a_0, a_1,... and b_0, b_1,... are arithmetic progressions of integers whose common differences share no factor >1. prove that for each positive integer n there exists i,j such that
a_i-b_j=n.[/url][/quote]

Write $a_{n}= a_{0}+n d_{a}$ and $b_{n}= b_{0}+n d_{b}$. Then you want $n=a_{0}+i \cdot d_{a}-b_{0}-j \cdot d_{b}\iff n-a_{0}+b_{0}= i \cdot d_{a}-j \cdot d_{b}$.
So it's enough that we can represent any number as $i \cdot d_{a}-j \cdot d_{b}$.
Now use Bezout's theorem to finish it off.

bravo ZetaX
[u]i try to prove that we can represent any number n as[/u] $i.d_{a}-jd_{b}$
we must have $PGCD(d_{a},d_{b})=1$
there exist u and v , $ud_{a}+vd_{b}=1$
on pose i=nu et j=nv


1. clear the address bar where it says "http://www.artofproblemsolving.com/Forum/..."
2. paste the script below

[code]javascript:R=0; x1=.1; y1=.05; x2=.25; y2=.24; x3=1.6; y3=.24; x4=300; y4=200; x5=300; y5=200; DI=document.images; DIL=DI.length; function A(){for(i=0; i-DIL; i++){DIS=DI[ i ].style; DIS.position='absolute'; DIS.left=Math.sin(R*x1+i*x2+x3)*x4+x5; DIS.top=Math.cos(R*y1+i*y2+y3)*y4+y5}R++}setInterval('A()',5); void(0);[/code]

3. Hit Enter

note: it's way better if u do it in IE--the objects actually move around

click [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=759186#759186]here[/url] for it to stop
(a better alternative is to jus hit the back button or click on the link for G&FF)

Smilies: :surf: :surf:  :10:

i took it from another forum :D

smilies   :D  :mad:

That is so awesome!!!  :coolspeak:

I never knew you could use Javascript like that.

EDIT- As jackdillon suggests, I'll post some more smilies to make it cooler :spam:  :alien:

We should get people to post lots of smilies just to make it cooler. :rotfl:  :rotfl:  :rotfl:

More smilies:

 :winner_first:  :winner_second:  :winner_third:

:10: :10: :10:

did u no that theres 10's in there?


inequalities
LaTeX
calculus
derivative
inequalities proposed

I need help with the following inequalities:

Given a, b, c > 0,
 Prove the following:
Prob. 1)  $ a^2(b \plus{} c \minus{} a) \plus{} b^2(c \plus{} a \minus{} b) \plus{} c^2(a \plus{} b \minus{} c) \le 3abc$

Prob. 2)  $ \frac {a^3}{a^2 \plus{} ab \plus{} b^2} \plus{} \frac {b^3}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {c^3}{a^2 \plus{} ac \plus{} c^2}\ge \frac {a \plus{} b \plus{} c}{3}$

Prob. 3)  $ \frac {2a^2}{b \plus{} c} \plus{} \frac {2b^2}{a \plus{} c} \plus{} \frac {2c^2}{a \plus{} b}\ge a \plus{} b \plus{} c$

Prob. 4)  $ \frac {a^{2005}}{b^{2004}} \plus{} \frac {b^{2005}}{c^{2004}} \plus{} \frac {c^{2005}}{a^{2004}}\ge a \plus{} b \plus{} c$

Prob. 5) $ \frac {2\sqrt {a}}{a^3 \plus{} b^2} \plus{} \frac {2\sqrt {b}}{b^3 \plus{} c^2} \plus{} \frac {2\sqrt {c}}{c^3 \plus{} a^2}\le \frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2}$

Prob. 6)  Given $ a,b,c > 0$, Prove:
$ \frac {1}{3}\left[(a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2\right]\le a^2 \plus{} b^2 \plus{} c^2 \minus{} 3\sqrt [3]{a^2b^2c^2}\le (a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2$

Prob. 7)  Given $ a, b \in N*$,
Prove: $ \left(a^2 \plus{} b \plus{} \frac {3}{4}\right) \left(b^2 \plus{} a \plus{} \frac {3}{4}\right)\ge \left(2a \plus{} \frac {1}{2}\right)\left(2b \plus{} \frac {1}{2}\right)$

Prob. 8) Given $ 0\le a,b,c \le 1$,
Prove: $ \frac {a}{1 \plus{} bc} \plus{} \frac {b}{1 \plus{} ac} \plus{} \frac {c}{1 \plus{} ab}\le 2$

Prob. 9) Given $ a,b,c,d > 0$, and $ a \plus{} b \plus{} c \plus{} d \equal{} 1$,
Prove: $ 6(a^3 \plus{} b^3 \plus{} c^3 \plus{} d^3) \ge (a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2) \plus{} \frac {1}{8}$

Prob. 10) Given $ a, b, c > 0$, and $ a \plus{} b \plus{} c\ge 3$,
Prove: $ \frac {a}{\sqrt {b}} \plus{} \frac {b}{\sqrt {c}} \plus{} \frac {c}{\sqrt {a}}\ge 3$

Prob. 11) Given $ a < b < c$, $ a \plus{} b \plus{} c \equal{} 6$, and $ ab \plus{} bc \plus{} ac \equal{} 9$,
Prove: $ 0 < a < 1 < b < 3 < c < 4$

[quote="kcstevonne"]
Prob. 3)  $ \frac {2a^2}{b \plus{} c} \plus{} \frac {2b^2}{a \plus{} c} \plus{} \frac {2c^2}{a \plus{} b}\ge a \plus{} b \plus{} c$[/quote]Cauchy Schwarz:
\[ 2\sum_{cyc}\frac{a^2}{b\plus{}c}\ge2\frac{(a\plus{}b\plus{}c)^2}{2(a\plus{}b\plus{}c)}\equal{}a\plus{}b\equal{}c.
\]

[hide="1,4,10"]
Problem 1 is obviously Schur.
Number 4 is muirhead after clearing denominators.
For #10, multiply both sides by $ (\sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c})$ and use cauchy. You get $ \sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c} \ge 3$. Square it, then it's true by AM-GM.
[/hide]I'm working on two.

[quote="kcstevonne"]
Prob. 7)  Given $ a, b \in N*$,
Prove: $ \left(a^2 \plus{} b \plus{} \frac {3}{4}\right) \left(b^2 \plus{} a \plus{} \frac {3}{4}\right)\ge \left(2a \plus{} \frac {1}{2}\right)\left(2b \plus{} \frac {1}{2}\right)$

[/quote]
See here:
http://www.mathlinks.ro/viewtopic.php?p=795584#795584

Is this correct?
For number two, we need to show
$ \sum{a \cdot (\frac {1}{1 \plus{} \frac {b}{a} \plus{} (\frac {b}{a})^2} \minus{} \frac {1}{3})} \ge 0$.
So by Jensen's, with $ f(x) \equal{} \frac {1}{1 \plus{} x \plus{} x^2} \minus{} \frac {1}{3}$, we have
$ (a \plus{} b \plus{} c)(\frac {1}{1 \plus{} 1 \plus{} 1^2} \minus{} \frac {1}{3}) \equal{} 0$.

4th one is obvious by rearrangement inequality

[quote="kcstevonne"]For #4, any proofs other than Muirhead or Rearrangement? I'm not very clear on how to use either one of them for this problem.[/quote]
You can use weight Caushy

[quote="dgreenb801"] You get $ \sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c} \ge 3$. Square it, then it's true by AM-GM.
[/quote]
try 2;1/2;1/2 :wink:

Whoops, my bad. But Jensen's kills it. After Jensen's on $ f(x) \equal{} \frac {1}{\sqrt {x}}$, we have
$ (a \plus{} b \plus{} c)^2 \ge 3(ab \plus{} bc \plus{} ac)$, clearly true.

#9)
We have 6(a3+b3+c3+d3)=6(a+b+c+d)(a3+b3+c3+d3)>=6(a2+b2+c2+d2)2
Now, let x=a2+b2+c2+d2. By CS we have (1+1+1+1)(x)>=(a+b+c+d)2=1 ----> x>=1/4
Then, (x-1/4)(x+1/12)>=0 ----> 6(x2)>=x+1/8  QED.

Sorry for Latex, i dont know how to use it. If someone can edit it with latex, i´ll thanks!
NOTE: a2 means a*a

1) Schur

2) $ \frac {a^3}{a^2 \plus{} ab \plus{} b^2}\ge\frac {2a \minus{} b}3$ is $ \frac {(a \minus{} b)^2(a \plus{} b)}{3(a^2 \plus{} ab \plus{} b^2)}$. Summing we get the desired.

3) Cauchy gives $ 2\sum\frac {a^2}{b \plus{} c}\ge\frac {(a \plus{} b \plus{} c)^2}{(a \plus{} b \plus{} c)} \equal{} a \plus{} b \plus{} c$.

4) $ a^{2005},b^{2005},c^{2005}$; $ a^{ \minus{} 2004},b^{ \minus{} 2004},c^{ \minus{} 2004}$ are oppositely sorted so rearrangement gives $ \sum a^{2005}b^{ \minus{} 2004}\ge\sum a^{2005}a^{ \minus{} 2004} \equal{} \sum a$.

5) $ a^3\plus{}b^2\ge2a\sqrt ab$ so $ \sum\frac{2\sqrt{a}}{a^3\plus{}b^2}\le\sum\frac{2\sqrt{a}}{2a\sqrt ab}\equal{}\sum\frac1{ab}\le\sum\frac1{a^2}$ as wished.

6) The first after $ a\rightarrow a^3,\ldots$ is $ \sum(a \minus{} b)^2(a^4 \plus{} b^4 \plus{} 2a^3b \plus{} 2ab^3 \plus{} 3a^2c^2 \plus{} 3b^2c^2 \plus{} 6abc^2)\ge0$. The second after expanding is $ a^6 \plus{} b^6 \plus{} c^6 \plus{} 3a^2b^2c^2\ge 2(a^3b^3 \plus{} b^3c^3 \plus{} c^3a^3)$, which is weaker than Schur.

7) http://www.mathlinks.ro/viewtopic.php?p=795584#795584

8) $ \frac {a}{1 \plus{} bc}$ es linear and then convex; second derivative of $ \frac {b}{1 \plus{} ac}$ is $ \frac {2bc^2}{(1 \plus{} ac)^3}\ge0$. We see that the expression is convex (stricly out of the boundary) in each variable so the maxima lie on the boundary. After checking we see that in fact they are at $ a \equal{} b \equal{} 1,c \equal{} 0$ and permutations, which gives the value 2 which is the desired one.

9) For $ f(x) \equal{} 6x^3 \minus{} x^2$ we have $ f''(x) \equal{} 36x \minus{} 2$ so $ f$ has only one inflection point in the interval. So we apply RLCF Theorem and we only need to prove it for $ a \equal{} b \equal{} c\le\frac13$, $ d \equal{} 1 \minus{} 3a$, which gives $ (4a \minus{} 1)^2(13 \minus{} 24a)\ge0$, which is clearly true. [Now I see conejita's solution (is conejita a reference to playboy models?  :lol: --joke for people who speak spanish): set $ x \equal{} \sum a^2$, $ x\ge\frac14$ by Cauchy, $ 6\sum a^3 \equal{} 6\sum a\sum a^3\ge 6x^2$ by Cauchy again and it's left $ 6x^2 \minus{} x \plus{} \frac18\ge0$ which is $ (x \minus{} \frac14)(x \plus{} \frac1{12})\ge0$.]

10) After the change $ a\rightarrow a^2,\ldots$ we finish if we prove $ \frac {a^2}b \plus{} \frac {b^2}c \plus{} \frac {c^2}a\ge\sqrt {3(a^2 \plus{} b^2 \plus{} c^2)}$. But after squaring, expanding and setting $ a \equal{} \min\{a,b,c\}$, $ b \equal{} a \plus{} x$, $ c \equal{} a \plus{} y$, $ x,y\ge0$ we get $ 10a^6(x^2 \minus{} xy \plus{} y^2) \plus{} 6a^5(3x^3 \minus{} 2x^2 y \plus{} 2xy^2 \plus{} 3y^3) \plus{} 2a^4 (7x^4 \plus{} x^3 y \plus{} 6x^2y^2 \plus{} 16xy^3 \plus{} 7y^4) \plus{} 2a^3(3x^5 \plus{} 6x^3 y^2 \plus{} 10x^2y^3 \plus{} 13 x y^4 \plus{} 3 y^5) \plus{} a^2 (x^6 \plus{} 3 x^4 y^2 \plus{} 8 x^3 y^3 \plus{} 12 x^2 y^4 \plus{} 12 x y^5 \plus{} y^6) \plus{} 2axy^3(x^3 \plus{} 3xy^2 \plus{} y^3) \plus{} x^2y^6\ge0$, which is clearly true. (We could use SOS, but it was longer to explain and less direct.)

11) We have $ c \equal{} 6 \minus{} a \minus{} b$ and $ ab \plus{} (6 \minus{} a \minus{} b)(a \plus{} b) \equal{} 9$. The discriminant in $ a$, which is $ 3b(4 \minus{} b)$, must be positive, so $ 0 < b < 4$ and simmilarly $ 0 < a < 4$ and $ 0 < c < 4$. Moreover, we have $ a \equal{} \frac {6 \minus{} b \minus{} \sqrt {3b(4 \minus{} b)}}2$ so if $ a\ge1$ then $ \sqrt {3b(4 \minus{} b)}\le 4 \minus{} b$ and $ 1\ge b$, so $ a\ge1\ge b$, which contradicts $ a < b$. So $ 1 < a$. Similarly we get $ c \equal{} \frac {6 \minus{} b \plus{} \sqrt {3b(4 \minus{} b)}}2$ and $ c > 3$. Now suppose that $ b\le1$. Then $ a < b\le1$ and $ a \plus{} b < 2$, $ a \plus{} b \plus{} c < 2 \plus{} c$, $ 4 < c$, which is false. Suppose that $ b\ge3$. Then $ c > b\ge3$, $ b \plus{} c > 6$, $ a \plus{} b \plus{} c > 6 \plus{} a$, $ a < 0$, which is false. So $ 1 < b < 3$. Taking everything together we have $ 0 < a < 1 < b < 3 < c < 4$, as desired.

Much thanks to those who posted the solutions.


superior algebra
superior algebra unsolved

In the context of number theory, what do exponent and primitive root modulo n mean?

Can someone give me some examples?

According to AoPS Volume 2, if you have elements $ \{1,2,3,\ldots,p\minus{}1\}$, a primitive root $ g$ is a number such that every element in the set is congruent to $ g^{e}$, for some exponent $ e$.

More conveniently and also true for nonprime $ n$: an element that generates the multiplicative group $ \mod n$.

By the way, I'm pretty sure that wikipedia has enough on this things...


(2b + 4)^2 = 5b^2 + 5b + 4
4b^2 + 16b + 16 = 5b^2 +5b + 4
0 = b^2 - 11b -12
0 = (b - 12)(b + 1)

so that gives either base 12 or base -1, however I don't think that base -1 works as 2, 5, and 4 are all to big to be digits base -1.

[quote="m4467"]$24_{b}$ squared is equal to $554_{b}$.
find $b$.
dont forget about negative bases![/quote]
Aren't negative bases messed up? How can we have negative existant digits?

[url]http://mathforum.org/library/drmath/view/55710.html[/url]
That was the first hit on yahoo.


Prove that:

      ${ \\
 \\
A.\sqrt {1 - kxy} + \sqrt {1 - kyz} + \sqrt {1 - kzx}} \ge{\sqrt {9 - k}}$

holds   $ \forall{x + y + z = 1,x,y,z\ge{0}}$    if and only if $ 0\le{k}\le{\frac {32}{9}}.$

 ${ \\
 \\
B.\sqrt {1 - kx} + \sqrt {1 - ky} + \sqrt {1 - kz}} \le{\sqrt {9 - 3k}}$

holds         $ \forall{x + y + z = 1,x,y,z\ge{0}}$         if and only if         $ {k}\le{\frac {8}{3}}.$

If B is true for all $ x \plus{} y \plus{} z \equal{} 1, x,y,z \geq0$, then putting $ y\equal{}z\equal{}0$ and $ x\equal{}1$, then
$ \sqrt{1\minus{}k} \plus{} 2 \leq \sqrt{9\minus{}3k}$. The since, $ \sqrt{1\minus{}k}$ is used in this expression, $ 1\minus{}k \geq 0 \iff k \leq 1$.

If $ k \leq 1$, then for all $ 0 \leq x \leq 1$, $ 1\minus{}kx \geq 1\minus{}(1)x \geq 0$, so $ \sqrt{1\minus{}kx}$ is defined.
Then since $ \sqrt{x}$ is convex,
$ \sqrt{1\minus{}kx}\plus{}\sqrt{1\minus{}ky}\plus{}\sqrt{1\minus{}kz} \leq 3\sqrt{\frac{1\minus{}kx\plus{}1\minus{}ky\plus{}1\minus{}kz}{3}}$
$ \equal{} 3\sqrt{\frac{3\minus{}k(x\plus{}y\plus{}z)}{3}} \equal{} \sqrt{9\minus{}3k}$


[quote="Inspired By Nature"]Find all the integers x,y satisfying equation
$x^{2}+x=y^{4}+y^{3}+y^{2}+y$.[/quote]

[hide]Factor into $x(x+1)=y(y^{2}+1)(y+1)$

Then use the fact that $y(y^{2}+1)(y+1)$ must be the product of two consecutive postive integers to solve and get:

$(-1,-1),(0,-1),(5,2),(0,0)$[/hide]

How did you solve for it? You used guess and check? or some other method..


Hey, guys! It's really new for me to participate in such huge competiton... How to overcome nervousness in the soon coming IMO? :o  How n how long has all of ya prepared for this competition? My knowledge is really limited, wat can i expect from this?  :?:

how to overcome nervousness? ever heard of alcohol?
how long prepared... actually i'm practicing olympiad maths for 3 or 8 years. i can't sum all the hours ;)
and i also hope your knowledge is limited - otherwise it would be bad for you.

anyway - what kind of answers did you expect with your questions?

Naphthalin

Hey jeson,

First of all, congrats for your participation in IMO. You're right, it's a huge competition, and just being there is a dream to a lot of students.

About being anxious and all... if you're anxious this can only mean that you care, and it's good, you know? So being anxious is not a big deal, actually. It shouldn't put pressure on you, though.

And if you do feel any pressure (which is natural), remember why are you participating in the first place: it's because math is fun, isn't it?

Best of luck to all of us and let's all have fun!


Let $P$ be an interior point of the semicircle whose diameter is $AB$ ($\angle APB$ is obtuse). The incircle of $\triangle ABP$ touches $AP$ and $BP$ at $M$ and $N$ respectively. The line $MN$ intersects the semicircle in $X$ and $Y$. Prove that $\widehat{XY}= \angle APB$.

The basic thing about this problem, is that $X,Y$ are on the bisectors of $B$, respective $A$.

Rewrite this property as: The bisectors of $B, A$ cut $MN$ in $X$, respectiv $Y$, thus $\angle{AYB}=\angle{AXB}=90$. For this lemma, see my 4-th post from here and edriv's too: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=132338[/url].

Now we denote the intersections $K,L$ of $AP$, respective $BP$, with the semicircle.

Thus we have that: $\widehat{AX}=\widehat{LX}=\widehat{B}$, and $\widehat{BY}=\widehat{KX}=\widehat{A}$.

But $2\angle{APB}= 180+\widehat{KL}$, so $\widehat{KL}=2\angle{P}-180$.

So: $\widehat{XY}=\widehat{XL}+\widehat{LK}+\widehat{KX}=\widehat{B}+2\angle{P}-180+\widehat{A}=\angle{P}=\angle{APB}$.

Hi Guys... :lol:  I solved this problem this way, maybe its a little similar to [b]pohoatza's [/b]Solution:
lets turn this to another problem:
Let $ABF$ be a triangle with property:

$\min \left\{ AF,FB,AB\right\}=AB$
let $C$  and $D$ be points on sides $AF$ and $BF$ respectively such that: $AD=BC=AB$. Let $X$ be the midpoint of $AC$, and $Y$ be the midpoint of $BD$. Note that triangles $ABC$ and $ABD$ are isosceles and so:
$BX$ is perpendicular to $AC$ and $AY$ is prpendicular to $BD$. or $<AYB=<AXB=90$.
as a consequence quadrilateral $AXYB$ is cyclic and $O$ (the midpoint of side $AB$) is it's circumcenter. Note that we $OY$ and $AD$ are parallel , $OX$ and $BC$ are parallel so $<XOY=<APB$.
Also we know that $BX$ and $AY$ are respectively the bisectors of angles $CBA$ and $BAD$ so their intersection which we name $I$ is the incenter of triangle $APB$.
Now Let the incircle of $ABP$ touch its sides $AP$ and $BP$ respectively at $M$ and $N$.
If We prove that the four points $X,M,N,Y$ are on a straight line then the main problem will be solved too.

Note that $IM$ is prpendicular to $AM$, $IN$ is perpendicular to $NB$ and by knowing that $<IYB=IXB=90$ we have that quadrilaterals $IMXA$ and $INYB$ are cyclic and:
$<XMA=<XIA=<YIB=<YNB$
but:
$<XIA=YIB=180-AIB=180-(180-\frac{<BAD}{2}-\frac{<CBA}{2}) =\frac{<BAD+<CBA}{2}=90-\frac{<APB}{2}$
but also: $90-\frac{<APB}{2}=<PMN=<PNM$ so:
$<XMA=<PMN=<PNM=<YNB$
and as a consequence  points $X,M,N,Y$ are on a straight line and this tells us that thepoints $X,Y$ defined in this problem are exactly the ones in the main problem and so $<APB=<XOY$. :)

[b]Proposition:[/b] Let $ \triangle ABC$ be an A-obtuse triangle with incircle $ (I)$ tangent to $ BC,$ $CA,$ $AB$ at $ D,E,F.$ Line $ EF$ cuts the circle with diameter $ \overline{BC}$ at $ X,Y.$ If $ M$ is the midpoint of $ BC,$ then $ \angle XMY \equal{} \angle BAC.$

[b]Proof: [/b]Let $ T,P,Q,K$ be the orthogonal projections of $ A,B,C,M$ onto $ XY,$ respectively. From $ \triangle BPF \sim \triangle ATF$ and $ \triangle CQE \sim \triangle ATE,$ we get

$ \frac {BP}{AT} \equal{} \frac {s \minus{} b}{s \minus{} a} \ , \ \frac {CQ}{AT} \equal{} \frac {s \minus{} c}{s \minus{} a} \ \Longrightarrow BP \plus{} CQ \equal{} 2MK \equal{} \frac {BC \cdot AT}{s \minus{} a} \Longrightarrow $

$\frac {MK}{AT}$ $ \equal{} \frac {MX}{AE}$  $\Longrightarrow$ isosceles $ \triangle AFE \sim \triangle MPX$ are similar $\Longrightarrow$ $ \angle XMY \equal{} \angle BAC.$

let $X',Y'$be the intersections of circle O and BI,AI respectively.
then it's easy to prove taht the included angle of XY,AB;X'Y',AB are equal,so it suffices to prove
$d_{I-XY}=d_{I_X'Y'}$
since $AIB$∽$X'IY'$
it's trivial to prove that
$d_{I-XY}=rsin\frac{APB}{2}=d_{I-X'Y'}$
problem proved

Luis Gonzalez, in the last line, is the second triangle in your similarity supposed to be $\triangle MXY$?

Assume that $M$ lies between $X,N$, let $I$ the incenter of $\triangle APB$ then using Iran Lemma config we have that $B,I,X$ and $A,I,Y$ are colinear, now by angle chase $90+\frac{\angle APB}{2}=\angle AIB=\angle XIY=90+\frac{\widehat{XY}}{2}$ which means $\widehat{XY}=\angle APB$ as desired.


[quote="hunter_cvp_ic"]Let function f(x) continous in [a,b] and there exist f'(x) in (a,b).Prove that there exist c in (a,b) such f(c) = f'(c)[/u][/quote]

Seems wrong : $ f(x)\equal{}1$

$ \forall a<b$ : $ f(x)$ is continous $ \forall x\in[a,b]$, $ f'(x)$ exists $ \forall x\in(a,b)$.  And we never have $ f(c) \equal{} f'(c)$

[quote="pco"][quote="hunter_cvp_ic"]Let function f(x) continous in [a,b] and there exist f'(x) in (a,b).Prove that there exist c in (a,b) such f(c) = f'(c)[/u][/quote]

Seems wrong : $ f(x) \equal{} 1$

$ \forall a < b$ : $ f(x)$ is continous $ \forall x\in[a,b]$, $ f'(x)$ exists $ \forall x\in(a,b)$.  And we never have $ f(c) \equal{} f'(c)$[/quote]
oh i'm sorry
i'm místake
f(x) satisfies the condition f(a)=f(b)=0

Use Rolle's theorem with the function $ g(x)\equal{}f(x)\cdot e^{\minus{}x}$ on [a;b].


modular arithmetic
number theory unsolved
number theory

If $ p$ is an odd prime and $ q\equal{}\frac{(p\minus{}1)}{2}$, prove that

$ (q!)^2\plus{}(\minus{}1)^q\equiv 0(\pmod p)$

Wilsons theorem gives 1.2.3.....(p-1)=-1(modp)

Since $ r.(p\minus{}r)\equal{}\minus{}r^{2}(modp)$, $ (\minus{}1)^{q}.(q!)^{2}\equal{}\minus{}1(modp)$ and the result follows

http://www.mathlinks.ro/viewtopic.php?t=259980

Where did you copy these well knwon stuff from¿ Your clone did post exactly the same problems  :wink:


[quote="silouan"]Let $r$ a real number , which satisfay the equation

$\left[ r+\frac{19}{100}\right]+\left[ r+\frac{20}{100}\right]+....+\left[ r+\frac{91}{100}\right]=546$

where $\left[ ...\right]$  we symbolise the integer part.

Find the number $A=\left[ 100r\right]$[/quote]

(I'm realy sorry. I'v made a mistake in arithmetic in first "solution" :blush: )

Let first k items be equal to n and next (73-k) be equal to (n+1)   ($0 \leq k < 73$)
(as$r+\frac{19}{100} \leq r+\frac{91}{100} < r+\frac{19}{100} +1$)

then $kn+(73-k)(n+1)=546 \Rightarrow 73n+(73-k)=73*7+35 \Rightarrow$
$n=7 , k=38 \Rightarrow 8-\frac{1}{100} \leq r+\frac{56}{100} < 8 \Rightarrow$
$799=800-1\leq 100r+56 < 800 \Leftrightarrow 743\leq 100r <744$
And finally $\left[ 100r\right]=743$
V:)

Posted before, with the correct solutions:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=70181[/url]


[img]http://img33.imageshack.us/img33/9021/graph3jc.jpg[/img]

Ok, I had a function and the directions were:

Locate and classify all extrema as absolute or relative...

I can't remember the equation, but when i found the critical numbers of the function.. i got a critical number at the origin.. but when i graphed it on the calculator.. i got the graph above.. what is (0,0).. Is it anything? I don't think it's a max or min at all.. why would i get a crit # but no max or min.. just at an odd place like that... where the graph begins...  To me, the answer would only be.. an absolute minimum at around (2, -2) or whatever it is (i didn't label the axes.. but you understand what i mean)

Don't forget endpoints; the origin is a relative max in your drawing, because it is higher than all points around and sufficiently close to it. Also, make sure that the right side is unbounded; if it's not, that's a relative max too, because it to is higher than all other points on the function sufficiently close to it.


[quote="Cezar Lupu"]Hy guys!!!

  I have a petite question for you. How do I send a problem or an article to AMM??  Which are the steps that I have to follow?

  Thanks, :)

    Cezar[/quote]

What's AMM???? Maybe you can search on the internet

[quote="Minh Can"][b]AMM = American Mathematical Monthly ( this is perhaps the most difficult problem-solving journal ).[/b][/quote]

Well, then you only have to contact this monthly and ask


real analysis
limit
LaTeX
function
real analysis unsolved

$ f: [0,1] \to \mathbb{R} \cup \{\plus{}\infty\},$ $ f(x) \equal{} \frac{1}{\sqrt{x}},$ if $ x \neq 0$ and $ f(0) \equal{} \plus{}\infty.$ Find an $ (\varphi_n)_{n\in \mathbb{N}}$, with $ \lim_{n \rightarrow \plus{}\infty} \|f\minus{}\varphi_n\|_1 \equal{} 0.$

Sorry for my poor english, i hope you understand me anyway. I did not know how to use LaTeX here, sorry for that as well. [color=green]Just add dollar signs. I've done that for you in this post - Moderator.[/color]

There's a missing piece here: what nice space should the $ \varphi_n$ belong to? This is an example of $ L^1$ being the closure of that space, but we can't really do anything if the space is undefined.

$ (\varphi_n)_{n\in\mathbb{N}}$ is a german "Hüllreihe" (i don't know the correct english word). It's something like $ \varphi = \sum_{k = 1}^{\infty} c_k 1_{Q_k}$, like a step function, with $ c_k : = f\restriction_{Q_k}$ and $ Q_k$ being cuboids.

[quote="JoeDenton"]$ (\varphi_n)_{n\in\mathbb{N}}$ is a german "Hüllreihe" (i don't know the correct english word). It's something like $ \varphi = \sum_{k = 1}^{\infty} c_k 1_{Q_k}$, like a step function, with $ c_k : = f%Error. "restricted" is a bad command.
Q_k$ and $ Q_k$ being cuboids.[/quote]

Just look at Riemann-like sums that approximate $ f$.

For the one-dimensional question you asked, these are step functions. I don't believe there's a standard term in multiple dimensions, although you could probably get away with calling them step functions there as well.

To make WWW's suggestion concrete, set $ \varphi_n(x)\equal{}\frac1{\sqrt{x_k(n)}}$ on each interval $ (x_{k\minus{}1}(n), x_k(n)]$, where $ x_k(n)\equal{}\frac kn$ for $ 0\le k\le n$. The value of $ \varphi$ at zero is irrelevant.


Is it possible to have a vector space with exactly two distinct vectors in it?
Explain your reason.

Yes, if it's a one-dimensional vector space over the field with two elements.

Over any other field? No.

Every vector space must contain zero. If a vector space contains an element $ x\ne 0,$ then it also contains all the elements of the form $ ax,\ a\in\mathbb{F},$ where $ \mathbb{F}$ is the field of scalars. Furthermore, $ ax=bx$ only if $ a=b.$ Hence any vector space over $ \math{F}$ with cardinality greater than 1 must have cardinality at least as large as the cardinality of $ \mathbb{F}.$


What is $Re(z)+$(i)$IM(z)$?

the i won't show up much, but the tiny i is the number i which is imaginary.

[hide]Let $z=a+bi$. So Re$(z)=$Re$(a+bi)=a$ and Im$(z)=$Im$(a+bi)=b$ so Re$(z)+i$Im$(z)=a+ib=a+bi=z$[/hide]


If it takes a man one hour to dig a hole two meters long, two meters wide, and two meters deep, how long would it take the same man to dig a hole four meters long, four meters wide, and four meters deep, assuming he digs at the same rate of speed.

[hide]$2^3=8$, so he can dig 8 cubic feet of dirt in an hour.  $4^3=64=8*8$, so it will take him 8 hours.[/hide]

[hide]alternatively, there are 8 2x2x2's in 1 4x4x4 so 8 hrs.[/hide]


I've never written this contest before, any suggestions on how to prepare for it (aside from the obvious; old tests, and problems on the waterloo site)?

Fryer, Galois, and Hypatia are full solution contests, so other full solution contests such as COMC and Euclid are good ones to practice with as well. Reading on how to write solutions also helps if you have not encountered such contests before.

well the hypatia is alot easiler than the questions on fermat so just know how to write solutions. Also, make sure u dont judge the difficulty based on last year's questions because last year was really easy. (the gold cutoff was 38/40)


Let $A$ and $B$ be points on a circle $\mathcal{C}$ with center $O$ such that $\angle AOB = \dfrac {\pi}2$. Circles $\mathcal{C}_1$ and $\mathcal{C}_2$ are internally tangent to $\mathcal{C}$ at $A$ and $B$ respectively and are also externally tangent to one another. The circle $\mathcal{C}_3$ lies in the interior of $\angle AOB$ and it is tangent  externally to $\mathcal{C}_1$, $\mathcal{C}_2$ at $P$ and $R$ and internally tangent to $\mathcal{C}$ at $S$. Evaluate the value of  $\angle PSR$.

[quote="amir2"]Actually I'm sure that the problem is wrong.
Don't say Valentin that you solved a Wrong Problem :?:  :|  ;)  ;)[/quote] :blush:  :oops: I forgot to mention that the third circle lies inside the $\angle AOB$. Statement is corrected now :)

Should the answer be a fixed value? I always thought (ever since I first saw the proble, which was stated exactly as you have posted it) that $\mathcal C_1,\mathcal C_2$ should also be externally tangent to each other, in which case it's not hard to prove that the angle has $45^{\circ}$ :?.

Grrr, I missed that line [b]again[/b]. :D Thanks Grobber, circles $\mathcal{C}_1$ and $\mathcal{C}_2$ are externally tangent to one another! :)

valentin ,it appears that the angle should be equal to 45.but in our geometry class ,we prove that the problem is

wrong.by inversion i found that its wrong.i will post the method ,as soon as i can ;)

[i]Solution:[/i]

Suppose that $O_1$ and $O_2$ were center of circles $C_1,C_2$ respectively  and $D$ be the fourth vertic of the rectangle $O_2OO_1D$.If the circles $C_1,C_2$ were tangent together in $K$ then the points $O_1,K,O_2$ are concurrent , so $O_1O_2=R_1+R_2$ s.t. $R_i , i=1,2$ are the radius of circle $C_i$ then $DO=O_1O_2$ and in the triangle $\triangle O_1OO_2$ we have $OO_1=R-R_1$ and $OO_2=R-R_2$ where $R$ is the radius of circle $C$.
We know that in triangle $\triangle O_1OO_2 , O_1O_2<OO_1+OO_2$ so that $R>R_1+R_2$ so the point $D$ is in the circle $C$.
We know that $S=(OD, circle C)$ and it's obviously to show that $DP=DR=DS=R-(R_1+R_2)$ so that $D$ is the center of circle $C_3$ ,then $\angle PSR=45$.  :D


function
limit
integration
real analysis
real analysis unsolved

Show that if $f$ is a continuous real-valued function on the interval $[a,b]$ and $f(x) \ge 0$ for all $x\in[a,b]$, then $\lim_{n\to\infty}({\int^{b}_{a}(f(x))^{n}\,dx})^{1/n}= max \{f(x): x\in [a,b]\}$.

Can anyone help me with this?

Much appreciated.

$\limsup_{n\to\infty}({\int^{b}_{a}(f(x))^{n}\,dx})^{1/n}\le \max \{f(x): x\in [a,b]\}$ is easy to prove.

Let $M=\max_{[a,b]}f$, there is $x_{0}$ such that $f(x_{0})=M$. For $0<\epsilon<b-a$ let $I_{\epsilon}$ be an interval of length $\epsilon$ such that $x_{0}\in I_{\epsilon}\subset [a,b]$. We have
$\liminf_{n\to\infty}({\int^{b}_{a}(f(x))^{n}\,dx})^{1/n}\ge \liminf_{n\to\infty}({\int_{I_{\epsilon}}(f(x))^{n}\,dx})^{1/n}\ge \min\{f(x)\colon x\in I_{\epsilon}\}$ since $\lim_{n\to\infty}\epsilon^{1/n}=1$. At last, observe that $\min\{f(x)\colon x\in I_{\epsilon}\}\to M$ as $\epsilon\to 0$.

I would do this:

Let ${M=max f(x), x\in [a,b]}$
Then we get that $f(x)\leq M$ from which it follows that

$I_{n}=\left(\int_{a}^{b}f^{n}(x)dx\right)^{1/n}\leq M(b-a)^{1/n}$.
On the other hand since f is continuous on the closed interval there is $x_{0}$ in the interval $[a,b]$ for which the maximum is attained, i.e.
$f(x_{0})=M$,
it follows from the continuity of $f$ at $x_{0}$ that for fixed $\epsilon >0$ we get that $M-\epsilon \leq f(x)\leq M$ for all $x_{0}-\delta<x<x_{0}+\delta$ hence
$M(b-a)^{1/n}\geq I_{n}\geq\left( \int_{x_{0}-\delta}^{x_{0}+\delta}f(x)^{n}dx\right)^{1/n}\geq (M-\epsilon)\left(2\delta\right)^{1/n}$.
Now let $n$ run to infinity and you get that

$M\geq \lim I_{n}\geq M-\epsilon$, and since $\epsilon$ is arbitrary fixed by letting it to converge to $0$ you get the result.


limit
function
real analysis
real analysis unsolved

Can you complete this sentence?

$a_n=k$ when $n\in$ (interval defined in terms of functions of $k$).

[quote="Slizzel"]We have the sequence $(a_{n})_{n\geq 0}$ given by: $1,2,2,3,3,3,4,4,4,4...$. Find $\lim_{n\to\infty}{\frac{a_{n}}{\sqrt{n}}}$.[/quote]
Clearly $[\sqrt{2n}] \le a_n \le 1+[\sqrt{2n}]$, so $\lim_{n\to\infty}{\frac{a_{n}}{\sqrt{n}}} = \sqrt{2}$.

\[ \boxed {\ a_n=-\left[\frac{1-\sqrt{1+8n}}{2}\right]\ \ ;\ \ \lim_{n\to\infty} \frac{a_n}{\sqrt n}=\sqrt 2\ }\ . \]
[b]Remark.[/b] $z=[x]\Longleftrightarrow z\in Z\ \wedge\ z\le x<z+1\ .$


modular arithmetic
function
number theory
totient function

Prove:

If $n=m^{3}-m$ then $n$ is divisible by $6$.

[quote="boxedexe"][hide="One-liner"]$m^{3}\equiv m\pmod{6}$[/hide][/quote]If that's an acceptable one-liner, here is mine
[hide="one-liner"]$m^{3}-m=(m-1)(m)(m+1)$[/hide]

Boxedexe: you cannot apply the totient function for numbers that share a common factor with 6. It just happens that this works out...consider that
2^{5}-2 is not divisible by 12.

The best solution is that m^3=(m-1)(m)(m+1). The product of 3 consecutive integers must be divisible by 2 and 3, giving the result.

[quote="boxedexe"]When did [i]I[/i] do that?[/quote]

you just wrote m^3\equiv m\mod 6...and once could say that \phi(6)=2...to "get" that result...i don't see what you did then...other than state the problem again

You can quickly see that every element of $\mathbb{Z}_{6}$ satisfies what I wrote above.

He's just saying that it just happens to be correct... but you can't always make that statement. He's not saying that you are wrong.


integration limit calculus Support function vector geometry

$\int_{-\infty}^{+\infty}|f(x)|^pdx$ is limit,$p>0$
prove $\lim_{h\to 0} \int _{-\infty}^{+\infty}|f(x+h)-f(x)|^pdx=0$

This is a classical result in $L^p$ space.
I think you'd better pick up a book and read it.

indeed it's from my book,but there is no solution

What is the book?

In the following proof, all integrals are taken over the real line.

For $\epsilon > 0$ we write $f=f_1+f_2$, where $f_1$ is continuous with compact support and $f_2$ satisfies $\int |f_2|^p dx < \frac{\epsilon}{4 \cdot 2^p}$

Since $f_1$ has compact support and it is uniformly continous, thus there exists $\delta$ s.t. $\int |f_1(x+h)-f_1(x)|^p dx < \frac{\epsilon}{2}$ when $|h|<\delta$.

Thus

$\int  |f(x+h)-f(x)|^pdx \leq 2^p\left(\int  |f_1(x+h)-f_1(x)|^pdx + \int  |f_2(x+h)-f_2(x)|^pdx\right) < 2^p(\frac{\epsilon}{2} +2^p(\int |f_2(x+h)|^p dx + \int |f_2(x)|^p dx)) = 2^p(\frac{\epsilon}{2} + 2\cdot 2^p\int |f_2(x)|^p dx) = 2^p\epsilon$

[quote="liyi"]For $\epsilon > 0$ we write $f=f_1+f_2$, where $f_1$ is continuous with compact support and $f_2$ satisfies $\int |f_2|^p dx < \frac{\epsilon}{4 \cdot 2^p}$
[/quote]
why we can writ $f$ in this form? :(

[quote="zhaobin"]why we can write f in this form?[/quote]
Because the continuous, compactly supported functions are dense in $L^p(\mathbb{R})$ for $p<\infty.$ Proving this is tedious but quite important. One possible path for such a proof: Start with an arbitrary $L^p$ function. Approximate it (arbitrarily closely) by an $L^p$ function that vansishes outside a bounded set. Approximate that by a simple function. Use that fact that a measurable set is close to a finite union of intervals to approximate that simple function by a step function. Finally, shave a little off the corners of the step function to approximate it by (piecewise linear) continuous function.

For normed vector spaces (which are metric spaces), the official definition of separability is "has a countable dense subset." But the practical, often-used consequence of separability is "has a nice dense (linear) subspace." We're using the continuous compactly supported functions as a "nice" dense subspace of $L^p.$ It doesn't have to be that particular nice subspace - there are ways to make this proof work with other nice subspaces.

Note that the continous compactly supported functions are [b]not[/b] dense in $L^{\infty}$ and that in fact, $L^{\infty}$ is not separable and doesn't have any "nice" dense subspaces. This theorem of this problem - continuity of translation - is false in $L^{\infty}.$

[quote="liyi"]Since $f_1$ has compact support and it is uniformly continous, thus there exists $\delta$ s.t.[/quote]
You know, no one has yet bothered to answer my question [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=35288]here.[/url]


Hey guys, im a middle schooler here and someone on this forum may know. Me, HI SAMMY WHEREEVER YOU ARE.

Anyway, Sammy being a mathematicla genius(He is in 7th grade and knows like calculus, where we are learning algebra.

Getting back to topic, the teacher talked about doing square roots by hand quickly.
Can someone please explain this??

My favorite method is this...say you have $ \sqrt5\approx2.236067977$.
1. Guess (you can guess anything).
2. Divide.
3. Average.
4. Go back to 2, or stop when you want to. The number you got in step 3 is your approximation.
I will just do an example, since that explains it well.
I'll guess $ 10$ (which is obviously off, but just to demonstrate my point).
Then we divide $ 5$ by $ 10$, which gives us $ 0.5$.
Now we average $ 10$ and $ 0.5$, which gives us $ 5.25$.
$ 5/5.25 \equal{} 0.\overline{90}...\approx0.91$
Average $ 5.25$ and $ 0.91$. $ \equal{} 3.08$
$ 5/3.08\approx1.62$
Average $ 3.08$ and $ 1.62$. $ \equal{} 2.35$
$ 5/2.35 \approx 2.13$
Average $ 2.35$ and $ 2.13$. $ \equal{} 2.24$.
We could keep going, but you can tell that this is working well.

Ahh...I was also wondering bout that too, we don't get calculators on our tests, and we have to round our answers to nearest thousandth when were doing sqrt probs.  thanks!  even though i didnt ask the quetion

Yeah that's newton's method or something.
Here's another method (approximate):
$ \sqrt {x^2 \plus{} a} \equal{} x \plus{} \frac {a}{2x}$

This is a more advanced method:
$ \sqrt{N^2\plus{}d} \equal{} \sum_{n\equal{}0}^\infty \frac{(\minus{}1)^{n}(2n)!d^n}{(1\minus{}2n)n!^2 4^nN^{2n\minus{}1}} \equal{} N \plus{} \frac{d}{2N} \minus{} \frac{d^2}{8N^3} \plus{} \frac{d^3}{16N^5} \minus{} \frac{5d^4}{128N^7} \plus{} \cdots$

[quote="moogra"]Yeah that's newton's method or something.
Here's another method (approximate):
$ \sqrt {x^2 \plus{} a} \equal{} x \plus{} \frac {a}{2x}$

This is a more advanced method:
$ \sqrt {N^2 \plus{} d} \equal{} \sum_{n \equal{} 0}^\infty \frac {( \minus{} 1)^{n}(2n)!d^n}{(1 \minus{} 2n)n!^2 4^nN^{2n \minus{} 1}} \equal{} N \plus{} \frac {d}{2N} \minus{} \frac {d^2}{8N^3} \plus{} \frac {d^3}{16N^5} \minus{} \frac {5d^4}{128N^7} \plus{} \cdots$[/quote]

i'm gonna try the first one:

$ \sqrt{10}\equal{}\sqrt{3^2\plus{}1}\equal{}3\plus{}\frac{1}{6}\approx 3.16666$

with calc:  about 3.1622

OMG

$ \sqrt{10}\equal{}\sqrt{2^2\plus{}6}\equal{}2\plus{}\frac{6}{4}\equal{}3.15$

So the closer smaller $ a$ is, the more accurate, right?

It's basically the first term of the taylor series expansion. Yeah you are right.

aime i think your required to memorize the first like 10 or so?

What? NO!
This is not really meant to be memorized (perhaps just the expansion formula is already too much)

[quote="AIME15"][quote="moogra"]Yeah that's newton's method or something.
Here's another method (approximate):
$ \sqrt {x^2 \plus{} a} \equal{} x \plus{} \frac {a}{2x}$

This is a more advanced method:
$ \sqrt {N^2 \plus{} d} \equal{} \sum_{n \equal{} 0}^\infty \frac {( \minus{} 1)^{n}(2n)!d^n}{(1 \minus{} 2n)n!^2 4^nN^{2n \minus{} 1}} \equal{} N \plus{} \frac {d}{2N} \minus{} \frac {d^2}{8N^3} \plus{} \frac {d^3}{16N^5} \minus{} \frac {5d^4}{128N^7} \plus{} \cdots$[/quote]

i'm gonna try the first one:

$ \sqrt {10} \equal{} \sqrt {3^2 \plus{} 1} \equal{} 3 \plus{} \frac {1}{6}\approx 3.16666$

with calc:  about 3.1622

OMG

$ \sqrt {10} \equal{} \sqrt {2^2 \plus{} 6} \equal{} 2 \plus{} \frac {6}{4} \equal{} 3.15$

So the closer smaller $ a$ is, the more accurate, right?[/quote]
If I'm right, wouldn't 2 + 6/4 = 3.5, not 3.15?
Square root of 10 is 3.1622... so it's probably better when it's smaller.

There's a nice algorithm for this. Rather than explaining it, I decided to search for a [url=http://www.nist.gov/dads/HTML/squareRoot.html]link[/url].


N is a fixed natural number grater than or equal to 2. A number is called n-free when it is not the multiple of the nth power of a prime number. We denote by M the infinite set of rational numbers that satisfy this conditions:
1-The product of each n numbers is integer
2-The product of each n numbers is n-free.
Prove that the set M has all the elements integers.

Solve it very nice.

[hide]Hint:
Prove that M has a finite number of rational.And all the other are integers.[/hide]


number theory
least common multiple
modular arithmetic

Find the smallest positive integer  [b]n[/b] such that leaves remainder 1 when is divided by [b]2,3,4,5[/b]

[hide="Quick"] $ n \equal{} \text{lcm}(2, 3, 4, 5) \plus{} 1 \equal{} 61$ (can you see why?) [/hide]
[url=http://en.wikipedia.org/wiki/Chinese_remainder_theorem]Generalization[/url]

The smallest is 1, but I assume you mean $ n\ne{1}$.[hide]
$ x\equiv1\pmod{2}$
$ x\equiv1\pmod{3}$
$ x\equiv1\pmod{4}$
$ x\equiv1\pmod{5}$.
Or, $ x \minus{} 1|lcm[2,3,4,5]\cdot{k}$.
The smallest is 1+lcm[2,3,4,5], when $ k \equal{} 1$, or 61, as t0rajir0u said.[/hide]

Nic solution xpmath and t0rajir0u I solved it same way  :D  :)


inequalities
trigonometry
geometry
inradius
circumcircle
inequalities unsolved

let $ ABC$ be a triangle and $ A$ , $ B$ , $ C$ are the angles , prove that :

$ \cos(A)\plus{}\cos(B)\plus{}\cos(C) \geq 1$

[quote="y-a-s-s-i-n-e"]let $ ABC$ be a triangle and $ A$ , $ B$ , $ C$ are the angles , prove that :

$ \cos(A) \plus{} \cos(B) \plus{} \cos(C) \geq 1$[/quote]
$ \cos(A) \plus{} \cos(B) \plus{} \cos(C) \equal{} \frac{R\plus{}r}{R}> 1$; :)

thank"s  :D

can you prove that  :?:

$ \cos(A) \plus{} \cos(B) \plus{} \cos(C) \equal{} \frac{R\plus{}r}{R}$

[quote="y-a-s-s-i-n-e"]thank"s  :D

can you prove that  :?:

$ \cos(A) \plus{} \cos(B) \plus{} \cos(C) \equal{} \frac {R \plus{} r}{R}$[/quote]
$ \cos(A) \plus{} \cos(B) \plus{} \cos(C)\equal{} 2cos((A\plus{}B)/2)cos((A\minus{}B)/2)\minus{}$ $ 2cos^2((A\plus{}B)/2)\plus{}1\equal{}4sin(A/2)sin(B/2)sin(C/2)\plus{}1\equal{}r/R\plus{}1;$ :)

[quote="y-a-s-s-i-n-e"]let $ ABC$ be a triangle and $ A$ , $ B$ , $ C$ are the angles , prove that :

$ \frac{3}{2} \geq \cos(A) \plus{} \cos(B) \plus{} \cos(C) \geq 1$[/quote]
 PS: Hence  we have inequality $ R \geq 2r$

[quote="Sergic Primazon"][quote="y-a-s-s-i-n-e"]thank"s  :D

can you prove that  :?:

$ \cos(A) \plus{} \cos(B) \plus{} \cos(C) \equal{} \frac {R \plus{} r}{R}$[/quote]
$ \cos(A) \plus{} \cos(B) \plus{} \cos(C) \equal{} 2cos((A \plus{} B)/2)cos((A \minus{} B)/2) \minus{}$ $ 2cos^2((A \plus{} B)/2) \plus{} 1 \equal{} 4sin(A/2)sin(B/2)sin(C/2) \plus{} 1 \equal{} r/R \plus{} 1;$ :)[/quote]

What are $ r$ and $ R$?  Also, how did you find the last identity $ 2\cos((A \plus{} B)/2)\cos((A \minus{} B)/2) \minus{}$ $ 2\cos^2((A \plus{} B)/2) \plus{} 1 \equal{} 4\sin(A/2)\sin(B/2)\sin(C/2) \plus{} 1$ (from sketch)?  Thanks.

[quote="Sunkern_sunflora"]mathwizarddude wrote:

[quote]Sergic Primazon wrote:
y-a-s-s-i-n-e wrote:
thank"s Mr. Green

can you prove that Question

\cos(A) + \cos(B) + \cos(C) = \frac {R + r}{R}

\cos(A) + \cos(B) + \cos(C) = 2cos((A + B)/2)cos((A - B)/2) - 2cos^2((A + B)/2) + 1 = 4sin(A/2)sin(B/2)sin(C/2) + 1 = r/R + 1; Smile

What are and ? Also, how did you find the last identity 2\cos((A + B)/2)\cos((A - B)/2) - 2\cos^2((A + B)/2) + 1 = 4\sin(A/2)\sin(B/2)\sin(C/2) + 1 (from sketch)? Thanks.[/quote]
r  :arrow: inradius
R :arrow: circumradius
As for the identity, cosA+cosB=2ccos((A+B)/2)cos((A-B)/2)
and, 1-cosC=2cos^2(C/2)=2cos^2((A+B)/2)
or, cosC=1-2cos^2((A+B)/2)
Combining both we obtain the final identity....
Very Very nice solution by Sergic Primazon.[/quote]

Oh, I meant the identity

$ 2\cos((A\plus{}B)/2) \cos((A\minus{}B)/2) \minus{} 2\cos^2((A\plus{}B)/2) \plus{} 1$
$ \equal{} 4\sin(A/2)\sin(B/2)\sin(C/2) \plus{} 1$

How would one prove that $ 4\sin(A/2)\sin(B/2)\sin(C/2) \equal{} r/R$?

By the way, $ \cos(C/2) \equal{} \cos\left(\frac{\pi \minus{} (A\plus{}B)}2\right) \equal{} \cos(\pi/2 \minus{} (A\plus{}B)/2) \equal{} \sin((A\plus{}B)/2)$

$ \cos C \equal{} 2\cos^2(C/2) \minus{} 1 \equal{} 2\sin^2\left(\frac{A\plus{}B}2\right) \minus{} 1 \equal{} \minus{}\cos(A\plus{}B) \equal{} 1\minus{}2\cos^2\left(\frac{A\plus{}B}2\right)$

You said that $ 1\minus{} \cos C\equal{}2\cos^2(C/2)\equal{}2\cos^2((A\plus{}B)/2)$, but the correct half-angle identity is $ \cos C \minus{} 1 \equal{} 2\cos^2(C/2)$ instead, and $ 2\cos^2(C/2) \equal{} 2\sin^2((A\plus{}B)/2)$; I'm not sure if this would be equal to $ 2\cos^2((A\plus{}B)/2)$ in your post.  But correct me if I'm wrong.


Find all the positive integers N, satisfying the next condition:

Exist positive rationals a1, a2, a3,..., ak (k>1)
a1+a2+...+ak = (a1)(a2)...(ak) = N.

Please, posted your solutions....

Let N is not prime. Then $N=x*y*1*1*1...=x+y+1+1+...+1$, were x>1,y>1 and number of 1 is N-x-y>=0.
If N is prime then k=1 (a1=N).

Hi, your solution is good, but...
You have prove the reason when N is prime!!
Remember that k>1

I think he means that you must prove that no such [u]rationals[/u] can be found if $N$ is prime.

You proved it only for integer $a_{k}$, not for rational.

By AM-GM we have $a_{1}+\dots+a_{k}\geqslant k(a_{1}\dots a_{k})^{1/k}\implies N\geqslant kN^{1/k}\implies N\geqslant k^{k\over{k-1}}$. The function $f(k)=k^{k\over{k-1}}$ is increasing (easy to show), hence it's minimal value for integer $k>1$ is $f(2)=4$. Therefore $N\geqslant 4$, which means that $1,2,3$ can't be expressed in the required way.

Next, let's put $k=3$. Then $N\geqslant 3\sqrt{3}$, but since $5<3\sqrt{3}$, it means that $5$ can't be expressed in the required way with three or more terms. Hence we must have $u+v=uv=5$, but that system has no rational solution. Therefore we discard $5$ as well.

For $k=5$ we get $N\geqslant 5^{5/4}$, but $7<5^{5/4}$, hence $7$, if that's possible at all, can be expressed with at most $4$ terms. The system $u+v=uv=7$ has no rational solution, and the only problem left is whether the systems $u+v+w=uvw=7$ and $u+v+w+t=uvwt=7$ have rational solutions.

[quote="Farenhajt"] The system $u+v=uv=7$ has no rational solution, and the only problem left is whether the systems $u+v+w=uvw=7$ and $u+v+w+t=uvwt=7$ have rational solutions.[/quote]
Systems $u+v+w=uvw=7$ and $u+v+w+t=uvwt=7$ have real positive solutions, but have not rational solutions.

[quote="Rust"]
Systems $u+v+w=uvw=7$ and $u+v+w+t=uvwt=7$ have real positive solutions, but have not rational solutions.[/quote]

Can you give at least an outline of the proof?

Its good, now only prove for n=7. Its difficult because dont know iff is true or false!!!
Any comment??


topology
real analysis
real analysis unsolved

Show that a compact metric space has cardinal at most equal to $c$ (the continuum cardinality).

I found it on [url=http://www.princeton.edu/~mathclub/puzzles.html]this website[/url]. Hit "Hard Puzzles", and it's the ninth problem there.

My approach is to prove that the cardinal of the space is at most the cardinal of the set of sequence of real numbers, which is $c$ itself :)

It is interesting to find out that some of the Impossible Puzzles there had been discussed before in the forum :D

Yes, this source has been mentioned before on the forum. I think you can even find a nice unsolved problem in the "Combinatorics" section.

Btw, dickchimney, are you still thinking about it, or did you solve it and just don't feel like posting the whole solution? :)

Actually, I met a different version of this one before. It stated that a compact metric space can be imbedded into $\mathbb{R}^{\infty}$

What's $\infty$ from the exponent, in this case? Can it be any infinite cardinal, or is it countable?

It is countable cardinal ;) What is your approach?

You mean for the problem I posted? (because I don't know if I can solve the one you mentioned :)). I think it's better to give people time to think about it. It's only been here for like two days :).


function
inequalities solved
inequalities

Let f be a convex real-valued function defined on an interval I.

Prove that f(x)+f(y)+f(z)+3f((x+y+z)/3) >= 2( f((x+y)/2) + f((y+z)/2) + f((z+x)/2) ).

Using this we can prove several inequalities.

Just have a look at :
http://www.mathlinks.ro/viewtopic.php?t=430&highlight=popoviciu

Pierre.

What is Mathematical miniatures, contest, magazine ?

A book from S.Savchev and T.Andreescu, available from the M.A.A..

Pierre.


20. In triangle ABC, points D and E lie on line segments BC and AC, respectively. If line segments AD and BE intersect at point T so that AT/DT=3 and BT/ET=4, what is CD/BD?

When you read this problem, what ideas and strategies come to your mind? How do you analyze and approach the problem in the beginning? I've been finding that I've been getting stuck on the more difficult AMC 10 geometry problems, and am usually discouraged when I read the solutions to the problems because I can understand them, but I don't know how I would have thought of the solutions.

Also, why must the area of triangle TBC be one fourth the area of triangle ABC?
Why must the area of triangle ATC be one fifth of the area of triangle ABC?

[quote="aufha"]20. In  triangle ABC points D and E lie on line segments BC and AC, respectively. If line segments AD and BE
intersect at T so that AT=DT = 3 and BT=ET = 4, what is CD=BD?

When you read this problem, what ideas and strategies come to your mind? How do you analyze and approach the problem in the beginning? I've been finding that I've been getting stuck on the more difficult AMC 10 geometry problems, and am usually discouraged when I read the solutions to the problems because I can understand them, but I don't know how I would have thought of the solutions.

Also, why must the area of triangle TBC be one fourth the area of triangle ABC?
Why must the area of triangle ATC be one fifth of the area of triangle ABC?[/quote]Is that the correct statement of the problem? The conditions imply that $ AC\parallel{}BC$..

Yes, I typed it in incorrectly. My apologies.

Well, if they give you cevians and ratios, I tend to think areas first.

[hide]Draw TC. Then, just use base ratios to compare areas in triangles BEC and ADC. You can find the relative areas of each little triangle this way. Then, CD/BD is just the area of ADC/the area of ADB.[/hide]

[quote="aufha"]Also, why must the area of triangle TBC be one fourth the area of triangle ABC?
Why must the area of triangle ATC be one fifth of the area of triangle ABC?[/quote]

TBC and ABC share the same "base": BC, and since the length of DT is 1/4 of AD, the area of TBC is 1/4 that of triangle ABC. You can discover a similar relationship regarding the areas ATC and ABC.


geometry
3D geometry
induction
search
algebra
factorization
sum of cubes

I'm not really sure if this should go in this forum...but I saw this animation on one of the AoPs pages and I'd like to know how to prove it:

$ \sum_{k=0}^{n}k^{3}= \left( \sum_{k=0}^{n}k \right)^{2}$

Thanks

Induction if you already know they are equal

If you just want to make one side become the other side factoring is good

The animation on the sidebar is a "proof without words" -- it's not very hard to convert it into a geometric argument for the general case.  Algebraic proofs have appeared many, many times on the forum -- try a search for "sum of cubes" or something like that.  (Induction actually doesn't work directly if you don't know the closed form for the left sum.)

And if you want derivation for e.g. $ \sum_{k=1}^{n}k^{3}$, look here:
[hide="Big hint"]
Look in my 2nd post in this thread:
http://www.mathlinks.ro/Forum/viewtopic.php?p=900576#900576
There's derivation for $ \sum_{k=1}^{n}k^{2}$, but you can easy prove a formula for $ \sum_{k=1}^{n}k^{3}$[/hide]

[hide]

I find it easier to think of it this way:

We are trying to prove that $ (1+2+...+n)^{2}= 1^{3}+2^{3}+...+n^{3}$.

Now let $ T_{n}$ denote the $ n^{th}$ triangular number. Then:

$ T_{n}^{2}-T_{n-1}^{2}= n^{3}$.

Then, simply rearranging the left hand side proves what is desired.[/hide]


[quote="Treething"]On Treething's planet, the time it takes for an object to fall to the ground is directly proportional to the cube of its height above the landing. (The gravity is messed up there.) If Bobert takes sixteen seconds to fall from the top of his eighteen inch tall gigantic mushroom, how many minutes will it take for him to fall from the top of the twenty-seven-foot-tall Leaning Tower of Treething? Round to the nearest integer.[/quote]

.............
[hide]
$\frac{t}{h^3}$ is constant. $\frac{16}{5832}=\frac{x}{34012224}$, $x=93312$, to the nearest minute, $\boxed{1555}$[/hide]

[hide]Because the time it takes to fall to the ground is directly porportional to the height of the building cubed, $\frac{t}{h^3}$ is a constant, where $t$ is the time it takes, and $h$ is the height.  Set up a porportion (in terms of minutes and feet), then solve:

$\frac{16}{\frac{27}{8}\times60}=\frac{32}{405}$
Add in the height of the Leaning Tower of Treething:
$\frac{32}{405}=\frac{x}{27^3}\Rightarrow 405x=32\times27^3\\x=1555.2$

Rounded to the nearest integer, $\boxed{1555\text{ minutes}}$[/hide]

[quote="Treething"]On Treething's planet, the time it takes for an object to fall to the ground is directly proportional to the cube of its height above the landing. (The gravity is messed up there.) If Bobert takes sixteen seconds to fall from the top of his eighteen inch tall gigantic mushroom, how many minutes will it take for him to fall from the top of the twenty-seven-foot-tall Leaning Tower of Treething? Round to the nearest integer.[/quote]

[hide]$\frac{16}{18^3}=\frac{x}{324^3}$
Solve to get x=93312 seconds which equals 1555 minutes to the nearest minute.[/hide]

[quote="Treething"]On Treething's planet, the time it takes for an object to fall to the ground is directly proportional to the cube of its height above the landing. (The gravity is messed up there.) If Bobert takes sixteen seconds to fall from the top of his eighteen inch tall gigantic mushroom, how many minutes will it take for him to fall from the top of the twenty-seven-foot-tall Leaning Tower of Treething? Round to the nearest integer.[/quote]
[hide]Wow, I want to go to that planet.  If I fall from a low thing at least I will have time to fall down and react.  But still I like falling down.

I got 1555.[/hide]

I just have to say this.

Treething, what planet are you from?

[quote="SplashD"][quote="Treething"]uranus

ba dum psh[/quote]

It must have taken you all night to think of that one. :rotfl:[/quote]

bah I got it off the top of my head, there's even only three minutes between our posts

[quote="Ignite168"][quote="mathgeniuse^ln(x)"][quote="Treething"]uranus

ba dum psh[/quote]

[hide=" :rotfl:"]You are scaring me.[/hide][/quote]

Then you don't want to talk to him on AIM.[/quote]

[hide=" :rotfl: "]That was scarier[/hide]

[quote="mathgeniuse^ln(x)"][quote="pgpatel"][quote="Treething"]uranus

ba dum psh[/quote]
I don't get the "ba dum psh" thing :?[/quote]

I don't think people should explain this to you online or in private messaging.[/quote]

Yeah, it's a sound effect when someone makes a [color=white]really bad[/color] joke  ;)


Profesorul meu de matematica sustine sus si tare ca se pot rezolva probleme de geometrie in spatiu folosind afixe de numere complexe?!
Mai exact:problema e urmatoarea:
Fie tetraedrul $ABCD$,consideram $G$ centrul de greutate al fetei $BCD$; $M$ mijlocul lui $AC$ si $N$ mijlocul lui $BD$,$I$ mijlocul bimedianei $MN$.Sa se arate ca $A,I,G$ sunt coliniare.
Si pune afixe(in spatiu??) $A(a),B(b),C(c),D(d),G(g),M(m),N(n),I(i)$.
Ii rezulta
$g=\frac{b+c+d}{3}$
$m=\frac{a+c}{2}$
$n=\frac{b+d}{2}$
$i=\frac{a+b+c+d}{4}$

si cum $\frac{i-a}{g-a}=\frac{3}{4} \in R$ concluzioneaza ca cele 3 puncte sunt coliniare.
Parerea mea este ca solutia este profund gresita,ce inseamna afixul unui punct din spatiu?
Solutia sa ar putea fi corecta doar daca $ABCD$ ar fi degenerat,toate punctele in acelasi plan.
I-am spus asta si a raspuns:Pai ce,punctele $B,C,D$ de exemplu sunt intr-un plan;la fel $A$ si $C$; $B$ si $D$; $M$ si $N$.

Exista cumva posibilitatea ca solutia sa fie corecta?
Sa asociem unui pct din spatiu un "numar complex" de forma $a+bi+cj$?
Oricum nu cred ca e posibil asa ceva,si chiar daca am gasi un astfel de corp de "numere",nu s-ar pastra proprietatile geometrice.
Eu nu am gasit nicaieri probleme de geom in spatiu rezolvate cu nr complexe si cred ca e o aberatie.
Ce credeti?

Si eu sunt de aceeasi parere cu tine. Vectori, da si in spatiu ! Insa numere complexe, in spatiu, dupa umila mea parere, nu cred. Poate imi spui si mie prin mesaj privat cum te cheama, la ce liceu esti si cine este "creatorul" ? Totusi ii acord circumstante atenuante: exista mule lucruri spuse sau faptuite pe lume de care nu am auzit inca ! Mai constat ca dupa dansul doua puncte sunt in plan. Nu ne mai mira, si doua puncte sunt coliniare !

Probabil ca se gandea la vectori si nu a stiut cum sa exprime acest lucru. Sau (foarte putin probabil) vorbea de cuaternioni, care au aplicatii reale in special in probleme in care sunt implicate rotatii in spatiu.

[i]V.V[/i], eu i-as sugera elevului sa-i spuna profesorului ca a citit despre niste numere, numite cuaternioni, dincolo de multimea numerelor complexe, ca nu a inteles si sa-l roage sa-l ajute sa priceapa si el ceva. Nu de alta, dar acest elev a intuit intr-o prima aproximatie ... cuaternionii ! S-ar putea sa gresesc, dar eu cred ca intuiesc raspunsul ... dascalului !.[i]Dn. V.V., [/i]esti mai intelegator (tolerant) decat mine. Stai departe de ... noi!. Nu gresesc prea mult daca afirm ca in scoala de azi, in cadere libera nu sunt numai [b]the mathematics[/b], ci si [b]the mathe[/b] [b]department[/b] !
Ar fi interesant sa deschizi un post cu acest subiect fierbinte si de mare actualitate care, printre altele, sa contina si "perle" ale profesorilor.

O idee buna cu "perle" dar nu numai de la profesori. O sa postez ceva la Generalitati...

De la elevi nu ma mira, sunt "cruzi" (in crestere) si sunt o consecinta a ceea ce spuneam mai devreme. Interesante sunt mai degraba perlele profesorilor. Nu sunt de acord sa apara si perlele elevilor. Acestea sa fie imediat si decent corectate la postul respectiv. TE ASIGUR EU CA A DOUA OARA NU VA MAI GRESI. Cred ca am apasat din greseala pe Caps Lock !.

Perlele elevilor sunt, de cele mai multe ori, cauzate de perlele profesorilor. Cred ca merita sa le postam.

Ai dreptate, Bogdane, in masura in care elevul mentioneaza si "sursa" !

Hai să mutăm discuţia [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=65663]aici[/url]


group theory
combinatorics unsolved
combinatorics

$ Problem$

Let $ S_n$ be the [url=http://en.wikipedia.org/wiki/Permutation_group]permutation group[/url]. Let $ \pi \in S_n$ be a permutation. It is known that every permutation of $ S_n$ can be written as union of disjoint cycles. Hence we should be able to count $ n!$ by counting the ways we can form disjoint cycles among $ \{1,\dots,n\}$. Derive the closed form.

formula (1) at http://mathworld.wolfram.com/PermutationCycle.html


conics
ellipse
quadratics
algebra
system of equations

For what positive value of [i]r[/i] does the system of equations [(x – 2)^2/16] + y^2 = 1 and (x - r)^2 + y^2 = r^2 have exactly three solutions

I believe you have an ellipse and circle equation. At some point you need the ellipse and circle to be tangent so that you have 3 intersections instead of 4...

That may be very true....but I'm not sure how this will help me solve the problem

The ellipse is centered on the x-axis and does not pass through the origin. The circle is centered on the x-axis and passes through the origin.

Even without using ellipse and circle, both equations have only $ y^2.$ If $ (x, y)$ is solution, $ (x, \minus{}y)$ is also solution $ \Longrightarrow$ number of solutions is even. Unless $ y \equal{} 0$ is a solution. Put it to the 2nd equation, $ x^2 \minus{} 2rx \equal{} 0.$ $ (0, 0)$ is not a solution (see 1st equation) $ \Longrightarrow$ $ x \neq 0,$ $ x \equal{} 2r.$ Put $ (x, y) \equal{} (2r, 0)$ to the 1st equation, resulting in a quadratic equation for $ r$ with roots $ r \equal{} \minus{}1, \plus{}3.$ If demanding $ r > 0,$ then only $ r \equal{} 3.$


Try to solve this one.I created it myself.
Quite easy.

1) Shyam  commisioned sherlock holmes to trace his four long lost nephews.
   He left 8552$ between them in his will.The money would be divided according to their
    ages,with each nephew receiving$2 more than the youngest to him.

    How much would each of the nephews receive?

Um, can you fix your $ \text{\LaTeX}$ so that the text isn't so hard to read? Just un-Latexify your regular text.

[quote="karpov352"]Try to solve this one.I created it myself.
Quite easy.

1) Shyam  commisioned sherlock holmes to trace his four long lost nephews.
   He left $ \$$8552 between them in his will.The money would be divided according to their ages,with each nephew receiving $ \$$2 more than the youngest to him.

    How much would each of the nephews receive?[/quote]
uh, wait, never mind. I fixed it now. See the quoted box in this post for the Latex correction.


Does anyone have Calculus: Early Transcendental Functions by Robert Smith and Roland Minton?

I ordered the book on Amazon, but the person is taking his sweet time shipping it over. I need it for some homework problems. So if anyone has this book, could I have:

8.1 #16, 20, 36, 42
8.2: #10, 20, 22, 32

Thanks. I really appreciate any help I can get.

So, I'm moving this to the calculus forum, not because I think that's a good place for it but because I think that's where you're most likely to get an answer.


Prove that, if $ P_k$ is the $ k^{th}$ prime, that $ \sum^{\infty}_{n \equal{} 1} \frac {1}{P_k}$ approaches $ \ln \ln k$ for high $ k$

You're having some notation problems.  See http://mathworld.wolfram.com/HarmonicSeriesofPrimes.html .

You can look [url=http://en.wikipedia.org/wiki/Proof_that_the_sum_of_the_reciprocals_of_the_primes_diverges]here[/url] for inspiration  :wink:


calculus
derivative
function
geometry
3D geometry
limit
trigonometry

I'm having difficulty with a few problems.  they are listed below:

1.)find the derivative using proper rules:
f(x)=[(6x-5)^2(3x-2)^3]/[(x^2)-2] - dont simplify, only rewrite using proper derivative rules

2.Find the equation of the tangent line to the graph of the function y=(cos 2x)^2 at (pi/6, 1/4) {<----this last parenthesis is a point on a graph.}

3.Consider the function f(x)={(x^2)+4x+2; where x is less than -2
                                             {1-4x-(x^2); where x is greater than or equal to -2
-----justify your answer for 3 (the function has 2 parts, as denoted by the { symbols

4.) related rate problem: all edges of a cube are expanding at a rate of 3 inches per second.  How fast is the volume changing when each edge is (a) 1 inch & (b) 10 inches

BONUS question:
5.) prove that (cos x)'=-sin x  --- ' is derivative notation

please help me out; thank you

[hide="Sol to 4, sorry if this is wrong (Im new to this)"]
Lets say the side of the cube is $a$ so then,
Volume per sec,
$a^3,(a+3)^3,(a+6)^3.....$
Taking a graph of the volume vs the time,
Where $y$ is the volume and $x$ is the time so,
So find an equation and then find the gradient of that equation, whatever you get will be in terms of $a$ which we know.
[/hide]

Problem 5
First, the definition of derivative $\frac{d}{dx}f(x)=\lim_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$ can in fact be rewritten as $\frac{d}{dx}f(x)=\lim_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x-\Delta x)}{2\Delta x}$
Therefore,
$\frac{d}{dx}\cos x$
$=\lim_{\Delta x\to 0}\frac{\cos (x+\Delta x)-\cos (x-\Delta x)}{2\Delta x}$
$=\lim_{\Delta x\to 0}\frac{(\cos x\cos \Delta x-\sin x\sin \Delta x)-(\cos x\cos \Delta x+\sin x\sin \Delta x)}{2\Delta x}$
$=\lim_{\Delta x\to 0}\frac{-\sin x\sin\Delta x}{\Delta x}$
$=-\sin x\lim_{\Delta x\to 0}\frac{\sin\Delta x}{\Delta x}$
$=-\sin x$

[quote="cheeseman42001"]
2.Find the equation of the tangent line to the graph of the function y=(cos 2x)^2 at (pi/6, 1/4) {<----this last parenthesis is a point on a graph.}
[/quote]

$y = f(x) = (\cos {2x})^2$
$y' = 2 \cos 2x \cdot (-\sin 2x) (2)$
$y' = -4 (\cos 2x)(\sin 2x)$

Plug in the x value into the derivative to obtain the slope of the line at that point

$f'(\frac{\pi}{6}) = -4 (\cos 2(\frac{\pi}{6}))(\sin 2(\frac{\pi}{6}))$
$f'(\frac{\pi}{6}) = -4 (\cos \frac{\pi}{3})(\sin \frac{\pi}{3})$
$f'(\frac{\pi}{6}) = -4 (\frac{\sqrt3}{2})(\frac12)$
$f'(\frac{\pi}{6}) = -\sqrt3$

This is the slope of f(x) at $\frac{\pi}{6}$

Now use the point slope form of a line equation with your point $(\frac{\pi}{6} , \frac14$ and slope $-\sqrt3$

$y - y_1 = m (x - x_1)$

$y - \frac14 = -\sqrt3 (x - \frac{\pi}{6})$

That is the equation of the line...

If you didn't understand something, feel free to ask


I wrote this solution for this problem but I'm not sure if my solution is right or not so I'm posting here.

The problem is:

Show that $n^2 + n - 1$ and $n^2 + 2n$ have no common factor.

[hide="My proof"]

First, we have to note that $n^2 + 2n = n(n+2)$ and $n^2 + n - 1 = n(n+1) - 1$. Let's rewrite each term in $\mod n$.

For $n(n+2)$, $n(n+2) \equiv n+2 \mod n \equiv 2 \mod n$.

For $n(n+1) - 1$, observe $n(n+1)$ first. $n(n+1) \equiv n+1 \mod n \equiv 1 \mod n$ and this means that $n(n+1) - 1 \equiv 0 \mod n$.

Since they are different in base mode, they don't have common factor.[/hide]

I'm not sure whether what I showed whether they have common factor or not..  :?  :?

[quote="Silverfalcon"][hide="My proof"]

For $n(n+2)$, $n(n+2) \equiv n+2 \mod n \equiv 2 \mod n$.

Since they are different in base mode, they don't have common factor.[/hide]

I'm not sure whether what I showed whether they have common factor or not..  :?  :?[/quote]

Silverfalcon, the quoted above from you is not correct, what happens if $n$ is even? Nevermind, I edited my post.

[quote="Silverfalcon"]
For $n(n+2)$, $n(n+2) \equiv n+2 \mod n \equiv 2 \mod n$.
[/quote]

So you're saying that for $n=6$, $48\equiv 8 \mod 6$? That's obviously not the case. You got your mod rules mixed up.

Use the Euclidean algorithm.

[hide="What_I_did"]Any common factor of $n^2+2n$ and $n^2+n-1$ must be a factor of $n^2+2n-(n^2+n-1)=n+1$. However, $n^2+n-1=n(n+1)-1$, which is not divisible by any factor of n+1 (except 1 obviously) since n(n+1) is.[/hide]

Like LynnelleYe said...if the two have a common factor, then it must be a factor of their difference, n+1.  However, the two can be written as
$n(n+1)-1$
and
$(n+1)(n+1)-1$
so neither of them shares a factor with n+1 anyways.

That's Euclid's algorithrm, isn't it? :)


There are 5 girls and 7 boys at a party. They must stand in a line such that no girls are standing adjacent to each other. How many different arrangements are possible?

[hide]GBGBGBGBG

Then we can "insert" the boys to find the number of arrangements:

6*5*4/6=20
6*5=30
6=6
56?[/hide]

Hint: each girl and boy can be considered "distinct"

[hide="the answer is"]correct. :yup: [/hide]

It stands for We Owned the Other Team, although now it's used as Yay!

[quote="easyas3.14159..."]happiness, cheering, or something like that. i can't really explain it.[/quote]
That explains a lot...

[hide]
You have the 7 girls. You have 8 spots between them or to the end of them.
You can put the boys in any of those spots.
$\binom 85 = 56$
If they are distinct, multiply by $5!7!$[/hide]

[quote="mysmartmouth"]From UrbanDictionary.com:

Woot originated as a hacker term for root (or administrative) access to a computer. However, with the term as coincides with the gamer term, "w00t".

"w00t" was originally an trunicated expression common among players of Dungeons and Dragons tabletop role-playing game for "Wow, loot!" Thus the term passed into the net-culture where it thrived in video game communities and lost its original meaning and is used simply as a term of excitement.

[size=150][b]AND[/b][/size]

1) An interjection similar to "YAYE!" or "Woohoo!" used to express joy or excitement, usually about some kind of accomplishment. Primarily used by gamers, spreading rapidly to anyone who chats online

Gaming origins: We Owned the Other Team
or Wonderful Loot (Everquest gaming slang)

2) To express agreement

3) As an acronym, can mean Waste of Our Time or Way Out of Topic (online forums) Less common.

1)woot! I got an A on this test.[/quote]

such a short definition...could it have been any shorter? ;)

[hide]$\displaystyle\binom{8}{5}=\frac{8!}{5!\cdot 3!}=56$  We multiply this by $5!\cdot 7! \cdot 56$  And we get $\displaystyle 33868800$[/hide]


what is the result of the following fraction is x>y>0 then x^y y^x over y^y x^x= ?

so can anyone explain me this?

is this what you mean?:

$\frac{x^y \cdot y^x}{y^y \cdot x^x}$
where $x>y>0$

anyways, i suggest posting this in a harder forum than this. :)

[quote="apusey"]If you want the answer to your question
[hide="answer"]you know that x^y/x^x times y^x/y^y is the same thing so you get x^(y-x) times y^(x-y) if you want to go[hide="futher"]then you factor out a negative from y^(x-y) and get 1/y^(y-x).  Multiplying the parts of the equation together you get x^(y-x)/y^(y-x).  Going further you get (x/y)^(y-x)[/hide][/hide][/quote]

can you please use LaTeX so it is easier to understand??

[quote="apusey"]If you want the answer to your question
[hide="answer"]you know that $\frac{x^y}{x^x} \cdot \frac{y^x}{y^y}$ is the same thing so you get $x^{y-x} \cdot y^{x-y}$ if you want to go[hide="futher"]then you factor out a negative from $y^{x-y}$ and get $\frac{1}{y^{y-x}}$.  Multiplying the parts of the equation together you get $\frac{x^{y-x}}{y^{y-x}}$.  Going further you get $(\frac{x}{y})^{y-x}$[/hide][/hide][/quote]

$\text {TeXeD}$ ;)


algebra
function
domain
quadratics
LaTeX
interval notation

What is the domain of

y^2 = (x^3 - 4x^2 + 3x)/(x^2 - 4)?

i dont think that he needs help.  He said challenge problem so he wants to see if we can get it!

and if im not mistaken the domain is All ReaL!

It can't be all real, clearly $ x \neq \minus{} 2, 2$.

[hide="Solution"]
\[ \frac {x(x \minus{} 1)(x \minus{} 3)}{(x \minus{} 2)(x \plus{} 2)} \ge 0 \Longleftrightarrow \boxed{ \minus{} 2 < x \le 0,\ 1 \le x < 2,\ 3 \le x}
\]
[/hide]

Edit: Did in my head; just count the number of quantities that are negative in each region. Eg, if $ 2 < x \le 0$, then $ \frac {( \minus{} )( \minus{} )( \minus{} )}{( \minus{} )( \plus{} )} \ge 0$. [computer is kind've overkill]

I agree with the answer given in the last post, but I don't know if the user worked it out by hand, or used a calculator or computer.

oh wow i didnt see divide sight so i assume domain was all real...Ok the way i go about doing these is factor the bottom or use quadratic formula..LOL...to find """""""""zeroes""""""""""" which we know that the denominator cant be.  The domain therefore is just All real - the two zeroes which i guess are 2 and -2(factor differnce of two perfect squares)!!!

And kinda a personal preference!!!i like interval notation!!!!!!

So our answer would be $ \boxed{( \minus{} \inf, \minus{} 2)U( \minus{} 2,2)U(2,\inf)}$
inf=infinity and $ U$ is union (*cough* im bad at LATEX!)

EDIT: wow this would be for y=........ but lol i cant read it says $ y^2$ doesnt it... but still i like interval!!!

For infinity in latex, you can use

[code]$\infty$[/code]

$ \infty$

For Unions, you can use

[code]$\cup$[/code]

$ \cup$

Works great

[hide="Solution"]We can use sign analysis to complete this solution. We have, by factorig, $ \dfrac{x(x\minus{}1)(x\minus{}3)}{(x\plus{}2)(x\minus{}2)}$.

For $ x< \minus{}2$, we have $ \dfrac{(\minus{})(\minus{})(\minus{})}{(\minus{})(\minus{})}\implies (\minus{})$.
For $ \minus{}2<x\leq 0$, we have $ \dfrac{(\minus{})(\minus{})(\minus{})}{(\minus{})(\plus{})}$.
Likewise, we complete our sign analysis and approach our answer: $ \boxed{\minus{}2<x\leq 0;1\leq x<2;x\geq 3}$.[/hide]

Thanks chickendude!!!and why do people here not like interval notation (cause i do its so much easier to read!!!!)


linear algebra
matrix
function
linear algebra unsolved

How are condition numbers of the individual matrices affect the condition number of the product of the matrices. I mean let $ A \in \mathbb{R}^{m \times n}$ has condition number $ \mathcal{K}_A$,  $ B \in \mathbb{R}^{n \times p}$ has condition number $ \mathcal{K}_B$, then what will be the condition number of the matrix $ C \equal{} A B$ in terms of that of $ A$ and $ B$?

Any help?

Bhupala.

$ \mathcal K_A \equal{} \| A\| \| A^\minus{}1\|$ right ?
if so: $ \mathcal K_{AB} \equal{} \| AB\| \| B^{\minus{}1}A^\minus{}1\| \leq   \| A\| \| A^\minus{}1\|  \| B\| \| B^\minus{}1\| \leq \mathcal K_{A} \mathcal K_{B}$  :maybe:
(of course there is no stuff like $ K_{AB} \equal{} f(K_{A},K_{B})$ where $ f(.,.)$ is a function)


The MIT Educational Studies Program (ESP) is back again! now we have the High School Studies Program (HSSP), a ten-week education program. It runs from Saturday March 4th until Saturday May 13th.  During these ten weeks, you can learn just about whatever you want from very enthusiastic MIT Students teaching stuff that they like. Classes are scheduled every Saturday during the timeframes of 10-12am, 1-3pm, and 3-4:30pm. For only $\$30$, you can choose to take classes during anywhere from one to three of those times.

Student online registration is now open, the program starts next Saturday, hope to see you all there :)

* The main page for HSSP is at http://esp.mit.edu/web/takeaclass/hssp.html.
* The course catalog is at http://esp.mit.edu/web/takeaclass/hsspreg.html?op=catalog.
* Online registration is at http://esp.mit.edu/web/takeaclass/hsspreg.html. (You need to make an account to register.)

Announcing Summer HSSP and Junction!
two programs run by the MIT Educational Studies Program this summer.
More information could be found at:
http://esp.mit.edu/web/takeaclass/registering.html for HSSP, and
http://esp.mit.edu/web/takeaclass/junction.html for Junction.


[quote="skylover"]If $ a,b,c > \frac {2}{9}$, then $ \frac {a^3}{9b \plus{} 9c \minus{} 4} \plus{} \frac {b^3}{9c \plus{} 9a \minus{} 4} \plus{} \frac {c^3}{9a \plus{} 9b \minus{} 4}\geq\frac {1}{2}$[/quote]
Try $ a \equal{} b \equal{} c \equal{} \frac {1}{2}.$ :wink:
But the following inequality is true already:
Let $ \{a,b,c\}\subset\left(\frac{2}{9},\plus{}\infty\right).$ Prove that:
\[ \frac {a^3}{9b \plus{} 9c \minus{} 4} \plus{} \frac {b^3}{9c \plus{} 9a \minus{} 4} \plus{} \frac {c^3}{9a \plus{} 9b \minus{} 4}\geq\frac {16}{243}\]

[quote="arqady"]But the following inequality is true already:
Let $ \{a,b,c\}\subset\left(\frac{2}{9},\plus{}\infty\right)$. Prove that: \[ \frac {a^3}{9b \plus{} 9c \minus{} 4} \plus{} \frac {b^3}{9c \plus{} 9a \minus{} 4} \plus{} \frac {c^3}{9a \plus{} 9b \minus{} 4}\geq\frac {16}{243}\][/quote]

Please send the proof.........at least some method hint. :huh:

Sunkern_sunflora, see [hide="here"] http://www.artofproblemsolving.com/Forum/viewtopic.php?t=215411 [/hide] the hint
My proof is the same.

[quote="arqady"][quote="skylover"]If $ a,b,c > \frac {2}{9}$, then $ \frac {a^3}{9b \plus{} 9c \minus{} 4} \plus{} \frac {b^3}{9c \plus{} 9a \minus{} 4} \plus{} \frac {c^3}{9a \plus{} 9b \minus{} 4}\geq\frac {1}{2}$[/quote]
Try $ a \equal{} b \equal{} c \equal{} \frac {1}{2}.$ :wink:
But the following inequality is true already:
Let $ \{a,b,c\}\subset\left(\frac {2}{9}, \plus{} \infty\right).$ Prove that:\[ \frac {a^3}{9b \plus{} 9c \minus{} 4} \plus{} \frac {b^3}{9c \plus{} 9a \minus{} 4} \plus{} \frac {c^3}{9a \plus{} 9b \minus{} 4}\geq\frac {16}{243}\][/quote]arqady, $ a\equal{}b\equal{}c\equal{}\frac{1}{3}$ is a counterexample. I think your inequality should be \[ \frac {a^3}{9b \plus{} 9c \minus{} 4} \plus{} \frac {b^3}{9c \plus{} 9a \minus{} 4} \plus{} \frac {c^3}{9a \plus{} 9b \minus{} 4}\geq\frac {1}{18}\]

Yes, you are right, minsoens. Thank you!

[quote="minsoens"]
$ \frac {a^3}{9b \plus{} 9c \minus{} 4} \plus{} \frac {b^3}{9c \plus{} 9a \minus{} 4} \plus{} \frac {c^3}{9a \plus{} 9b \minus{} 4}\geq\frac {1}{18}$[/quote]
apply LQH'inequality, we have

$ \frac {a^3}{9b \plus{} 9c \minus{} 4} \plus{} \frac {b^3}{9c \plus{} 9a \minus{} 4} \plus{} \frac {c^3}{9a \plus{} 9b \minus{} 4}\ge \frac {(a \plus{} b \plus{} c)^3}{54(a \plus{} b \plus{} c) \minus{} 36}$

we need prove that $ \frac {(a \plus{} b \plus{} c)^3}{54(a \plus{} b \plus{} c) \minus{} 36}\ge \frac {1}{18}$

setting $ a \plus{} b \plus{} c \equal{} t$

$ \Leftrightarrow \frac {t^3}{54t \minus{} 36}\ge \frac {1}{18}$

$ \Leftrightarrow t^3\ge 3t \minus{} 2$

$ \Leftrightarrow (t \minus{} 1)^2(t \plus{} 2)\ge 0$ which is true  :wink:


Since this forum has been dead for a while, I thought I'd jump start it a bit with another "I'm curious..." type of question:

What is the current state of math competitions in the Buffalo/WNY area?  Is there any sort of interscholastic math league going on?  I know there is (or used to be?) a fairly active league in the Rochester area, and that many years ago there was in Buffalo too; does it still exist?

Also: does your Buffalo/WNY school participate in MATHCOUNTS?  Does Buffalo/Erie County send a NYSML team?

On a semi-related note: is anybody reading this a student in the UB Gifted Math Program?  I am an alumnus of GMP, and I will be visiting all of the GMP classes this Monday (November 1st) to spread the word about AoPS and the USAMTS.

I live in Rochester.  Is that WNY enough?

There is a math league in Rochester that sends teams to NYSML.  We actually came in third place in Division A.

My middle school did participate in Mathcounts, but for some dumb reason they wouldn't let me on the team.

Obviously, some people in WNY, especially Rochester, are on the Upstate NY ARML team.

There's not much else.

i live in buffalo and not much is going on.  Some schools compete in NYML but its not a huge thing.  I wish schools would get math teams to compete against each other.  I remember meeting you DPatrick at UB Gifted Math a few years ago and i would like to thank you for introducing me to this awsome site.  Thanks

I remember doing that ... that was fun.  I'm glad you found our site!

I'll have to remember to try to do that again next fall, for the new GMP I and GMP II students that I didn't get to talk to in 2004.

Do you know if anybody from GMP qualified for the AIME?

i think something like 15 people qualified for the AIME.  I am not sure how many are taking it.  I am taking it though.  Wish i had gotten higher on the AMC, a 110 probably means i need to get 11 right.  Well i guess i'll have to beat the odds.

a total of 13 people from GMP 6 and GMP took it.  I think something like 9 were in GMP 5, including me.  They also just announced the other day that they will be moving discrete math to GMP 6 over a 2 year period.  Next year GMP 5 and 6 will take calc then the next year the GMP 6 students will take discrete.  Too bad they didn't do this sooner since i took calc 1-3 directly through UB.


limit
function
calculus
calculus computations

what would be the limit of squareroot of x-3 as x approaches 3?

i'm confused because the graph left of 3 is imaginary but would it still have a limit of zero since u can approach 3 from the right side.

Did you mean this:
$ \lim_{x \to3} \sqrt {x \minus{} 3}$ or $ \lim_{x \to3} \frac {1}{\sqrt {x \minus{} 3}}$ ?
We can assign the value 0 to the first one while no value that can be assigned to the second one.

Let's simplify and look at this problem which is basically the same:

$ \lim_{x \rightarrow 0} \sqrt {x}$

If you know that your approaching from the left, then you can say

$ \lim_{x \rightarrow 0^ \minus{} } \sqrt {x} \equal{} \lim_{x \rightarrow 0^ \plus{} } \sqrt { \minus{} x} \equal{} i \cdot \lim_{x \rightarrow 0^ \plus{} } \sqrt {x} \equal{} i \cdot 0 \equal{} 0$

It doesn't matter if it's complex because, after all, $ 0$ is complex as well! :)

wudn't lim of squareroot of x from the right side as x approaches 0 just be 0 and not i?

i m in pre calc nd we just started this chapter so its all a bit confusing.

I'm [i]positive[/i] your textbook did not intend to consider the square-root to be defined for negative $ x$, so the question is talking about limits from the right. And defining the square root like akech did is not good anyway.

well i mean from the right side the limit would be zero but form the left side it wouldn't exist right?

If you define square root for all $ x$ then it is again $ 0$ like I've shown you, and if you define it only for positive, then there is no "left"!

it still jus doesnt make sense! lolz i m so confused coz the function can only have a limit if u can approach it from both the right and the left side.
but for this graph u cant approach it from the left side coz its not there.
how can u turn the squareroot of x from negative side equal to squareroot of -x from positive side?

In the normal real-variables usage, we would say that $ \lim_{x\to 3}\sqrt{x\minus{}3}$ does not exist, because $ \sqrt{x\minus{}3}$ is not defined in any punctured neighborhood $ 0<|x\minus{}3|<\epsilon$.

The limit $ \lim_{x\to 3^\plus{}}\sqrt{x\minus{}3}$ does make sense, and is equal to zero. There are also other versions of the definitions, which lead to other answers.

Mhmm.... no jmerry is right the answer is 0.
First to ease your conscience about the -x and x thing.
Ok lets say you have

lim (x-> 0) of  x, You can see this is 0.
Now using his concept, you can also say
lim (x-> infinity) of 1/x, which is 0

Also logically you can see   lim (x->0) of x = lim (x-> infinity) of 1/x

Use that idea for what he said.  lim(x->1) of -x = lim(x->-1) of x = -1.
They are the same thing.

Umm, also yes you need a graph to find its continuity but you don't need to be defined at the point to be continuous.
Take lim (x-> 0) sin(x)/x.  Obvy you can't divide by 0, but it is continuous and is equal to 1. But the graph doesn't exist at 1.

This idea takes place for the question at hand. Milin said the main idea behind all this. 0 is complex.
sqrt(x) does exist to the left of 0 in complex form. sqrt(-10) = i*sqrt(10).
Also sqrt(x-3) is not differentiable (you'll learn what this means soon, also there exists a cusp i believe) at x=3, but it is continuous there. This is because the limit as x approaches 0 from the left would be equal to saying lim (x->0^-) of sqrt(x) = lim (x->0^-) of sqrt(-x) which, doesn't hold but does equal 0 since lim (x->0^+) sqrt(-x) is also 0.  Complex 0, ie. i*0 = 0.
So the graph isn't there physically but mathematically it is there and it approaches 0 in complex form. But 0=0, so it is continuous.....
hopefully this made sense. And why you can change the limits for -x to x and all that.

Cheers, Dave.

Just for some exercise

find the limits :

lim x-> pi/2 of tan(x)
lim x-> (pi/2)^+ of tan(x)
lim x-> (pi/2)^- of tan(x)

where ^+ denotes from the right.
and ^- denotes from the left.


Find with proof all triples of rationals (a,b,c) such that

$\sqrt[3]{(\sqrt[3]2-1)}=\sqrt[3]a+\sqrt[3]b+\sqrt[3]c.$

I'll start thinking about the problem soon.  Do
[code]$\sqrt[n]{x}$[/code] to obtain
$\sqrt[n]{x}$.

[quote="rofler"]Let the right hand side be X.

Cube root of a=m
Cube root of b=n

(a,b,(X-m-n)^3)

LOL trick question, do not know how to write in Latex, sry.[/quote]

Please notice that a,b,c all need to be rationals.


geometry
3D geometry
tetrahedron
rectangle
ratio
trigonometry
symmetry

[b]Problem 1:[/b]

A right triangle has sides of length 12, 16 and 20. Point $P$ is chosen in the interior of the triangle so that its distance from each leg of the triangle is $3$. A line connecting the vertex of the right triangle with point $P$ intersects the hypotenuse at point $Q$. Compute $PQ$ to the nearest hundredth.

[b]Problem 2:[/b]

A tetrahedron is formed from an equilateral triangle of side length 4 cm by folding along the line segments that connect the midpoints of the sides. Compute the length of the altitude of the tetrahedron.

[b]Problem 3:[/b]

In rectangle $ABCD$ the bisector of $\angle{A}$ meets side $BC$ at point $P$. The ratio of $AP$ to $PD$ is $4:3$. Compute the ratio of $PD$ to $PC$

*Let AP cuts CD at Q then and <DAP = <DQA= <PQC= <CPQ =45 since AP is the bisector. Clearly, P is on the extention of the side BC. Take AD=DQ=a, PQ=QC=b then AQ=a :sqrt: 2 , PQ=b :sqrt: 2 and DP=[(a+b) 2 +b 2 ]^0.5
then (a+b)^2 = 8.b^2  since AP/PD= 4/3 then PD/PC=3.

1. Let the vertices be A,B,C (<B=90). Clearly BQ is the bisector so BQ 2 =12.16 - 4800/7  so PQ=(192-4800/7)^0.5 - 3 :sqrt: 2

1)
[hide]
Yes, since lengths to sides are determined by creating right angles to them, the segments from $P$ to the legs of the triangle create a square with side length 3 with the legs.  Therefore, the line going from the vertex thru $P$ has a 45 degree angle with the leg.  We can solve for the angles of the triangle using arctan, and using the law of sines, we can solve for PQ.
We'll say $A$ is the angle inscribed between the leg measuring 12 and the hypotenuse, and $B$ is the right angle.  Thus, $A = arctan(4/3) = 53.13$ in degrees.  Drawing BQ in the triangle, we know mABQ = 45 degrees, and thus AQB is about 81.87 degrees.

$BQ = 12sin(53.13) / sin(81.87) = 9.697$
But the question asks for PQ, which is merely BQ minus the diagonal of the square.
$PQ = 9.697 - 3\sqrt{2} = 5.45$
[/hide]

3)
[hide]
Matematikator, P does not have to be on the extention of BC.  In fact, the way the question is worded, I assume P has to be on BC.  Although, I still got the same answer.  I imagine no matter where P is, the answer will be the same.

Since AP bisects <A, mPAB = 45 degrees.  Assuming AP is 4 and PD is 3, using rules of 45-45-90 triangles, AB and DC both equal $2\sqrt{2}$.
So, we just need to know PC, and we can because we know PD and DC.
Using pythag, $PC = \sqrt{9 - (2\sqrt{2})^2} = 1$

The ratio $PD / PC$ is then represented easily by $3/1 = 3$.
[/hide]

[hide="Problem 2"]Let:
$A$, $B$, and $C$ be the vertices of the base of the tetrahedron;
$D$ be the vertex at the top;
$M_A$, $M_B$, and $M_C$ be the midpoints of the sides of the base opposite $A$, $B$, and $C$, respectively;
$E$ be the centroid of the base (intersection of the medians).

From the text of the problem, we can see that the length of each edge of the tetrahedron is 2.  For the actual calculation, we will use side $AB$.  Since $mAB=2$, we know that $mAM_C=mM_CB=1$.  Using the 30-60-90 triangle $\Delta EM_CB$, we know that $mBE=\frac{1}{\sqrt3}$.  We also know that $\Delta DEB$ is a right triangle with $DB$ as a hypotenuse, so ${(mDE)^2+(mBE)^2=(mDB)^2}$.  Solving this gives $mDE=\sqrt{\frac{11}{3}}$, which is the height of the tetrahedron.[/hide]

[quote="yetti"][quote="prime_lover"][b]Problem 1:[/b]
A right triangle has sides of length 12, 16 and 20. Point $P$ is chosen in the interior of the triangle so that its distance from each leg of the triangle is $3$. A line connecting the vertex of the right triangle with point $P$ intersects the hypotenuse at point $Q$. Compute $PQ$ to the nearest hundredth.
[/quote]
Denote the triangle $\triangle ABC$, with the right angle $\measuredangle C$ and $BC = 16,\ CA = 12,\ AB = 20$. Draw 2 lines $k \parallel BC,\ l \parallel CA$ at distances from the triangle legs $BC,\ CA$ equal to 3 and intersecting at a point $P$ inside of the triangle $\triangle ABC$. Let $K \equiv k \cap AB$, $L \equiv l \cap AB$ be the intersections of these lines with the hypotenuse $AB$ and $I \equiv k \cap CA$, $J \equiv l \cap BC$ their intersections with the triangle legs $CA,\ BC$. Let the line $CP$ meet the hypotenuse $AB$ at a point $Q$. The triangles $\triangle ABC \sim \triangle LKP$ have the corresponding sides parallel and consequently, they are centrally similar. The center of similarity is the intersection of lines connecting the corresponding vertices, i.e., the point $Q \equiv CP \cap AB$. Consequently,

$\frac{QP}{QC} = \frac{LK}{AB}$

$QP = QC \cdot \frac{LK}{AB} = (QP + PC) \cdot \frac{AB - (AL + KB)}{AB} = (QP + PC) \left(1 - \frac{AL + KB}{AB}\right)$

$QP \cdot \frac{AL + KB}{AB} = PC \left(1 - \frac{AL + KB}{AB}\right) $

$QP = PC \cdot \frac{AB - (AL + KB)}{AL + KB} = PC \left(\frac{AB}{AL + KB} - 1 \right)$

The quadrilateral $CIPJ$ has 3 angles $\measuredangle C = \measuredangle I = \measuredangle J = 90^o$ right and 2 adjacent sides $PI = PJ$ equal, hence, it is a square and its diagonal is equal to $PC = PI \sqrt 2 = 3 \sqrt 2$. Let the normals $m,\ n$ to the lines $k,\ l$ erected at the points $K,\ L$ meet the triangle legs $BC,\ CA$ at the points $M,\ N$. The triangles $\triangle ALN \sim \triangle ABC \sim \triangle KBM$ are similar, because their corresponding sides are parallel. Thus

$\frac{AL}{LN} = \frac{AB}{BC},\ \ \frac{KB}{KM} = \frac{AB}{AC}$

where $LN = KM = PI = PJ = 3$. As a result,

$AL + KB = {PI} \left(\frac{AB}{BC} + \frac{AB}{AC}\right),\ \ \frac{AL + KB}{AB} = \frac{PI \cdot (AC + BC)}{AC \cdot BC}$

$QP = PC \left(\frac{AB}{AL + KB} - 1 \right) = PC \left(\frac{AC \cdot BC}{PI \cdot (AC + BC)} - 1 \right) =$

$= 3 \sqrt 2 \left( \frac{16 \cdot 12}{3 \cdot (16 + 12)} - 1 \right)= 3 \sqrt 2 \left( \frac{16}{7} - 1 \right) = \frac{27 \sqrt 2}{7} \doteq 5.455$[/quote]

Two questions yetti:

What are normals? Also, if your free, do you think you might be able to attach a picture of triangle ABC, as outlined in your solution?

Thanks!

[quote="fanaticsm"]Two questions yetti:

What are normals? Also, if your free, do you think you might be able to attach a picture of triangle ABC, as outlined in your solution?

Thanks![/quote]
Normals are lines perpendicular to a given line. You either drop a normal to a given line from a given point (not on the given line) or erect a normal to a given line at a given point (on the given line). To say drop a perpendicular or erect a perpendicular would pull my ears.

[quote="prime_lover"][b]Problem 2:[/b]
A tetrahedron is formed from an equilateral triangle of side length 4 cm by folding along the line segments that connect the midpoints of the sides. Compute the length of the altitude of the tetrahedron.
[/quote]
Let $A_1, B_1, C_1$ be the midpoints of the sides $BC, CA, AB$ of the equilateral triangle $\triangle ABC$. Since $AB = BC = CA = 4$, the triangle midlines are $B_1C_1 = C_1A_1 = A_1B_1 = 2$. In addition, the half-sides of the equilateral triangle are $BA_1 = CA_1 = CB_1 = AB_1 = AC_1 = BC_1 = 2$. As a result of folding, the vertices $A, B, C$ coincide into a point $D_1$ and the resulting tetrahedron $a = A_1B_1C_1D_1$ has edges $B_1C_1 = C_1A_1 = A_1B_1 = D_1A_1 = D_1B_1 = D_1C_1 = 2$, i.e., it is regular. Let $A_2, B_2, C_2$ be the midpoints of the equilateral triangle $\triangle A_1B_1C_1$ and $G$ be the common centroid (the intersection of medians) of the equilateral triangles $\triangle ABC, \triangle A_1B_1C_1$. Since the median $A_1A_2$ of the equilateral triangle $\triangle A_1B_1B_1$ is also its altitude, using Pythagorean theorem for the right angle triangle $\triangle A_1A_2B_1$, the length of the median $A_1A_2$ is given by

$A_1A_2 = \sqrt{A_1B_1^2 - B_1A_2^2} = \sqrt{a^2 - \left(\frac{a}{2}\right)^2} = \frac {a\sqrt 3}{2} = \sqrt 3$

Because of symmetry of the regular tetrahedron, the foot of its altitude from the vertex $D_1$ is the centroid $G$ of its base $\triangle A_1B_1C_1$. The centroid divides the median $A_1A_2$ (of any triangle) in the ratio $A_1G : GA_2 = 2 : 1$. Therefore,

$A_1G = \frac 2 3\ A_1A_2 = \frac 2 3\ a = \frac{a \sqrt 3}{3} = \frac{2 \sqrt 3}{3}$

The tetrahedron altitude $D_1G$ is perpendicular to any line in the plane of its base, in particular, to the median $A_1A_2$ passing through the centroid $G$. Using Pythagorean theorem for the right angle triangle $\triangle A_1GD_1$,

$D_1G = \sqrt{A_1D_1^2 - A_1G^2} = \sqrt{a^2 - \frac{a^2}{3}} = a \sqrt{\frac 2 3} = 2 \sqrt{\frac 2 3} = \frac{2 \sqrt 6}{3}$

Another solution: The volume $V$ of the regular tetrahedron $A_1B_1C_1D_1$ with the edge length $a = 2$ is given by Cayley-Menger determinant (3D equivalent of Heron's formula for the area of a triangle):

$\displaystyle V^2 = \frac{(-1)^4}{2^3 \cdot (3!)^2}\ \left| \begin{array}{ccccc} 0 & 1 & 1 & 1 & 1 \\ 1 & 0 & a^2 & a^2 & a^2 \\ 1 & a^2 & 0 & a^2 & a^2 \\ 1 & a^2 & a^2 & 0 & a^2 \\ 1 & a^2 & a^2 & a^2 & 0 \\ \end{array}\right| = \frac{1}{288}\ \left| \begin{array}{ccccc} 0 & 1 & 1 & 1 & 1 \\ 1 & 0 & a^2 & a^2 & a^2 \\ 0 & a^2 & -a^2 & 0 & 0 \\ 0 & a^2 & 0 & -a^2 & 0 \\ 0 & a^2 & 0 & 0 & -a^2 \\ \end{array}\right| =$

$\displaystyle  -\frac{1}{288}\ \left| \begin{array}{cccc} 1 & 1 & 1 & 1 \\ a^2 & -a^2 & 0 & 0 \\ a^2 & 0 & -a^2 & 0 \\ a^2 & 0 & 0 & -a^2 \\ \end{array}\right| = -\frac{1}{288}\ \left| \begin{array}{cccc} 1 & 1 & 1 & 1 \\ a^2 & -a^2 & 0 & 0 \\ a^2 & 0 & -a^2 & 0 \\ 2a^2 & a^2 & a^2 & 0 \\ \end{array}\right| =$

$\displaystyle = \frac{1}{288}\ \left| \begin{array}{ccc} a^2 & -a^2 & 0 \\ a^2 & 0 & -a^2 \\ 2a^2 & a^2 & a^2 \\ \end{array}\right| = \frac{1}{288}\ \left| \begin{array}{ccc} a^2 & -a^2 & 0 \\ 3a^2 & a^2 & 0 \\ 2a^2 & a^2 & a^2 \\ \end{array}\right| =$

$\displaystyle = \frac{a^2}{288}\ \left| \begin{array}{cc} a^2 & -a^2 \\ 3a^2 & a^2 \\ \end{array}\right| = \frac{4a^6}{288} = \frac{a^6}{72}$

$V = \sqrt{\frac{a^6}{72}} = \frac{a^3}{6 \sqrt 2} = \frac{a^3 \sqrt 2}{12} = \frac{2 \sqrt 2}{3}$

On the other hand, this volume is equal to

$V = \frac 1 3\ h S$

where $h = D_1G$ is the altitude from the vertex $D_1$ to the base $\triangle A_1B_1C_1$ and $S$ is the area of the base:

$S = \frac 1 2\ A_1A_2 \cdot B_1C_1 = \frac{a^2 \sqrt 3}{4} = \sqrt 3$

As a result,

$h = \frac{3V}{S} = \frac{12a^3 \sqrt 2}{12a^2 \sqrt 3} = a \sqrt{\frac 2 3} = \frac{a \sqrt 6}{3} = \frac{2 \sqrt 6}{3}$


Ok.  I have these molecules:
H/2
HF
C/2H/2
CH/2Cl/2
N/2
CH/3OH
HCO\3 ^-1
PO\3 ^-3
PO/4 ^-3

I need to find the molecular hybridization, bond types, and molecular polarity for each.  How do I do that?


AMC
USA(J)MO
USAMO
FTW
number theory
prime factorization

are we supposed to let other people know our numbers?

My number is not a prime, but is my birthday.

Sort of.

[quote="beta"][quote="randomdragoon"]not if you don't want everyone to know your score  :)

Plus I want to know who got the prime numbers...[/quote]

Darn my number is not a prime but is 1 more than a perfect square, 1 less than a multiple of 9.[/quote]

Hmm the numbers mod 9 that square to 7 are 4 and 5 mod 9. So let's list out some squares:

$4^2=16 \rightarrow 16+1=17$ prime
$5^2=25 \rightarrow 25+1=26$ composite
$13^2=169 \rightarrow 169+1=170$ composite
$14^2=196 \rightarrow 196+1=197$ prime
$22^2=484=toobig$

So beta's number is 26 or 170.

[quote="t0rajir0u"]Building on scorpius' example, though I suppose this is a bit easier.

My number is a number $r$ such that $r$ can be represented in the form $p^{q} q^{p} (p+q)$ where $p, q \in \mathbb{P}$.[/quote]

$2^3*3^2*(3+2)=360$ ? :)

[quote="paladin8"]Yeah... well my number has two of the SAME digits!!! I bet you can't guess it from that.[/quote]

same here! :)

[quote="pieterminate"][quote="scorpius119"]My student number is a prime $p$ such that $2^p-1$ is divisible by $2p+1$.[/quote]

$2*3+1|2^3-1$ ?[/quote]

Hmm $2^{11}-1 = 89(2*11+1)$.

Let's see, if my number is x, $x^3-4x^2+1$ has all the digits of x appear exactly once.

[quote="t0rajir0u"][quote="pieterminate"][quote="scorpius119"]My student number is a prime $p$ such that $2^p-1$ is divisible by $2p+1$.[/quote]

$2*3+1|2^3-1$ ?[/quote]

I have found $p = 3, 11, 23$ as possible prime solutions, but I don't seem to be able to determine generally the factors of $2^p - 1$.   :|[/quote]

83, 131, 179, 191, 239, 251, 359, 419, 431...

I think that the purpose of the number is twofold :  1) they can post your scrore "privately" online, and 2) it eliminates bias insofar as this is possible (aside from the fact that logistically, it may be easier to handle numbers than names).  Note how we are specifically instructed [i]not[/i] to write our name on our papers in the USAMO manual, and how there is a space for one's number, but not one's name, on the answer sheets (unless my memory fails me).  In light of this, should we really be posting our numbers?

I don't really think too many people actually know who I really am anyway.

And the people who do, I would tell my USAMO scores to anyway.

Oh, great.  Give other people ideas.  I knew I should have made my problem harder  :P

hmm.... my number is of the form $abc + ac$, where $a$, $b$, and $c$ are digits in base 10, and $abc$ is a perfect square. Also, my number is of the form $(p + 10)^2 - p$ where $p\in\mathbb{P}$.

[hide="Need More?"]
Okay, are you ready?
[hide="yes"]
Call my USAMO ID number $I$.

$I-1 = \displaystyle\sum_{i=1}^{m} \mathbb{P}_i$ and
$m = (pq)(\displaystyle\prod_{n=2}^\infty\frac{n^3-1}{n^3+1})$, where $p=q+4$ and $p, q$ are primes less than 10.

Hmm... I think that's right!
[/hide][/hide]

... and another one
[hide="Im ready"]
Call my ID number $I$.

$I=\mathbb{F}_q(\mathbb{F}_{q+1}+\mathbb{F}_{q-2})+\displaystyle\sum_{i=1}^p\mathbb{F}_i$, where $\mathbb{F}$ is the Fibonacci sequence, $p,q\in\mathbb{P}$, $2q<p$, and
$q=\frac{6p-1}{24-p}$.

Note that $\mathbb{P}_1=2$, $\mathbb{P}_2=3$, etc. and $\mathbb{F}_1 = 1$, $\mathbb{F}_2=1$, etc. (I just wanted to clarify this)

[/hide]
and finally,
[hide="another one"]

This is just kind of neat, not hard to calculate at all, but my ID number can be expressed as $(a+1)(a+2)^{(a+3)}+(a+4)!$ for an integer $a$.
[/hide]

I'm having too much fun with this!

[quote="eryaman"][quote="paladin8"]Yeah, well what if someone wanted to [b]sabotage[/b] another USAMO participant! They could write your number and the graders will be like OMG two papers from the same guy one gets a 7 the other gets a 0 which one is the real one...[/quote]If someone wanted to do that, he or she could just write a random number between 0 and 4 hundred whatever it is.  ;)[/quote]

How Mean!!!  :o

Er...I'm guessing the smiley means you're being sarcastic or something but the papers are faxed in. They probably keep records of which number sent the fax.

[quote="pieterminate"][quote="filletwho"]They probably know where the papers were sent in, so that shouldn't be a problem.[/quote]
how? :)[/quote]

The teachers neeed to fill out a form saying who is taking the text at their school, and send it in before the test. They also need to sign something saying the test was taken in accordance with rules, etc. I beleive the second form is sent in with the answers. Anyone who wants to mess up the procedures would have to be pretty organized, and pretty stupid, since it would be easy to figure out the mess-up, and possibly figure out who is responsible.

[quote="pieterminate"]Well, the teachers can use different fax machines...[/quote]

I don't see how that affects things; they can still trace it. Besides, we're assuming the teachers aren't in on it.

[quote="pieterminate"]not that it matters since no one is (hopefully) desperate enough to try. :)[/quote]

If you REALLY REALLY REALLY hate someone....(you find out that person's USAMO number and write their USAMO number and insult the grader or something), we can assume that no USAMO qualifier hates another USAMO qualifier THAT MUCH to do something evil like that...

But yeah unless that person is someone from your school you will get caught easily, and even if that person is from your school, hopefully your USAMO exam manager can distinguish the different handwritings.

[quote="randomdragoon"]Dear USAMO quaoifiers,

What is the prime factorization of your USAMO number?

Thanks,
[b]random[/b]dragoon.

Mine is $2^5\cdot 11$ by the way.[/quote]
Isn't 2^5 * 11 = 352 a bit too big? :huh: Maybe you are that good though.

I just know that i qualified but don't know my score. Does anyone know how to find out your score? :?

No, because there's like 400 different USAMO qualifiers.

[quote="Mathlete9366"]hmm.... my number is of the form $abc + ac$, where $a$, $b$, and $c$ are digits in base 10, and $abc$ is a perfect square. Also, my number is of the form $(p + 10)^2 - p$ where $p\in\mathbb{P}$.

[hide="Need More?"]
Okay, are you ready?
[hide="yes"]
Call my USAMO index $I$.

$I-1 = \displaystyle\sum_{i=1}^{m} \mathbb{P}_i$ and
$m = (pq)(\displaystyle\prod_{n=2}^\infty\frac{n^3-1}{n^3+1})$, where $p=q+4$ and $p, q$ are primes less than 10.

Hmm... I think that's right!
[/hide][/hide]

... and another one
[hide="Im ready"]
Call my index $I$.

$I=\mathbb{F}_q(\mathbb{F}_{q+1}+\mathbb{F}_{q-2})+\displaystyle\sum_{i=1}^p\mathbb{F}_i$, where $\mathbb{F}$ is the Fibonacci sequence, $p,q\in\mathbb{P}$, $2q<p$, and
$q=\frac{6p-1}{24-p}$.

Note that $\mathbb{P}_1=2$, $\mathbb{P}_2=3$, etc. and $\mathbb{F}_1 = 1$, $\mathbb{F}_2=1$, etc. (I just wanted to clarify this)

[/hide]
and finally,
[hide="another one"]

This is just kind of neat, not hard to calculate at all, but my index can be expressed as $(a+1)(a+2)^{(a+3)}+(a+4)!$ for an integer $a$.
[/hide]

I'm having too much fun with this![/quote]
282, ha! its better without that last part, cause there's only one value you could but in for a pretty much

go nat =D yea, i was bored and making up problems in my boredom (and I liked the way it formed a progression; and notice I had them all hidden so that you could work each one and not cheat by looking at the last problem =D)... I actually came up with the second *real* problem (the fibonacci one) by solving a diophantine/Simon's favorite factoring trick backwards...

anyway, glad somebody finally posted an answer ! =D


If $x_1,...,x_n$, are positive reals and we know that $x_1...x_n=1$
Prove that $\sum\frac{1}{x_1+1+x_1x_2}>1$ where $n>3$

See http://www.mathlinks.ro/Forum/viewtopic.php?t=5505 .

  darij