geometry
circumcircle
search
geometry unsolved

In cyclic quadrangle $ ABCD$, $ M$ and $ N$ are midpoints of $ AC$ and $ BD$ respectively.$ AC$ is bisector of $ \angle BMD$.prove that $ BD$ is the bisector of $ \angle ANC$ too.

Let $ P\equiv DA \cap CB , \ Q \equiv AB \cap DC$ and $ E \equiv AC \cap BD.$ It's known that $ E$ is the pole of $ PQ$ WRT the circumcircle $ (O)$ of $ ABCD$ $ \Longrightarrow$ $ OE \perp PQ.$ We have $ OM \perp AC$ and let $ X \equiv OM \cap PQ.$ Since $ (B,D,E,X) \equal{} \minus{} 1$  and $ CA$ bisects $ \angle BMD,$ then it follows that $ X \equiv DB \cap OM \cap PQ.$ Hence if $ Y\equiv CA \cap PQ,$ then $ E$ is the othocenter of $ \triangle OXY$ $ \Longrightarrow$ $ DB \perp OY$ and due to $ (C,A,E,Y) \equal{} \minus{} 1,$ we deduce that $ DB$ bisects $ \angle ANC.$

See here:
http://www.mathlinks.ro/viewtopic.php?search_id=2088370956&t=6557

Best regards,
sunken rock


geometry ratio circumcircle inradius rotation

[quote="xxxyyyy"]Here are $3$ geometry related questions.

Question $1$
    My pedal Bicycle has the front wheel smaller than the rear. The dimensions of the wheels; the pedal Sprocket and the other sprocket are : $24, 36, 12, 4$ ; all in inches. Arrange their linear speeds in descending order.

Question $2$
    Following is a figure of a symmetrical star (see figure 1)

    Find the angle of each of the corners $1$ to $5$ . Can you generate the general formula for the corner angles for a $n$ cornered Star?

Question $3$
    You have an equilateral triangle (see figure 2).
     a) Find the ratio of the areas of  $\triangle ABC$ & $\triangle DEF$
     b) Which is larger in area : $\triangle DEF$ or , the $EBCF$?

While providing the answer, do mention the method as well.[/quote]

[hide="2"]Inscribing the star inside a circle, there are 5 arcs in 360 degrees; each arc is 72 degrees, and each angle is $\boxed{36}$ degrees.
For an $n$ cornered star, there are $n$ arcs, so the formula is $\frac{360}{2n}=\frac{180}{n}$.
[/hide]

[hide="3a"]Call the radius of the circle $x$. For the small triangle, the circumradius is $\frac{\sqrt{3}s}{3}$ where $s$ is the triangle's side length. So $s=\sqrt{3}x$ and the area of the triangle is $\frac{3\sqrt{3}x^2}{4}$.
For the larger triangle, the inradius is A/s, where A is the area and s is the semiperimeter. So if the side length of the triangle is y, then $x=\frac{\frac{y^2\sqrt{3}}{4}}{3y}=y\sqrt{3}/12$. So then $y=4\sqrt{3}x$ and the area is $12\sqrt{3}x$. So the ratio is $\frac{12\sqrt{3}x^2}{\frac{3\sqrt{3}x^2}{4}}=\boxed{16}$? [/hide]
[hide="3b"]
If the radius of the circle is x, the area of the large triangle is $12\sqrt{3}x^2$ and the small triangle's area is $\frac{3\sqrt{3}x^2}{4}$. Subtract large from small and divide by 4: $\frac{45\sqrt{3}x^2}{16}$, ...which is larger than the triangle?[/hide]

3a doesn't look reasonable to me..

[hide="my idea"]Let the side of the big triangle be $6$.

Then the inradius of the equilateral triangle is 1/3 the altitude or $\sqrt3$.

We can see a 30-60-90 triangle in the smaller equilateral triangle, whose sides are the radius of the circle, inradius of the smaller triangle, and half the side of the smaller triangle itself (this is very difficult to describe). The length of a side of the smaller triangle is thus $3$.

Since the side of the bigger triangle is twice the side of the smaller triangle, and they are similar, their area ratio (big:small) is 4:1. [/hide]

Hm. I probably made an error somewhere...which means 3b is also incorrect. >.>

[hide="2"]The triangles have sum of interior angles =180*n. The base angles are all exterior angles. So, each corner angle
$=\frac{180n-2*360}{n}=180-\frac{720}{n}$. For n=5, the answer is 36.[/hide]

[hide="3a"]Rotate the smaller triangle so that its corners lie on the midpoints of the sides. 1/4.[/hide]

[hide="3b"]Draw AD. We know that the space surrounding DEF is 3/4 the area of the bigger triangle. AD, BE, and CF divide the that space into thirds. So EBCF is 1/4 of the area of the bigger triangle. So EBCF and DEF have the same area.[/hide]


real analysis
calculus
integration
limit
function
real analysis unsolved

Hi, can we write $ \displaystyle \lim_{r\rightarrow\infty}\int_{B(x,r)}f(x)dx\equal{}\int_{R^n}f(x)dx$ ?

If $ f \in L^1(\mathbb{R}^n)$ then write :
\[ \int_{B(0,r)}f(x) dx = \int_{\mathbb{R}^n} f(x) \chi_{B(0,r)}(x)dx \longrightarrow \int_{\mathbb{R}^n}f(x) dx
\]
by virtue of lebesgue dominated convergence theorem.

Indeed $ \chi_{B(0,r)} \longrightarrow 1$ and $ 0\leq \chi_{B(0,r) \leq 1}$ where $ \chi_B$ is the characteristic function of the set $ B$.

Anyway keyword is lebesgue theorem and characteristic function

But note that popolux invoked the condition of $ f\in L^1.$ Without either that hypothesis (so as to invoke the DCT) or the hypothesis of $ f\ge 0$ (so as the invoke the MCT), the original statement would be false.

In particular, we can have $ \lim_{r\to\infty}\int_{B(0,r)}f$ exist but have $ \int_{\mathbb{R}^n}f$ be undefined as a Lebesgue integral.


AMC USA(J)MO USAMO monovariant induction ARML

so yeah, the title pretty much says it all...i was considering putting up a poll but since only like 15 people qualified from NC and since only like 10 of the above actually use this forum, just post.

EDIT: SORRY DID NOT MEAN TO TALK ABOUT USAMO!!! WHOOPS!!!  Thanks for catching my error, Jeremy.

EDIT:plus i wanted to be the first to start the topic.

Do not talk about the test before tomorrow

I hope I can get 12-16... (can you get garbage points for more than one problem?)

The AMC forum is now unlocked, so we can now discuss it.

I have no idea how difficult it was compared to other years, but it was probably easier because I think I solved 2 of them  :trampoline:  :clap:  :D

#1:  I didn't think it was hard, but I wrote 5 different versions of my solution  :D   I used a monovariant, but I don't think it was necessary.

#2:   :) ; I had no idea how to solve it so I drew a smiley face.

#3:  I might get partial credit here.  I wrote up about induction and the base case and a neat lemma, but the computation and stuff near the end I skipped.

#4:  This was the easiest problem to solve, but harder than #1 to write up.  I'm pretty sure the answer is 8025.

#5:  You cannot factor this as far as I can tell.  Induction also did not seem to work  :(
$7^{7}+1=2^{3}\times113\times911$

#6:  I simplified what needed to be proved some with AM-GM and got that $\frac{(x+y+z)^{3}}3=R^{3}$ (I think that was it, anyway), but didn't prove it.  I also drew a nice picture.

ooh nice job bryce

i couldn't factor 102943 without a calculator though... i factored it after the test and was like WHY DOES 113+911=1024!??

very coincidential (or not?)

i solved 1, 4, maybe 1 point on 5/6 if I'm lucky

yeah - i put the sets on a graph but then proceeded to make an incredibly dumb mistake - edges on a graph can belong to more than one complete graph...woo hoo.

but yeah, i got 1,2,4 although 2 and 4 were not completely airtight...w/e.  but depending on whether or not i get a nice grader i could get a point or two on #6, i got the equality, AM-GMed it down to a sum of PQ's, and got best possible lower bounds on AP, BP, CP but couldn't get workable upper bounds on AQ,BQ,CQ so i couldn't finish...but maybe a point.

blue mop cutoff better be < 20.  i would like for it to be 19 so that i could probably make it if they like my #2...

Could one of you post the USAMO questions 2007?
EDIT:  This is Noah.

What do you mean by "not completely airtight?"  In my experience this is a good candidate for a zero  :(

I solved only two problems, and will not make MOP.  I know the scores of several other people but it's probably best not to post that kind of information without permission.  All in all, NC did not have a good showing.

what i mean is that at one or two points in my proofs (and no more) i was not happy with how i argued a certain point.

i got 1 completely, but not nicely.

actually, jeremy, you are probably right...f*. that means like a 2-3 point improvement from last year...which hurts.

I may as well post what I got...

I got #1 easily. Half a page and very clear and concise, so I can't possibly miss full-credit.

#2 is just silly. I proved that the triangles had to have different radii and could not smooth them into a case where they had the same radii. 1-2 points?

#3... yikes!

I mostly got #4, but not a 7 and possibly not even a 6.

On #5, I got the factorization for $7^{7}+1$ and wrote it nicely in terms of 7. Unfortunately, I could not make it generalize, though I discussed how I thought you could do so in my paper. 1 point? Maybe?

On #6, I incorrectly thought that a bunch of centers were colinear, thereby trivially reducing it to 2r<=R, which of course is well-known. Luckily for me, it apparently does reduce to my falsity eventually, after proving that this case is optimal, so maybe 1 point if I'm lucky?

So 10-11 plus whatever I get on #4. Which could make MOSP iff I get 6, which I probably won't.

After reading through what everyone else has said about the USAMO, I have no idea what I will get.  I still think my solutions are correct, if not very nice/concise, but they may not be what the USAMO people are looking for.  My #4 wasn't very rigorous at all...

Don't worry b-flat... you still have 4 more years  :)

When scores come out I am pretty sure that 80-95% of "solvers" of #2 and/or of #4 will wind up with about a point :/

I know of [i]several[/i] past USAMO winners (top 12) that got no more than 2 problems this year.  Also, some IMO people couldn't get #4.

If you got even one problem fully correct this year you should be in great shape for red group next year.

you and everybody else except for Reid Barton and maybe Altheman/JSteinhardt...no offense, but a lot of people who thought they had sound solutions are quickly finding that the boat is sinking. (edit: including me)

jb05, i dunno about yours - your outline gave me some idea that it is unique, but i cant really say anything about how rigorous it is.  but maybe MOP this year will be really low: it is looking that way.  :(

I think I can fairly predict a black MOP cutoff of $21 \pm 3$, a blue MOP cutoff of $16 \pm 3$, and a red MOP cutoff of $7 \pm 2$

Here are my predictions, with 85% confidence:

Red: 7-9
Blue: 15-18
Black: 24-27

argh - no blue unless somebody figures out my proof for #4.  which i think works, but maybe not.

EDIT: Zhai, nice to have you on the ARML team.  :rotfl:

EDIT: Noah, to paraphrase Dorothy.  I don't think you're in North Carolina anymore. (i.e. cough cough your location profile)

EDIT: Noah, to paraphrase Dorothy.  I don't think you're in North Carolina anymore. (i.e. cough cough your location profile)[/quote]

Once from north carolina, always from NC.  You learn that very quickly when you hang around northerners and they all start telling NC jokes

I know of several past USAMO winners (top 12) that got no more than 2 problems this year. Also, some IMO people couldn't get #4.  [/quote]

Then no need to worry, jeremy, you can still make blue.  Seriously, its possible.  good luck.

oh i'm sorry i don't hang around with northerners the dirty carpetbaggers...

argh - hope i make blue then too...as well as EVERYBODY ELSE FROM NC!

EDIT: Ignite168, is '14' your blue cutoff prediction or what?  And what's with the new avatar?

Ignite got his avatar from this link, another website :

http://www.mafiascum.net/forum/profile.php?mode=viewprofile&u=3596

dude zhai wtf that is an unfortunately high cutoff for black wtf man

but then that leads to USAMO being a popularity contest, assuming that AoPSers care so much about USAMO that they would not grade fairly (which is pretty reasonable...)

but you can't rate your own posts...

and arnav is not sure that he scored that high.

[quote="probability1.01"]dude? you probably scored at least that high? also ummm dude this is pretty awesome you get to rate posts... everyone rate this post 6/6

edit--also they should make it out of 7 so then you can be all like dude here are my usamo solutions and people can just rate them and then we can grade usamo democratically from now on[/quote]

dude that would be awesome I bet we'd be able to call the IMO team correctly then

also my poster rating seems to be 2.3 arrgh

dude that was an awesome post I rated it a 6

[quote="morpheus"]Well, since i need at least ten posts to vote, i am going to be spamming this forum for a bit.
Then, MY AVERAGE SHALL REIGN!!!!!!!!!!
do posts count if they get deleted?[/quote]

I thought you needed 25...

@PI-Dimension:  Please stop spamming and reviving old topics to point out minor details.

Thanks,
[color=red]b-flat[/color]

This topic is locked.


geometry
trigonometry
inequalities
geometry proposed

Suppose that $O$ is a point inside a convex quadrilateral $ABCD$ such that \[ OA^2 + OB^2 + OC^2 + OD^2 = 2\mathcal A[ABCD] , \] where by $\mathcal A[ABCD]$ we have denoted the area of $ABCD$. Prove that $ABCD$ is a square and $O$ is its center.

[i]Yugoslavia[/i]

From AM-GM we have that $OA^2 + OB^2 \geq 2OA \cdot OB$.
On the other hand we have that $OA \cdot OB \geq OA \cdot OB\sin \alpha = 2\left[OAB\right]$ because the $\sin$ is on $(0, 1)$.
So $OA^2 + OB^2 \geq 4\left[OAB\right]$ and $OA^2 + OB^2 + OC^2 + OD^2 \geq 2\left[ABCD\right]$.
The equality holds iff the two that we've considered hold, i.e $OA = OB$ for the first and $\alpha = 90^{\circ}$ for the second one.
Hence $O$ is the center of a square.

[quote="frenchy"]Andreas i think your solution is a little bit mistaken
because you can asume that the area of AOB is bigger then the COB or DOB but you cannot asume that the area of triangle AOB and COD are simmultainesly the bigest in the quadrilateral
P.S.Sorry if i didnt understood smt. wrong[/quote]

Well, actually these assumptions haven't been made in the solution proposed by Andreas, which is actually correct.
What he did is the following:
he used the inequality that he proved for AOB for the other 3 triangles (AOD, DOC, BOC) and then he added all 4 of them concluding that 2(OA^2 + OB^2 + OC^2 + OD^2) >= 4( [AOB] + [AOD] + [DOC] + [COB] ) = 4 [ABCD]


For which integers $ n$ are there integers $ x,y,z$ such that
\[ \frac{x}{y}+\frac{y}{z}+\frac{z}{x}=n?\]

I think this is an open problem. You should see

Two more representation problems.  Proc. Edinburgh Math. Soc. (2)  40  (1997),  no. 1, 1--17 by Bremner & Gyu.

They have solved the cases $ |n|\leq 100$. Unfortunately I can see only a review of the paper via MathSciNet. The solutions corresponds the rational points on the elliptic curve $ Y^{2}=n^{2}X^{3}+(nX+4)^{2}$. For fixed $ n$, the number of integer points of the equation equals the rank of the curve.


Find the number of units in the length of diagonal $ DA$ of the regular hexagon shown. Express your answer in simplest radical form.
[asy]draw((1,0)--(3,0)--(4,1.732)--(3,3.464)--(1,3.464)--(0,1.732)--cycle);
draw((1,0)--(1,3.464));
label("10",(3.5,2.598),NE);
label("$A$",(1,0),SW);
label("$D$",(1,3.464),NW);[/asy]

The answer is $ 10\sqrt{3}$.

However, I can't explain it because I don't really understand Asymptote...

Draw the altitude from the point between $ D$ and $ A$ to $ \overline{DA}$.  Then this altitude measures $ 5$ by $ 30\minus{}60\minus{}90$ triangle ratios, so $ DA\equal{}2(5\sqrt{3})\equal{}\boxed{10\sqrt{3}}$.


I wish to create multiple numbered lists; but not have the counter continue numbering. e.g.
some text
1
2.
3.
some more text
1
2
3
4

is there something like the
\setcounter{equation}{0}
command that I can use?

I'm not sure I understand you, as the numbering from start as you wanted is what happens by default.
[code]\begin{enumerate}
\item line 1
\item line 2
\end{enumerate}
Text goes here
\begin{enumerate}
\item line 3
\item line 4
\end{enumerate}[/code] will give
1. line 1
2. line 2
Text goes here
1. line 3
2. line 4
If you do want to change the numbering, the counter you are after is enumi so
[code]\begin{enumerate}
\item line 1
\item line 2
\end{enumerate}
Text goes here
\begin{enumerate}
\setcounter{enumi}{12}
\item line 3
\item line 4
\end{enumerate}[/code]will give
1. line 1
2. line 2
Text goes here
13. line 3
14. line 4

sorry about that... I realized that it already did what I wanted it to do, immediately after i posted


calculus
integration
inequalities
algebra proposed
algebra

Find all integral solutions of

$ 1\plus{}x\plus{}x^2\plus{}x^3\plus{}x^4\equal{}y^4$

$ 1 \plus{} x \plus{} x^{2} \plus{} x^{3} \plus{} x^{4} \equal{} y^{4} \Leftrightarrow 1 \plus{} x \plus{} x^{2} \plus{} x^{3} \equal{} y^{4} \minus{} x^{4} \equal{} (y^{2} \minus{} x^{2})(y^{2} \plus{} x^{2}) \equal{} (y \minus{} x)(y \plus{} x)(y^{2} \plus{} x^{2})$.
By the other hand, $ 1 \plus{} x \plus{} x^{2} \plus{} x^{3} \equal{} (1 \plus{} x)(1 \plus{} x^{2})$ so we get the equality
$ (y \minus{} x)(y \plus{} x)(y^{2} \plus{} x^{2}) \equal{} (1 \plus{} x)(1 \plus{} x^{2})$. Suppose now that $ (x,y)$ is an integer solution to this equation
and $ (y^{2} \plus{} x^{2},1 \plus{} x) \equal{} 1$, then $ 1 \plus{} x^{2} \equiv 0$ $ ($ $ mod$ $ y^{2} \plus{} x^{2})$ and it implies $ y \equal{} \pm 1$. So
$ x(1 \plus{} x)(1 \plus{} x^{2}) \equal{} 0$ so we get the pairs $ (0,\pm 1),( \minus{} 1,\pm 1)$.
But I don't know what to do when $ (y^{2} \plus{} x^{2},1 \plus{} x) \equal{} d > 1$.
Maybe it may help to consider this factorisation $ x(1 \plus{} x)(1 \plus{} x^{2}) \equal{} (y \minus{} 1)(y \plus{} 1)(y^{2} \plus{} 1)$.
wich gives us trivially the pairs $ (0,\pm 1),( \minus{} 1,\pm 1)$.

What do you mean by d > 1 and also what do you mean by (y^2+x^2,1+x)=1?

$ (a,b)$ denotes the greatest common divisor of $ a$ and $ b$.

Some of these dastardly Diophantine equations are better solved by algebraic, rather than number theoretical, methods! It is classical to try and surround one of the powers by consecutive similar powers.

For $ x\equal{}0$ we have the solution $ y\equal{}\pm 1$.
For $ x>0$ we see that $ (4y)^4 \equal{} 256(x^4 \plus{} x^3 \plus{} x^2 \plus{} x \plus{} 1)$ lays strictly between $ (4x\plus{}1)^4$ and $ (4x\plus{}2)^4$. The first inequality is equivalent to $ 160x^2 \plus{} 240x \plus{} 255 > 0$, true, while the second is equivalent to $ 16x^3 \plus{} 8x^2 \minus{} 8x \minus{} 15 \equal{} x^3 \plus{} 15(x^3\minus{}1) \plus{}8x(x\minus{}1) > 0$, also true. Therefore there are no solutions.
For $ x<0$ denote $ z\equal{}\minus{}x$; we see that $ (4y)^4 \equal{} 256(z^4 \minus{} z^3 \plus{} z^2 \minus{} z \plus{} 1)$ lays strictly between $ (4z\minus{}1)^4$ and $ (4z\plus{}1)^4$. The first inequality is equivalent to $ 160z^2 \minus{} 240z \plus{} 255 > 0$, true, while the second is equivalent to $ 512z^3 \minus{} 160z^2 \plus{} 272z \minus{} 255 > 0$, also true. Therefore there is only left $ (4z)^4$, when the equation for equality is $ z^3\minus{}z^2\plus{}z\minus{}1 \equal{} (z\minus{}1)(z^2\plus{}1) \equal{} 0$, with only solution $ z\equal{}1$, i.e. $ x\equal{}\minus{}1$, and $ y\equal{}\pm 1$.


If the length of each edge in a cube is increased by the same percent and the result was a cube that had 25% more volume than it originally had, by what percentage were the edges increased?


[quote="sunchips"]cubic formula :lol:[/quote]

Unless of course cubic formulas are not allowed :wink:

I think rewriting in following way may help :roll:
\[(x+4)^{3}-(167\cdot12+1)x^{3}=0\]


inequalities
geometry
three variable inequality
geometric inequality

I think we should transform the inequality into as follow  :maybe:
Let $ a,b,c\ge 0$ such that: $ ab\plus{}bc\plus{}ca\plus{}abc\equal{}4$. Prove that:
\[ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\ge a\plus{}b\plus{}c\]
or

\[ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\le a\plus{}b\plus{}c\]

I have ever tried, but I think the first is correct  :) Using Mixing Variable, maybe we can solve your one ... I will try!

My proof:
\[ f(a,b,c) \equal{} \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \minus{} a \minus{} b \minus{} c
\]
We will prove that:
\[ f(a,b,c)\ge f\left(a,\frac {2}{a \plus{} 1},\frac {2}{a \plus{} 1}\right)
\]

\[ \iff \frac {[4 \minus{} (a \plus{} 1)(b \plus{} c)][bc \minus{} a \minus{} 1]}{bc(a \plus{} 1)}\ge 0
\]
\[ \iff (b\minus{}c)^2\left(\frac{1}{bc(b\plus{}c)}\minus{}\frac{1}{4\plus{}b\plus{}c}\right)\ge 0\]
Note that:
\[ a \equal{} \frac {4 \minus{} bc}{bc \plus{} b \plus{} c}
\]
Remain, we proved:
\[ f\left(a,\frac {2}{a \plus{} 1},\frac {2}{a \plus{} 1}\right) \equal{} \frac {(a \minus{} 1)^2}{a(a \plus{} 1)}\ge 0
\]

Hope it is true
$ a \equal{} \frac {2x}{y \plus{} z},b \equal{} \frac {2y}{z \plus{} x},c \equal{} \frac {2z}{x \plus{} y}$.

[quote="arqady"]Let $ x,$ $ y$ and $ z$ are positive numbers such that $ x \plus{} y \plus{} z \equal{} \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z}.$ Prove that $ xyz \plus{} yz \plus{} zx \plus{} xy\geq4.$[/quote]
$ (x \plus{} y \plus{} z)(yz \plus{} zx \plus{} xy)\geq 9xyz\Longrightarrow$

$ \sqrt {\frac {xyz(x \plus{} y \plus{} z)}{yz \plus{} zx \plus{} xy}} \plus{} x \plus{} y \plus{} z \minus{} \frac {4xyz(x \plus{} y \plus{} z)^2}{(yz \plus{} zx \plus{} xy)^2}$

$ \geq\frac {3xyz}{yz \plus{} zx \plus{} xy} \plus{} x \plus{} y \plus{} z \minus{} \frac {4xyz(x \plus{} y \plus{} z)^2}{(yz \plus{} zx \plus{} xy)^2}$

$ \equal{} \frac {x^3(y \minus{} z)^2 \plus{} y^3(z \minus{} x)^2 \plus{} z^3(x \minus{} y)^2}{(yz \plus{} zx \plus{} xy)^2}\geq 0\Longrightarrow$

$ 1 \plus{} \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z}\geq \frac {4}{xyz}\Longleftrightarrow xyz \plus{} yz \plus{} zx \plus{} xy\geq4$.

Ji Chen's solution is very ugly  :D Because the inequality is very very simple  :D  We have:
\[{ \frac{4x}{2(y+z}}\le \frac{x}{2y}+ \frac{x}{2z}\]
Using above result and adding up them we have the inequality. Only one line  :D

[quote="yutian_yzh"]If $ x,y,z$ are positive numbers, then

$ \frac {2x}{y + z} + \frac {2y}{z + x} + \frac {2z}{x + y}\leq\frac {y + z}{2x} + \frac {z + x}{2y} + \frac {x + y}{2z}$.[/quote]
Substitute $ x = s - a,y = s - b,z = s - c$, we sharpen the above inequality as follows :

$ \sum\frac {2x}{y + z} = \sum{\frac {h_a}{r_a}\leq\sum{\frac {w_a}{r_a}\leq\sum{\frac {m_a}{r_a}\leq\sum{\frac {r_a}{h_a} = \sum{\frac {y + z}{2x}}}}}}$,

where $ s$ is the semi-perimeter of a triangle, $ a,b,c$ are the sides, $ r_a,r_b,r_c$ are the radius of excircles,  $ h_a,h_b,h_c$ are the altitudes,
$ w_a,w_b,w_c$ are the angle-bisectors and $ m_a,m_b,m_c$ are the medians.

Note $ \Delta$ is the atea of the triangle, then

$ m_a = \frac {\sqrt {2b^2 + 2c^2 - a^2}}{2} = \frac {\sqrt {4x(x + y + z) + (y - z)^2}}{2}$,

$ r_a = \frac {\Delta}{s - a} = \sqrt {\frac {yz(x + y + z)}{x}}$,

$ h_a = \frac {2\Delta}{a} = \frac {2\sqrt {xyz(x + y + z)}}{y + z}$,

the last geometric inequality is equivalent to the following algebraic inequality

$ \sum\sqrt {\frac {x\left[4x(x + y + z) + (y - z)^2\right]}{yz(x + y + z)}}\leq \sum{\frac {y + z}{x}}$.

Using Cauchy inequality,

$ \sum\sqrt {\frac {x\left[4x(x + y + z) + (y - z)^2\right]}{yz(x + y + z)}}\leq\sqrt {\sum\frac {2x + y + z}{yz}}\sqrt {\sum\frac {x\left[4x(x + y + z) + (y - z)^2\right]}{(x + y + z)(2x + y + z)}}$,

we have to prove

$ \sum\frac {2x + y + z}{yz}\sum\frac {x\left[4x(x + y + z) + (y - z)^2\right]}{(x + y + z)(2x + y + z)}\leq\left(\sum\frac {y + z}{x}\right)^2$.

[hide="Proof."]$ \left(\sum\frac {y + z}{x}\right)^2 - \sum\frac {2x + y + z}{yz}\sum\frac {x\left[4x(x + y + z) + (y - z)^2\right]}{(x + y + z)(2x + y + z)}$

$ \equiv\frac {F(x,y,z)}{x^2y^2z^2(x + y + z)(2x + y + z)(2y + z + x)(2z + x + y)}$.

By Muirhead's theorm,

$ F(x,y,z) = 2\sum{x^8\left(y^2 + z^2\right)} - 4\sum{x^8yz}$

$ + 13\sum{x^7\left(y^3 + z^3\right)} - 13\sum{x^7yz(y + z)}$

$ + 34\sum{x^6\left(y^4 + z^4\right)} - 10\sum{x^6yz\left(y^2 + z^2\right)} - 32\sum{x^6y^2z^2}$

$ + 46\sum{y^5z^5} + 3\sum{x^5yz\left(y^3 + z^3\right)} - 21\sum{x^5y^2z^2(y + z)}$

$ - 24\sum{x^2y^4z^4} - 2\sum{x^4y^3z^3}$

$ = 2\sum{x^8\left(y^2 + z^2\right)} - 4\sum{x^8yz}$

$ + 13\sum{x^7\left(y^3 + z^3\right)} - 13\sum{x^7yz(y + z)}$

$ + 26\sum{x^6\left(y^4 + z^4\right)} - 10\sum{x^6yz\left(y^2 + z^2\right)} - 32\sum{x^6y^2z^2}$

$ + 42\sum{y^5z^5} - 21\sum{x^5y^2z^2(y + z)}$

$ + 8\sum{x^6\left(y^4 + z^4\right)} + 4\sum{y^5z^5} + 3\sum{x^5yz\left(y^3 + z^3\right)}$

$ - 24\sum{x^2y^4z^4} - 2\sum{x^4y^3z^3}\geq 0$.[/hide]

[quote="zaizai-hoang"]arquady, can you post your hard solution?  :roll:[/quote]
After beautiful Ji Chen's proof, my proof seems very ugly:
Let $ a + b + c = 3u,$ $ ab + ac + bc = 3v^2$ and $ abc = w^3.$ Then
$ a + b + c = \frac {1}{a} + \frac {1}{b} + \frac {1}{c}\Leftrightarrow uw^3 = v^2$ and
${ abc + ab + ac + bc\geq4\Leftrightarrow w^3 + 3v^2\geq4\Leftrightarrow w^3 + 3v^2\cdot\sqrt {\frac {uw^3}{v^2}}\geq4(\sqrt\frac {uw^3}{v^2}})^3\Leftrightarrow$
$ \Leftrightarrow4u\sqrt {u}w^3 - v^3\sqrt {w^3} - 3v^4\sqrt {u}\leq0\Leftrightarrow\sqrt {w^3}\leq\frac {v^3 + \sqrt {v^6 + 48v^4u^2}}{8u\sqrt {u}}.$
$ (a - b)^2(a - c)^2(b - c)^2\geq0\Leftrightarrow w^6 + 2(2u^3 - 3uv^2)w^3 - 3u^2v^4 + 4v^6\leq0.$
It gives: $ w^3\leq3uv^2 - 2u^3 + 2\sqrt {(u^2 - v^2)^3}.$
Thus, it remains to prove that $ \sqrt {3uv^2 - 2u^3 + 2\sqrt {(u^2 - v^2)^3}}\leq\frac {v^3 + \sqrt {v^6 + 48v^4u^2}}{8u\sqrt {u}}.$
Let $ u^2 = tv^2.$ Then $ t\geq1$ and $ \sqrt {3uv^2 - 2u^3 + 2\sqrt {(u^2 - v^2)^3}}\leq\frac {v^3 + \sqrt {v^6 + 48v^4u^2}}{8u\sqrt {u}}\Leftrightarrow$
$ \Leftrightarrow\sqrt {192t^2 - 128t^3 + 128\sqrt {(t - 1)^3t^3}}\leq1 + \sqrt {1 + 48t}\Leftrightarrow$
$ \Leftrightarrow8t^3 - 12t^2 + 3t + 1 + \frac {6(t - 1)}{7 + \sqrt {1 + 48t}}\geq8\sqrt {(t - 1)^3t^3}\Leftrightarrow$
$ \Leftrightarrow8t^2 - 4t - 1 + \frac {6}{7 + \sqrt {1 + 48t}}\geq8\sqrt {(t - 1)t^3}\Leftrightarrow$
$ \Leftrightarrow1 + 8t + \frac {12(8t^2 - 4t - 1)}{7 + \sqrt {1 + 48t}} + \frac {36}{(7 + \sqrt {1 + 48t})^2}\geq0,$ which true for $ t\geq1.$
The original inequality is equivalent to the following my inequality:
Let $ \{\alpha,\beta,\gamma\}\subset( - \frac {\pi}{2},\frac {\pi}{2})$ such that $ \tan\alpha + \tan\beta + \tan\gamma = 0.$ Prove that:
\[ \frac {1}{3 - \tan{\frac {\alpha}{2}}} + \frac {1}{3 - \tan{\frac {\beta}{2}}} + \frac {1}{3 - \tan{\frac {\gamma}{2}}}\geq1
\]

it is hard to think out as your solution   :wink: Anyway, we have a lot of ways to solving above inequality. In this topic, we have four solution...
Using contraction method as my proof (using mixing varialbe method) and using substitution like yutian_yzh's proof. Remain proofs are not easy to think out  :wink: Have any solution for it?

[quote="tardi"][quote="arqady"]
The original inequality is equivalent to the following my inequality:
Let $ \{\alpha,\beta,\gamma\}\subset\left( \minus{} \frac {\pi}{2},\frac {\pi}{2}\right)$ such that $ \tan\alpha \plus{} \tan\beta \plus{} \tan\gamma \equal{} 0.$ Prove that:
\[ \frac {1}{3 \minus{} \tan{\frac {\alpha}{2}}} \plus{} \frac {1}{3 \minus{} \tan{\frac {\beta}{2}}} \plus{} \frac {1}{3 \minus{} \tan{\frac {\gamma}{2}}}\geq1
\]
[/quote]

Dear arquady, can you post your solution?[/quote]
Let $ a\equal{}\frac{1\plus{}\tan{\frac{\alpha}{2}}}{1\minus{}\tan{\frac{\alpha}{2}}},$ $ b\equal{}\frac{1\plus{}\tan{\frac{\beta}{2}}}{1\minus{}\tan{\frac{\beta}{2}}}$ and  $ c\equal{}\frac{1\plus{}\tan{\frac{\gamma}{2}}}{1\minus{}\tan{\frac{\gamma}{2}}}.$
Then $ a>0,$ $ b>0,$ $ c>0,$  $ \tan\alpha \plus{} \tan\beta \plus{} \tan\gamma \equal{} 0\Leftrightarrow a\plus{}b\plus{}c\equal{}\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}$ and
$ \frac {1}{3 \minus{} \tan{\frac {\alpha}{2}}} \plus{} \frac {1}{3 \minus{} \tan{\frac {\beta}{2}}} \plus{} \frac {1}{3 \minus{} \tan{\frac {\gamma}{2}}}\geq1\Leftrightarrow abc\plus{}ab\plus{}ac\plus{}bc\geq4.$ :wink:

基本是齐次化手段的炸弹不等式

[quote=arqady]Let $ a,$ $ b$ and $ c$ are positive numbers such that $ a\plus{}b\plus{}c\equal{}\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}.$ Prove that
\[ abc\plus{}ab\plus{}ac\plus{}bc\geq4\][/quote]

Very beautiful !

Just Rolle .[quote=Nguyenngoctu]Dear, leonardg Why $p \le {\left( {s - t} \right)^2}\left( {s + 2t} \right)$?[/quote]


Start with circle C1 with radius 6. Circles C2 and C3 with equal radius are inscribed in C1, each tangent to each other and C1. C4 and C5 are inscribed on opposite sides of C2 and C3, both internally tangent to all three previous circles. Finally there is circle C6, inscribed in the region between C2, C3 and C4. Find the radius of C6.

[quote="Hexaditidom"]C4 and C5 are inscribed on opposite sides of C2 and C3, both internally tangent to all three previous circles.[/quote]

You mean they're internally tangent to $C_1$ but externally tangent to $C_2$ and $C_3$ like a four leaf clover kind of thing? Or do you mean something else?

Yes, it's internally tangent to C1 and externally tangent to C2 and C3. There are two such regions in the figure.
Anyone want to try it?

[hide]
Let $O_k$ denote the center of circle $C_k$ and $r_k$ denote the length of the radius of $C_k$.
Clearly, $r_2=r_3=3$.
Then we want to find $r_4=r_5$.
Consider the radius of $C_1$ passing through $O_4$.
Now connect $O_2$ and $O_4$.
Consider the right triangle $O_2O_1O_4$.

$O_2O_1 = 3$ because it is a radius of $C_2$.
$O_1O_4 = r_1-r_4 = 6-r_4$.
$O_4O_2 = r_2+r_4=3+r_4$.

So we have $3^2+(6-r_4)^2=(3+r_4)^2 \Rightarrow r_4=2$.

Similarly, construct the right triangle $O_2O_1O_6$.

$O_2O_1 = 3$ again.
$O_1O_6 = r_1-2r_4-r_6 = 2-r_6$.
$O_6O_2 = r_2+r_6 = 3+r_6$.

So we have $3^2+(2-r_6)^2=(3+r_6)^2 \Rightarrow r_6=\frac{2}{5}$.[/hide]


geometry
geometric transformation
reflection
3D geometry
pyramid
algorithm

They say that we can live for 5000 years with having the same youthul look:  http://www.popsci.com/popsci/medicine/article/0,20967,929447,00.html

My question to you is, would you? Can you imagine living 5 millenniums. It will be a psychological challenge. It could be a good thing, like having the opportunity of having 500 different careers or learning everything about math  :)  ! Or it can be a bad thing, over population, suicidal death rate will go up, global corruption?...  :(

I like to hear your opinion... I personally like to live for 1000 years ;)  but what about you?

Considering the current human aging rate, 80 years are enough for me.
Plus, there might be heaven waiting for me after death.. hmm.

It's hard to live 5000 years. It would be boring.

But 500 years would be nice.  :D

80 years for me.  I'm not sure why, but living for hundreds of years doesn't appeal to me.
Perhaps its the thought of everything growing older but me.
Would [i]everybody[/i] live this long, or just us?

Hello,

[quote="sky9073"]Considering the current human aging rate, 80 years are enough for me.[/quote]

Well Lady, I would like to know if you'll think the same when you'll be 79... but, given that you are a woman and that I think I am older than you (just a little), acconding to the stadistics, chances are that I'll be dead by then. Acconding to most of the stadistics I have read since I was a little kid, women use to live longer and healthier than man, at least in average  :D

[quote="Derek"]It's hard to live 5000 years. It would be boring. [/quote]

I don't know, I am just 24, and I still think that there are that many things that I would like to do that all my life ten times wouldn't be time enough to do it. I don't think that to live forever would really be wise, eventually you will get tired to live, but eitherway, I think I would like to see 5 millenia pass by and enjoy them as much as I could.

Of course, it would be nice if you could live that without aging after certain age, I don't know, may be stay with your body around 20 to 30 as long as you live after that. There is a proverb in my country that, translated as well as I can it says "If the young man know... or the old mand could..." and that is write, life is sometimes not fair, because we finally learn something when it is not useful anymore, probably the 70 years old guy around the corner has the experience (and the money) and the decire to live very nice adventures, but unfortunatelly, because of the state of his body, he has to settle down for something less demanding, enjoy live anyway, but not with that intencity.

Aging is something that everyone suffer, and I believe that no matter how well is your physical condition, when you are 70 you are not able to do the same thing that 40 or 50 years before that.

Best,

[quote="shobber"]I choosed 500 years...
I think that when i am 500 years old, i will have done most of the things that i want to do, and also by that time, i should have been bored with life......[/quote]After your answer, I presume the first digit in your age is still 1, so wait until it turns to 2. Then you will quickly change your opinion :)

How long I live isn't as important as how I'm living. I'd rather have a happy life doing stuff until 70 than stay in a hospital bed from 60-90 and live until 90.

Yup.  :D  Surprisingly, when I was as old as my lucky number, it wasn't a really good year for me at all.  :)

Well to be honest, I'd rather have another 20-30 years for [img]http://artofproblemsolving.com/Forum/images/avatars/45961890041e4a6671adbd.jpg[/img]

I was just thinking about this topic.  Most young people think, "Oh, I want to live a lot longer!  I don't ever want to die."  But if you think about it, would it really be fun to live until you're 500 or so?  By age 200 for sure, you would have a ton of medical problems and I doubt you would have much fun living (you'd probably rather die).

Now, this is all assuming the only thing changing between right now and the future is that you can live longer.  However, medicine will improve dramatically by the time it's possible to live 500 years, so maybe you [i]could[/i] live a fun life until you die?

Bah... 100 year is good enough. I don't want to be some old idiot. I want to live young and actually be able to do stuff - I don't want to sit around for a few millenia - I think though, that the world would be reduced to dust by then, so it shouldn't matter.  :?

It seems like the last two posters didn't read the first message,

The idea is to live long, but healthy and with a young apparience! That is almost paradise here on Earth.

Best,

[quote="Valentin Vornicu"]You also have the digit 1 as your first digit in your age, right? :D[/quote]

I guess you have it as second digit. So let us wait for three month.  ;)

[quote="aznness"]Think about it..

Even with today's technology you'd be a lump of fat in a wheelchair by 150. I'd rather die at 80 and be happy with life. I don't have to see my great-great-great-great-great-great-great-great-great-great-great-great-grandchildren( if i could) :P[/quote]

aznness,

You didn't read the full thread, did you? The idea is basically, what if you can live healthy, looking and feeling like in the 20s or 30s as long as you want!

Best,

Wouldn't you think how long you would like to live would depend on the conditions under which you were living.  I know I wouldn't like to live 500 years as a comatose vegetable.  ;)

Do we get to live for 5000 years no matter what? I mean, are we invincible until we turn 5000, then we instantly die? That might be kinda cool, but to be honest, I'd rather be in heaven those 4920 years than stuck here . . .

Invincible is quite different from immortal.  I feel that we could suffer from some sort of extremely painful disease for several thousand years but simply not die if we were immortal over that time frame.  If this were the case, I don't think I would want to live that long.

I would not want to live long, even if I could. I just feel that sometimes going through life in a stagnant phase is just dumb, and (gasp) I want to die (in the future)

i chose 100 cuz i feel you would get bored after so long. and besides, think of how bad the ozone layer will be in 5000 years? the environment would be a complete disaster with a lot less land (global warming) and a bunch more people (hmm....how does that happen?).  besides either robots or aliens are going to take over in the next 1500 years, so who wants to be one of the last humans? i don't want to be around when that goes down.

not too long... maybe 40 years, 45 at most  :?
i think i'll be bored by then already (partly bored already now  ;) )

I chose for infinity. Hey guys, if life gets boring, you can always stem out right? :?

And as long as it doesn't, just keep living I'd say :D

[quote="kentliu"]if living long means living youthfully.. then its not a bad deal.. but i think there should be a law tat says if u wanna live long, u gotta get a vasectomy.. lol[/quote]

You know, I agree. There are too many people in the world already.  :D . You know how many sons and dauthers you can have in 500 years of health and youth? :?  :huh: . That is why in Highlander they couldn't have descendence.  :D .

[quote="kentliu"]if living long means living youthfully.. then its not a bad deal.. but i think there should be a law tat says if u wanna live long, u gotta get a vasectomy.. lol[/quote]

Now who wants to live forever? lol...

I see no absolute reason in life.  Once your dead... thats it... Its like playing an RPG, completing it to only an infintesimal amount and then having the game stolen from you...

It would be nice to be like the energizer bunny so that you could complete more that an infintesimal, to keep going on [size=92]and on[/size] [size=84]and on[/size] [size=75]and on[/size] [size=67]and on[/size] [size=59]and on[/size] [size=50]and on[/size] [size=42]and on[/size] [size=34]and on[/size] [size=25]and on[/size] [size=17]and on[/size] [size=9]and on...[/size]

I would want to live for 5000 years or more.  I would spend the the first 1000 years strictly learning new stuff here on earth.  Imagine being asked how many languages you speak, and replying "All Of Them."

As for overpopulation, since our life spans would be longer prolonged space travel and space colonization would be viable options.  No matter how old you get there is always something new to do or learn (if you are willing).  So many people stop enjoying life in the short life spans that we have now.

To create a new term, it would be great to say " I remember being one of the first space explorers to use the UNIVERSAL FOLD (a folding of time and space as a means of travel). There is so much adventure waiting, and I want a FRONT ROW SEAT!

No, I wouldn't want to live to be 5,000, probably not even 80. Life just gets to be tiresome, and I would just want a break. I don't get this whole heaven thing: why would you want to continue living after you're dead? Don't you just want to get some rest?


Notice that $ (9)(9)\equal{}81$ and $ 9\plus{}9\equal{}18$. Also notice that $ 3\plus{}24\equal{}27$ and $ (3)(24)\equal{}72$. Find another pair of integers such that their sum and product are equal to two digit numbers with their digits reversed.

Hi Elefont.

[hide=" Click to reveal hidden solution"]

$ 2\plus{}47\equal{}49$ and $ 2\times 47\equal{} 94$

[/hide]

[quote="Elefont"]How did you get that answer? Can somebody explain why my solution doesn't work.

$ xy \equal{} 10M \plus{} N, x \plus{} y \equal{} 10N \plus{} M$ for integers $ M$ and $ N$ $ 1 \minus{} 9$.
$ xy \plus{} x \plus{} y \equal{} 11(M \plus{} N)$.
$ (x \plus{} 1)(y \plus{} 1) \equal{} 11(M \plus{} N) \plus{} 1$
$ xy < 100, x \plus{} y < 100, so (x \plus{} 1)(y \plus{} 1) < 201$
$ (x \plus{} 1)(y \plus{} 1) \equal{} {23, 34, 45 ,56 ,67 ,78, 89, 100, 111, 122, 133, 144, 155, 166, 177, 188, 199}$

This is clearly wrong, though, because $ (x \plus{} 1)(y \plus{} 1) \equal{} 34$ so $ x \equal{} 1, y \equal{} 16,$ however $ xy \equal{} 16, x \plus{} y \equal{} 17$. Why don't all integer solutions for $ x$ and $ y$ in the above equation work?[/quote]

Because even though $ xy\equal{}10M\plus{}N$ and $ x\plus{}y\equal{}10N\plus{}M$ implies $ xy\plus{}x\plus{}y\equal{}11(M\plus{}N)$, it doesn't work the other way around (the converse is not necessarily true).

Ok thank you. so then how would you solve this problem?

Hi Elefont.

I just tried numbers that looked like they might work.

Alternatively, you could test the seventeen possibilities you have for $ (x\plus{}1)(y\plus{}1)$, some of which you can ignore immediately. It only takes a few minutes to see that the only solutions are given by $ 100$ and $ 144$.

$ 100 \equal{} (1)(100)$ has factors $ (1)(100), (2)(50), (4, 25), (5, 20), (10, 10)$ which give $ (1, 49), (3, 24), (4, 19) (9,9)$ as possibilities for $ (x, y)$. Of these, only $ (3, 24)$ and $ (9, 9)$ work.

Similarly the only other solution comes from $ 144 \equal{} (3, 48)$, giving $ (x, y) \equal{} (2, 47)$.

I hope this is helpful as I can't see any other way of doing this at the moment.


There are $ n$ houses numbered consecutively on a street from $ 1$ to $ n$. There is a house in between, numbered $ x$, where the sum of the house numbers on the left of the house is equal to the sum of the house numbers on the right. If $ 50<n<500$, find $ n$ and $ x$.

[hide="Answer"]We have that $ (x \minus{} 1)^2 \plus{} (x \minus{} 1) \equal{} n^2 \plus{} n \minus{} (x^2 \plus{} x) \iff 2x^2 \equal{} n^2 \plus{} n \iff \frac {n(n \plus{} 1)}{2} \equal{} x^2$.

Suppose $ n$ is odd. Then, since $ n$ and $ n \plus{} 1$ are relatively prime, $ n$ must be a square. This means $ n \in \{9^2 \ldots 21^2 \}$, a mere $ 7$ cases, none of which work.

Suppose $ n$ is even. Then $ n \plus{} 1$ and $ \frac {n}{2}$ are both squares. Thus, $ \frac {n}{2} \in \{ 5^2, \ldots 15^2 \}$ and $ n \plus{} 1 \in \{8^2 \ldots 22^2 \}$. We find that $ n \equal{} 288$ works, since $ \frac {n}{2} \equal{} 144 \equal{} 12^2$ and $ n \plus{} 1 \equal{} 289 \equal{} 17^2$.

Then, we see that $ x \equal{} 12 \cdot 17 \equal{} 204$, and $ (n, x) \equal{} (288,204)$.
[/hide]


geometry
parallelogram
geometry unsolved

problem : there is a point P inside a  parallelogram such that angle ABP = angle ADP. Show that angle PAB = angle PCB

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

I'm korean. so , that sentence is all right??...... Please ~

korean version>> 평행사변형 ABCD의 내부에 점 P가 있고, 각 ABP=각ADP이다. 각PAB=각PCB임을 증명하여라.

혹시.. 한국인 계시면, 저 제일 위에 문장을 제대로된 영문장으로 바꿔주세요....

Please, Solve the problem

Let $ Q$ be a point such that $ ABPQ$ is parallelogram. Since $ \angle AQP\equal{}\angle PBA\equal{}\angle PDA$ we obtain that $ APDQ$ is cyclic hence $ \angle PAB\equal{}\angle APQ\equal{}\angle ADQ\equal{}\angle PCB$ qed


Let a is a real number a>1
Find max, min of
$ \sqrt {a \plus{} \sin x}  \plus{} \sqrt {a \plus{} \cos x}$

I think that the maximum must occur when $ x\equal{}\frac{\pi}{4}$ and is $ 2\sqrt{a\plus{} \frac{\sqrt{2}}{2}}$.

Can you prove more detail ? please :lol:


iet x,y be positive real numbers satisfying the inequality
$x^{3}+y^{3}\leq x-y$
prove that
$x^{2}+y^{2}\leq 1$

You actually "foiled" wrong at the end of your solution, but I think that your idea still works.  This inequality doesn't seem very sharp at all.


dude I will be pissed if OSU becomes number one just because florida lost yesterday and we lost last week

it will be all like dude quit hogging the number one spots dude

[quote="junggi"][quote="now a ranger"]dude I bet OSU will lose to the number 2 team in the BCS championship  :roll:

And that OSU will lose in the NCAA tournament.[/quote]
DUDE NO WAY
OSU is like the BEST EVER! s
HOW CAN OSU LOSE TO ANYONE? THEY PWN! s[/quote]

um...easy they'll lose to duke  :lol:

um, yeah

that's going to happen

especially as . . . they don't seem to be playing each other

y'know?

but they will lose this wednesday so it's all good

haha good one. you're just loaded with great jokes aren't you?  :rotfl:

um, dude

seriously

osu and duke are not playing?

(wtf)

[quote="MysticTerminator"]
but they will lose this wednesday so it's all good[/quote]
dude NO WAY
OSU has the best chinese food in their caf
GOOD FOOD=COMPLETE PWNAGE

$\mathfrak{QED}$

[quote="Inspired By Nature"][quote="MysticTerminator"]dude![/quote]
[quote="MysticTerminator"]uhh ok then[/quote]
Your posts need more meaning or atleast more words. :rotfl:[/quote]
dude that's for non-special ppl only
arnav is like unique so he can spam

You do, but you are exempt. Lucky Arnav...

 :spam:

Why does MysticTerminator get to be above the law? :(
EDIT: Him being special doesn't count as a good reason. :D

dude... did anyone go to the Buncombe County Math Circle today? :yup:

eh? what you say?

me no understand you!

i confuse!

this poem is about not trig's grammar  :lol:

your grammar is like a rotten egg

like a person in a waste factory who likes to beg

canada for which you like so much

is horrendous and pig-poopy as Michael Jackson and such

so go to english class and say to your teacher

My Milkshake Brings all the girls to the yard!!!!

wtf

It's called engrish, you know

purposely written incorrectly

come on

get with the program, eh

why, eh? i liked the poem. this chap's got some good on-demand poetry :moose:

yo man go around the tree dawg write a poem about me too man

okay

this one is about chess64.

the luckiest man in the world

 when he  touched this amazing spinner

it reached the lottery mark and he won 2 Billion! what a winner!

then he went to the christmas tree

singing oh say can you see.... ( you won't believe how he sounded)  :wink:

then he received a Pumped up High 5 from SANTA! ! !

the luckiest man in the world

no doubt about it

that's chess64 for you  :)

yo man that was amazing

let me just add a few lines:

canada is amazing

go canada!!!!!!!!

hahaha getting crazy, eh?
EDIT: i mean we are all crazy lol i am getting crazy srry man people keep telling me that and thought transferrance occursss and my friennd iss all like YO LET'S SAY MEAN THINGS ABOUT CANADIANS and i am all like are you reading my emails?
AND THEY ARE All LIKE HEHEHE NO I AM NOT I JUST LIKE SAYING MEAN THINGS ABOUT CANADIANS YO and I am all like woah i can't believe it really happens

[quote="funcia"]hahaha getting crazy, eh?[/quote]

yo man funcia man wtf are you talking about man stfu ok


abstract algebra
superior algebra
superior algebra unsolved

Let $A$ be a simple $R$-module. Prove that $Hom_R (A,A)$ is a division ring.

Clearly, $Hom_R(A,A)$ is a ring with pointwise operations.

Now consider $f \in Hom_R(A,A)$. The kernel of $f$ is a submodule of $A$, hence $0$ or $A$. So, if $f \neq 0$, the kernel must be $0$, i.e. $f$ is injective. Analog, we can conclude, that $f$ is surjective. So, $f$ is bijective, and has a multiplicative inverse in $Hom_R(A,A)$.


The radius of the larger of the concentric circles is 5 units longer than the radius of the smaller. How many times greater is the circumference of the larger circle?
[asy]draw((1,0)..(0,1)..(-1,0)..(0,-1)..cycle);
draw((.5,0)..(0,.5)..(-.5,0)..(0,-.5)..cycle);
draw((0,0)--(1,0));
dot((0,0));
label("$O$",(0,0),S);
label("5",(.25,0),N);[/asy]

Okay, we can tell that the radius of the smaller circle is $ 5$, and thus, the circumference is $ 10\pi$.  The radius of the larger circle is $ 5\plus{}5$, or $ 10$.  Thus, the circumference of that circle is $ 20\pi$.  Finally, our answer is $ 20\pi \minus{} 10\pi$, or $ \boxed{10\pi}$.

Note: The circumference of a circle is equal to $ 2\pi$ times the radius.

it asks how many times greater.
the circumfrence is a linear function of the radius, so the ratio of circumfrences is equal to the ratio of the radii.
big radius=10
small radius=5
10/5=2
2 is  our answer.

The diagram was misleading. It wasn't fair

Don't concentrate on the diagrams unless you truly need them. Just use them as a small guide.


in how many ways can 6 similair rings be worn on 4 fingers of 1 hand?

Please explain the thought flow behind solving this... i know i can use the formulae for partitioning of n non-distinct things into r distinct groups... but i'm not able to get the given answer.. so whats the flaw in my logic?

heres my thought process

This is more like possible arrangements for 10 items of which 6 are similair minus the arrangements where the fingers are all empty(E)... ie: (10!/6!)-E

Now there are 2 x 4! ways of them being empty.. so the possible arrangements are (10! / 6!) - (2x4!)

Well now my text book has a different explanation... whats wrong here?

have you tried the hockey stick identity?

whtever that it. anyways, ill give it a go.

we have $ \binom{5}{4}$ ways to chose the fingers. thats 5. then we have we must have atleast one finger one each hand(otherwise we wouldnt be using four hands) and then two others. so we have $ \binom{4}{2}$ ways to arrange the other two rings. so we have 6*5=30. then we have 2 different hands to choose from. so the answer is 60.

EDIT: i see where you got your answer. your saying were using 10 fingers. atleast, thats what i think your saying. we have to have atleast one ring on one hand so basically were doing $ x mod4$ where x is the number of rings were working with. in this case x is 6 so instead of doing $ \binom{6}{4}$, we do $ \binom{4}{6 mod4\equal{}2}$ for how to arrange the rings. the other parts, you just solve normally with combinations.

To clarify what i was saying was... there are 10 items totally.. 4 fingers + 6 rings... where 6 items are alike...so total arrangements possible = 10! / 6!. Now this includes the arrangements where all the fngers are empty...which is not possible so the dedcution from this... I'm still not clear. :(

[quote="math_music"]If we let fingers be empty then the problem turns out to find the number of solutions of $ x_{1}\plus{}x_{2}\plus{}x_{3}\plus{}x_{4}\equal{} 6$ with $ x_{i}\geq0$. You said that there is no difference between rings... so no matter the relative order of them. The solution is $ \text{CR}_{4}^{6}\equal{}\text{C}^{6}_{4\plus{}6}\equal{}\text{C}^{6}_{10}\equal{} 210$. If the rings were different we need to multiply by $ 6!$. If you need non-empty fingers then make a substitution like $ y_{i}\equal{} x_{i}\plus{}1$...

[url]http://en.wikipedia.org/wiki/Combinatorics#Combination_with_repetition[/url][/quote]
Yes, that's what I was thinking... I think that might be right

I'm so sorry for reviving but can't you just do balls and urns and then multiply that by 5C4?

What math_music did [i]is[/i] balls and urns, just a more algebraic way to think about it.  It is also known as stars and bars.

If it is balls and urns, then can someone elaborate on exactly why math_music did it that way?

They are the same thing, so he did use balls and urns; just different terminology.  He simply took a [i]slightly[/i] more general approach by mapping it to an algebraic equation rather than applying the balls and urns formula.  I'm not entirely sure what you are asking here or how you are confused about what he did...

Just confused on  this:

http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/4/4/0/4401e0a2b4214443f8e4b20ea64c4bb10a03dc73.gif

OK.  Explaining this particular example should make the general case clear, but I can explain the general case as well if necessary.

We wish to find the number of solutions to $ x_1\plus{}x_2\plus{}x_3\plus{}x_4\equal{}6$ where each $ x$ is a non-negative integer.  Consider lining up 6 1s with 3 dividers in arrangement like the one below:

11|111|1|

This corresponds to $ x_1\equal{}2$, $ x_2\equal{}3$, $ x_3\equal{}1$, and $ x_4\equal{}0$ (we are just counting the number of ones between each pair of consecutive dividers).  No matter how we place the dividers, we will always create a solution in the $ x$'s.  Conversely, we can represent any solution to $ x_1\plus{}x_2\plus{}x_3\plus{}x_4\equal{}6$ using 1s and |s, so we have a bijection.  Hence, the number of solutions to the equation $ x_1\plus{}x_2\plus{}x_3\plus{}x_4\equal{}6$ is the same as the number of ways to rearrange:

111111||| (we only need 3 dividers to create 4 regions)

Hence, there are actually $ \binom93\equal{}84$ solutions to this system of equations.

@mathemagician1729: I see your logic, but I don't think that's what the problem is asking for.

If you still want the general case, it's [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1323456#1323456]here (case 4)[/url].

So the problem is assuming that these 4 fingers are pre-picked?

[quote="mathemagician1729"]Yes I understand that and I basically did that but shouldn't you multiply by five because the problem said to use 4 of the 5 fingers on the hand?[/quote]

You cannot simply multiply by 5 if you assume that the fingers are not pre-picked because you will overcount when multiple fingers have no rings.

In this case:
We will simply put the rings on any of the 5 fingers and subtract out the cases where every finger has a ring.  As before, we are rearranging:
OOOOOO||||
which can be rearranged in $ \binom{10}4 \equal{} 210$ ways.

Finding the number of arrangements in which every finger has a ring is the same as placing a ring on every finger and then placing the remaining ring on any finger, which can be done in 5 ways.

Hence, in this case, there are $ 210 \minus{} 5 \equal{} 205$ ways to rearrange the rings on the fingers.


Evaluate $ 24\minus{}(2x\minus{}y)$ if $ x\equal{}4$ and $ y\equal{}3$.


[b]Problem 1[/b] : What is the value of $ x$ if $ 1^x \equal{} 1$? $ 1$ or $ 0?$ Please give a bit explanation with your answer.

[b]Problem 2 [/b]: How can I find out the $ a^x\equal{}y$ if  $ a$ and $ x$ are non-integers or if $ a$ is non-integer or if $ x$ is non-integer? Please show the solution for $ ({1.1}^{0.1})$.

Please help. Thank you.

[hide="1"]It's all real numbers. 1 to any power is 1.[/hide]


geometry
function
integration
logarithms
analytic geometry
graphing lines
slope

Anybody come across this result:

Find the equation of the tangent at any point on y =1/x. Draw the tangent and find the area of the triangle formed by the intersection of the tangent with the x and y axes. It turns out that the area is always constant!

The thing i don't get is why this works for y=1/x and are there a family of functions for which this result holds true? What properties does y=1/x have for this work?

Any thoughts?

[hide]The tangent line at the point $(x_{o},y_{o})$ has equation: $y-y_{o}= f'(x_{o})(x-x_{o})$. This line intersects the axes at $A(0,y_{1})$ and $B(x_{1},0)$

For point $A: \ x=0 \Longrightarrow y_{1}-y_{o}=-x_{o}f'(x_{o}) \Longrightarrow y_{1}= y_{o}-x_{o}f'(x_{o})$

For point $B: \ y=0 \Longrightarrow-y_{o}= f'(x_{o})(x_{1}-x_{o}) \Longrightarrow x_{1}=\frac{x_{o}f'(x_{o})-y_{o}}{f'(x_{o})}$

It is necessary that $f'(x_{o})\neq 0$

We want $x_{1}y_{1}=-\frac{(y_{o}-x_{o}f'(x_{o}))^{2}}{f'(x_{o})}\text{ = constant}$

Setting $x,y,y'$ instead of $x_{o}, y_{o}, f'(x_{o})$ we have: $\frac{(y-xy')^{2}}{y'}\text{ = constant}$

$\forall x$ such that $\exists y''$ we have: $\left(\frac{(y-xy')^{2}}{y'}\right)'=0$

$\frac{2(y-xy')(-xy'')-y''(y-xy')^{2}}{(y')^{2}}= 0$

$2(y-xy')(-xy'')y'-y''(y-xy')^{2}=0$

$-y''(y-xy')(2xy'+(y-xy')) =0$

$-y''(y-xy')(xy'+y) =0$

So, at least one of the 3 factors is zero

Each of $y''=0$ and $y=xy'$ leads to $\boxed{y=ax+b}, \ \ a\neq 0$
This function has the desired property, because the tangent line is the function itself, so it determines a fixed triangle (however, if $b=0$ there is not a triangle)

From $y=-xy'$ we get: $y=-x\frac{dy}{dx}\Rightarrow$

$\frac{dy}{y}=-\frac{dx}{x}\Rightarrow$

$\int{\frac{dy}{y}}=-\int{\frac{dx}{x}}\Rightarrow$

$\ln{y}= \ln{a}-\ln{x}\Rightarrow$

$\boxed{y = \frac{a}{x}}$

[color=brown]Note:
I set just the constraint that the area of $\triangle OAB$ doesn't change in an [u]interval[/u] in which $y',y''$ exist.
I mean, we could get a function which is twice differentiable in $(-\infty,m) \cup (m,+\infty)$ and giving different areas in each of the intervals.
Fortunately, the function $y = \frac{a}{x}$ gives equal areas in $(-\infty,0)$ and $(0,+\infty)$ (accurately equal products, because I didn't take absolute values)
[/color][/hide]

A most thrilling solution :yup:

Here's how I would look at it:

1. We can find that $\frac{d}{dx}\frac{1}{x}=-\frac{1}{-x^{2}}$
2. We notice that $-\frac{1}{x^{2}}=-\frac{y}{x}$
3. Consider the picture below:
[img]http://img.photobucket.com/albums/v194/mukmasterxps/picturex-1.gif[/img]
4.We know the slope from the origin to a point is $\frac{y}{x}$, and that the slope of the function at the point, and therefore the tangent line, is $-\frac{y}{x}$. We also know that they have a common point, which is the vertex of the green triangles shown.
5. From this, it is easily shown that the two lowermost green triangles have the same height ($y$ value of the chosen point) and the same width $y/|m|=x$
6. This means the lower two green triangles are congruent.
7. Because $y=\frac{1}{x}$ is symmetric about $y=x$, we can apply the same logic to show that the other two green triangles are congruent, and that in fact all four green triangles are congruent.
5. Since each of these triangles has a height of $y$ and a width of $x$, and since $y=\frac{1}{x}$, the total area is $4\frac{xy}{2}=2$, a constant.

Weird. It's showing on my computer. Must be a problem with photobucket. Or else its not loading fast enough. Just in case, I've attached it to this post.


integration
trigonometry
calculus
calculus computations

I must show with Fubini's theorem that $ \int_0^{\pi}xf(\sin{x})dx\equal{}\int_0^{\pi/2}f(\sin{x})dx$, where $ f: [0,1]\minus{}>\mathbb{R}$ is a continuous function. Do you have any idea?

I have got no idea how Fubini should help here. However there is an easy proof
using the substitution $ y\equal{}\pi\minus{}x$.


definitely...happy chinese new yr....DOG!!!

[quote="dragongrl6"]???????????? [/quote]I think what tokenadult wrote is in chinese ;)

I knew that.

I'm the year of the rooster!! :lol:

I'm the year of the monkey! How about you guys? ;)

[quote=lakecomo224]I THINK THAT IS THE LONGEST BUMP EVER![/quote]

I will bump it again when it’s Chinese New Year! It’s amazing to learn about different cultures, in my scenario my own!

OMG it's my year! Year of the dog! I am turning 12 this year. What is really sad is that Chinese New Year is the day before my piano test.

[quote=lakecomo224]I THINK THAT IS THE LONGEST BUMP EVER![/quote]

Actually...

[quote=lakecomo224]I THINK THAT IS THE LONGEST BUMP EVER![/quote]

This is the longest bump ever. https://artofproblemsolving.com/community/c7h5_wierd_problem_in_linear_algebra
EDIT: I just set the record. https://artofproblemsolving.com/community/c48h2024_how_to_read_the_names


Hint: we have

$ \frac {x^4}{1 \minus{} 4x(1 \plus{} x^2) \plus{} 6x^2 \plus{} 2x^4} \equal{} \frac {x^4}{1 \minus{} 4x \plus{} 6x^2 \minus{} 4x^3 \plus{} x^4 \plus{} x^4} \equal{} \frac {x^4}{(1 \minus{} x)^4 \plus{} x^4}$

Now try again. If you can't solve...

[hide]Let $ g(k) \equal{} \frac{f(k)^4}{(1 \minus{} f(k))^4 \plus{} f(k)^4}$. Then, since $ f(k) \plus{} f(2009 \minus{} k) \equal{} 1$, we have $ g(2009 \minus{} k) \plus{} g(k) \equal{} 1$. Then

$ \sum_{k \equal{} 0}^{2009} g(k) \equal{} \sum_{k \equal{} 0}^{1004}g(k) \plus{} g(2009 \minus{} k) \equal{} 1005.$[/hide]


function
integration
calculus
real analysis
real analysis theorems

Given some sequence of  functions $ f_n$ that admit antiderivates ,such that  $ f_n \to \ f$, I want to ask if there are some minimum conditions to impose in order to establish the fact that $ \int f_n \to \int f$.

For instance if I have $ f_n(x)\equal{}\sum_{k\equal{}0}^{n} x^k \to \frac{1}{1\minus{}x}$ (as $ n \to \infty$) , for $ x \in (0,1)$, I know that $ \int \sum_{k\equal{}0}^{\infty} x^k  \to \int \frac{1}{1\minus{}x}$.

But I want to ask : does this happen all the time or there are some restrictions regarding that sequence of functions ?

If integration is taken in the sense of [url=http://en.wikipedia.org/wiki/Lebesgue_integration]Lebesgue[/url], then two theorems relevant to your question are the [url=http://en.wikipedia.org/wiki/Monotone_convergence_theorem]Lebesgue Monotone Convergence Theorem[/url] and the [url=http://en.wikipedia.org/wiki/Dominated_convergence_theorem]Dominated Convergence Theorem[/url].


inequalities
set theory
calculus
calculus open

This is not an unsolved problem, it's actually an [b]unsolvable[/b] problem. See information about the Continuum Hypothesis.

Thanks! I was actually thinking about it !:D.

I can't help asking this: is it really equivalent to the Continuum Hypothesis?

Yes, it's equivalent to the Continuum Hypothesis. Because the answer to the question depends on whether you assume the Continuum Hypothesis to be true or not.

If you assume that there exists an uncountable set who's cardinality is less than that of the reals, then the answer to your question becomes no. Because the cardinality of any set who's power set has cardinality less than that of the reals would have to be less than that of the integers - which is not possible as the cardinality of the integers is the smallest infinite cardinal. It's like an infinite descent arguement.

However, if you assume that there do not exist any such sets, then the answer to your question becomes yes - as the only possibile cardinalities are those of the iterated power sets of the integers. So any uncountably infinite set would have a bijection to one of these sets, and hence a bijection to the power set of some smaller set.

The key here, is that that cardinality of the integers is the smallest infinite cardinality (see if you agree with this). Ance once you have any infinite set, you have a discrete chain of increasing cardinalities obtained from it's repeated power sets. Furthermore, this process of repeated "power setting" preserves the order of the original cardinalities of the sets (meaning if |A| < |B|, then |P(A)| < |P(B)|). See if you can prove this.

I don't want to annoy anybody (:D), but I have one more question:

By |P(A)|<|P(B)| do we mean |P(A)| =/= |P(B)| or just |P(A)|<=|P(B)|? I don't know if the notations are exactly the same for cardinal numbers. I can't prove that if |A|< (=/=) |B| then |P(A)|<|P(B)|, just that it's <=. So from what I understand, I have to show that if |P(A)|=|P(B)| then|A|=|B|, and I don't know how to do that.

Anyway, I think that we don't need strict inequality in order to show the equivalence between the continuum hypothesis and the question I originally asked.

I'm sure the inequality is strict. The key is to note that the power set of a set can be thought of as a set of sequences of the set's members. So we just need to find a way to use a bijection between those sequences to get a bijection between the sets.

Well, I'm sure you're right, but I was just saying that we don't really need it to be strict in order to show the equivalence. However, I'd like to see a proof of the fact that |P(A)|=|P(B)|=>|A|=|B|. Could you help me with that? I couldn't find it anywhere (I've searched the Internet for various topics related to this, but could find nothing).

Yes, I believe you're right about strict inequality not being necessary for equivalence to the CH. To prove |P(A)| = |P(B)| => |A| = |B|,

Suppose there is a bijection f: P(A) -> P(B). Consider f({a}) for each a in A. This is some unique subset of B. Use the axiom of choice to choose some member b of this subset for each a. Then let g: A -> B be such that g(a) = b.

We are assured of being able to pick such a b uniquely for each {a}, by Zorn's Lemma (again axiom of choice), making g injective. Analogously we can find an injection from B to A. So |A| = |B|.

This hits on another interesting point of set theory. Without the axiom of choice I don't think this proposition is true. That's just an impression though, don't ask me to prove that!  :D

I had the exact same idea, but I didn't know that Zorn's lemma allowed us to choose an element from each set such that they are all different. I thought it just allowed us to choose an element in each set, so I wasn't sure about that g you defined being injective.

[quote="grobber"]Assume A is an uncountable set. Is there a set B s.t. |A|=2<sup>|B|</sup>?[/quote]

The anser is NO! (and is independent from CH)

Take A_0 the set of positive integers. And A_(n+1)=P(A_n).
Take A=A_0 U A_1 U A_2 U ....
Then there is no B such that |A|=2^|B|

Yes A is a set from Axiom of Union. (it is the union of a family of sets, and the family is also a set, is the set of nonnegative integers)

Hmm. Actually, I believe you are right. A is a set. But it is a very unsual set (even for uncountable sets). You have to be careful about what you assume is true about it. I don't believe that it can be proven (from regular set theory) that it satisfies our condition. For example, how do you know that |A| =/= |P(B)| for some B /subset A?

The type of sets we have been talking about here have what are called strongly inaccessible cardinalities. They are (non-trivial) cardinalities that cannot be reached by the power set of a smaller cardinality. I believe the existence of such sets cannot be proven in ZFC set theory. I thought that it required the GCH, but it actually turns out there there is more to it than that.

While it is equivalent to GCH for "small sets", there is an axiom called the Axiiom of Inaccessible Cardinals which is needed to deal with the cases of these "very large" sets.

This set A is very interesting though. It has many many structures which are interesting to study. Here is a very good paper about it - http://philsci-archive.pitt.edu/archive/00001620/01/How_Set_Theory_Impinges_on_Logic.pdf


Hey well since my last "game show" pretty much failed  :( . I've decided to start a new one. As you can guess it would be Survivor. So here is how were going to do this and this time I won't add rules because a whole bunch of people tell me to unless its acctually good.
Rules:
1 Teams will be made of 8 people on ea team (there will be 2 teams)
2 If a team loses a "challenge" then the team will have to vote one member off there team!
3 The team that wins is the team that is the last one standing
4 There will be no stealing players from another team like in the real Survivor
5 Every three weeks I will post an Immunity Question! The first person who gets it right is automatically safe from getting voted out and they can keep immunity until they need to use it. Also other players will not be informed that, that certain person won immunity. HOWEVER if that player wishes to give his immunity to another player then simply PM the player and me that you are giving the immunity to them.
6 There will be one elimination a week
7 There will be one challenge a week (Immunity questions do not count as a challenge!)
8 You may play dirty stratagies among your team about eliminating a certain person from your team and I do not need to know about it.
9 Really this game is about surviving so play stratagies to stay alive and not get eliminated
10 There is absolutely no posting in this forum about strategy! If you do it is an automatic elimination!
11 The challenges are teamwork challenges not individual challenges. Each week I will post the challenges in a new forum pst and have two seperate topic pages for each team. However the challenges will be the same for both teams.
12 When a player is disqualified they may not discuss strategy among the remaining people! If I get a legitimate post from someone about this that person will be banned from the next 3 survivor games.

Hints: Disqualify members you think are hurting your team
Hint: The object of the game is to stay alive so build allies among your team! In the real survivor you will usually see a team split into really two teams!

If you wish to join then just post in this forum saying you want to join

i'm in if it doesn't push into end of june.

[quote="perfect628"]I (rather) recently tried this and it failed dismally. Given that I (no offense) was more organized than this in approaching it, I don't see this getting off the ground(as you've stated the rules), but good luck anyway.[/quote]

I agree. But it was pretty funny, I remember I was going to join then went out at the last minute because of our feud about Roy vs Marth, lol.

er ah....so only like one person is joining o fun...come on people!

So are the challenges going to be math, random trivia, what?

challenges are teamwork objectives like you have to work together to win...So it cant be math or trivia

Well, I never saw Survivor, but I'll join if someone fills me in

err.. survivor is a game where you ahve two teams and the teams fight to keep from having to vote one of their players out and having the other team vote someone out

Just a note of sympathy to crazychinesemaniac about the failure of this game  :lol: .

This is a BUMP.

A BUMP is when someone makes a useless post so that the topic goes to the top

of the Forum, when viewed.

A BUMP is entirely useless.

A BUMP is when you want people to see the topic.

You would want someone to see your topic if

1. Its a game where you sign up, and you need people to see this.

2. You just like being on top of other people.

3. You're bored

The first option applies to this game

and many more.

The first option is because applications may be a little slow

Or it may be people just don't scroll down.

The second option is a problem. You are obsessed with power.

The third option applies to me. I am making a long post because I'm bored.

Yay.

This is another BUMP.

A BUMP is when someone makes a useless post so that the topic goes to the top

of the Forum, when viewed.

A BUMP is entirely useless.

A BUMP is when you want people to see the topic.

You would want someone to see your topic if

1. Its a game where you sign up, and you need people to see this.

2. You just like being on top of other people.

3. You're bored

The first option applies to this game

and many more.

The first option is because applications may be a little slow

Or it may be people just don't scroll down.

The second option is a problem. You are obsessed with power.

The third option applies to me. I am making a long post because I'm bored.

Yay.

Well, I don't think that Survivor will work here... =(

Can we just lock this? It's getting spammy now.

[quote="DonkeyKingKong"]yo homie wazzup?

dont lock this[/quote]

My point exactly.

This is the FINAL BUMP.

A BUMP is when someone makes a useless post so that the topic goes to the top

of the Forum, when viewed.

A BUMP is entirely useless.

A BUMP is when you want people to see the topic.

You would want someone to see your topic if

1. Its a game where you sign up, and you need people to see this.

2. You just like being on top of other people.

3. You're bored

The first option applies to this game

and many more.

The first option is because applications may be a little slow

Or it may be people just don't scroll down.

The second option is a problem. You are obsessed with power.

The third option applies to me. I am making a long post because I'm bored.

Yay.

Apparently, not many people are interested.

If this BUMP does not work I call for a LOCK.

Stop spamming. This obviously isn't happening, so just let it fall to the bottom and be forgotten.

I'll join, but it doesn't look like it's going to work.


Let ABC be an acute triangle, and w a circle tangent to AB at X and AC at Y. If k is a circle that passes through B and C and is tangent to w at Z, prove that XZIB is a
ciclic quadrilateral (where I -incenter).
(I guess this problem was presented as a lemma to an Indian problem , 2001)

This lemma follows from another lemma, which was proved by Fedor Bakharev in http://www.mathlinks.ro/Forum/viewtopic.php?t=33 . His proof is a kind of proof I really hate: it is written up in such a brief and messy way that one has to think for an hour before one is able to understand it, then a dumb typo disturbs the understanding (I've corrected it now), and finally, for the motivation of the reader, at the very beginning of the post, the writer calls the lemma "trivial".

  Darij

I don't think I have the time to get into details right now, but I inverted the figure wrt the incircle, then looked at it backwards (i.e. tried to prove a converse from which the problem itself would follow), and got this mugh easier problem:

Assume $H$ is the orthocenter of $ABC$. Take points $M,N$ on the circles $(AHB),(AHC)$ respectively, s.t. $MN\|BC$, and $M,N$ are on opposite sides of $AH$. Let $P=MB\cap NC$. The circle $(MNP)$ is tangent to both $(AHB)$ and $(AHC)$.

It comes from the fact that $I$ is the incenter of the triangle formed by the images of $A,B,C$ under the inversion, and so on (like I said, I don't have that much time right now).

Well some details will certainly be needed here...

Ok, I think I have some spare time right now :).

Invert, as I said, wrt the incircle of $ABC$. Let $\mathcal F'$ be the image of $\mathcal F$ ($\mathcal F$ is a general figure, which might be a point, or whatever we want). $I$ is now the orthocenter of $A'B'C'$. $w'$ is a circle tangent to $(AB)',(AC)'$ in $X',Y'$ respectively. Consider now a circle through $B'C'$, tangent to $w'$ in $Z'$. We must show that $X,B',Z$ are collinear (or, similarly, $Y,C',Z$).

Now we look at it backwards: clearly, in the construction above $X'Y'$ is parallel to $B'C'$, so assume we take $X',Y'$ on the circles $(AB)',(AC)'$ respectively, s.t. $X'Y'\|B'C'$. Now take $Z'=X'B'\cap Y'C'$. It is true that the circle $w'=(X'Y'Z')$ to be tangent to $(B'C'Z)$ (actually, this is what the problem I posted above asks for), but because of the uniqueness of the circle $(B'C'Z')$ defined as the circle passing through $B',C'$ and tangent to $w'$, we see that this is also sufficient, so the two problems are actually equivalent.

Now, I suppose you have already recognized the problem I've posted in my previous post, although hidden behind different notations.

Dear Mathlinkers,
this problem is also solved as a lemma.
see http://perso.orange.fr/jl.ayme  vol. 2 Un remarquable résultat de Vladimir Protassov.
Sincerely
Jean-Louis


This problem requires the use of [b]combinations[/b]. Here's a good explanation of what they are:  [url]http://www.artofproblemsolving.com/Wiki/index.php/Combinations[/url]

For any four points you choose on the circle, there will be exactly 1 convex quadrilateral.  Since it doesn't matter what order you pick the four points (for example, choosing A, B, C, and D is the same as choosing B, A, D, and C),  this problem is a combination problem and not a a permutation problem.

Now, computate C(10, 4). This is 10!/(6! * 4!) =210
Therefore, you can connect the points in [b]210 [/b]to create concave quadrilaterals.

I hope that helps :)

thanks a lot for this help i think the convex part threw me off. i did not realize this before.

[quote="shoobish"]could you guys please explain how to solve this problem.

Ten points are randomly placed on the circumference of a circle all a different distance from each other apart.
In how many ways can you connect four of these points to create a convex quadrilateral.

Could you pelase explain this.
I may be a noob at math but help anyway!!!!![/quote]

There are $ 10\cdot9\cdot8\cdot7\=\frac{10!}{6!}$ ways to choose 4 points [b]with regard to order[/b] But we divide by the ways to arrange the points, $ 4!$ so our total is $ \frac{10!}{4!6!}=210$ The expression $ \frac{n!}{r!(n-r)!}$ is generally written as $ nCr,\binom{n}{r},$ or $ C(n,r)$

[quote="shoobish"]i think the convex part threw me off.[/quote]

You can't make a concave cyclic quadrilateral, so it must be convex.


function
linear algebra
linear algebra theorems

Can I ask something about Linear Programming?
This is really elementary i thinks.

if we have an objection function $f(x,y)$ and we want to minimize or maximize this function subject to any given condition (usually line equation with inequality). (as we do in Linear Programming Lesson).
Then drawing a graph will help us. The intersection points of any line is condition are called critical points. And  max or min value often occurs at this point?

My Question: Why we only need to check the value of objection function  in the extreme points to get the max/min value. Is there any theorem told us about that?

thanks in advance.

If both the function and the constraints are linear, the vertices of the region are all that matters.

Every point in the interior is a convex combination of the vertices: $x=a_1v_1+a_2v_2+\cdots+a_nv_n$ with $\sum_i a_i=1$ and $a_i\ge0$ for each $i$. Then $f(x)=\sum_i a_if(v_i)\le\max_i f(v_i)$.

This assumes that the region is bounded- if not, you need some extra virtual vertices to handle the behavior in unbounded directions.


Please help me find a solution and a detailed explanation on how to solve this problem, I have the answer but I don't get it ill put a hidden text below the problem that is the answer. Thanks.

In the multiplication problem below, A,B,C,and D are different digits. What is A+B?

ABA*CD=CDCD
(I don't know latex its the one where the ABA is on top then you line up CD below it and there is the line then CDCD)

A)1            B)2                    C)3                        D)4                        E)9

[hide]The answer is A[/hide]

Sorry occupied doing the other stuff forgot to put the problem up lol

[hide="solution"] ABA is the same as 100A+10B+A=101A+B (just like how 213 is 200+10+3). Similarly, CD is 10C+D, and CDCD is 1000C+100D+10C+D. Now substituting back in, we have $ (101A \plus{} B)(10C \plus{} D) \equal{} 1000C \plus{} 100D \plus{} 10C \plus{} D \equal{} 101(10C \plus{} d)$ so $ 101A \plus{} B \equal{} 101$. If A was bigger than 1, $ 101A\plus{}B$ would be bigger than 101. Thus, A=1, B=0, so A+B=1.[/hide]

BTW, I just happened to solve 2006 AMC 8 for the 3rd time yesterday and this problem is from 2006 AMC 8's # 24

[hide="easy solution"]ABA must be 101, since 101 is the only number that, when multiplied by a two digit number, ends up having a product that is the two digit number followed by itself(ABA*CD=CDCD). so, 1+0=1. No LaTeX cuz im lazy.[/hide]

whoa, this is a #24? either im getting smarter or amc8 is getting easier.

AMC 8 is trivial, therefore like sometimes number 17 can be harder than number 25. Whereas for AIME number 1 is way more easier than AIME number 15


[quote="Omid Hatami"]aslan $2^{\sqrt 2}$ rooye $\mathbb Q$ jabri nist.

[hide]Adade $a$ ra rooye meidan $F$ jabri gooyim agar yek chandjomleyi ba zarayebe $F$ vojood dashte bashe ke $a$ risheye an bashad[/hide]

Har addade gheire jabri rooye $\mathbb Q$ (estelahan motaali) be vozooh hong ast. Albate esbate inke $2^{\sqrt 2}$ rooye $\mathbb Q$ motaali ast asan nist.[/quote]
khob nabashe mage chi mishe?

Tavajjoh kon har adade gooya jabri ast chon $ax=b$ dar har meidani javab darad. Pas aadade gheire jabri gong hastand

salam
az koja mishe sabet kard ke roye $\mathbb Q$ jabri nist va bad chetori sabet mishe adadi ke roye $\mathbb Q$ jabri nistand gongand.

Man tanha ghaziyei ke midoonam baraye esbate gheire Jabri boodane yek adad ghaziyeye "Liovil"e.

salam
aghaye hatami age momkene ghazize ro post post konid ya inke ketabi ke in ghaziye toshe ro be ma moarefi konid.mamnoon misham

shoma hokmo kamel saabet nakardin dar vaghe saabet kardin ke adade mazboor be ezaye hame ye $k$ ha gooya nist

[quote="BaBaK Ghalebi"][quote="jensen"]sabet konid $2^{\sqrt{2}}$ adady gong ast.[/quote]
man ye rahe hal daram bebinid doroste ya na
miaim hokm ro tamim midim
yani migim adade be shekle $2^{n\sqrt{2}}(n\in \mathbb N)$
farz konid intor nabashad va $\exists k\in \mathbb N : 2^{k\sqrt{2}} \in \mathbb Q$
hal koochiktarin $k$ ba sharayete fogh ra dar nazar begirid
yani $2^{(k-1)\sqrt{2}} \in \mathbb Q^c$
hal faz konid $2^\sqrt{2}$ gong nabashad
midanim ke zarbe va taghsime do adade gooya adadi gooyast va hasele zarbe gooya dar gong ham gonge
hal az anjai ke adade $2^{\sqrt{2}}$ adadi gooyast (tebghe farze kholf) pas agar adadi gong dar an zarb shavad hasel gong ast pas hasele ebarate $2^\sqrt{2}$ $*$ $2^{(k-1)\sqrt{2}}$adadi gong ast chon farz kardim ke adade $2^\sqrt{2}$  gooya va adade digar gong ast
amma hasele in ebarat barabare $2^{k\sqrt{2}}$ ke adadi gooyast
pas be tanaghoz resideim
pas chenin $k$ i vojood nadarad
pas $2^{\sqrt{2}}$ gong astage joof zadam begid
chon ino sare emtehane dini be dast ovordam :D[/quote]

Joobe mahz :D avvalan az hich khasiate $2$ va $\sqrt 2$ estefade nakardi, Dovvom inke to ye farzi kardi va baad az naghize farz estefade kasrdi khob maaloome ke be tanaghoz miresi.    :D

[quote="Omid Hatami"]aslan $2^{\sqrt 2}$ rooye $\mathbb Q$ jabri nist.

[hide]Adade $a$ ra rooye meidan $F$ jabri gooyim agar yek chandjomleyi ba zarayebe $F$ vojood dashte bashe ke $a$ risheye an bashad[/hide]

Har addade gheire jabri rooye $\mathbb Q$ (estelahan motaali) be vozooh hong ast. Albate esbate inke $2^{\sqrt 2}$ rooye $\mathbb Q$ motaali ast asan nist.[/quote]
doroste vali inike $2^\sqrt{2}$rooye $\mathbb Q$jabri nist hokme ghavitarie
maslan adade $\sqrt{2}$ rooye $\mathbb Q$jabrie va gong ham hast
dar natije to dari hokmi ghavitar ro esbat mikonio ye zare fekr konam karet gharare sakht beshe

ye masaleye kheili ghashang va kheili asoon.kheili jalaebe vaghti bare avval besh nega mikonid intori be nazar nemiyad:
saabet konid adadi gooya vojood darad ke agar be tavane $\sqrt2$ berese gooya mimoone.


dude, i double post like everything so shut up loll.  :)


inequalities function calculus derivative logarithms

We have n positive real numbers where their sum is 1976. What is the largest product of these n positive real numbers.

This type of problem has been discussed a couple of times before.

[hide="answer"]$2\cdot3^{658}$[/hide]

EDIT- I probably misinterpretted the problem...

I believe the problem should read "...n positive integers..."  As written, the largest possible product would be
$\displaystyle (\frac{1976}{n})^n$

I get a slightly different answer, though I am not entirely confident in it.  Let $a_1,a_2,...,a_n$ be the sequence.  From the AM-GM inequality, we know:
$\frac{a_1+a_2+...+a_n}{n} \geq \sqrt[n]{a_1 a_2 ... a_n}$
w/ equality iff $a_1=a_2=...=a_n$.  Raising both sides of the inequality to the nth power, we have:
$f(n)=\left(\frac{1976}{n}\right) ^n \geq a_1 a_2 ... a_n$
We want to find the maximum value $f(n)$, with $n$ an integer. Clearly, the function is continuous for all positive intervals.  Thus inorder to find the maximum we can use derivatives:
$f'(n)=1976^n \left(\ln \left(\frac{1}{n}\right)-1\right)
   \left(\frac{1}{n}\right)^n+1976^n \ln (1976)
   \left(\frac{1}{n}\right)^n$
This only has one root, $x=\frac{1976}{e} \approx 726.93$. Thus the maximum of $f(n)$ for integer $n$ is at either $n=727$ or $n=726$.  By inspection, its at 727, for which $f(727) \approx \boxed{5.027*10^{315}}$

I think the problem means $n$ is fixed, but unknown. So it's just AM-GM to give the $f(n)$ you have.


Janelle averages 40 kilometers per hour biking on level ground. She averages $ 60\%$ of her level-ground speed riding uphill, and she averages $ 120\%$ of her level-ground speed riding downhill. The course is level for 5 kilometers, uphill for 6 kilometers, and then downhill for 6 kilometers. How many kilometers per hour is her average speed for the entire course?

She goes 24 km/hr uphill, 48 km/hr downhill.
She takes 1/8 of an hr going level, 1/4 of an hr going uphill, and 1/8 of an hr going downhill, totaling 1/2 an hour for 17 miles, or 34 km/hr total.


ratio
geometry
circumcircle
incenter
geometry proposed

I like this problem, is not so hard, but the solution is very beautiful.

there are 2 questions

1 - $\frac{DH}{BH}= ?$  The ratio of lengths

2 - $\frac{a(BCDH)}{a(BDH)}= ?$  the ratio of Areas

Kouichi Nakagawa I like all your geometric images and solutions

you noticed that H is fermat point , hehe, i thought nobody would see that

how much time did you spent on it?

I spent around 30 minutes to create and solve it.

but you solved it in a better way than me

you are sugoi

thanks

Thank you for yagaron :)
I was able to make smoothly answer in itself unexpectedly.
However, I noticed following the evolution problem, but there is not yet proof.

About triangular BCD

F:Fermat point
H:Orthocenter
O:Circumcenter
I:Incenter
$J_{C}$:Excenter corresponding to C
K:New point when we made an equilateral triangle to assume side BD one side for triangular BCD outward
L:Point of intersection to have when I extended side AB and side ED

There are nine points of $FHOIJ_{C}GMBD$ on the same circumference.

$K, G_{3}, O$ is in the same straight line.
$J_{C}, G_{3}, I$ is in the same straight line.
$L, G_{3}, H$ is in the same straight line.

I understand that, besides, I am interesting about these nine points.
There are parallelism and a perpendicular affiliated thing in them when I bind a certain points together in a line.
A lot of various interesting properties seem to be covered in this figure.

Thank you for a splendid problem!!

woow.. You are really inspired on this problem :)

I fell very very happy when you said my problem is good.

I began in Geometry recently and is a pleasure "heard (actually Read)" it from you

thank you


geometry
incenter
circumcircle
geometry proposed

Given a triangle $ABC$, we construct a circle $\Gamma$ through $B,C$ with center $O$. $\Gamma$ intersects $AC, AB$ at points $D$, $E$, respectively($D$, $E$ are distinct from $B$ and $C$). Let the intersection of $BD$ and $CE$ be $F$. Extend $OF$ so that it intersects the circumcircle of $\triangle ABC$ at $P$. Show that the incenters of triangles $PBD$ and $PCE$ coincide.

It seems like I did not include $F$ in my original post! I have posted a correction. Sorry for your inconvenience.

nice problem! It is well known that P is the Miquel point and thus the inverse of point F in circle O; thus inverting line BFD in circle O, we find that BPDO is cyclic, so if I is the intersection of line OPF and circle O, PBD = POD = IOD = 2IBD, which suffices to show that BI bisects PBD; everything else follows analagously and we are done.

let me expand on that well-known bit: basically, what we want to show is that P is the inverse of F in (O). Well, define Q as the inverse of F in (O) and we shall show that Q = P. Define G = BC ^ DE. Consider a tangent GT from G to (O), and consider the circle with center G passing through T. Suppose the inverse of O in (G) is X and let the inverse of G in (O) be G'. Then GX*GO = GT^2 = GO^2 - OT^2 = GO^2 - GO*OG' = GO*GG', so GX = GG' and thus X = G'. Thus, the pole of O in (G) is the line through X perpendicular to GO, which we know is the line through G' perpendicular to GO, which we know is the pole of G in (O), which we know is line AF. Thus F lies on the pole of O in (G), so O lies on the pole of F in (G), i.e. the inverse F' of F in (G) must be the foot of the perpendicular from O to GF. Thus GF' * GF = GT^2 = GB * GC, so F'BCF is cyclic. Now, invert this circle in (O) and see what circle it goes to. B and C stay fixed, and by definition, F goes to Q. As F' is the projection of O onto GF and GF is the pole of A in (O), F' must be the inverse of A, so F'DEF goes to AQBC, i.e. Q lies on the circumcircle of BC. But Q, being the inverse of F in (O), must also lie on ray OF. These two conditions clearly determine a unique point and as P also satisfies these conditions, Q = P, i.e. P is the inverse of F in (O).

I believe you are right! However, there are a lot of those who do not know inversion. Can anyone find an alternative solution?