \lbrace {x|g(x)=0;x \neq 0} \rbrace \in R-Q [/tex]
Prove that [tex]  f(x)=g(x)+\lbrace {x} \rbrace [/tex] is not cycle. :)

Wouldn't that be very easy?

If we assume $T$ is a period, since we have $g(0)+\{0\}=0$, we must also have $g(T)+\{T\}=0$. On the other hand, we also have $g(-T)+\{-T\}=0$, so $\{-T\}=-\{T\}$, meaning that $T\in\mathbb Z$. We thus get $g(x+T)=g(x)$ (because $\{x+T\}=\{x\},\ \forall x$), so $g(T)=0$, contradiction.

Am I missing something here?

I am very sorry anout that but I had some mistake in my typing. It must be [tex] g(x)=-g(-x), \forall x \in R- \lbrace{0}\rbrace [/tex] :(. However, it is still very easy.It is one of the easiest problems in the test. :)


In the rectangle $ ABCD$, $ E$ is the projection of $ C$ on $ BD$, $ M$ is the projection of $ E$ on $ AB$ and $ N$ is the projection of $ E$ on $ AD$. If $ AB\equal{}a,BC\equal{}b,EN\equal{}c,EM\equal{}d,$ then prove that $ ab\plus{}bc\equal{}ad.$

I think we must prove that $ ab\equal{}bc\plus{}ad$ .

We have $ [BCD]\equal{}[ABD]\equal{}[ADE]\plus{}[ABE]\Longleftrightarrow \frac{BC\cdot CD}{2}\equal{}\frac{EN\cdot AD}{2}\plus{}\frac{EM\cdot AB}{2}$

$ \Longleftrightarrow ab\equal{}bc\plus{}ad$ .


[asy]int i, j;
    for(i=0; i<10; i=i+1)
    {
    for(j=0; j<5; j=j+1)
    dot((i,j), red);
    }[/asy]
How can I label these dots, so that the top left dot is labeled 1001, one to the right is 1002, etc, all the way to the bottom right is 1050?  Also, how can I space the dots more (if I labeled a dot with a 4-digit number, it wouldn't fit in such a small space).

Thanks in advance.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=226578v

look at the 7th post

[asy]size(11cm);
for(int i=0; i<10; ++i)
for(int j=0; j<5; ++j)
dot(scale(0.5)*Label("",align=NW),(i,j), blue);[/asy]

Now how do I change the labels to 1001, 1002, 1003, etc?

And why is the bottom cut off...

[quote="AIME15"][asy]size(11cm);
for(int i=0; i<10; ++i)
for(int j=0; j<5; ++j)
dot(scale(0.5)*Label("",align=NW),(i+1000,j+1000), blue);[/asy]

Now how do I change the labels to 1001, 1002, 1003, etc?

And why is the bottom cut off...[/quote]

How do i mkae it just the number, not an ordered pair though?  So on top of 0,4 would be just 1000, instead of (1000,1004)?

edit: the bottom is still cut off..