Hi,
Last year I scored 104.5 on AMC 12A and a 2 (lol) on the AIME I. My goal is to at least make USAMO this year, and since the AMCs are coming up soon, would past USAMOers or equivalent please explain the best thing for us newbies to do in order to improve our problem solving skills in 3 weeks?

Also, does anyone know of websites where you can get past AMC 12 questions? I always hear about people doing AMC questions to get into the problem solving 'mood' but I can't find any.

Also, I have the Art and Craft of Problem Solving by Paul Zeitz and briefly used AoPS Vol. 1 since last year as resources. Thanks for your help.

I'd say the best thing to do is just practice old contests. Look at the button up at the top that says 'contests,' this site has tons of old problems there (Go into USA, AMC 12, etc). AoPS V. 1 and 2 would also be good for you at this point.

If you're looking for an AIME-level "crash course", we'll be having our annual weekend [url=http://www.artofproblemsolving.com/Classes/AoPS_C_Enroll.php#class2]Special AIME Problem Seminar[/url] on March 3-4.

Thank you. For past USAMOers, was practicing problems, not reading problem solving books, what mainly improved your scores?

[quote="cooljoe"]For past USAMOers, was practicing problems, not reading problem solving books, what mainly improved your scores?[/quote]

Well, I haven't even been to AIME before(aiming for it this year), but I think that doing BOTH will improve your score.

[quote="cooljoe"]Also, does anyone know of websites where you can get past AMC 12 questions? I always hear about people doing AMC questions to get into the problem solving 'mood' but I can't find any. [/quote]

[url]http://www.artofproblemsolving.com/Books/AoPS_B_Item.php[/url]

I recommend the books.

[url]http://www.kalva.demon.co.uk/aime.html[/url]

AIME's are good, too.

[quote="cooljoe"] Also, I have the Art and Craft of Problem Solving by Paul Zeitz and briefly used AoPS Vol. 1 since last year as resources. Thanks for your help.
 [/quote]

That's a great start. Try The AoPS volumes 1 and 2 for AIME, and only 2 for USAMO.

As a MOsPer, I ONLY do old problems (especially in books like Engel's Problem Solving Strategies that are essentially problem sets grouped by subject) and read the solutions after sufficient work. Of course, if you come across a solution with ideas you don't understand, certainly read about it. However, I suggest trying to read about it on Wikipedia (or better yet, Mathworld if you understand it) so that you can get a lot of in depth knowledge and truly understand what you are doing, rather than reading about it in a book that just briefly touches on it.

If you want books, Zeitz is excellent for USAMO and hard AIME level material. Though plenty of people seem to disagree with me, I think the AoPS books only help with AMC and easier AIME problems (they're still great books for that, though). They certainly won't help you on the USAMO.

So that was more what to do eventually if you want to get to the USAMO and do well on it. At your current level, you may prefer to read through all of AoPS and do old AMC's until you can consistently get at least 5 on AIME's, and then go to AIME's.

You could also improve your problem solving skills by trying to generalize everything you solve, if possible.  That'll make you understand the material better.

I think that this: [url]http://www.artofproblemsolving.com/Forum/resources.php?c=182[/url]

is probably what you're looking for.

And, yes, the best way to improve your problem-solving skills is by solving problems.

If your goal is to make USAMO and you have a relatively low score on the AIME, I'd advise that you work on problems 21-25 on any AMC 12 and on problems 5-12 on any AIME.

[quote="cooljoe"]

Also, is it true that they've changed the blank answer points from 2.5 to 1.5 this year on the AMC 12?[/quote]

This is the only question which I can answer.   :P The answer is yes.  This means that you need to get 14 correct and 0 wrong to qualify with an 100.5.  Previously you could qualify with 11 correct, 0 wrong.

I think AoPS Vol. II is useful on the AIME to an extent, but it isn't the panacea for lousy scores on the AIME. Doing past and mock tests helps more, I think.

[quote="cooljoe"]I have a question for you naturals: Do solutions really come to you naturally, or is it usually because you've seen a certain idea in a previous problem?

For example, Mr. Patrick, and others, do the solutions to the AIME problems during Math Jams come to you immediately?

Also, is it true that they've changed the blank answer points from 2.5 to 1.5 this year on the AMC 12?[/quote]
Once you do enough problems, you will see the same themes being repeated over and over. It takes a lot of practice before you reach that level, but it is a great feeling to be able to get through the first 2/3 of an AMC without any sweat - "yeah, I saw something like this before!"

The other things that don't come naturally, you just fiddle around with until you notice a pattern that you've seen before, and then you can solve it.

For this reason, when you are doing practice tests, if you must look at the solution for one of them, don't just look at it -- be so familiar with it that you are able to show the solution by heart to someone else. Then it will show that you will be able to use a similar idea on a similar problem that may come out during the actual test.

So do most people here (like you) post their solutions for really hard problems copious experimentation? Is it just me or does Mr. Patrick instantly generate solutions for intractable problems during the AMC/AIME Math Jams?

So my theory is that you MOSPers and such have reached a level where the average 'problem' now looks like an exercise.

Btw, I got 100.5 on AMC 12B :(

bump...

I keep doing problems, and my problem solving isn't getting any beter (refer to above post)!

Strangely and funnily enough, this year I scored a 0 on the AIME--both of my scores DROPPED from last year.

Now a senior, my son has a most satisfying journey in math problem solving. Here's what have worked for him:

[u]MOST HELPFUL:[/u]
The AoPS classes my son took were definitely his shortcuts to major improvement. The short and sweet Special AIME Problem Seminar probably helped him solidify his path to USAMO (and Red MOP) in 9th grade.

[u]QUITE HELPFUL:[/u]
Attending math circles

[u]ALSO HELPFUL:[/u]
Working old problems and solutions

[u]HELPFUL WHEN HE USED THEM:[/u]
As far as books go, he has only worked with AoPS Vol. 1 & 2 in the early going, then later the Zeitz and Engel books as assigned by AoPS classes.

Yeah, the AIME Seminar this year probably raised my score about 2 points.
And I'm taking WOOT next year, so that should definitely help.  :P

Personally, I found this link helpful for old contests (AMC/AIME).

http://www.math.ksu.edu/main/handbook/ProblemSets

Also, beware that some of the files are a couple of megabytes (although not many), so some files may download slowly on dialup.[/youtube]


calculus
integration
trigonometry
calculus computations

$ \int_{\theta_1}^{\theta_2}\frac{1}{1\plus{}\tan\theta} d\theta$

where $ \theta_2\equal{} \frac{1003\pi}{2008} ,\theta_1\equal{} \frac{\pi}{2008}$
:D

[quote="kunny"]Let $ f(\theta) = \frac {1}{1 + \tan \theta}\ d\theta$,

$ \int_{\theta _1}^{\theta _2} f(\theta)\ d\theta = \int_{\theta _1}^{\theta _2} f\left(\frac {\pi}{2} - \theta \right)\ d\theta$.

Since $ \int_{\theta _1}^{\theta _2} f(\theta)\ d\theta + \int_{\theta _1}^{\theta _2} f\left(\frac {\pi}{2} - \theta \right)\ d\theta = 1$.

$ \therefore \int_{\theta _1}^{\theta _2} f(\theta)\ d\theta = \frac {1}{2}$. :lol:[/quote]

$ \boxed{\int_{\theta _1}^{\theta _2} f(\theta)\ d\theta = \int_{\theta _1}^{\theta _2} f\left(\frac {\pi}{2} - \theta \right)\ d\theta=1}$.
I think it might be not true.[color=red]corrections[/color]
On the other hand,

$ f(\theta) + f\left(\frac {\pi}{2} - \theta \right)=1$. :)

[hide="ans."]
$ \frac {501\pi}{2008}$
[/hide]

$ \boxed{\int_{\theta _1}^{\theta _2} f(\theta _1 \plus{} \theta _2 \minus{} \theta)\ d\theta \equal{} \int_{\theta _1}^{\theta _2} f(\theta)\ d\theta }$

[quote="kunny"][quote="vinskman"]$ \int_{\theta_1}^{\theta_2}\frac {1}{1 \plus{} \tan\theta} d\theta$

where $ \theta_2 \equal{} \frac {1003\pi}{2008} ,\theta_1 \equal{} \frac {\pi}{2008}$
:D[/quote]

Let $ f(\theta) \equal{} \frac {1}{1 \plus{} \tan \theta}\ d\theta$, we have $ \int_{\theta _1}^{\theta _2} f(\theta)\ d\theta \equal{} \int_{\theta _1}^{\theta _2} f\left(\frac {\pi}{2} \minus{} \theta \right)\ d\theta$. Since $ \int_{\theta _1}^{\theta _2} f(\theta)\ d\theta \plus{} \int_{\theta _1}^{\theta _2} f\left(\frac {\pi}{2} \minus{} \theta \right)\ d\theta \equal{} \theta _2 \minus{} \theta _1 \equal{} \frac {1002}{2008}\pi$,

$ \therefore \int_{\theta _1}^{\theta _2} f(\theta)\ d\theta \equal{} \boxed{\frac {501}{2008}\pi}$. :lol:[/quote]


inequalities
linear algebra
matrix
linear algebra solved

If we have 2 square matrices $X$ and $Y$ of order $n$ then $rank(X)+rank(Y) - n <=rank(XY)$.
I suppose it must be ease for people that are good at linear algebra, but I don't know hot to prove it. Please Help!!! :?

First of all, we can find invertible $P_X,P_Y,Q_X,Q_Y$ s.t. $P_X\cdot X\cdot Q_X=I_X$ and $P_Y\cdot Y\cdot Q_Y=I_Y$, where, in general, for a matrix $Z$, $I_Z$ is the matrix having $1$ on the first $r(Z)$ positions of the main diagonal and $0$ everywhere else.

After using that we are left with this: $r(X)+r(Y)-n\le r(I_X\cdot U\cdot I_Y)$, where $U$ is some invertible matrix. $I_X\cdot U\cdot I_Y$ is the matrix having the corresponding elements of $U$ at the intersections of the first $r(X)$ lines and $r(Y)$ columns, and $0$ everywhere else. We must show that its rank is at least $r(X)+r(Y)-n$. This isn't hard at all. Just draw a picture.

Hi, grobber! Sorry, but I think I don't understand something.
From $P_X\cdot X\cdot Q_X = I_X$ we get $X=P_X^{-1}\cdot I_X\cdot Q_X^{-1} $. Same thing for $Y$. Why then does it reduces to that expression when we multiply them ? Or is it because $r(AX)=r(X)$ if A is invertible??? :?

Yup, you got it. The matrix in the RHS is $A\cdot I_X\cdot BV\cdot A'\cdot I_Y\cdot B'$ for some invertible matrices $A,B,A',B'$. We can disregard $A,B'$ because of the property you have already mentioned, and we make the notation $U=BA'$. We're left with $I_X\cdot U\cdot I_Y$.


geometry
circumcircle
trigonometry
ratio
power of a point
radical axis
cyclic quadrilateral

1) Let $ M$ be a point inside the $ \triangle ABC$. The line $ l_a$ is perpendicular to $ MA$ at $ M$. $ l_a$ cuts line $ BC$ at $ A_1$. We define $ B_1,C_1$ similarly. Prove that $ A_1,B_1,C_1$ are collinear.

2) Let $ M$ be a point inside the $ \triangle ABC$. Denote by $ H$ the orthocenter of $ \triangle ABC$. The line $ d_a$ passes through $ H$ and is perpendicular to $ MA$. $ d_a$ cuts $ BC$ at $ A_0$. We define $ B_0,C_0$ similarly. Prove that $ A_0,B_0,C_0$ are collinear.

[quote="KDS"]1) Let $ M$ be a point inside the $ \triangle ABC.$ The line $ l_a$ is perpendicular to $ MA$ at $ M.$ $ l_a$ cuts line $ BC$ at $ A_1.$ We define $ B_1,C_1$ similarly. Prove that $ A_1,B_1,C_1$ are collinear.[/quote]
Let $ H$ be the orthocenter of $ \triangle ABC$ and $ H_a,H_b,H_c$ the feet of the altitudes onto $ BC,$ $CA,$ $AB.$ Circles $ (A') \equiv \odot(MAA_1),$ $ (B') \equiv \odot(MBB_1)$ and  $ (C') \equiv \odot(MCC_1)$ with diameters $ AA_1,BB_1,CC_1$ pass through $ H_a,H_b,H_c,$ respectively and since $ AH \cdot HH_a \equal{} BH \cdot HH_b \equal{} CH \cdot HH_c,$ it follows that $ (A'),(B'),(C')$ are coaxal with common radical axis the straight line $ HM$ $ \Longrightarrow$ $ A',B',C'$ lie on a perpendicular to $ HM.$ Since $ A',B',C'$ are the midpoints of $ AA_1,BB_1,CC_1,$ then $ A',B',C'$ lie on the sidelines $ EF,FD,DE$ of the medial triangle $\triangle DEF$ of $ \triangle ABC,$ respectively. Therefore

$ \frac {\overline{A'F}}{\overline{A'E}} \equal{} \frac {\overline{A_1B}}{\overline{A_1C}} \ , \ \frac {\overline{B'D}}{\overline{B'F}} \equal{} \frac {\overline{B_1C}}{\overline{B_1A}} \ , \ \frac {\overline{C'E}}{\overline{C'D}} \equal{} \frac {\overline{C_1A}}{\overline{C_1B}}$

$ \frac {\overline{A_1B}}{\overline{A_1C}} \cdot \frac {\overline{B_1C}}{\overline{B_1A}} \cdot \frac {\overline{C_1A}}{\overline{C_1B}} \equal{} \frac {\overline{A'F}}{\overline{A'E}} \cdot \frac {\overline{C'E}}{\overline{C'D}} \cdot \frac {\overline{B'D}}{\overline{B'F}} \equal{} 1$

By the converse of Menelaus' theorem, the points $ A_1,B_1,C_1$ are collinear.

[quote="KDS"]2) Let $ M$ be a point inside the $ \triangle ABC.$ Denote by $ H$ the orthocenter of $ \triangle ABC.$ The line $ d_a$ passes through $ H$ and is perpendicular to $ MA.$ $ d_a$ cuts $ BC$ at $ A_0.$ We define $ B_0,C_0$ similarly. Prove that $ A_0,B_0,C_0$ are collinear.[/quote]
Let $ X \equiv d_a \cap MA$ and define cyclically $ Y,Z.$ Using the same notations above, since $ AXH_aA_0$ is a cyclic quadrilateral on account of the right angles at $ X$ and $ H_a,$ we have $ \overline{HX} \cdot \overline{HA_0} \equal{} \overline{AH} \cdot \overline{HH_a}$ and by cyclic exchange:

$ \overline{HY} \cdot \overline{HB_0} \equal{} \overline{BH} \cdot \overline{HH_b} \ \ , \ \ \overline{HZ} \cdot \overline{HC_0} \equal{} \overline{CH} \cdot \overline{HH_c}$

$ \Longrightarrow \overline{HX} \cdot \overline{HA_0} \equal{} \overline{HY} \cdot \overline{HB_0} \equal{} \overline{HZ} \cdot \overline{HC_0} \equal{} k^2$

Since $ X,Y,Z$ lie on the circle with diameter $\overline { HM},$ it follows that $ A_0,B_0,C_0$ lie on the inverse line of the circle with diameter $\overline{ HM}$ under the negative inversion with center $ H$ and radius $ k.$

[quote]1)Let $ M$ be a point inside the $ \triangle ABC$. The line $ l_a$ is perpendicular to $ MA$ at $ M$. $ l_a$ cuts line $ BC$ at $ A_1$. We define $ B_1,C_1$ similarly. Prove that $ A_1,B_1,C_1$ are collinear.[/quote]
[i]Solution[/i]

Consider the inversion through pole $ M$, with any power $ k$. Let $ A',B',C'$ and $ A'_1,B'_1,C'_1$ respectively be the image of $ A,B,C$ and $ A_1,B_1,C_1$ through $ \mathcal {I}(I,k)$. Therefore, we have $ \mathcal {I}(I,k):$ $ \overline {ABC_1}\mapsto (A'B'C'_1M)$, $ \overline {ACB_1}\mapsto (A'C'B'_1M)$ and $ \overline {BCA_1}\mapsto (B'C'A'_1M)$. Then $ MC'\perp MC'_1$, $ MB'\perp MB'_1$ and $ MA'\perp MA'_1$. We claim that $ M,B'_1,C'_1,A'_1$ are concyclic.

Indeed, let $ O_1,O_2,O_3$ respectively be the circumcenters of $ (A'C'B'_1M)$, $ (B'C'A'_1M)$ and $ (A'B'C'_1M)$. Since $ O_2O_3\perp MB'$, $ MB'_1\perp MB'$, thus $ O_2O_3\| MB'_1$. Therefore, the perpendicular bisector of $ MB'_1$ will be the line $ d_1$ passes $ O_1$ and $ d_1\perp O_2O_3$. With the same argument for $ O_1O_2$ and $ O_1O_3$. As the consequence, $ d_1,d_2,d_3$ repsectively will be the perpendiculars from $ O_1,O_2,O_3$ wrt $ \triangle O_1O_2O_3$, obviously, they are concurrent at the orthocenter of $ \triangle O_1O_2O_3$, which implies that $ M,B'_1,A'_1,C'_1$ are concyclic. Therefore, $ A_1,B_1,C_1$ are collinear. $ \square$

problem 1)
in fact we have $ \frac{BA_1}{A_1C}\equal{}\frac{MB}{MC} . \frac{\sin \angle BMA_1}{\sin \angle CMA_1}$

if we write the same thing for the other ratios and then multiply them (menelaus ' theorem)
we'll see that each of the sides $ MA,MB,MC$ will appear one time as the denominator and another time as the numerator so we have to prove the equality which is consisted of some sinuses of the angles,this part is the same as the previous one, i mean by using the equalities :
$ \sin \angle BMA_1\equal{} \sin \angle AMB_1$ and similarly for the other four angles for which their sin become equal with one of the other and u can see that in this way we'll obtain what we wanted.
so we are done.
problem 2)
let $ d_a \cap AM \equal{} M_a$ and similarly for the other lines.
for this part we do the same thing for computing the ratio $ \frac{CA_0}{BA_0}$ in fact we have

$ \frac{CA_0}{BA_0}\equal{} \frac{CH}{BH} . \frac{\sin \angle CHA_0}{\sin \angle BHA_0}$

by the same reasoning for the problem one and by using the equality (use the fact that $ AFHM_aE$ is cyclic wher $ F,E$ are the feet of altitudes)
$ \sin \angle BHA_0 \equal{}  \sin \angle M_aAC\equal{} \sin \angle MAC \ \ ; \ \ \sin \angle CHA_0\equal{} \sin \angle M_aAB\equal{}\sin \angle MAB$

and similarly for the other angles we must prove that :
$ \frac{ \sin \angle MAB}{\sin \angle MAC} . \frac{\sin \angle CBM}{\sin \angle MBA} . \frac{\sin \angle MCA}{\sin \angle MCB} \equal{} 1$

which is obvious since the lines $ AM,BM,CM$ are concurrent at $ M$ (ceva's theorem,sinus form)
so we are done!
in fact in this way we have also proven this (the converse statement):
suppose that $ A_1,B_1,C_1$ are some points on $ BC,CA,AB$ respectively.
let the perpendicular from $ H$ to $ AA_1$ meet $ BC$ at $ A_0$ and similarly for the other sides.
then prove that $ A_0,B_0,C_0$ are collinear if and only if $ AA_1,BB_1,CC_1$ are concurrent.
here is an added problem similar to problem 1:
added problem (it may be posted before):
let $ l$ be a line which intersects the sides of $ ABC$ at $ A_1,B_1,C_1$ respectively.
consider a point $ P$ on the plane of $ ABC$ and let the perpendicular from $ P$ to $ PA_1$ intersects $ l$ at $ A_2$.
define similarly $ B_2,C_2$.
prove the concurrency of $ AA_2,BB_2,CC_2$.
[hide="hint"]
try to find the ratios $ \frac{\sin \angle C_1BB_2}{\sin \angle B_2BA_1}$ and $ \frac{C_1B_2}{B_2A_1}$ in an appropiate form ! and the rest is the same as above!
[/hide]

[b]Problem 2:[/b] Denote $ A'', B'', C''$ the projections of $ A,B,C$ on $ BC, CA, AB$, respectively. $ C'H\cap CM at I$.
We have $ C'M^2 \minus{} HM^2 \equal{} C'I^2 \minus{} HI^2 \equal{} 2C'H.HI \plus{} C'H^2$
$ \Rightarrow C'M^2 \minus{} C'H^2 \equal{} HM^2 \plus{} 2C'H.HI \equal{} HM^2 \plus{} 2HC.HC''$ (because $ C'C''IC$ is cyclic)
Similarly $ A'M^2 \minus{} A'H^2 \equal{} HM^2 \plus{} 2HA.HA'', B'M^2 \minus{} B'H^2 \equal{} HM^2 \plus{} 2HB.HB''$.
Note that $ HA.HA'' \equal{} HB.HB'' \equal{} HC.HC''$
$ \Rightarrow A'M^2 \minus{} A'H^2 \equal{} B'M^2 \minus{} B'H^2 \equal{} C'M^2 \minus{} C'H^2$.
So $ A',B',C'$ lie on a line which is perpendicular to $ HM$.


I keep seeing recursion problems in this forum, and I don't quite get them. For example, the Fibonnaci Series. The general term is something quite complicated involving the golden ratio, and it must be proved with induction. Any hints/books/sites?

recursion problems are basically formulas where the current term is expressed as a relationship of previous terms... kinda confusing but stare at it long enough and you should be able to get it, that's what i did

[quote="vishalarul"]and it must be proved with induction. [/quote]

Certainly not.  The explicit formula can be derived in one go using the theory of generating functions.

You can find a [b]lot[/b] of posts on this subject with the Search function.

Here a recursion problem..read my post about $Induction$...

$F_{2n}=F_{n}*L_{n}$

[quote="t0rajir0u"][quote="vishalarul"]and it must be proved with induction. [/quote]
Certainly not.  The explicit formula can be derived in one go using the theory of generating functions.
[/quote]
That would be killing flies with cannonballs. The easiest standard proof is using induction:
If $x^{2}=x+1$ then by induction $x^{i}=f_{i}x+f_{i-1}$, where $f_{i}$ is the $i$-th number in Fibonacci's series.
Since there are two solutions to $x^{2}=x+1$, $x=\phi$ and $x=\phi'$, we have a system:
$\phi^{i}=f_{i}\phi+f_{i-1}$,
$\phi'^{i}=f_{i}\phi'+f_{i-1}$,
we can solve it for $f_{i}$: $f_{i}=\frac{\phi^{i}-\phi'^{i}}{\phi-\phi'}$.

The thing is, induction is non-constructive. Suppose someone told you to find the formula for say $a_{i}$ where $a_{0}=a_{1}=1$ and $a_{n}=2a_{1}+2a_{2}$ for $n>1$. You don't have the formula already so you can't induct. This goes well with generating functions and, in this case, linear algebra.


[quote="Inequalities Master"]Let $ a,b,c > 0$. Prove that:

$ \sqrt {\dfrac{a \plus{} b}{c} \plus{} \dfrac{b \plus{} c}{a} \plus{} \dfrac{c \plus{} a}{b}} \plus{} 2\cdot \sqrt {\dfrac{ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}}\ge \sqrt {6} \plus{} 2$[/quote]
It follows from $ \frac{ab\plus{}ac\plus{}bc}{a^2\plus{}b^2\plus{}c^2}\cdot\sum_{cyc}\frac{a\plus{}b}{c}\geq6.$ :wink:

The soln had alreadyy been posted. i didn't see.

[quote="arqady"][quote="Inequalities Master"]Let $ a,b,c > 0$. Prove that:

$ \sqrt {\dfrac{a \plus{} b}{c} \plus{} \dfrac{b \plus{} c}{a} \plus{} \dfrac{c \plus{} a}{b}} \plus{} 2\cdot \sqrt {\dfrac{ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}}\ge \sqrt {6} \plus{} 2$[/quote]
It follows from $ \frac {ab \plus{} ac \plus{} bc}{a^2 \plus{} b^2 \plus{} c^2}\cdot\sum_{cyc}\frac {a \plus{} b}{c}\geq6.$ :wink:[/quote]
My argument is true because the inequality $ \frac {ab \plus{} ac \plus{} bc}{a^2 \plus{} b^2 \plus{} c^2}\cdot\sum_{cyc}\frac {a \plus{} b}{c}\geq6$ is wrong.
I'll try to prove it more.

It's an old problem of Nguyen Anh Cuong.
I tried to solve it but I think it's not easy to find a nice solution  :wink:

[quote=Inequalities Master]Let $ a,b,c>0$. Prove that:

$ \sqrt{\dfrac{a\plus{}b}{c}\plus{}\dfrac{b\plus{}c}{a}\plus{}\dfrac{c\plus{}a}{b}}\plus{}2\cdot \sqrt{\dfrac{ab\plus{}bc\plus{}ca}{a^2\plus{}b^2\plus{}c^2}}\ge \sqrt{6}\plus{}2$[/quote]
By [url=http://www.artofproblemsolving.com/community/c6h278791]uvw[/url] it's enough to check one case only $b=c=1$, which is easy.


analytic geometry
trigonometry
geometry
circumcircle
rotation
graphing lines
slope

Let $ k$ be a circle with radius $ r$, let $ A$ be any point on $ k$, and let $ t$ be the tangent line to $ k$ at $ A$. Let $ B$ and $ C$ be points on $ t$ on opposite sides of $ A$ such that $ AB=mr$ and
$ AC=nr$ for some positive real numbers $ m$ and $ n$. Let $ P$ be any point of $ k$ different from $ A$. Show that $ \cot\angle APB+\cot\angle APC$ is a constant for all such points $ P$, and determine this constant value in terms of $ m$ and $ n$.

We apply a system of coordinates with $ A$ as origin and $ O(r,0)$

Let $ (A)$ be the circle centered at $ A$ with radius $ 2r$
$ B$ is at positive $ y$-axis and $ C$ at negative

Let $ F(2r,0)$. Let $ \theta = \angle FAP$ such that $ \theta \in \left(-\frac{\pi}{2},\frac{\pi}{2}\right)$. The ray $ AP$ intersects the line $ \boxed{FF: x=2r}$ at $ P_{2}$. Then $ P_{2}(2r,2r\cdot \tan\theta)$

The circle $ (B)$ centered at $ B$ and passing through $ A$ intersects $ (A)$ at $ L(x_{L},y_{L})$

$ L\in(B)\cap k \Rightarrow \left\|\begin{array}{l}x_{L}^{2}+(y_{L}-mr)^{2}= m^{2}r^{2}\\ (x_{L}-r)^{2}+y_{L}^{2}= r^{2}\end{array}\right\}\Rightarrow x_{L}=m\cdot y_{L}$

Setting $ M=\frac{1}{m}$ we get $ y_{L}= Mx\ \ (1)$

The line $ AL: \ y=Mx$ intersects $ FF$ at $ B_{0}$. Then $ B_{0}\left(2r,2rM\right)$

[img]10006[/img]

Now let $ B_{1}$ be the 2nd intersection point between $ BP$ and $ k$. Clearly $ B_{1},P$ are inverse wrt $ (B)$
Let $ B_{2}$ be the inverse of $ B_{1}$ wrt $ (A)$. Of course $ B_{2}\in FF$ (because $ FF$ and $ k$ are inverse wrt $ (A)$)
We have $ AB_{1}\cdot AB_{2}= (2r)^{2}= AP \cdot AP_{2}\Rightarrow PB_{1}B_{2}P_{2}$ is cyclic, let $ W_{B}$ be its circumcenter
$ \hline$

Let $ \mathbb{P}_{q}X$ denote the power of the point $ X$ wrt the circle $ q$

$ \mathbb{P}_{(W_{B})}A=AP\cdot AP_{2}= AF^{2}\Rightarrow (W_{B})\perp (A)$
$ \mathbb{P}_{(W_{B})}B=BB_{1}\cdot BP = BA^{2}\Rightarrow (W_{B})\perp (B)$

So we have

$ \mathbb{P}_{(A)}W_{B}= \mathbb{P}_{(B)}W_{B}\Rightarrow W_{B}$ is on the radical axis of $ (A)$ and $ (B)$
But the circles $ (A),(B)$ are fixed, so their radical axis is also fixed. We can easily find that, the radical axis is the line $ y=\frac{2r}{m}=2rM$

So $ B_{0}W_{B}\perp B_{2}P_{2}$ and this means that $ B_{0}$ is the midpoint of $ B_{2}P_{2}$
$ \hline$

So we have
$ P(2r,\ 2r\cdot\tan\theta)$
$ B_{0}(2r,2rM)$ and consequently
$ B_{2}(2r,\ 2r(2M-\tan\theta))$

$ AB$ is tangent to $ k$, so $ \angle APB = \angle B_{2}AB \Rightarrow \cot(\angle APB) = \cot(\angle B_{2}AB)$ for any position of $ P$

If $ 2M>\tan\theta$ then the $ y$-coordinate of $ B_{2}$ is positive and $ \cot(\angle B_{2}AB) = \tan(\angle B_{2}AF) = \frac{B_{2}F}{2r}= 2M-\tan\theta$

If $ 2M<\tan\theta$ then $ y<0$ and $ \angle B_{2}AB= \angle B_{2}AF+90^{o}\Rightarrow$

$ \cot(\angle B_{2}AB) =-\tan(\angle B_{2}AF)=-\frac{|B_{2}F|}{2r}= 2M-\tan\theta$

So it is always $ \boxed{\cot(\angle B_{2}AB) = 2M-\tan\theta}$

[color=red](* We can see the diagram to be rotated and inversed such that $ A$ remains at the origin, $ F$ goes to $ (0,2r)$ and $ B$ to the positive x-axis, and then it is clear, that the line $ FF$ is the axis of the cotangents)[/color]
$ \hline$

We define the respective points $ C_{1},C_{2},W_{C}$ for the point $ C(0,-nr)$
Now the circle $ (C)=(C,nr)$ intersects $ k$ at $ Q$ which satisfies the equation $ y=-Nx$, where $ N=\frac{1}{n}$

Then $ W_{C}$ is the point $ (2r,-2rN)$. Again we show that $ W_{C}$ is the midpoint of $ P_{2}C_{2}$

So it is $ C_{2}(2r,-2r(2N+\tan\theta))$

[img]10007[/img]

Again $ \cot(\angle APC) = \cot(\angle CAC_{2})$ but it is $ \frac{-C_{2,y}}{AF}$, where $ C_{2,y}$ is the y-coordinate of $ C_{2}$ (rotate the diagram 90 degrees left to see it)

So we have:

$ \cot(\angle CAC_{2}) = 2r\frac{2N+\tan\theta}{2r}= 2N+\tan\theta\Rightarrow \boxed{\cot(\angle CAC_{2})=2N+\tan\theta}$

and finally $ \cot\angle APB+\cot\angle APC = 2M+2N = 2\left(\frac{1}{m}+\frac{1}{n}\right)=$
[color=red]Correction[/color]$ = 2\frac{m+n}{mn}$

It took you a lot of time to writ these and thank you very much for this solution.

[quote="delegat"][color=darkred]Let $ A$ be a fixed point on the given circle $ k=\mathcal C(r)$. Let $ B$ and $ C$ be two fixed points on the tangent $ t=AA$ such that $ A\in (BC)$ and $ AB=mr$ and $ AC=nr$, where $ m$, $ n$ are given positive real numbers. Let $ P\ne A$ be a mobile point of $ k$. Show that $ \cot\widehat{APB}+\cot\widehat{ APC}$ is constant.[/color][/quote] [color=darkblue][b]Proof I (synthetic).[/b] Define the second intersections $ U$, $ V$ of the circle $ k$ with the lines $ PB$, $ PC$ respectively. Denote $ \{\begin{array}{c}m(\widehat{APB})=x\ ,\ m(\widehat{APC})=y\\\ m(\widehat{AUB})=\phi\ ,\ m(\widehat{AVC})=\psi\end{array}$.

Observe that $ \phi+\psi=180^{\circ}$, i.e. $ \sin\phi=\sin\psi$. Suppose w.l.o.g. that $ r=\frac{1}{2}$, $ m=2b$, $ n=2c$. Thus, $ AB=b$, $ AC=c$, $ AU=\sin x$, $ AV=\sin y$.

Apply the [b]Sinus' theorem[/b] : $ \{\begin{array}{ccc}\triangle ABU\ : & \frac{BU}{\sin x}=\frac{AB}{\sin \phi}\\\\ \triangle ACV\ : & \frac{CV}{\sin y}=\frac{AC}{\sin\psi}\end{array}$ $ \implies$ $ AB\cdot CV\cdot\sin x=AC\cdot BU\cdot \sin y\ \ (1)$.

Apply the [b]generalized Pythagoras' theorem[/b] : $ \{\begin{array}{ccc}\triangle ABU\ : & BU^{2}=b^{2}+\sin^{2}x-2b\sin x\cos x=\sin^{2}x\cdot[b^{2}+(b\cdot\cot x-1)^{2}]\\\\ \triangle ACV\: & CV^{2}=c^{2}+\sin^{2}y-2c\sin y\cos y=\sin^{2}y\cdot[c^{2}+(c\cdot\cot y-1)^{2}]\end{array}$.

Therefore, the relation $ (1)$ becomes : $ b^{2}\cdot\sin^{2}x\cdot\sin^{2}y\cdot[c^{2}+(c\cdot\cot y-1)^{2}]=c^{2}\cdot\sin^{2}y\cdot\sin^{2}x\cdot[b^{2}+(b\cdot\cot x-1)^{2}]$ $ \Longleftrightarrow$

$ |b\cdot (c\cdot\cot y-1)|=|c\cdot (b\cdot\cot x-1)|$ $ \Longleftrightarrow$ $ \{\begin{array}{c}\cot x-\cot y=\frac{c-b}{bc}\\\\ \mathrm{OR}\\\\ \cot x+\cot y=\frac{b+c}{bc}\end{array}$, i.e. $ \boxed{\cot x-\cot y=\frac{2(n-m)}{mn}\ \ \vee\ \ \cot x+\cot y=\frac{2(m+n)}{mn}}$.

[b]Proof II (analytic).[/b] Suppose w.l.o.g. that the circle is $ \mathcal C(O,1)$, where the center $ O$ is the origin of the analytical system and $ \overline{BAC}\parallel Ox$.

Therefore, $ \{\begin{array}{c}P(\cos\phi ,\sin \phi )\\\\ A(0,-1)\ ,\ B(-m,-1)\ ,\ C(n,-1)\end{array}$ and the slopes of the lines $ PA$, $ PB$, $ PC$ are the following :

 $ \{\begin{array}{c}s_{PA}=\tan\alpha =\frac{\sin\phi+1}{\cos\phi}\\\\ s_{PB}=\tan\beta =\frac{\sin\phi+1}{\cos\phi+m}\\\\ s_{PC}=\tan\gamma=\frac{\sin\phi}{\cos\phi-n}\end{array}$. Since $ \{\begin{array}{c}x=\alpha-\beta\\\ y=\gamma-\alpha\end{array}$ obtain : $ \{\begin{array}{c}\tan x=\frac{\tan\alpha-\tan\beta}{1+\tan\alpha \tan\beta}=\frac{m(1+\sin\phi )}{2+m\cos\phi+2\sin\phi}\\\\ \tan y=\frac{\tan\gamma-\tan \alpha}{1+\tan\gamma\tan\alpha}=\frac{n(1+\sin\phi)}{2-n\cos\phi+2\sin\phi}\end{array}\|$ $ \implies$

$ \cot x+\cot y=$ $ \frac{2+m\cos\phi+2\sin \phi}{m(1+\sin\phi )}+\frac{2-n\cos\phi+2\sin\phi}{n(1+\sin\phi)}=$ $ \frac{2(m+n)(1+\sin\phi )}{mn(1+\sin\phi )}=$ $ \frac{2(m+n)}{mn}$ $ \implies$ $ \boxed{\cot x+\cot y=\frac{2(m+n)}{mn}}$.

[b]Remark.[/b] $ \frac{UV\cdot 2r}{AU\cdot AV}=\frac{2r\sin (x+y)\cdot 2r}{2r\sin x\cdot 2r\sin y}=\cot x+\cot y$. Therefore,  the conclusion is equivalently with "[b]the ratio $ \boxed{\frac{UV}{AU\cdot AV}}$ is constant[/b]". [/color]

Most probably problem is not well defined.
Maybe result depends on position of point $ P$ so sometimes it is $ +$ and sometimes $ -$ sign :?:


Hi I'm trying to set up Aysmptote on my computer, but I'm having some problems. I managed to get it to work, but now I'm trying to get it to work with the TeXnicCenter, which is where I'm stuck. I'm using this guide: http://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_Getting_Started/Windows/Interactive_Mode

I did everything it said but when I run the Asympote tool, I get the DOS-like screen giving me an error. It says

[code]
Cannot execute 'latex' \scrollmode
Please put in config.asy:

import settings;
textpath="PATH";

where PATH denotes the correct path to the directory containing your latex engine (latex).

Alternatively, set the environment variable ASYMPTOTE_TEXPATH or use the command line option -textpath="PATH"
[/code]

I've looked for this config.asy, but I couldn't find it, and even if I did, I'm not quite sure exactly what I need to put in it. Can anybody help me?

I'm assuming you are using MikTeX on Windows XP as you don't say.

1. Is TexnicCenter able to compile normal latex documents?

2. Go to Start, Run then type cmd and press OK. This will give you a DOS box. Type [i]path[/i] then press Enter. In amongst the mess is there something like c:\program files\miktex 2.5\bin? If not, that means miktex hasn't told windows where it should be found.

This can be changed but I am reluctant to advise this as it is too easy to make a mistake. So we'll do as Asymptote says and alter config.asy. It can be found in C:\Documents and Settings\[i]your_name[/i]\.asy. Make a copy of it and then open it in Notepad. Add the line
texpath="c:\program files\miktex 2.5\bin";
using whatever your path is.

Okay I did that; then when I tried it again it began to output a bunch of stuff like it was working. But then it closed without showing anything. Then I couldn't even get it to do that again and it seems to be unable to execute something called "dvips". I think it's unable to find that, but I don't know where it is either...

Can you tell me if
1. Texniccenter compiles ordinary tex files correctly?
2. Go to Start, Run then type cmd and press OK. This will give you a DOS box. Type path then press Enter. Is there anything about miktex?
3. Is Ghostscript installed?

1. Regular LaTeX works fine
2. No, there is nothing about MiKTeX. I added the path to config.asy but now there's another error.
3. Yes I have Ghostscript.

[quote="tjhance"]2. No, there is nothing about MiKTeX.[/quote]That's the source of all your problems. You can add it to the path if you look up how to do it (it's called Environmental Variable) but I will give you an alternative risk-free method which will add this info when you run asymptote.
Follow the instructions at [url=http://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_Advanced_Configuration#Showing_Asymptote_error_messages_in_TeXnicCenter]Showing Asymptote error messages in TeXnicCenter[/url]. But in asymptote.bat after the line [code]echo Asymptote is working ... please wait[/code]add the lines
[code]set opath=%path%
set path=%path%;"c:\program files\miktex 2.5\bin"[/code]
And on the line after [code]:end [/code]add
[code]set path=%opath%
set opath=[/code]I hope this will work.

Hm well okay so I did that, and now I've seemed to get rid of this dvips error, but there's still nothing. When I run the Asymptote tool from TeXnicCenter, the black DOS screen comes up and instantly disappears without displaying anything. I do have the path in config.asy and stuff and I'm not sure what's wrong now.

I really have no clue what I'm doing.

You are, I hope, using asymptote on file called somefile.asy NOT somefile.tex I hope?
Put
pause
at the end of the aymptote.bat file and the DOS window shouldn't close until you press a key.
If that doesn't help delete the first line @echo off from asymptote.bat then see what errors it gives you.

My file is called "test.asy"

Okay I'm getting output now... but still nothing that looks like an error. Here's a screenshot of my output.

I still don't quite understand how this tool is supposed to work...

Believe it or not that's looking good!
1. Is there a test_asy.log and is there anything in it (I suspect it's empty)?
2. In asymptote.bat can you make a couple of changes (my fault :oops: )
a) remove the quotes from c:\program files\miktex 2.5\bin (are you absolutely sure this is the correct path to MiKTeX - it should contain lot of files including latex.exe, please double-check)
b) Move
set path=%opath%
set opath=
to after :end

If still no joy try saving your test.asy into a different directory which has no spaces in it and open it in TeXnicCenter from there.

Okay I made your changes; first of all, no, i wasn't using the correct path; it's actually c:\program files\MiKTex 2.5\miktex\bin. I've been aware of that for a while, but I must've forgotten to change it when I made the .bat file. I made the other changes, with the quotes and stuff. I also tried a different directory.

It appears to be working but it still isn't producing an image  :(

Did you save test.asy inside TexnicCenter? If so, are you sure it hasn't been saved as test.asy.tex?
Is there a test_asy.log file?

Actually, I don't think test.asy was a pure asy file... the file appears with the TeXNicCenter icon. I created a new file, called supertest.asy, outside of TeXNicCenter to make sure it was just .asy, but it still doesn't work :wallbash:

As for the test_asy.log file (or supertest_asy.log), where is it supposed to be? If it is supposed to appear in the same folder as test.asy and supertest.asy, then I see files called "test_asy" and "supertest_asy" there, but I don't see the .log extension. They appear to be plain text files, and are both empty.

Looks like it's all Microsoft's fault - you haven't got extensions turned on so you can't see what they are or are fooled by double extensions.

In My Computer, Tools, Folder Options, View, Untick the box that says Hide extensions for known file types, OK out and look again. I bet you find supertest_asy.log and also supertest.asy.tex which you will need to rename to supertest.asy.

wow i can see extensions now... that was bugging me for so long before i even started this.....  :lol:

okay well there's test.asy.tex and supertest.asy. Then there's test_asy.log and supertest_asy.log, which are both empty.

Should all work now and using test.asy you should get test.pdf. You should also be able to use asymptote code inode a tex file as well.

Once it's working add @echo off back on the first line of asymptote.bat and get rid of pause at the end. Then when you call asymptote from TeXnicCenter you'll get a window with a message which will close automatically and any errors will be found in the test_asy.log file. If there are any errors, test_asy.log will open automatically in TexnicCenter.

When I open the pdf it says "the file was damaged and could not be repaired". And I'm not getting and pdf from the supertest one. I know I'm just missing something stupid now, but I can't figure out what.......

Here is the Asymptote code I'm using; maybe there's something wrong with it?

[code]
draw((0,0)--(100,100));
draw((0,100)--(100,0));
dot((50,50));
label("$P$",(50,50),S);[/code]

EDIT- also, while you say I should get a PDF, the guid http://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_Getting_Started/Windows/Interactive_Mode makes it sound like running the tool should just make the image appear on the screen.

What happens if you remove the dollar signs from the last line?
Is there a test.pdf file created? If so, it will be opened which is what "appear on the screen" means.

removing the dollar signs did nothing. And no, there is no pdf generated.

There appears to be a problem with asymptote. Does it work outside TexnicCenter?

Yes I did get it to work outside TeXNicCenter

Can you double-check that asy.exe is in C:\Program Files\Asymptote?

If asy.exe is there then try this:
1. In asymptote.bat remove the first part of
"C:\Program Files\Asymptote\redir.exe" -e %1 "C:\progra~1\Asymptote\asy.exe" -batchView -tex "pdflatex" -align C -f "pdf" %2
so it reads
"C:\progra~1\Asymptote\asy.exe" -batchView -tex "pdflatex" -align C -f "pdf" %2
2. From your screen shots, on the line
find /i ".asy" %3 >nul
there seems to be an extra 1 before the >nul which needs removing.

Now try again and show me the screen which will tell me if asy.exe is really running properly.

Hmm yeah asy.exe is there. I fixed that one line, but I have no idea where the extra 1 is coming from. I don't see it asymptote.bat. Here's the code for asymptote.bat:

[code]echo Asymptote is working ... please wait
set opath=%path%
set path=%path%;C:\Program Files\MiKTex 2.5\miktex\bin
"C:\progra~1\Asymptote\asy.exe" -batchView -tex "pdflatex" -align C -f "pdf" %2
find /i ".asy" %3 >nul
If errorlevel 1 goto end
"C:\Program Files\TeXnicCenter\TEXCNTR.EXE" /ddecmd "[open('%3')]"
pause
:end
set path=%opath%
set opath=[/code]

Do I have the settings for the Asymptote tool set correctly?

Command: "C:\Program Files\Asymptote\asymptote.bat"
Arguments: %tc_asy.log %tc "%dc\%tc_asy.log"
Initial Directory: %dc

Everything seems fine and asymptote is being found but isn't producing anything which is odd. (It turns out that the extra 1 is normal, it's just that normally you don't see it when @echo off is there).

Can you try re-installing Asymptote over the top of the one you have just in case there's a problem there?

What happens if I can't find config.asy in the path you've given?

Nothing. Under normal circumstances you do not need config.asy at all. It is just one of the files Asymptote executes before reading your source file. You can always create it if it is absent and put whatever you want in there but it is by no means necessary for the proper functioning of Asymptote.

I see. I use Foxit Reader for pdf files, and had to create a new config.asy in the .asy directory to change the pdfviewer path; it works now. Yes, it did not change Asymptote's proper functioning. Thank you for this board; it's really helpful.


Let Triangle ABC be an isosceles triangle(AB=AC) with Angle BAC=20 degrees. Point D is on side AC such that Angle DBC=60 degrees. Point E is on side AB such that Angle ECB=50 degrees. Find, with proof, the measure of Angle EDB.

Can you please try to find a solution not involving trigonometry? Thanks.  :roll:

It'll be very helpful to draw it out.

See here: http://www.cut-the-knot.org/triangle/80-80-20/index.shtml

(and here: http://www.mathlinks.ro/viewtopic.php?t=310088 )


Prove that for given $ a,b,c \ge0$ we have

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

EDITED : Sorry,It seemed to me that the sign was reversed !

If you take $ b$, $ c \to 0$, then the right-hand-side gets arbitrarily close to zero, but the left-hand-side not. So the ineuality holds true in this case.  :wink:

Yes , the inequality is true . See [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=106975]here[/url]

It is equivalent to $ \frac {1}{2}[(a \minus{} b)^6 \plus{} (b \minus{} c)^6 \plus{} (c \minus{} a)^6] \plus{} 2[ab(a \minus{} b)^4 \plus{} bc(b \minus{} c)^4 \plus{} ca(c \minus{} a)^4] \plus{} \frac {1}{2}(a^4(b\minus{}c)^2\plus{}b^4(c\minus{}a)^2\plus{}c^4(a\minus{}b)^2) \ge 0$.

Can someone tell me how to post cyclic sum and symmetric sum?
EDIT: Thank you, alex2008

[quote="dgreenb801"]
Can someone tell me how to post cyclic sum and symmetric sum?[/quote]

cyclic sum - $ \sum_{cyc}$

[code]\sum_{cyc}[/code]

symmetric sum - $ \sum_{sym}$

[code]
\sum_{sym}[/code]

i hope you was reffering to those .


In principle, the only thing that theoretically limits efficiency of heat machines is Carnot's inequality. But I don't know how to explain it to a 10 year old boy :?

[quote="aahmadi"]Please explain why us humans cannot obtain 100% efficiency with energy.  I would prefer an explanation that is so simple that even my 10-year-old cousin could understand.  Thanks[/quote]

I'll give it a try. First, there's the 1st law of thermodynamics: the amount of energy that a thermodynamic system can store is the difference between the amount of heat added to the system and the amount of work the system does on its surroundings.

Second, the amount of energy the system can store is equal to the greatest amount of heat that the system can add to its surroundings. Further, heat can only flow from "hot" to "cold."

This uni-directionality of heat flow indicates that the system draws heat from a hot reservoir, and rejects heat to a cold reservoir. Written as an equation, the first law of thermodynamics says: $ Q_C \equal{} Q_H \minus{} W$.

For a reversible process, which is the most efficient kind, $ \frac{Q_H}{T_H} \equal{} \frac{Q_C}{T_C}$  This implies $ \frac{T_C}{T_H} \equal{} \frac{Q_C}{Q_H}$

The efficiency $ \eta \doteq \frac{W}{Q_H}$.

If we then substitute $ W$ from the above equation into the definition of $ \eta$ we get:

$ \eta \equal{}\frac{Q_H \minus{} Q_C}{Q_H} \equal{} 1 \minus{} \frac{Q_C}{Q_H}$

We can now substitute $ \frac{T_C}{T_H}$ for $ \frac{Q_C}{Q_H}$ in the above expression.

$ \eta \equal{} 1 \minus{} \frac{T_C}{T_H}$

Now let's suppose that $ \eta\geq 100\%$.

Then $ 1 \minus{}\frac{T_C}{T_H} \geq 1$.

Subtracting 1 from both sides of the inequality,

$ \minus{}\frac{T_C}{T_H} \geq 0 \Rightarrow \frac{T_C}{T_H}\leq 0$

This is impossible according to the third law of thermodynamics; therefore, our supposition is false.

Ergo, $ \eta < 100\%$

I don't know if a ten year old would comprehend the algebra involved, but the conclusion is clear enough: One simply cannot get as much work out of a thermodynamic system as the amount of energy one puts into it.

I would just emphasize that, even if a human intends to do work,he can not consentrate 100% to it.So from this,it can be understood that there can't be 100% eficiency.

Yes, essentially the explanation of Carnot's inequality is exactly that

a) No matter how cunning we try to be, we cannot create a process the only result of which is passing heat from a cold body to a hot one.

b) We can be cunning enough to pass heat $ Q$ from the cold body to the hot one using work $ W$ much less than $ Q$ (namely, $ [1\minus{}(T_{\text{low}}/T_{\text{hot}})]Q$)

Ergo, we cannot make the heat machine that passes heat $ Q$ to produce work larger than $ W$ (otherwise run this machine, and use the resulting work to transfer more heat back than was transferred forward)

Unfortunately, neither (a), nor (b) is obvious. (a) has to be taken for granted, (b) is a bit easier to explain, but it still requires an explicit construction of such a machine and computation of the mechanical work done and the heat transferred. :(

[quote="Carcul"][quote="Dr. No"]One simply cannot get as much work out of a thermodynamic system as the amount of energy one puts into it.[/quote]

Wrong again: yes we can. The Kelvin-Planck statement of the second law simply states that such a conversion is not possible [i]for a cyclic process[/i]. There is no problem in converting completely in work a given amount of heat by a non-cyclic process.[/quote]

I wish you would stop finding fault with what I write merely for the sake of finding fault.

Your objection is pointless. I didn't appeal to the Kelvin-Planck statement of 2LoT, or any other statement of it for that matter, in making my argument. Starting with a reversible process, the Carnot inequality guarantees that $ \frac{Q_H}{T_H} \equal{} \frac{Q_C}{T_C}$; that said process is or is not cyclic notwithstanding.

Following which I proved that an efficiency greater than or equal 100% requires a ratio of absolute temperatures that is either negative or zero. By the 3LoT this circumstance is impossible. Thus the efficiency of [b]any[/b] thermodynamic process is less than 100%, cyclic or not.

Carnot's inequality applies only to cyclic processes: that's the point of using it to demonstrate the existence of a state function that we call Entropy. And an example of a thermodynamic process with 100% efficiency is the isothermal expansion of an ideal gas.

[quote="Dr. No"]I wish you would stop finding fault with what I write merely for the sake of finding fault.[/quote]

If you don't like people to make a critic about what you write, then think twice before posting. If you want to keep writing wrong things, be me guest. That's your problem and a problem for those who read your posts. And don't worry, I will not make any further comments about your posts.

The Loss of Heat
The reversible system of heat
The inreversible system of heat

[quote="Carcul"]Carnot's inequality applies only to cyclic processes: that's the point of using it to demonstrate the existence of a state function that we call Entropy. And an example of a thermodynamic process with 100% efficiency is the isothermal expansion of an ideal gas.

[quote="Dr. No"]I wish you would stop finding fault with what I write merely for the sake of finding fault.[/quote][/quote]

[i]Oh, good grief....[/i]  $ dS\geq \frac{\delta Q}{T}$ has nothing to do with cyclic processes necessarily. The case I cited was the equality, which occurs when $ dS \equal{} \frac{\delta Q}{T}$, i.e. a reversible process.

For an isothermal expansion of an ideal gas, the efficiency is given by:

$ \eta \equal{}\frac{W}{Q} \equal{} \frac{W}{\Delta U \plus{} W} < 1$

[quote="Carcul"]If you don't like people to make a critic about what you write, then think twice before posting. If you want to keep writing wrong things, be me guest. That's your problem and a problem for those who read your posts. And don't worry, I will not make any further comments about your posts.[/quote]

I don't mind criticism, especially when it's legitimate. But you have an annoying habit of trying to pull at loose threads in my arguments that don't exist. I don't intentionally post things that are wrong and I don't come here to get help so much as to give it. I don't have any need to prove myself to be the smartest kid in the room. I know that there are many here, including probably you Carcul, who are brighter than I. But that shouldn't matter since I don't come here with the intention of putting anyone down. I know the subject of physics, I know a fair bit about several fields of mathematics, and I enjoy figuring things out. If I can help anyone in the process of learning things myself then good. If not, then I'll leave but in the meantime I don't like being attacked by 'know-it-all' pups with ego malfunctions.

[quote="Dr. No"] I don't like being attacked by 'know-it-all' pups with ego malfunctions.[/quote]

Ouch. Carcul is a teacher. I think it's part of his job to pick out small things. So yeah, let's just stick to the academic stuff.

[quote="aahmadi"]I would prefer an explanation that is so simple that even my 10-year-old cousin could understand.[/quote]

I think a really simple, understandable explanation is that in any machine, parts rub together and release some of the use[i]ful[/i] energy put into them as use[i]less[/i] heat due to friction, or heat is released through other processes (e.g., a wire heating due to the electric current passing through it). Because some of the energy expended is always released as heat, the machine cannot be 100% efficient.

That's probably not the most "scientific" explanation and not easily derived mathematically and whatnot, but I do think a 10-year-old would be able to understand it.

A simple explanation is not an accurate one.  Pick what you value more.


[quote="2424"]With $ a,b,c > 0$ such that $ ab,ca,bc \in[0;\sqrt {2}]$    and  $ (ab)^2 \plus{} (bc)^2 \plus{} (ca)^2 \plus{} (abc)^2 \equal{} 4$
Prove that $ a \plus{} b \plus{} c \geq 3$[/quote]
You can see the inequality than stronger:
http://www.mathlinks.ro/viewtopic.php?t=289839


geometry
perpendicular bisector
geometry proposed

Let $ABCD$ be a convex quadrilateral with $AB=CD$ and not a parallelogram. $M,N$ are midpoints of diagonals $AC,BD$ respectively. Prove that the orthogonal projections of $AB$ and $CD$ onto $MN$ both have length equal to $MN$.

Let $U, V$ be the midpoints of $AD$ and $BC$. Prove the fact the $UV$ is the perpendicular bisector of $MN$. Thus, if $G$ is the centroid of $ABCD$ and $A', B', C', D'$ are the projections of $A, B, C, D$ on $MN$, then $G$ is the common midpoint of $MN, A'D', B'C'$. The result immediately follows.

[quote="treegoner"] then $G$ is the common midpoint of $MN, A'D', B'C'$. The result immediately follows.[/quote]

Why? I see only $A'D'=B'C'$, not $=MN$...

Let U be the midpoint of AD. We have 2*UM=CD=AB=2*MN so triangle MUN is isosceles.
Let MN meet AB at K, and PQ is the orthogonal projection of AB onto MN
We have PQ=AB*cosAKN=2*UN*cosUNM=MN (because triangle MUN is isosceles)
Similarly with CD we have the result above q.e.d

[quote="Huyền Vũ"]Let U be the midpoint of AD. We have $2 \cdot UM=CD=AB=2 \cdot UN$ so triangle $MUN$ is isosceles.
Let $MN$ meet $AB$ at $K$, and $PQ$ is the orthogonal projection of $AB$ onto $MN$
We have $PQ=AB \cdot \cos AKN=2 \cdot UN \cdot \cos UNM=MN$ (because triangle $MUN$ is isosceles)
Similarly with $CD$ we have the result above q.e.d[/quote]

 Please see the figure. (There was a  typo in  the above solution. I have corrected it.)


calculus
derivative
algebra
polynomial
function
domain
complex numbers

Solve it:

  ${\begin{array}{l}x^5-y^5=992\\x-y=2\end{array}}$

Please if somebody could help me with this :mad:

Well,
it is easy to see that
$4^5-2^5=992$.
And also
$(y+2)^2-y^2$ grows when $y$ is positive because the first derivative of $x^5$ is $5x^4$ and is growing. So this is only solution for positive $x$ and $y$ $(4,2)$. also $x$ and $y$ must have the same sign because otherwise $x^5-y^5\leq 2.2^5=64<992$.
And for negative $x,y$ we find the solution $(-2,-4)$.

Finally $(4,2)$ and $(-2,-4)$ are the solutions!

there isn't more solutions? Are you sure?

You mean aren't there more solutions?

Well, it sounds like you're quizzing instead of asking.

Actually, to the poster, imortal it is a throwaway statement "it is easy to see that..."

It certainly didn't occur to me to use those numbers!    From now on. please don't use that phrasing; it's counterproductive.

And imortal, because the problem never stated to find only integer solutions in x and y, your solutions are incomplete.

After getting (y+2)^5 - y^5 = 992 and subtracting off the y^5 terms
from each side, you have a 4th degree equation in y.  After setting
it to zero and dividing both sides by 10, you have

y^4 + 4y^3 + 8y^2 + 8y - 96 = 0

After dividing out the factors (y-2) and (y+4) to go along with the known solutions of y = 2 and y = - 4 from the earlier post, you are left with

y^2 + 2y + 12 = 0 which has the roots of

y = -1 - i*sqrt(11) and -1 + i*sqrt(11)

Other solutions to x^5 - y^5 = 992 and x - y = 2 are then

(1 - i*sqrt(11), -1 - i*sqrt(11)) and

(1 + i*sqrt(11), -1 + i*sqrt(11))

thank you, now i see that it is good solution of the problem

[hide="Here is a solution only with the polynominal  of the second degree."] $S=x+y$, $P=xy$ $\Longrightarrow$ $S_2\equiv x^2+y^2=S^2-2P$ $\Longrightarrow$ $S_4\equiv x^4+y^4=$ $\left(x^2+y^2\right)^2-2(xy)^2=$ $S^2_2-2P^2=$ $\left(S^2-2P\right)^2-2P^2=$ $S^4-4PS^2+2P^2$ $\Longrightarrow$ $\boxed {S_2=S^2-2P\ ,\ S_4=S_4-4PS^2+2P^2}\ .$
$x-y=2,\ x+y=S\Longrightarrow \boxed {x=\frac{S+2}{2}\ ,\ y=\frac{S-2}{2}}$ $\Longrightarrow S^2-4=4P\ \ (P\ge -1)\ .$
$x^5-y^5=992$ $\Longleftrightarrow$ $(x-y)\left[ \left(x^4+y^4\right)+xy\left(x^2+y^2\right)+(xy)^2\right]=998$ $\Longleftrightarrow$ $S_4+PS_2+P^2=496$ $\Longleftrightarrow$ $S^4-4PS^2+2P^2+P\left(S^2-2P\right)+2P^2=496$ $\Longleftrightarrow$ $\boxed {S^2=4(P+1)\ge 0\ ,\ S^2\left(S^2-3P\right)+P^2=496}$ $\Longrightarrow$ $4(P+1)(P+4)+P^2-496=0$ $\Longleftrightarrow$ $P^2+4P-96=0$ $\Longrightarrow$ $P=8$ $\Longrightarrow$ $S^2=36$ $\Longrightarrow$ $S\in \{6,-6\}$ $\Longrightarrow$ $\boxed {x=4\ ,\ y=2\ \ \vee\ \ x=-2\ ,\ y=-4}\ .$

[b]Remark.[/b] For $P=-12$ and $S^2=-44$, i.e. $S\in \{\pm 2i\sqrt{11}\}$ we obtain and the another solutions, but in the set of the complex numbers:
$t^2-2i\sqrt{11}\cdot t-12=0\Longrightarrow \boxed {x=1+i\sqrt{11}\ ;\ y=-1+i\sqrt{11}}\ .$
$t^2+2i\sqrt{11}\cdot t-12=0\Longrightarrow \boxed {x=1-i\sqrt{11}\ ;\ y=-1-i\sqrt{11}}\ .$

[color=darkred][b]Remark.[/b] I don't know the set in which we find the solutions of the proposed system !
[b]Convention.[/b] [u]In the case when the domain of the existence isn't specified, then it is the largest domain of the existence, i.e. in the our case, the set of the complex numbers.[/u][/color][/hide]

Hi!

Of course, when I saw the problem, I had no idea what kind of numbers x and y are so I decided that they are real. If I knew that x and y may be complex I would try to find them but it was not stated in the problem so I overlooked it :)
But if you read carefully my solution you will see that I solve the equation in real numbers (NOT ONLY IN INTEGERS) and my solution is full (for real numbers).

Actually, A cute angle, it is not really hard to see that (4,2) is a solution.
When I saw the problem I firstly thought: wow, the number 992 sounds very strange, maybe the reason it appers in this problem is that  INTEGERS are solutions.
.

Then I started trying with small integers if they satisfy the condition. Because the differnce between the perfect fifth degrees grows very fast I knew that can check if the equation have integer solutions or not. So I tried it and was lucky to find (4,2).

So, maybe it may sound as throwaway "it is easy to see that..." to you, but for me it was really easy to see. :D

imortal,

you do not have the leisure to decide that x and y are exclusively
real.

Just because the problem did not state that x and y may be complex does NOT mean you can ignore them as a possibility.

Your solution in real numbers is not complete enough.

No, your reasoning is again off, as pertains 992.  It only means
that perhaps integers CAN be the solutions, not ARE the sole solutions.

You'll have to get your story straight; "it is easy to see that..."
sounds like the solution jumps off the screen with little effort, BUT  you disclosed that you tried (4,2) and were lucky, which is altogether a different meaning.  You cannot have it both ways logically.

The main point is this: the problem has to have accompanying restrictions on the variable(s) if it stands alone.  If there are none, then you solve for as large an encompassing set of solutions that you can, taking into account, for example, real world restrictions.

You were wrong to discard complex solutions, for example, with this problem, because you stated you felt like overlooking it.
You simply don't have a choice in the matter.

Hi,
A cute angle, you are correct that I should have solved the problem for complex, not just for real. I agree with that. I won't argue about that.

But you are not correct when you are saying that my solution is not correct if we have the restriction $x,y \in R$. It is correct just like your solution is correct but it is different. Still for a problem there can be more than one solution, right?

You are bringing up a specific objection for you that I did not raise.
Look back over my sentences.

As long as you, the poster imortal, are having the restriction for x and y
belonging to the real numbers, then yes, your solutions are correct and consistent with that.  My past words are also consistent with that.

Read all, please, the my last reply ! I added there a theoretical comment. Finish this exchange of words, please ! Thanks.


[quote="nguyentrang"]In a graph with even number of vertices, each vertex has degree $ 3$. This graph has a Hamiltonian cycle. Show that it has at least two.[/quote]

Posted before (not quite answered) at http://www.mathlinks.ro/Forum/viewtopic.php?t=80281 .

  darij


trigonometry
quadratics
rational numbers

Suppose that $\tan \alpha =\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$. Prove the number $\tan \beta$ for which $\tan 2\beta =\tan 3\alpha$ is rational only when $p^2 +q^2$ is the square of an integer.

If $ \tan\alpha \equal{}\frac{p}{q}$ then the equation reads $ \frac{2\tan\beta}{1\minus{}\tan^{2}\beta}\equal{}\frac{p(p^{2}\minus{}3q^{2})}{q(3p^{2}\minus{}q^{2})}$, solving to $ \tan\beta$ gives a quadratic with discriminant $ (p^{2}\plus{}q^{2})^{3}$, so $ \tan\beta\in\mathbb{Q}$ iff $ p^{2}\plus{}q^{2}$ is a rational square. But $ p^{2}\plus{}q^{2}$ is an integer, so is the square of an integer, QED.


A homogeneous ball with decelerates during $4.0\,s$ to rest on a smooth,
horizontal plane. The braking distance is $s=1,5\,dm$. What inclination, $\theta$, should the
plane have if the ball is to roll with constant velocity?

[color=red]EDITED[/color]

assuming you made a mistake. because the plane is horizontal. so the angle is zero.

assuming misspelling.

the ball is rolling down to a inclined plane with constant velocity.

So there is no force on it. so the aceleration is zero.

So the component of the gravitational force is equal to the aceleration.

since you didnt say if there is friction or another thing and the problem and your statement.

I will be happy if you say it more clear :)

The only thing i have forgot to say is that it decelerates uniformly.

Yes of course friction must be taken into account, otherwise the ball
wont decelerate on a horizontal plane.

acctually, there is a lot of ways a particle decelerate without friction.

we can use eletromagnetic fields, of others fields.

i just wanted the cause of deceleration. thanks

but this problem is simple. we  can use

$s'' = a$ and $s'=v$

i put in diferential forme to become more simple, but is the law of movement on newtonian mechanics

I have already mentioned the cause: friction

and of course you have to rely
on Newtonian mechanics.

If you find it simple, then please post a solution.

the solutions is analogous of solution of http://www.artofproblemsolving.com/Forum/viewtopic.php?t=151971

I cannot lose time to make a lot of analogous problem dude.

color it would be helpful if you restate the entire problem . actually it is difficult to get you.

Please explain what is difficult. I am sorry for that, but the problem
text will not reveal more information, that is the point. It gets
more difficult to solve that way.

Well then the problem text is probably wrong.  it happens sometimes :wink:

[quote="color"]They are not that analogous.

You don't have to solve it if you don't want to.   :P[/quote]

er?? ahm?!!?

not!? are you sure?

since, like i said, there is no force acting on the particle. so $s = v t+d$  $a=0$

and your problem has wrongs things, but you dont want to correct it.

I just think you thought you made a good problem, and everybody is saying it is wrong and it's simples etc.. and you get angry. dont be mad man..

thats happen like taakisfox said.

if you'd be able to rewrite the problem correctly, we can try to solve it.

I cant see why the problem is wrong.
Also I have said that friction acts on the ball, so count that as a force.
And I am not mad :lol:

I will post my solution later, but I cant see why the problem should be changed.
---------------------------------

A ball with mass $m$ and radius $R$ decelerates uniformly during $4.0$s to rest
on a smoth horizontal table due to friction.
The breaking distance is $l$ (1.5 dm). If one instead want the ball to roll with constant velocity
what should the inclination  $\theta$ be?

no i assume roll her means roll without slipping but if it has to roll without slipping and with a constant speed .that implies the friction(which must act because it has to roll without slipping to produce the net torque) must balance the component of the gravity along the surface.But then this friction gives ise to torque which in turn provides angular accelaration and since the speed of c.m has to be constant the ball no longer rolls.

if you meant simple rolling not considering about sliding then

$mg \sin \theta$ $=$ $mg \cos \theta \frac{2s}{t^{2}g}$

$\ tan \theta$ $=$ $\frac{2s}{t^{2}g}$
this means there is no use of teh so given variables like $R$...

Sorry for giving extra information like $R\ldots$
That is edited.

[hide="Here is a solution"]
The force which causes the ball to decelerate, is the resulting force from the normal force
$N$ and friction force $f$. Assume the lever arm of the normal force $N$ is $d$ relative the center of mass, and $f$ has the arm $R$.

Let the braking distance be $s$ and braking time be $t$, then we have
$f=ma$
$s=\frac{at^{2}}{2}$ or $a=\frac{2s}{t^{2}}$
$Nd-fR=I\alpha=\frac{2}{5}mR^{2}\cdot \frac{a}{R}=\frac{2}{5}mRa$
all these give $s=\frac{14Rs}{5gt^{2}}$

If the plane is inclined with $\theta$ in braking is compensated. Braking torque will be zero as well as acceleration. (This means the resultant of $N$ and $f$ goes through the center of the ball).
This gives:
$f-mg\sin{\theta}=0$
$Nd-fR=0$ or $mg\cos{\theta}\cdot d-fR=0$
gives
$\tan{\theta}=\frac{d}{R}=\frac{14Rs}{5gt^{2}}=\boxed{\frac{14s}{5gt^{2}}}$

so $\theta\approx 0.15^{o}$
[/hide]

Come on man, The normal force gets exerted at the point where the ball touches the floor, and since the ball is homogenous, its c.m. is above this point so the normal force cant have torque on c.m. :wink:

[b]@Thaakisfox[/b]
So how would you solve it?

Please, explain more detailed than [b]pardesi[/b]s solution above

[b]@pardesi[/b]
Could you develop your conditions and the solution a bit more, please?
Also what do you think about my solution?

I would do it the same way Pardesi did:
The coefficient of friction from the given data:
$\mu =\frac{2s}{gt^{2}}$
since the ball will roll with constant velocity:
$mg\sin \theta =\mu m g \cos \theta$

So the needed angle:
$\theta =\arctan \frac{2s}{gt^{2}}$


Complementary Probability Part 3

We use complementary probability to find the probability that a geyser goes off in 150 minutes, given that it is 2/3 likely to go off at least once during any 50 minutes.

Umm...this is late at night, and I have to go to bed right after this

If it is 2/3 likely to go off during any 50 minutes, then it is 1/3 likely that it will not go off.
We can divide 150 minutes into 3 50-minute segments.
Thus the probablility is $ \frac{1}{3}*\frac{1}{3}*\frac{1}{3}*\equal{}\frac{1}{27}$

Nope, that's not complementary counting, it's multiplying probabilities. Oh well.

aren't we supposed to find the probability that it [i]does [/i]go off?

therefore, the probability that it doesn't go off is $ \frac{1}{27}$, so the probability that it [i]will[/i] go off is $ \frac{26}{27}$

since we found the part not wanted to find the part wanted, we used complementary counting

i don't know how to solve the question mentioned at the end of the video.
"If the probability that you will see it go off is 2/3 in 50 minutes, then what is the probability you will see in 20minutes"
 :|  :|  :|  ~~
Thanks, for anyone who replied

I have the same question as Fersolve. Can someone please answer it?


Solve the folowing diophantine equation:

$n!+1=m^{2}$


[quote="jeez123"]Let $p={(2,7), (-1,\pi), (4, \sqrt{3}), (0,7)}$.
and let $r={(-2, 1), (\pi, 4), (\sqrt{3}, 2), (6, 0), (2,9)}$. Find the sum of all distinct $x$ such that $p(r(x))=7$.

Can you please give me an example as to how the problem works? Thanks![/quote]

[hide]If $p(r(x))=7$, then you see where $p(\text{something})=7$ so you look in the set of points of $p$ to see where the y-coordinate is 7, which is at $(2,7)$ and $(0,7)$. Then you know that $r(x)=2$ or $r(x)=0$ so you look at the set of points in $r$ to see where the y-coordinate is $2$ or $0$ and you get $(\sqrt{3},2)$ and $(6,0)$ so the sum of the $x$ where $p(r(x))=7$ is $\sqrt{3}+6$[/hide]

Hope this helped...


analytic geometry
MATHCOUNTS
Pythagorean Theorem
geometry

[quote="mathnerd314"]mathnerd314 is at the center of a circular building of radius 127. He decides to start walking north (somehow he happens to know the compass directions even though he has no compass). Halfway to the wall, he stops and decides to go east. Again he only gets halfway to the wall before he changes his mind yet another time and walks south. Then, he walks halfway to the wall and turns to walk west. After going halfway to that wall, he stops, utterly lost. How far is he from the center?

I didn't understand the explanation on mathcounts.saab.org (where I got this problem).

Please tell me if my wording is too unclear.[/quote]

when you say west or south do you mean mathnerd314's  :lol: west and south or west and south in general?

Let the equation of this circle be $x^2+y^2=127^2$. Mathnerd314 starts at $(0,0)$ and walks North half of the radius to $(0, \frac{127}{2})$. To find how far away the circle is to the left, let $y=\frac{127}{2}$. $x^2+(\frac{127}{2})^2 = 127^2 \Rightarrow x = \pm \sqrt{127^2 - (\frac{127}{2})^2}$. Choose the positive because Mathnerd314 is walking east into the first quadrant. The new point is half of this distance, $x= \frac{\sqrt{127^2 - (\frac{127}{2})^2}}{2}$ So now with the point $(\frac{\sqrt{127^2 - (\frac{127}{2})^2}}{2}, \frac{127}{2})$ find the distance to the circle coordinate with the x-coordinate being $\frac{\sqrt{127^2 - (\frac{127}{2})^2}}{2}$ and the y-coordinate being negative, because it's in the fourth quadrant because Mathnerd314 is walking South. The numbers get increasingly ugly, but this will eventually give you the point where Mathnerd314 ends, and by using the distance formula on that from the origin you get the total distance Mathnerd314 is from the origin.
Hope this helped.

[quote="catcurio"][quote="ch1n353ch3s54a1l"]when you say west or south do you mean mathnerd314's  :lol: west and south or west and south in general?[/quote]I think it means west and south in general. Aren't west and south always the same anyway? ;)  I think you're confusing this with left and right.[/quote]

oh. i should have known..oops  :oops:  :?  :roll:

i got ........[hide]42? [/hide]

[quote="thebrokenlight"][quote="ch1n353ch3s54a1l"][hide]42? [/hide][/quote]

Ah, such an inexplicably beautiful number...  :P

How did you arrive at that answer?

-BLt[/quote]

well ACTUALLY, i was estimating half the time.  :D

so he goes up $\frac{127}{2}$. Forming a parallel line that goes through where mathnerd314 is and the diameter of the circle, you can find the length of the line, divide it by two, and then divide it by two again, and i found that to be $\approx54.9926$. this is how much he goes east. then the "halfway south" part i found to be $\frac{127}{2}+\frac{127}{4}$. then "halfway west" i found to be 88.9799, u take away 54.9926 and get 33.9873, use pythagorean theorem and get...........................................................46.51. hm, different answer............. I don't think 42's right anymore.. hehe
so [hide][size=200]46.51[/size][/hide]

[quote="Hokkage"]But 42 is the answer to life, the universe, and everything! How could it not be right?[/quote]

wait huh? so the answer's 42??  :lol:  :P  :P

[quote="Hokkage"]But 42 is the answer to life, the universe, and everything! How could it not be right?[/quote]
So long, so long so long, and thanks for all the fish!  :rotfl:

Please stop spamming, everyone.

[quote="Hokkage"]The numbers get increasingly ugly,[/quote]

That was the problem with the website's solution. I'm trying to find an easy solution to this problem.

[quote="gmslinger"]Maybe it's just me.....but don't you guys think this problem belongs under the intermediate forum and not the getting started one. Don't get me wrong the problem sounds fairly simple, but I have tried it and came up with some very ugly numbers.....and unless there's an easier way to do it, things like this should be posted somewhere else.  :?:[/quote]

At this point I agree with this (and it's been moved anyway). I just thought that since it was from mathcounts.saab.org, it would be around Mathcounts level.

I had hoped for a less ugly, fairly simple solution too.

Thanks guys,
   Sorry I'm new at this and some of this seems very overwhelming. I'm only a freshman in geometry.

                                           [b]THANKS[/b]

One simple question:

If the initial point is in somewhere on the first quadrant, and then go south, then go west, go north, go east, and so on....

What is the initial point, if at the end of each step, the distance from the origin is constant?

Angel "Java" Lopez

No no no... the initial point is the center of the circle, and the distance from the origin does not remain constant. Unless I've totally misunderstood your idea...

btw it's a mathcounts target round problem... IMO it should be in the MathCounts forum.

And BTW that's where I got it (I posted it in the Getting Started forum), but it was moved because of all the hard solutions submitted. Do you have the Mathcounts-level solution, Treething?

walks up half radius, then right rsin(60)/2
then its kinda good to know how far it is up to the wall from that point.
That would be sqrt(r^2 - (r^2)(sin^2(60))/4) which happens to be:
sqrt(r^2 - (3/16)r^2)=sqrt((9/16)r^2)=(3/4)r
Which means that the distance to the other wall is (3/4)r-(1/2)r+(3/4)r=r
He now walks r/2 from the coordinates where he first was, and end up at
the coordinates (relative the middle) (rsin(60)/2 ; 0)
the way to the (our) left wall is now (r+rsin(60)/2)
k divide by two..

r/2+rsin(60)/4... thats the way he walks

lets subtract that from the x-coordinate
aand get a better value for sin(60) again.. sqrt(3)/2 yess
rsqrt(3)/4 -  r/2 + rsqrt(3)/8 = r(3sqrt(3)/8-1,5).. according to me that should be negative, so i guess he passed origo.. yei
so take |127(3sqrt(3)/8 - 1,5)| for an answer. thats like 108 and something.

I may have done some counting mistakes, but the rest should due. ;P
Excuse me cant handle latex.. and btw sqrt(x) is supposed to mean squareroot


analytic geometry
pigeonhole principle
number theory unsolved
number theory

Find the largest integer $n$ for which there exist $n$ different integers such that none of them are divisible by either of $7,11$ or $13$, but the sum of any two of them is divisible by at least one of $7,11$ and $13$.

the answer is actually 8.

we may represent each integer by a triple $ (a, b, c)$, where $ a, b, c$ are the remainders upon division by 7, 11, 13, respectively. for each remainder $ a$, let $ a'$ denote the complementary remainder; e.g., the complementary remainder modulo 7 of 3 is 4. for two $ n$-tuples $ (a_1, ..., a_n)$, $ (b_1, ..., b_n)$ of remainders, we say that they "match" in the $ i$-th position if $ a_i \equal{} b_i'$. a set of $ n$-tuples will be called "matching" if any two of them match in at least one position; otherwise, we call them "mismatched." so the statement we want to prove is, that a set of matching triples must have size at most 8. first we prove the following lemma.

Lemma: a set of matching pairs can have size at most 4.

Proof: assume, for a contradiction, that we have a set of 5 matching pairs, and let $ a, b, c, d, e$ denote these pairs. assume for the moment that $ a$ matches 3 of the other pairs - WLOG we may assume they are $ b, c, d$ - in the first position. then $ b, c, d$ coincide in the first position, so any two of them must match each other in the second position - but this is impossible. so $ a$ can match at most 2 of the other pairs in the first position, and similarly it can match at most 2 of the other pairs in the second position. since there are four other pairs, a must match *exactly* 2 in the first position and 2 in the second position. now let $ a \equal{} (x,y)$. by what we've just proved, the other pairs must be of the form $ (x, a), (x, a'), (b, y), (b', y)$. now, for $ (x, a')$ to match $ (b, y)$, we must have either $ x \equal{} b'$ or $ y \equal{} a$. in both cases, we have 3 pairs which coincide in one position, which as before, is impossible. this completes the proof of the lemma.

now to the result we want to prove. assume, for a contradiction, that we have a set of 9 matching triples. take the 9 numbers in the first coordinates of these triples and partition them into two sets so that no two numbers in the same set are complementary. (this is possible because every number has a unique complement and no number is its own complement because the moduli are odd.) by the pigeonhole principle, one of these sets contains 5 numbers, and these numbers are pairwise non-complementary. thus there are 5 triples which are mutually mismatched in their first coordinates. their second and third coordinates then give us 5 matching pairs, contrary to the lemma.

to see that 8 is indeed possible, take 8 triples of the form $ (a, b, c), (a', b, d), (a, b', e), (a', b', f), (a', b', f'), (a, b, c'), (a', b, d'), (a, b', e')$.

thank you.

I realized 120 was too high, as I thought that every pair added up to a multiple of 7, 11, 13 in my set of 120, but I was wrong.

Nice solution.


function
limit
inequalities open
inequalities

Hello everybody.

I would like to know if there exists a linear function f so that for all positive integers n : $ (x\plus{}y)^n \le f(n) (x^n \plus{} y^n)$.

Thanks.

Take $ x\equal{}y$. Hence $ f(n) \ge 2^{n\minus{}1}$ for all $ n$. $ f(n)\equal{}An\plus{}B \ge 2^{n\minus{}1} \ge n^2$ for $ n \ge 7$. But $ \lim_{n \to infty}g(n)\equal{}\lim_{n \to \infty}\minus{}n^2\plus{}An\plus{}B \to \minus{} \infty$. Hence there exists no such function.


[quote="tranthanhnam"]let ABC be square triangle ($c>a \geq b$)
   find minimal of expression P
$P= \frac{a^2(b+c)+b^2(c+a)}{abc}$
and problem 2 :
find minimal of expression P
$P= \frac{a}{4+bc} + \frac{b}{4+ac} + \frac{c}{4+ba}$ where $a,b,c \in$ [$0;2$][/quote]
What is a square triangle? :?

I guess he means "right triangle".
http://mathworld.wolfram.com/SquareTrianglePicking.html for "square triangle"

[quote="tranthanhnam"]problem1: let ABC be square triangle ($c>a \geq b$)
   find minimal of expression P
$P= \frac{a^2(b+c)+b^2(c+a)}{abc}$
and proble m 2 :
find maximum of expression P
$P= \frac{a}{4+bc} + \frac{b}{4+ac} + \frac{c}{4+ba}$ where $a,b,c \in$ [$0;2$][/quote]
solution for problem 2:
let $f(a)=\frac{a}{4+bc} + \frac{b}{4+ac} + \frac{c}{4+ba}$
we have $\frac{d^2f}{da^2}\geq 0$
so P reach the maximum when $a,b,c$ is $0$ or$2$
then the maximum is 1


function
limit
algebra
domain
real analysis
real analysis unsolved

It's well-known that a function $f: E\subset\mathbb{R}^{n}\rightarrow F\subset\mathbb{R}^{m}$  continuous at almost every points in $E$ is measurable.
Can you find a function measurable on $E$ that is not continuous at almost every points in $E$?

take a non-trivial dense mesurable subset $S$, $\mathbb{Q}^{n}$ for example, and the function $1_{S}$.

Thanks :) And how can you say that $1_{S}$ is not continuous on $\mathbb{R}^{n}\backslash S$, it takes the constant value 0. So $\forall y \in \mathbb{R}^{n}\backslash S,\hspace{0,2cm}\lim_{\{x\rightarrow y; x\in\mathbb{R}^{n}\backslash S\}}1_{S}(x)=0=1_{S}(y)$. I'm maybe confused by the definition of "$f$ is continuous at point $y$" for $y$ in a strange subset :p...

Continuity of $f$ depends on what is understood by the domain of $f$. In alekk's post the domain ($E$ in your question) is the entire space $\mathbb R^{n}$


inequalities
geometry
circumcircle
analytic geometry
ratio
absolute value
inequalities proposed

Let ABC be a triangle. M and N are arbitrary points inside ABC. D, E, F are points that belong to BC, CA, AB, respectively such that: MD is parallel to NA, ME is parallel to NB, and MF is parallel to NC. Prove that:

area of DEF =< 1/4 area of ABC.

You may know the problem in some more specific cases :

1) AN is orthogonal to BC, so on...

or

2) M and N are the same points

This problem is little more general. ;)

This hangs around unsolved since June?

We begin with a lemma:

[b]Lemma 1.[/b] If M is an arbitrary point inside triangle ABC, then the area of the pedal triangle of the point M with respect to triangle ABC is less or equal to $\frac14$ of the area of triangle ABC.

[i]Note.[/i] The pedal triangle of a point M with respect to triangle ABC is the triangle formed by the orthogonal projections of the point M on the sides BC, CA, AB of triangle ABC.

[i]Proof of Lemma 1.[/i] After what I proved in [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=5658[/url], the area of T the pedal triangle of the point M with respect to triangle ABC equals $T=\frac{Sp}{4R^2}$, where S is the area and R is the circumradius of triangle ABC, and p is the absolute value of the power of the point M with respect to the circumcircle of triangle ABC. We have to prove $T \leq \frac14 S$. This simplifies to $\frac{Sp}{4R^2} \leq \frac14 S$, i. e. to $\frac{p}{R^2} \leq 1$. In other words, we have to show $p \leq R^2$. But this is immediately clear: The point M lies inside the triangle ABC, hence also inside the circumcircle of triangle ABC, and thus the power of the point M with respect to the circumcircle is negative. On the other hand, this power must be $\geq -R^2$ (in fact, if O is the center of the circumcircle of triangle ABC, the power of M with respect to the circumcircle equals $OM^2-R^2$, what is clearly $\geq -R^2$ since $OM^2 \geq 0$). Hence, the absolute value p of this power is $\leq R^2$, and Lemma 1 is proven.

Now I cite a lemma which I won't prove since the proof is straightforward:

[b]Lemma 2.[/b] If ABC is a triangle with the sidelengths a, b, c, then the homogeneous barycentric coordinates of the orthocenter of triangle ABC with respect to triangle ABC are $\left(\frac{1}{b^2+c^2-a^2}:\frac{1}{c^2+a^2-b^2}:\frac{1}{a^2+b^2-c^2}\right)$.

Now a more important lemma:

[b]Lemma 3.[/b] For every triangle ABC and every point N inside triangle ABC, there exists an affine transformation mapping the triangle ABC to a triangle A'B'C' and the point N to a point N' such that the point N' is the orthocenter of triangle A'B'C'.

[i]Proof of Lemma 3.[/i] Let $\left(x:y:z\right)$ be the homogeneous barycentric coordinates of the point N with respect to triangle ABC, such that all the numbers x, y, z are positive (this is possible, since the point N lies inside triangle ABC). Let A'B'C' be a triangle lying somewhere in the plane and having the sidelengths

$a^{\prime} = B^{\prime}C^{\prime} = \sqrt{\frac{1}{y}+\frac{1}{z}}$;
$b^{\prime} = C^{\prime}A^{\prime} = \sqrt{\frac{1}{z}+\frac{1}{x}}$;
$c^{\prime} = A^{\prime}B^{\prime} = \sqrt{\frac{1}{x}+\frac{1}{y}}$.

Then,

$b^{\prime}\ ^2 + c^{\prime}\ ^2 - a^{\prime}\ ^2 = \left(\frac{1}{z}+\frac{1}{x}\right) + \left(\frac{1}{x}+\frac{1}{y}\right) - \left(\frac{1}{y}+\frac{1}{z}\right) = 2\frac{1}{x} = \frac{2}{x}$,

and thus

$\frac{1}{b^{\prime}\ ^2 + c^{\prime}\ ^2 - a^{\prime}\ ^2} = \frac{x}{2}$.

Similarly,

$\frac{1}{c^{\prime}\ ^2 + a^{\prime}\ ^2 - b^{\prime}\ ^2} = \frac{y}{2}$;
$\frac{1}{a^{\prime}\ ^2 + b^{\prime}\ ^2 - c^{\prime}\ ^2} = \frac{z}{2}$.

Hence, after Lemma 2, the homogeneous barycentric coordinates of the orthocenter of triangle A'B'C' with respect to triangle A'B'C' are

$\left(\frac{1}{b^{\prime}\ ^2 + c^{\prime}\ ^2 - a^{\prime}\ ^2}:\frac{1}{c^{\prime}\ ^2 + a^{\prime}\ ^2 - b^{\prime}\ ^2}:\frac{1}{a^{\prime}\ ^2 + b^{\prime}\ ^2 - c^{\prime}\ ^2}\right)$
$=\left(\frac{x}{2}:\frac{y}{2}:\frac{z}{2}\right)=\left(x:y:z\right)$.

Now, it is well-known that any triangle can be mapped to any other triangle by an affine transformation. Consider the affine transformation mapping the triangle ABC to the triangle A'B'C'. Let N' be the image of the point N under this transformation. Since affine transformations preserve homogeneous barycentric coordinates, it follows that the homogeneous barycentric coordinates of the point N' with respect to triangle A'B'C' are the same as the homogeneous barycentric coordinates of the point N with respect to triangle ABC. In other words, the homogeneous barycentric coordinates of the point N' with respect to triangle A'B'C' are $\left(x:y:z\right)$. But we know that $\left(x:y:z\right)$ are the homogeneous barycentric coordinates of the orthocenter of triangle A'B'C' with respect to triangle A'B'C'. Hence, the point N' is the orthocenter of triangle A'B'C'. And Lemma 3 is proven.

Now, in order to solve the problem, we apply Lemma 3 and conclude that there is an affine transformation mapping the triangle ABC to a triangle A'B'C' and the point N to a point N' such that the point N' is the orthocenter of triangle A'B'C'. Let this affine transformation map the points D, E, F, M to the points D', E', F', M', respectively; of course, then, the points M' and N' lie inside the triangle A'B'C', and the points D', E', F' lie on the lines B'C', C'A', A'B', respectively. And since affine transformations preserve parallelisms, from MD || NA it follows that M'D' || N'A'. But since the point N' is the orthocenter of triangle A'B'C', we have $N^{\prime}A^{\prime} \perp B^{\prime}C^{\prime}$. Hence, $M^{\prime}D^{\prime} \perp B^{\prime}C^{\prime}$. In other words, the point D' is the orthogonal projection of the point M' on the side B'C' of triangle A'B'C'. Similarly, the points E' and F' are the orthogonal projections of the point M' on the sides C'A' and A'B' of triangle A'B'C'. Hence, the triangle D'E'F' is the pedal triangle of the point M' with respect to triangle A'B'C'. Now, since the point M' lies inside the triangle A'B'C', Lemma 1 tells us that the area of the pedal triangle of the point M' with respect to triangle A'B'C' is less or equal to $\frac14$ of the area of triangle A'B'C'. In other words, the area of triangle D'E'F' is less or equal to $\frac14$ of the area of triangle A'B'C'. But since affine transformations preserve ratios of areas, the ratio between the areas of triangles D'E'F' and A'B'C' equals the ratio between the areas of triangles DEF and ABC. Hence, we see that the area of triangle DEF is less or equal to $\frac14$ of the area of triangle ABC. Problem solved.

Note that, as a special case of your inequality for M = N, we get the following famous inequality: If N is a point inside a triangle ABC, and DEF is the cevian triangle of this point N with respect to triangle ABC (in other words, the points D, E, F are the points of intersection of the lines AN, BN, CN with the lines BC, CA, AB, respectively), then the area of triangle DEF is less or equal to $\frac14$ of the area of triangle ABC.

  Darij


geometry
circumcircle
inequalities unsolved
inequalities

[i]First Day - Budapest, April 9[/i]

Let $ABC$ be an acute-angled triangle. The tangents to its circumcircle at $A,B,C$ form a triangle $PQR$ with $C\in PQ$ and $B\in PR$. Let $C_1$ be the foot of the altitude from $C$ in $\triangle ABC$. Prove that $CC_1$ bisects $\angle QC_1P$.

Excuse me. There is a mistake. This shouldn`t be in this section  :( .

See http://www.mathlinks.ro/Forum/viewtopic.php?t=55918 or http://www.mathlinks.ro/Forum/viewtopic.php?t=59664 .

  darij


George is planning a dinner meeting for 150 people.  Each table will be set for nine people.  If there are enough tables to accommodate everyone, what is the fewest possible number of unused table settings?

Each table seats nine people, so 17 tables will seat 17*9 or 153 people.
However, there are only 150 people at the dinner meeting, so there will be 153-150=$ 3$ unused table settings.


Oh you are right I missed $ n$ and I am suprised at your solution, here is the solution I known.


Write the value of $ 6600 \div .006$ in scientific notation.

This is just $ 6.6\times 10^{3}\div 6\times 10^{\minus{}3}\equal{}\frac{6.6\times 10^{3}}{6\times 10^{\minus{}3}}\equal{}\frac{(6.6\times 10^{3})\times 10^{3}}{6}$.  Now just re-write it.


AMC USA(J)MO USAMO Putnam calculus algebra polynomial

[quote="djimenez"][quote="Scrambled"]Take the first n squares (where n is a variable) and insert plus or minus signs in between them
prove you can get any number
for example
1=1^2
2=-1^2-2^2-3^2+4^2
3=-1^2+2^2
4=1^2-2^2-3^2+4^2[/quote]

I may be wrong, and just loosing the whole point, but I don't think this is a trivial fact at all (If it is true, what I don't know). It sounds similar to perfect square dissection. Do you have any source for this problem or it was just something that poped up of your mind? I mean, may be it is hard, but is nevertheless very interesting![/quote]
i got this from a book, which is a collection of good problems from various sources (including, but not limited to USAMO and putnam, but theres no need for calculus)

I know this is in Zeitz, is that where you got it from?  Also, I started a discussion about this question a while back on the forum.  If you are asking about the question because you need help I'll find the link...if you are just posing a challenge I'll let people try it before giving the linik ;)

[quote="Myth"][quote="djimenez"][quote="Scrambled"]Take the first n squares (where n is a variable) and insert plus or minus signs in between them
prove you can get any number
for example
1=1^2
2=-1^2-2^2-3^2+4^2
3=-1^2+2^2
4=1^2-2^2-3^2+4^2[/quote]

I may be wrong, and just loosing the whole point, but I don't think this is a trivial fact at all (If it is true, what I don't know). It sounds similar to perfect square dissection. Do you have any source for this problem or it was just something that poped up of your mind? I mean, may be it is hard, but is nevertheless very interesting![/quote]
But it is really trivial, just apply well known trick $(n+3)^2-(n+2)^2-(n+1)^2+n^2=4$ !!! Therefore it sufficies to obtain representation for 1, 2, 3 and 4 as showed above ;)[/quote]

 :first:  :first:  :cool:

[quote="Myth"][quote="djimenez"]I may be wrong, and just loosing the whole point, but ... [/quote]
But it is really trivial, just apply well known trick $(n+3)^2-(n+2)^2-(n+1)^2+n^2=4$ !!! Therefore it sufficies to obtain representation for 1, 2, 3 and 4 as showed above ;)[/quote]

Upss!  :blush:  :blush:  I didn't see that!

It's also true for changing the square to another power, or even for a special type of polynomial:
http://www.mathlinks.ro/Forum/viewtopic.php?t=30998


It says we can't use outside resources.
Are we allowed:
1) Calculators?
2) To code computer programs to solve stuff?
3) Printed reference material (formula books, etc.)

1) Definitely
2) This I would say no. In USAMTS, they dont mind if you use programming, but I think programming would fall under an "outside resource." That, and they probably wouldn't give you much credit if you said "Plugging this into Mathematica gives me..."
3) Not sure about

man, the cmor people need to be more clear with their rules.


[geogebra]a5d1d921ed4d1930f0538e3cdbaa82aadaaa9ed3[/geogebra]

Explain why the figure above is impossible given that ...

A' on the same side of BC as A, A' in the interior of triangle ABC, AB = A'B, AC = A'C.

Note that BC is the common segment, and A and A' are the points that are above (it doesn't really matter which you pick to be A and A' (I didn't know how to label them on the Geogebra that this post allows for...)

NOTE: In that book Plane Euclidean Geometry: Theory and Problems, we have not "learned" the SSS Theorem yet, but we do know the SAS Postulate (i.e. prove the above assertion of impossibility using the SAS Postulate).

Any help in this matter is greatly appreciated. I thank you in advance.

Because a circle centered at B having radius AB and a circle centered C having radius AC can't intersect in 2 points on the same side of BC to each other (A and A'), since their intersections must be symmetrical about the line BC.

Cheers,

Rofler

okay.....

But I want to have a more "solid" proof (I totally understand that your intuitive proof works - and if we were to set up the formal geometry, it would totally be a great solution) in the sense that we start with the SAS Postulate (which then proves the Isosceles Triangle Theorem, which are about the only two things that the book has proved before that problem) and then prove the impossibility of this result.

Is anyone willing to take the challenge?

:)

I guess not...  :(

But luckily I found the solution in a footnote in the dover publication of euclid's elements (book one - footnote on proposition 8)

[hide="Hint"]
Introduce three auxiliary lines
[/hide]


Let p \geq 3 be a prime and n \in N. Written in the 'regular' base 10 p^n has 20 digits.

Prove that there is a digit which appears more than twice !

Since there are 10 digits (0 through 9) and p^n has 20 digits, if there is no digit which appears more than twice then all the digits must appear exactly twice. If p>3, p^n has digit sum =2(1+2+..+9), which is divisible by 9, so 9|p^n, which is impossible. I guess this only leaves the case p=3.


We have just revamped our [url=http://www.artofproblemsolving.com/Resources/AoPS_R_Contests.php]directory of contests[/url].  I will add NOML and AHSMIC within the next couple of days, but otherwise please email me at crawford@artofproblemsolving.com if you recognize any omissions from this list.

Kinda unrelated, but I would recommend adding SIMUW (http://www.math.washington.edu/~simuw/) to the list of summer programs. Yes it's only restricted to students from WA but adding it won't flood the list either. :)


hope my solution is correct:
let:$a=\frac y x,b=\frac z y,c=\frac w z,d=\frac x w$
we will find the inequality is equal to:
$\frac{x+z}{x+y}+\frac{y+w}{y+z}+\frac{z+x}{z+w}+\frac{w+y}{w+x} \ge 4$
but use caughy's,we have:
$\left((x+y)(x+z)+(y+z)(y+w)+(z+w)(z+x)+(w+x)(w+y)\right)\left( \frac{x+z}{x+y}+\frac{y+w}{y+z}+\frac{z+x}{z+w}+\frac{w+y}{w+x}\right) \ge 4(x+y+z+w)^2$
but $(x+y+z+w)^2=(x+y)(x+z)+(y+z)(y+w)+(z+w)(z+x)+(w+x)(w+y)$
 :)

[quote="nhat"]$a,b,c,d >0$ satisfy $abcd=1$.Prove that:
             $\frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+cd}{1+c}+\frac{1+da}{1+d} \geq 4$[/quote]
it is equivalent to :
$\sum \frac{2+a+ad}{1+a}\geq 8 \leftrightarrow 2\sum \frac{1}{1+a}+\sum \frac{a(1+b)}{1+a}\geq 8$
because of wellknown $\sum \frac{1}{1+x_i}\geq \frac{n}{2}$ when $\prod x_i=1$
and AM-GM

[quote="Albanian Eagle"]
...because of wellknown $\sum \frac{1}{1+x_i}\geq \frac{n}{2}$ when $\prod x_i=1...$
[/quote]
This is wrong. Try $n=3,$ $x_1=x_2=2,$ $x_3=\frac{1}{4}.$ ;)


please help me in attaching pdf and img files to the post ???? :?:  :(

When you're posting an attachment, you can click the "Add an Attachment" next to the "Message body" tab.


I found a past complete AMC 12 exam on this website, but cannot find the solutions to them.  I don't need complete worked-out solutions; I'd just like the letter-answers to the multiple choice questions.  Does anyone know where I can find this (specifically for the 2004 exam)?

Thank you.

[quote="Silverfalcon"]Here is for 2004:

http://www.unl.edu/amc/e-exams/e5-amc10/e5-1-10archive/2004-10a/04AMC10Aanswers.html

There are no "official" key or full solutions for AMC in this site yet. We hope to have one in future, however.

Please ask this type of question in AMC forum. Moderators, can you move this to AMC forum?[/quote]Woah! I can check my answers now! :wow:

Shouldn't this be in the AMC forum?  ;)


Se considera in plan familia de cercuri $ x^{2}+y^{2}-4x-2\alpha y+3=0$ unde $ \alpha$ este un parametru real.
i) Sa se arate ca exista doua puncte prin care trec toate cercurile familiei.
ii) Sa se determine cercul de raza minima din familie.
iii) Sa se scrie ecuatiile cercurilor din familie care sunt tangente axei $ Oy$.

1) Pentru $ y=0$ rezulta ecuatia $ x^{2}-4x+3=0$ cu solutiile $ x_{1}=1,x_{2}=3$. Deci punctele prin care trec toate cercurile din familie sunt $ A(1,0),B(3,0)$.
2) Ecuatia cercului se poate scrie
$ (x-2)^{2}+(y-\alpha )^{2}=\alpha^{2}+1$ Rezulta ca $ r^{2}=\alpha^{2}+1$, unde $ r$ este raza cercului. Atunci raza este minima pentru $ \alpha =0$.
3) Centrul cercului este $ C(2,\alpha )$. Cercul este tangent axei $ Oy$ daca distanta de la $ C$ la $ Oy$ este egala cu raza cercului. Atunci $ \alpha^{2}+1=4\Rightarrow \alpha =\pm\sqrt{3}$.


abstract algebra
superior algebra
superior algebra unsolved

suppose $ R$ be a commutative ring with unity and $ f: M \longrightarrow  M^{\prime}$ be a R-homomorphism .assume $ I$ be an ideal of $ R$ so that $ f^{*} : \dfrac{M}{IM} \longrightarrow \dfrac{M^{\prime}}{{IM}^{\prime}}$  is a R-homomorphism induced by $ f$ is onto
proove that if $ M^{\prime}$ is finitely generated and  $ I \subseteq J(R)$ then $ f$ is onto.

$ U: \equal{}im(f)$ is a submodule of $ M'$ such that $ I*M'/U \equal{} M'/U$. nakayama's lemma implies $ M'/U\equal{}0$, i.e. $ f$ is onto.

it is not simple to prove  $ I \dfrac{M^{\prime}}{U}\equal{}\dfrac{M^{\prime}}{U}$
how I can do this


[quote="scatteredclouds"]Please find all non-negative integers n, a, b,  n doesnt equal to 0, and (n^b-1) | (n^a+1).[/quote]
Supposition $ \rightarrow  a \ge b$.We have, $ (n^b\minus{}1) | (n^a\plus{}1) \leftrightarrow n^a\plus{}n^b \vdots n^b\minus{}1 \rightarrow n^{a\minus{}b}\plus{}1 \vdots n^b\minus{}1 (gcd(n^b,n^b\minus{}1)\equal{}1)$.With $ a\equal{}kb\plus{}p (0 \le p <b) \rightarrow n^{a\minus{}kb}\plus{}1 \vdots n^b\minus{}1 \rightarrow n^p\plus{}1 \vdots n^b\minus{}1 \rightarrow n^p\plus{}1\equal{}n^b\minus{}1\rightarrow n^b\minus{}n^p\equal{}2 \rightarrow n\equal{}2,p\equal{}1,b\equal{}2,a\equal{}kb\plus{}1$.

[quote="scatteredclouds"]Please find all non-negative integers n, a, b,  n doesnt equal to 0, and (n^b-1) | (n^a+1).[/quote]
[hide]
Since $ n\minus{}1|n^{b}\minus{}1$, we have that
$ n^{a}\plus{}1 \equiv 0 \mod n\minus{}1$
$ 2 \equiv 0 \mod n\minus{}1$
n=3 or 2. Now we just bash cases?[/hide]

If $ b\equal{}0$ or $ n\equal{}1$, the problem is nonsense. Note that $ n\minus{}1|n^a\plus{}1\implies n\minus{}1|1\plus{}1$, so $ n\equal{}2,3$.

Let $ a\equal{}bq\plus{}r$ where $ q,r$ are non-negative integers; $ 0\le r< b$. Then

$ n^b\minus{}1|n^a\plus{}1\implies n^b\minus{}1|n^r\plus{}1\implies n^r\plus{}2\ge n^b\ge n^r\plus{}1$ since $ b>r$.

Hence $ 2\ge n^b\minus{}n^r\ge 1$ and $ n^b\minus{}n^r\equal{}1,2$

If $ r\equal{}0$ and $ n^b\minus{}n^r\equal{}1$, then $ n\equal{}2$, $ b\equal{}1$.
If $ r\equal{}0$ and $ n^b\minus{}n^r\equal{}2$, then $ n\equal{}3$, $ b\equal{}1$.

If $ r>0$, $ n|1,2$, $ n\equal{}1$ is nonsense, so $ n\equal{}2$.
$ 2^{b\minus{}1}\minus{}2^{r\minus{}1}\equal{}1$, so $ r\minus{}1\equal{}0$ and $ b\minus{}1\equal{}1$.

Our solutions are $ (n,b,r)\equal{}(2,1,0), (3,1,0), (2,2,1)$ which correspond to

$ (n,a,b)\equal{}(2,q,1), (3,q,1), (2,2q\plus{}1,2)$ where $ q$ is a non-negative integer.

[one can easily check that these solutions work].


trigonometry
integration
calculus
calculus computations

messed up on previous post the real problem is
∫	(x^4/(1+x^2)^2) dx

Again, you should try to use the same methods on those you previously got responded and see if it works.

having trouble with the long division because equal powers on top and bottom

Remember that $\frac{dx}{1+x^2}$ has *something* to do with $\tan^{-1}x$.  :)

Some algebra, starting from kunny's expression:
$\left(\frac{x^2}{1+x^2}\right)^2=\left(1-\frac1{1+x^2}\right)^2= 1-\frac2{1+x^2}+\frac1{(1+x^2)^2}.$

You should be able to integrate that.

okay im having trouble integrating that i keep coming up with 1/ sec^2 x       please help

please help me get this to a point i can solve.  from the begining i may be making a mistake somewhere

i get

x-1/2 arctan x  + 1/2 x    +   1/4   sin x +  c

is that right for the whole problem

please help

Your answer is absolutely wrong... There would not be $\sin x$.

$\int \frac{x^4}{(1+x^2)^2}dx = \int \left(\frac{x^2}{1+x^2}\right)^2 dx = \int \left(1-\frac{1}{1+x^2}\right)^2 dx = \int 1 dx - 2\int \frac{dx}{1+x^2} + \int\frac{dx}{(1+x^2)^2}$.

From my post,
$\int\frac{dx}{(1+x^2)^2} = \int \frac{1}{1+x^2}dx - \int \frac{x^2}{(1+x^2)^2}dx = \int \frac{1}{1+x^2}dx + \frac{x}{2(1+x^2)} - \frac{1}{2} \frac{dx}{1+x^2}$.

Combining these results,

$\int \frac{x^4}{(1+x^2)^2}dx = x + \frac{x}{2(1+x^2)} - \frac{3}{2}\int \frac{dx}{1+x^2} = x + \frac{x}{2(1+x^2)} - \frac{3\arctan x}{2} + c$


group theory
abstract algebra
superior algebra
superior algebra solved

if all the sylow subgroups of a finite group G, are normal in G, then G is soluble?

It's even nilpotent. $G$ would be the direct product of its sylow subgroups, and since each sylow subgroup is a $p$-group (for various primes $p$) and thus nilpotent, $G$ itself must be nilpotent.

i think it is only true for groups that are abelian

No, it's also true in other cases as shown above. And not to agree with it is no reason to open a new topic, especially in Solved Problems!

i really can't see why it works for general groups
can someone help me?

As grobber proved and I proved again in http://www.mathlinks.ro/Forum/viewtopic.php?t=62847 , it's true for all groups of the type you gave.
Where exactly is your problem¿ Please be more specific at which point you have problems to understand.

i do understand that there is only one sylow p group of G for every prime p dividing the order of G, and clearly the intersection is empty(a part the identity)
but i cant find a prof that A x B is solvable iff A and B are solvable

Let $\{e\} = A_0 \subset A_1 \subset ... \subset A_m=A$ and $\{e\} = B_0 \subset B_1 \subset ... \subset B_n=B$ be normal series with $A_{k+1}/A_k$, respectivly $B_{k+1}/B_k$, being abelian.
Then the normal series
$\{e\} \times \{e\} = A_0 \times B_0 \subset A_1 \times B_0 \subset ... \subset A_m \times B_0 =A \times B_0 \subset A \times B_1 \subset ... \subset A \times B_n= A \times B$
should be a normal series with abelian factors.
The other direction isn't needed here, but can be shown in a similar way.

Another way to prove such facts is to consider the $i$-th commutators.

yes but the problem is the quotient group, how can I be sure it is abelian?

Which ones¿ $(A_{k+1} \times B_0)/(A_k \times B_0)$ is isomorphic to $A_{k+1}/A_k$, and the last one is given to be abelian (we want to conclude from "$A,B$ are solvable" that "$A \times B$ is solvable"). Similar then for the other quotients.

now i am getting really lost.

the only thing i know by hypothesis is that every sylow p-group is normal in G
i cant follow your proof

All what I gave is the prove for $A \times B$ to be solvable iff $A,B$ are. Not more. It has at first nothing to do with the given $G$.

So what is left to prove now is that every Sylow subgroup is solvable. That follows more or less directly from the following: the center of a nontrivial $p$-group is not trivial.


If $ a,b,c>1$ and $ a\plus{}b\plus{}c\equal{}6$ , then prove that : $ \log_a(\sqrt[3]{bc}\plus{}a)\plus{}\log_b(\sqrt[3]{ca}\plus{}b)\plus{}\log_c(\sqrt[3]{ab}\plus{}c)\ge \frac{11}{2}$ .

Please don't  post a solution until 10 a.m. This problem is proposed in an online competition.