[quote="Fraenkel"]$ (ABCD)*4$ = $ DCBA$.

Find $ ABCD$[/quote]

[hide]
A must be even, since DCBA is even.  Also, A must be 1 or 2, or the product will be a 5-digit number.  Therefore, A=2.  If A=2, then D=3 or 8.  Since D can't be 3, or ABCD would only have 3 digits, D=8.  Also, we must have B<3.  Then, $ 4D\equal{}32$, so when the 3 is carried over in the product, we have $ 4C\plus{}3\equal{}B$, so B is clearly odd and thus is 1.  Finally, C=2 or 7, but 2 doesn't make sense, so C=7.  So the answer is $ \boxed{2178}$ (if that's what you're asking)
[/hide]

[quote="Nerd_of_the_Ages"]Well, technically, he never said that A, B, C, and D were digits, so the only answer would be 0.  :D[/quote]

the above answer is correct!

[hide] it could also be 8712[/hide]

[quote="Fraenkel"][hide] it could also be 8712[/hide][/quote]
Reverse the digits. :wink:

I may not be right, but I saw this in the other topics forum, and I believe the title was called "Fun Problem" and it was dealing with pallidromes and went off topic to this problem...

[quote="Nerd_of_the_Ages"]Actually, it was about ABCD * 9, I think[/quote]
Yeah.. I think it was too... Sorry..


function
induction
algebra proposed
algebra

I guess you mean

$\left(e_0-e_2+e_4\pm ...\right)^2+\left(e_1-e_3+e_5\pm ...\right)^2=\prod_{i=1}^n\left(1+x_i^2\right)$.

Anyway, why is this in Combinatorics? It's algebra IMHO...

  dg

Right. It's just that I've been used to dealing with symmetric functions most often in combinatorical contexts. Never mind :roll:

spanferkel, this is obviously true since, $(x^2+y^2)(z^2+t^2)=(xz+yt)^2+(xt-yz)^2.$ ;)

arqady, for me it doesn't look directly obvious from that identity. :? Don't you need some sort of induction? (my proof is direct) :)

Maybe you mean to $\left(|\prod_{k=1}^n(i+x_k)|\right)^2$?

[quote="spanferkel"]Yes, exactly. Really (well, let's say "complexly" :) ) easy with the generating function of the $e_k$, evaluated at $i$  :)
$LHS=\|\sum_{k=0}^ni^ke_k\|^2=\|\prod_{k=1}^n(1+ix_k)\|^2$ :D[/quote]

  Sorry for my curiousity. I didn't know much about complexity. May you elaborate your sols clearly? The idea of using complex number is so new to me.  :)

OK.
The nice fact is that $e_0-e_2+e_4\pm ...$ and $e_1-e_3+e_5\pm ...$ happen to be the real and imaginary parts of the complex number $\sum_{k=0}^ni^ke_k=e_0+ie_1-e_2-ie_3++--...$ and so the LHS is the square of its modulus (norm).
Now, if we define the generating function of the elementary symmetrical functions $e_k$'s by the formal power series $f(t)=\sum_{k\ge0}t^ke_k$, it is well known (and/or easy to see) that $f(t)=\prod_{k\ge1}(1+x_kt)$. (You can find more [url=http://www-igm.univ-mlv.fr/~ace/ACE/1.0/BACKGROUND/STRUCT/manual/node9.html]here[/url], and by the way, Macdonald's book, a classic, is really delightful to read. :) )
Now the proof consists in applying that for $t=i$ and taking norms, that's all.
Hope that helps  ;)


Suppose that a set , say $X$ is partitioned by different methods to $X_{1}, ..., X_{n}$ and $Y_{1}, ..., Y_{n}$ and ,... . Is it possible to find a mutual SDR for these partitions?
Note: as a first try consider only two different patitions.
Example: $X=\{1, 2, 3, 4\}, X_{1}=\{1, 2\}, X_{2}=\{3, 4\}. Y_{1}=\{1, 3\}, Y_{2}=\{2, 4\}.$ Then clearly we have an SDR $\{1, 4\}$.

There are counterexamples: $\{1,2,3,4\}=\{1,2\}\cup\{3\}\cup\{4\}$, and also $\{1\}\cup\{2\}\cup\{3,4\}$. It is true, however, if in both partitions $X_{i},\ i=\overline{1,n}$ and $Y_{j},\ j=\overline{1,n}$ all sets $X_{i},Y_{j}$ have the same cardinality (namely $\frac{|X|}n$). This is a simple application of Hall's Marriage Theorem.


When will the scores for this test be posted?  Many comprehensive and ALG II did most of it, while geometry people like me basically sat there and tried to do the easy ones assigned. which wasnt much

Hahaha

Yeah, I was running around helping the geometers like half the time...

It was fun  :rotfl:

Our math club is pretty sad... we had about 16 people, and assigned 4 people to each 25 problems... and we did... pretty badly. (Even though we have pretty strong individuals)

well...if the group assigned the first 25 problems did better.... :P  i think the group doing the 3rd set also got owned

btw the results are up at [url]http://www.natassessment.com[/url]

yeah the results have been out several days already.  :D

what school are you guys from? I didn't take part in it much, but the older kids did.

hmmm i think my school got DQ'd for undisclosed reasons

Or we suck very bad... on second thought, that's a much more viable alternative :)


geometry
parallelogram
Functional Analysis
advanced fields
advanced fields unsolved

Let $(x_n)_n$ be a bounded sequence in a Hilbert space $\mathcal H$. Prove that it has a subsequence $(x_{n_k})_k$ such that the sequence $(y_m)_m$ defined by $y_m=\frac 1m\sum_{k=1}^mx_{n_k}$ is convergent.

The $x_n$ live in a bounded ball $B(O,M)$ of a Hilbert space. Such a ball is weakly sequentially compact
(because a Hilbert is reflexive). Hence one can suppose that the $x_n$ tends weakly towards a limit $x$.
Moreover one can suppose that this limit is $0$, otherwise one can study $x_n-x$ instead.

  Now, the $y_n$ will strongly converge towards $0$ iff $|y_n|$ tends towards $0$.(it is for example
an easy consequence of the parallelogram formula) But:
$|y_n|^2=\frac{1}{n^2}(\sum_1^n |x_{n_k}|^2 + \sum_{1 \leq i < j \leq n} (x_{n_i}|x_{n_j}) )$
$|y_n|^2 \leq \frac{M^2}{n} + \frac{1}{n^2}(\sum_{1 \leq i < j \leq n} (x_{n_i}|x_{n_j}) )$.
Hence it suffices to make $(x_{n_i}|x_{n_j})$ small, which is easy because $x_n$ tends weakly
towards $0$. For example one can choose the $n_j$ such that $|(x_{n_i}|x_{n_j})| < \frac{1}{j}$ for every $i<j$.


MATHCOUNTS
inequalities
geometry
linear algebra
matrix
HMMT
AMC

We are curious to know a little bit about the makeup of the AoPS community.  If you are beyond high school in your education, please refrain from answering.

Somewhat average non-magnet high school with little math opportunities.

Just a plain old, non-magnet public high school  :|  :)

I'm pretty sure I'm a magnet public school, you gotta wait on a list since birth to get into the school (unless you go into the IB program ^^). But its free.

i go to a magnet school too; we have the IB program as well

just a regular public high school. although it is brand new, first year it opened.

so no math opps, no math club, no amc's, etc.

[quote="236factorial"]I'm from a public one.

What's a magnet school?  :huh:[/quote]

from answers.com:
magnet school
n.
A public school offering a specialized curriculum, often with high academic standards, to a student body representing a cross section of the community.

Shoot, I answered before I read the thing about not answering if you're beyond high school.  Sorry.  I went to a public school until last spring.

[quote="Rushil"]What's a Magent School??? Sorry I live in India - we have different systems here![/quote]

read what I said above

I'm homeschooled. Been so for all my life. For something like, 10 years? Just as all my siblings too.

[quote="Treething"]*starts cracking up*

Never... mind...[/quote]

Nonon, be consistent, explain your "joke"

I attend a college-preparatory private day school for 7-12 grade. The high schoolers are having their finals now so I don't have to attend school most of this week. Only one more official day of school this year! (Thursday, we get our finals back)

Stop quoting everybody! Anyways, I have a friend from Argentina and he would shoot himself if he was sent to an all-boy school.  :D All-boy school doesn't sound good to me either.

This thread has been edited to remove:
1) Repeated abuse of the quote button.
2) Something that was evidently supposed to be humorous.

In the future, please:
1) Use a quotation only if it would be otherwise unclear what you are responding to, and trim away irrelevant parts of the quoted material to avoid the creation of a "ziggurat."
2) Grow up.

public. I wasn't exposed to middle school Mathcounts or AMC or anything which is sad. But I am realllllllllly hooked up in geometry this year.

I got to a [i]very[/i] good and high-ranking public non-magnet school.

Public school.

Private schools are too expensive!!!!

I see no point in going to private schools. I go to a public school and we have 7 ph.D science teachers.

But I think the best system is magnet public schools. I wish my district had magnet schools.

normal public middle school with mathcounts and other good extra-curriculars.

I go to a private school that is somewhat but not intensely religious (Episcopalian). I am liking it for many reasons. (good science and math programs)

I go to a corrupt public middle school in a corrupt public school system in a corrupt town. The only thing good about it is that we have, in my opinion, one of the 5 best middle school math coaches in the country.

Evening School
and i have to work and pay them.


function
algebra
domain
complex analysis
complex analysis unsolved

My friend told me that this is a very easy problem, but I seem couldn't solve it. Wuld u mind helping me?
f is a analytic function with domain D and codomain:Complex field. If Re( f'(z))not equal to 0 for every z in D.
Prove that f is univalent!


Prove that

$n!\geq n^{\frac{n}{2}}$ for all $n\in \mathbb{N}$

(without using calculus of course :P )

[quote="Peter VDD"]Cou count stirling's formula as calculus?  :whistling:

$(n!)^2 > n^n\\$

$n*1\ge n\\
...\\
(n-k)*k > n\\
...\\
(n/2)*(n/2) > n \text{(or floor/ceil of it)}\\
...\\
1*n \ge n$

so it's certainly valid for n>=4. For n<=3 we can check it. :)[/quote]

What does stirling's formula say ? Anyway Im looking for another method (maybe using those standard inequality like AM-GM , cauchy ... )

http://mathworld.wolfram.com/StirlingsApproximation.html -> this is stirling.

Anyone have a different proof ? I have a proof for this inequality , but I want to see whether theres any other way  :?

[quote="Magnara"][hide]Methinks you have to use some level of calculus.  Well, typically calculus results.  Try proving it by induction for $n\geq{2}$ by using the fact that (1+1/n)^n is bounded above by $e$ (and more importantly, by 3).  It can be done using this typically calculus result and no others.  If you'd like to, it can be proven using elementary, but not fun, methods that (1+1/n)^n<3 for every integer n.[/hide][/quote]

I have a method using AM-GM and telescoping   :)  . I will show it if no one else is going to attempt .

[quote="shyong"]I have a method using AM-GM and telescoping   :)  . I will show it if no one else is going to attempt .[/quote]We are waiting. :)

Ok here is how I get it  ;)

[hide]
We use AM-GM on

$\frac{\sum^{n-1}_{k=1}\frac{1}{k(k+1)}}{n-1}\geq \left(\frac{1}{\Pi^{n-1}_{k=1}k(k+1)}\right)^{\frac{1}{n-1}}$

But the RHS is actually $\frac{1-\frac{1}{n}}{n-1}=\frac{1}{n}$

So we get $\frac{1}{n}\geq \left(\frac{n}{(n!)^2}\right)^{\frac{1}{n-1}}$

which simplify to our desired inequality  :)

[/hide]


function
geometry
geometric transformation
reflection
inequalities
number theory
complex numbers

Hi,

I am at a stage, where I am done with the introducion series of AoPS (execpt for Geometry). And I am doing intermediate Algebra (Chapter 6 right now) and Intermediate CP (chapter 3 right now), and also, the book Number Theory: Structures, Examples, and problems.

I am also very much interested in contest mathematics, as well as research mathematics (mainly in the field of Number Theory, more specifically, right now I am very interested about the Riemann Hypothesis, although I am far from actually completely understanding it.) I also love Problems and Puzzles.

So, based on that, what would be a very good Mathematical Magazine/Journal for me?

Thank you very much for your help :)

BTW, maybe a little bit off topic, but I'll try to explain Riemann's the best I can

You have the function $ f(x) \equal{} \frac {1}{1^x} \plus{} \frac {1}{2^x} \plus{} \ldots$ It has infinite fractions, the more you compute, you are closer to its value.

Let's find the zeros of $ f(x)$, in other words, we must find the numbers $ x$ for which $ f(x) \equal{} 0$.

In real numbers, such numbers are $ \minus{} 2, \minus{} 4,\ldots$

But if we turn to complex numbers, which are $ a \plus{} bi$, where $ a$ and $ b$ are real numbers, and $ i$ is the imaginary unit, $ i^2 \equal{} \minus{} 1$, then the hypothesis says:

[i]If $ a \plus{} bi$ is a zero of the function $ f$, then $ a \equal{} \frac {1}{2}$[/i]

On topic: Mathematical Reflections is an olympiad resource, but NT: SEP is also.

Bugi, thank you very much for that explanation.

[quote]Let's find the zeros of f(x), in other words, we must find the numbers x for which x=0. [/quote]

You mean: we must find the number x for which for which $ f(x)\equal{}0?$ :P

Actually, shouldn't $ a$ be $ \plus{}\frac{1}{2}?$

So, basically the question is that:

Prove that all the zeroes of the zeta function lie of the line $ \Re(s) \equal{} \frac{1}{2}?$

So doesn't this make it more of an algebra problem than Number Theory?

And also, I saw the movie "The Music of the Primes" an they said that there were several such lines along the Complex Plane. So why do people only focus on the line $ \Re(s) \equal{} \frac{1}{2}?$

Thanks again Bugi :)

IDK about magazines... But MR's problem section is useful. You can do Juniors' problems

[quote="limac"]About the magazines, the one you suggested, Bugi, is right now I guess a little too tough for me. Is there any at my level?[/quote]

I know this a revival, but the magazine Crux Mathematicorum has a lot of good problems, ranging from, say, AMC 10-ish problems to Olympiad problems.  Unfortunately, many of their inequalities require a lot of bashing.

Crux seems to be OK for me. I am gonna check it out. Thank you very much AIME15 :)


geometry
parallelogram
trapezoid
geometry proposed

Let a right-angled triangle $ABC$. We constuct the squares $CADF$  and $ABEP$ at the exterior of the triangle . .Let $K$ the intersection of the lines $EC$ and $BF$ .Prove that $AK$ passes through the midpoint $M$ of $PD$ .

If right angle is in $A$ then there is a very simple (and short!) analytic proof. Otherwise, this fact is quite obvious.

this sentence isnt the proof.

(it is equal to prove that $AK$ is perpendicular to $BC$)

First we can prove $CE\bot BQ, AQ\bot BC, BF\bot CQ$.

Then $CE, AQ, BF$ meet at $K$, that is, $AK\bot BC$.

Construct the srquares at the exterior of the triangle .

this is true for any angle A, please look at:
http://agutie.homestead.com/files/index.html
Theorems and problems
Triangle and squares case 2 and case 5.

It is also true for not right angle at vertex A and for similar rectangular parallelograms instead of squares.

     If we draw two lines Lb, Lc, through vertices B, C respectively and perpendicular to the sideline BC, then we will have the configuration of the ''isogonic'' theorem ( the lines AE, AF, are isogonal with respect to the angle A, the lines BE, Lb, are isogonal with respect to the angle B and the lines CF, Lc, are isogonal with respect to the angle C ).

     So, the lines BF, CE and the altitude of ABC through vertex A ( as the line connecting the vertex A with the intersection point of Lb, Lc, at infinity ), are concurrent.

     Let D', P' be, the orthogonal projections of D, P respectively, on the sideline BC. It is easy to prove that the triangle AP'D', is isosceles ( from cyclic quadrilaterals ABP'P, ACD'D and the similarity of the parallelograms CADF, ABEP ).

     Hence the line AK is midperpendicular of the segment D'P',and it is clear that passes through the midpoint of the segment PD ( from the trapezium DD'P'P ).

     The ''isogonic'' theorem states that if we draw two lines through each vertex of a given triangle and isogonal with respect to the corresponded angle, then the lines connecting each vertex with the intersection point of the isogonal lines through the other vertices of the given triangle, are concurrent.

     This theorem is a more general formulation of Jacobi theorem and you can find two it’s simple synthetic proofs in Hyacinthos forum.

     http://groups.yahoo.com/group/Hyacinthos/message/13680

     http://groups.yahoo.com/group/Hyacinthos/message/13685

     Kostas Vittas.


trigonometry geometry circumcircle rectangle

[hide="solution"]
We have $  PT=\frac{1}{2},ST=\frac{\sqrt{3}}{2},SR=\sqrt{3},PR=2,TR=\frac{3}{2}$.
Let $ RU\perp PQ$ and $ {\widehat{URQ}=\alpha}$
Then $  \frac{PQ}{\sin(90+\alpha)}=\frac{PT}{\sin(60-\alpha)}$ (1)
$  \frac{PQ}{\sin(60+\alpha)}=\frac{PR}{\sin(90-\alpha)}$ (2)
From (1) and (2) $  \Rightarrow \cos^{2}\alpha=4\sin(60-\alpha)\sin(60+\alpha)\Rightarrow \cos\alpha=\frac{1}{\sqrt{3}}\Rightarrow\sin\alpha=\sqrt{\frac{2}{3}}$
Then $ PQ=\sqrt{3}+\sqrt{2}$
[/hide]

[quote="wonki_noh"]How did you solve $ \cos^{2}(\alpha)=4\sin(60-\alpha)\sin(60+\alpha)$?[/quote]

$ \cos^{2}\alpha=2(\cos 2\alpha-\cos 120)\Leftrightarrow\cos^{2}\alpha=2\cos 2\alpha+1\Leftrightarrow$
$ \Leftrightarrow\cos^{2}\alpha=4\cos^{2}\alpha-1\Leftrightarrow \Leftrightarrow \cos^{2}\alpha=\frac{1}{3}$
$ \alpha\in(0,90)\Rightarrow\cos\alpha=\frac{1}{\sqrt{3}}$

[quote="wonki_noh"]$ PQRS$ is a quadrilateral right angled at $ P$ and at $ S$. $ T$ is a point on the diagonal $ PR$ and $ ST$ is perpendicular to $ PR$. The angles $ TPQ$, $ TSP$, $ TRS$, and $ TQR$ are 30 degrees. $ PS$ = 1 unit. Then, find the length of $ PQ$.[/quote]
[hide]$ PT = \frac{1}{2}, \ PR = 2,\ SR = \sqrt{3}$

$ \angle TQR = \angle TRS$, the circumcircle of $ \triangle TQR$ touches $ SR$ at $ R$

Let $ O$ be the circumcenter of $ \triangle TQR$

$ OR\perp SR \Rightarrow OR \perp PQ$. Let $ K\in PQ \cap OR$. Then $ PKRS$ is a rectangle, so $ PK = SR = \sqrt{3}$

Suppose that the circumcircle intersects $ PQ$ at two points $ Q,Q'$

Let $ x=KQ=KQ'$ (if we find $ x=0$ this would mean that the solution is only one point $ Q\equiv Q'$)

$ PT\cdot PR = PQ\cdot PQ' \Rightarrow$

$ \frac{1}{2}\cdot 2 = (\sqrt{3}-x)(\sqrt{3}+x) \Rightarrow 1 = 3-x^{2}\Rightarrow x=\sqrt{2}$

So we have two possible points $ Q,Q'$ with respective distances $ PQ=\sqrt{3}+\sqrt{2}$ and $ PQ'=\sqrt{3}-\sqrt{2}$[/hide]


geometry rectangle incenter circumcircle trigonometry vector search

In a triangle $ ABC$ (non rectangle), $ H$ and $ I$ are the orthocenter and incenter respectively. Let $ E$, $ F$ be the projections of $ B$, $ C$ on $ AC$, $ AB$  respectively. Prove that the lines $ BC$, $ EF$, $ HI$ are concurrent if and only if
\[ r_ah_a\equal{}r_br_c
\]
where $ r_a$, $ r_b$, $ r_c$ are the exradii and $ h_a$ is the $ A$-altitude.

$ A$ is the internal similarity center of the B-, C-excircles $ (I_b), (I_c),$

$ h_a \equal{} r_b \plus{} (r_c \minus{} r_b)\ \frac {r_b}{r_b \plus{} r_b} \equal{} \frac {2r_br_c}{r_b \plus{} r_c}.$

Let $ A'$ be the midpoint of $ BC.$ The external bisector $ I_bAI_c$ of the $ \angle A$ cuts the circumcircle $ (O, R)$ again at the midpoint $ M$ of the arc $ BC$ containing $ A.$ Thus

$ \frac {r_b \plus{} r_c}{2} \equal{} MA' \equal{} R(1 \plus{} \cos A)$

Thus $ r_br_c \equal{} r_ah_a\ \Longleftrightarrow\ r_a \equal{} \frac {r_b \plus{} r_c}{2}\ \Longleftrightarrow\ r_a \equal{} R(1 \plus{} \cos A).$

Substituting $ \cos B \plus{} \cos C \minus{} \cos A \equal{} \frac {r_a}{R} \minus{} 1,$

$ r_br_c \equal{} r_ah_a\ \Longleftrightarrow\ \cos B \plus{} \cos C \equal{} 2 \cos A.$
______________________________________

$ D \in BC$ is foot of the A-altitude $ AH$ and $ X \in BC$ foot of the internal bisector $ AI$ of the $ \angle A.$ $ EF$ cuts $ BC$ at $ P$ and $ IH$ cuts $ BC$ at $ Q.$

$ 1 \plus{} \frac {\overline{BC}}{\overline{PB}} \equal{} \frac {\overline{PC}}{\overline{PB}} \equal{} \minus{} \frac {\overline{DC}}{\overline{DB}}\ \Longleftrightarrow$

$ \overline{PB} \equal{} \minus{} \frac {\overline{DB} \cdot \overline{BC}}{\overline{DB} \plus{} \overline{DC}} \equal{} \frac {\overline{BD} \cdot \overline{BC}}{\overline{DC} \minus{} \overline{BD}} \equal{} \frac {ca \cos B}{b \cos C \minus{} c \cos B}.$

By Menelaus theorem for the $ \triangle ADX$ cut by the transversal $ QHI,$

$ \frac {\overline{QD}}{\overline{QX}} \cdot \frac {\overline{IX}}{\overline{IA}} \cdot \frac {\overline{HA}}{\overline{HD}} \equal{} 1\ \Longleftrightarrow$

$ \frac {\overline{QD}}{\overline{QX}} \equal{} \frac {\overline{QB} \plus{} \overline{BD}}{\overline{QB} \plus{} \overline{BX}} \equal{} \frac {\overline{IA}}{\overline{IX}} \cdot \frac {\overline{HD}}{\overline{HA}}\ \Longleftrightarrow$

$ \frac {\overline{QB} \plus{} c \cos B}{\overline{QB} \plus{} \frac {ca}{b \plus{} c}} \equal{} \frac {b \plus{} c}{a} \cdot \frac {\cos B \cos C}{\cos A}\ \Longleftrightarrow$

$ \overline{QB} \equal{} \frac {ca \cos B\ (\cos C \minus{} \cos A)}{a \cos A \minus{} (b \plus{} c) \cos B \cos C}$

Since $ \angle B, \angle C \neq 90^\circ,$ we have $ \cos B, \cos C \neq 0.$ Combining the two expressions, $ P \equiv Q$ are identical $ \Longleftrightarrow$ $ \overline{PB} \equal{} \overline{QB}\ \Longleftrightarrow$

$ \frac {ca \cos B}{b \cos C \minus{} c \cos B} \equal{} \frac {ca \cos B\ (\cos C \minus{} \cos A)}{a \cos A \minus{} (b \plus{} c) \cos B \cos C}\ \Longleftrightarrow$

$ a \cos A \minus{} (b \plus{} c) \cos B \cos C \equal{} (b \cos C \minus{} c \cos B)(\cos C \minus{} \cos A)\ \Longleftrightarrow$

$ (a \minus{} c \cos B) \cos A \equal{} b \cos C\ (\cos B \plus{} \cos C \minus{} \cos A) \ \Longleftrightarrow$

(using the cosine theorem formulas on the left side)

$ b \cos C \cos A \equal{} b \cos C\ (\cos B \plus{} \cos C \minus{} \cos A) \ \Longleftrightarrow$

$ 2 \cos A \equal{} \cos B \plus{} \cos C.$

Denote by $ D$ the projection of $ I$ onto $ BC$ and let $ X\equal{} EF \cap BC$.
We know that $ XI \perp AD$. Then  $ HI, BC, EF$ are concurrent iff $ IH \perp AD$,
so quadrialteral $ AHDI$ must be orthogonal. Hence $ AH^2\plus{}ID^2\equal{}AI^2\plus{}HD^2$.
Now all upper lengths can be found and only tedious calculation remains.

[quote="prowler"]Denote by $ D$ the projection of $ I$ onto $ BC$ and let $ X \equal{} EF \cap BC$.
We know that $ XI \perp AD$. Then  $ HI, BC, EF$ are concurrent iff $ IH \perp AD$,
so quadrialteral $ AHDI$ must be orthogonal. Hence $ AH^2 \plus{} ID^2 \equal{} AI^2 \plus{} HD^2$.
Now all upper lengths can be found and only tedious calculation remains.[/quote]

Two problems with this solution. First, $ IX$ or $ IH$ are not perpendicular to the A-Gergonne line $ AD,$ because $ E, F$ are feet of the B-, C-altitudes and not the incircle tangency points with $ CA, AB.$ Second, you skipped the tedious calculation, which would have raised a warning flag.

This problem can be solved by the vector method.
See at [url]http://www.mathlinks.ro/viewtopic.php?p=1178726[/url]

Sorry yetti, I read too quickly the statement. I thought $ E$ and $ F$ are projections from $ I$.

But anyway the idea is the same. Let $ E$ and $ F$ be the projections of $ B$ and $ C$
onto $ AC$ and $ AB$ and let $ M$ be the midpoint of $ BC$.
If $ X \equal{} EF \cap BC$, then  $ HX \perp AM$, because $ XH$ is the polar of $ A$
with respect to the circle $ BEFC$, whose center is $ M$.
Thus $ HI \perp AM$ and $ AHMI$ is orthogonal quadrilateral.
So $ AH^2 \plus{} IM^2 \equal{} AI^2 \plus{} HM^2$. It follows that

$ AH^2 \plus{} \frac {BI^2}{2} \plus{} \frac {CI^2}{2} \equal{} AI^2 \plus{} \frac {BH^2}{2} \plus{} \frac {CH^2}{2}$.

Because $ AH^2 \equal{} 4R^2 \minus{} a^2$ and $ AI^2 \equal{} bc \minus{} \frac {abc}{p}$, we get

$ ac \plus{} ab \minus{} 2bc \equal{} 2a^2 \minus{} b^2 \minus{} c^2$

which is equivalent to $ a(b \plus{} c \minus{} a) \equal{} (a \plus{} c \minus{} b)(a \plus{} b \minus{} c)$ or $ a(p \minus{} a) \equal{} 2(p \minus{} b)(p \minus{} c),$ that is

$ \frac {S}{p \minus{} b}\cdot \frac {S}{p \minus{} c} \equal{} \frac {2S}{a}\cdot \frac {S}{p \minus{} a}$,

or $ r_br_c \equal{} r_ah_a$.

[quote="mr.danh"]This problem can be solved by the vector method.
See at http://www.mathlinks.ro/viewtopic.php?p=1178726[/quote]
Here is a vector solution.

[quote="Leon"] [color=darkred]In $ \triangle ABC$ (non right) with the orthocenter $ H$ and the incenter $ I$ denote $ \left\|\begin{array}{c} E\in AC\ ,\ BE\perp AC \\
 \\
F\in AB\ ,\ CF\perp AB\end{array}\right\|$ .

 Prove that $ \boxed {\ BC\cap EF\cap HI\ne\emptyset\ \Longleftrightarrow\ r_ah_a \equal{} r_br_c\ \Longleftrightarrow\ 2\cos A \equal{} \cos B \plus{} \cos C\ }$ .[/color] [/quote]

[color=darkblue][b][u]Proof (similarly with Prowler's).[/u][/b]

Observe that $ \left\|\begin{array}{ccc} HA^2 \equal{} \left(2R|\cos A|\right)^2 \equal{} 4R^2\cos ^2A \equal{} 4R^2\left(1 \minus{} \sin^2A\right) \equal{} 4R^2 \minus{} a^2 & \implies & \boxed {\ HA^2 \equal{} 4R^2 \minus{} a^2\ } \\
 \\
IA^2 \equal{} \frac {bc(p \minus{} a)}{p} \equal{} bc \minus{} \frac {abc}{p} \equal{} bc \minus{} 4Rr & \implies & \boxed {\ IA^2 \equal{} bc \minus{} 4Rr\ }\end{array}\right\|\ \ (1)$ .

Denote the power $ p(X)$ of the point $ X$ w.r.t. the circumcircle $ w$ of $ \triangle ABC$ . Construct the midpoint $ M$ of the side $ [BC]$ . From the [b]

Lagrange's identity[/b] $ 3\cdot XG^2 \equal{} XA^2 \plus{} XB^2 \plus{} XC^2 \plus{} 3\cdot p(G)$ (for any $ X$ ) obtain  $ \left\|\begin{array}{ccc} HG^2 & \equal{} & \frac 13\cdot\left(HA^2 \plus{} HB^2 \plus{} HC^2\right) \plus{} p(G) \\
 \\
IG^2 & \equal{} & \frac 13\cdot\left(IA^2 \plus{} IB^2 \plus{} IC^2\right) \plus{} p(G)\end{array}\right\|\ \ (2)$ .

Therefore, $ \boxed {\ BC\cap EF\cap HI\ne\emptyset\ }$ $ \stackrel{(*)}{\ \Longleftrightarrow\ }$ $ HI\perp AM\ \Longleftrightarrow\ HI\perp AG$ $ \Longleftrightarrow$ $ HG^2 \minus{} HA^2 \equal{} IG^2 \minus{} IA^2$ $ \stackrel {(2)}{\ \Longleftrightarrow\ }$

$ \minus{} 2\cdot HA^2 \plus{} HB^2 \plus{} HC^2 \equal{} \minus{} 2\cdot IA^2 \plus{} IB^2 \plus{} IC^2$ $ \stackrel {(1)}{\ \Longleftrightarrow\ }$ $ 2a^2 \minus{} b^2 \minus{} c^2 \equal{} \minus{} 2bc \plus{} ac \plus{} ab$ $ \Longleftrightarrow$ $ a^2 \minus{} (b \minus{} c)^2 \equal{} a(b \plus{} c \minus{} a)$ $ \Longleftrightarrow$

$ \boxed {\ 2(p \minus{} b)(p \minus{} c) \equal{} a(p \minus{} a)\ }$ $ \Longleftrightarrow$ $ 2\cdot\frac {S}{r_b}\cdot\frac {S}{r_c} \equal{} \frac {2S}{h_a}\cdot\frac {S}{r_a}$ $ \Longleftrightarrow$ $ \boxed {\ r_br_c \equal{} h_ar_a\ }$ $ \Longleftrightarrow$ $ \frac {a}{2p} \equal{} \frac {(p \minus{} b)(p \minus{} c)}{p(p \minus{} a)}$ $ \Longleftrightarrow$ $ a \equal{} (a \plus{} b \plus{} c)\tan^2\frac A2$ $ \Longleftrightarrow$

$ a\left(1 \minus{} \tan^2\frac A2\right) \equal{} (b \plus{} c)\tan^2\frac A2$ $ \Longleftrightarrow$ $ a\cos A \equal{} (b \plus{} c)\sin^2\frac A2$ $ \Longleftrightarrow$ $ \sin A\cos A \equal{} (\sin B \plus{} \sin C)\sin^2\frac A2$ $ \Longleftrightarrow$

$ 2\sin\frac A2\cos\frac A2\cos A \equal{} 2\cos\frac A2\cos\frac {B \minus{} C}{2}\sin^2\frac A2$ $ \Longleftrightarrow$ $ \cos A \equal{} \cos\frac {B \minus{} C}{2}\cos\frac {B \plus{} C}{2}$ $ \Longleftrightarrow$ $ \boxed {\ 2\cos A \equal{} \cos B \plus{} \cos C\ }$ .

$ \blacktriangleright^{(*)}$ [b]Remark.[/b] If $ L\in BC\cap EF$ , then the property $ LH\perp AM$ is well-known (see [b]P1.[/b] from[/color] [url=http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1695143161&t=46146][color=red][b]here[/b][/color][/url]). [color=darkblue]Thus, $ BC\cap EF\cap HI\ne\emptyset \Longleftrightarrow\ HI\perp AM$ .[/color]


A particular number in base 3 requires 3 digits(---). When the number is written in base three and base four, the digits are the reverse of each other. QWhat is the number in base 10

Assume the digits are $ a$,$ b$, and $ c$.(In base 3)
$ abc base 3 \equal{}  c \plus{} 3b \plus{} 9a$(In base 10)
So in base 4......the number will be $ cba\equal{} a\plus{} 4b \plus{} 16c$$ (In base 10)$
Now equate!
$ c \plus{} 3b \plus{} 9a$ = $ a \plus{} 4b \plus{} 16c$
So 15c + b = 8a.
So a=2, c=1, and b=1.
$ 211 base 3$ = $ 22 base ten$...so the answer here is $ 22$!!!!
$ 22$

AIME, you make a base, do this:

[code]$x_y$[/code]

Where $ x$ is the number and $ y$ is your base.  Like this: $ 211_3$

[quote="AIME"]Assume the digits are $ a$,$ b$, and $ c$.(In base 3)
$ abc_3 = c + 3b + 9a$(In base 10)
So in base 4......the number will be $ cba = a + 4b + 16c_{10}$
Now equate!
$ c + 3b + 9a$ = $ a + 4b + 16c$
So $ 15c + b = 8a.$
So $ a=2, c=1,$and$ b=1.$
$ 211_3$ = $ 22_{10}$...so the answer here is $ 22$!!!!
$ \boxed{22}$[/quote]

fixed your $ \text{\LaTeX}$  :)


I see the resources and stuff. I have a question. How do we add to that?

I think you have to be in the RManagers usergroup


function
complex analysis
complex analysis unsolved

Suppose that $ f(z) \equal{} z \plus{} \dots$ is an entire function and $ g \equal{} z \plus{} \dots$ is analytic in the unit disk and satisfies $ f(z) \equal{} g(g(z))$ in some open disk centered at the origin. Is it true that $ g$ can be analytically extended to an entire function?

Just put $ g(z)\equal{}\minus{}\frac{z}{z\plus{}1} \ldots$

Erm, this starts with $ {\color{red}\minus{}}z$ :roll:

There are no linear fractional transformation counterexamples. A linear fractional transformation that "starts with $ z$" could always be written as $ f(z)\equal{}\frac{z}{az\plus{}1},$ and $ f(f(z))\equal{}z$ then forces $ a\equal{}0.$

Yes, fedja, I am sorry  :blush:

I can show that there are no analytic in
the whole plane except isolated singularities
functions which satisfy the conditions,
but this does not solve the problem.


Suppose $ a,b,c$ are three sides of a triangle.

Prove that
$ a^2(b\plus{}c\minus{}a)\plus{}b^2(c\plus{}a\minus{}b)\plus{}c^2(a\plus{}b\minus{}c)\le 3abc$

this should be in the inequalities forum

[quote="aadil"]a+1=2(bc)^1/2
b+1=2(ac)^1/2
c+1=2(ab)^1/2
work from here[/quote]

I can't understand why you could get  $ a\plus{}1\equal{}2(bc)^{1/2},b\plus{}1\equal{}2(ac)^{1/2},c\plus{}1\equal{}2(ab)^{1/2}$ for start.

It seems that the most common way to solve it is using notation $ a\equal{}x\plus{}y,b\equal{}y\plus{}z,c\equal{}z\plus{}x$
So the question $ \Leftrightarrow \sum (x\plus{}y)^2(2z)\le 3(x\plus{}y)(y\plus{}z)(z\plus{}x)$
$ \Leftrightarrow 6xyz \le \sum xy^2\plus{}\sum x^2 y$, which is obviously true.

[quote="PKMathew"]Suppose $ a,b,c$ are three sides of a triangle.

Prove that
$ a^2(b \plus{} c \minus{} a) \plus{} b^2(c \plus{} a \minus{} b) \plus{} c^2(a \plus{} b \minus{} c)\le 3abc$[/quote]
$ a,b,c$ doesn't have to be sides of a triangle ;) They just have to be positive.
It's equivalent to $ a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc \ge a^2b \plus{} a^2c \plus{} b^2a \plus{} b^2c \plus{} c^2a \plus{} c^2b$ which is Schur....


Mario throws a fireball across a cliff that's 20 feet wide and 1000 feet deep.  It starts 2 feet above the edge of the cliff, travels in a parabolic path, and lands right at the corner of the other side of the cliff.  See the diagram.

If the fireball reaches its peak height when it has a horizontal distance of 8 feet from the edge of the cliff Mario is on, compute its maximum height (i.e., the purple line).

The diagram is not to scale.  It shows three places that the fireball passes through.

Sorry if this is a bit hard for HSB...

[hide] Let's plot this on a coordinate graph. The ball begins at $(0,2)$ and ends at $(20,0)$. Also the x-coordinate of the vertex is $8$ so $-\frac{b}{2a}=8\Longrightarrow b=-16a$.

So plugging in $(20,0)$, we get $a(20)^{2}-16a(20)+2=0\Longrightarrow a=-\frac{1}{40}\Longrightarrow b=\frac{2}{5}$

$-\frac{1}{40}(8)^{2}+\frac{2}{5}(8)+2=-\frac{8}{5}+\frac{16}{5}+2=3.6$

So $\boxed{1003.6}$. [/hide]


calculus
derivative
integration
function
calculus computations

I need to find the derivative of integral -2cos x(dx) on the interval [x,1] and the average value of f(x)=2x on the interval [-1,1]

I should have posted this here, instead of the other forum. But any help is appreciated

The average value of an odd function on an interval symmetric about zero is zero - you can say that without calculating.

If you don't feel comfortable without something written down, then that average value is

$\frac12\int_{-1}^12x\,dx.$

Why that $\frac12$ ? Because we divide by the length of the interval.

Your first question involves an abuse of notation; you're making the letter $x$ stand for too many things at once, and as a result, the question is meaningless.

By the way, two things: only post problems once; if they're in the wrong place, a moderator will move them. And the right place for this kind of problem is here, Calculus Computations and Tutorials.


limit
logarithms
real analysis
real analysis unsolved

Assume $ a_{i}>0$, for $ \forall$ $ i \in \mathbb{N}^\plus{}$,

show that:

$ \lim_{n \rightarrow \infty}n\left(\frac{a_{n}}{a_{n\plus{}1}}\minus{}1 \right)$

existed implies

$ \lim_{n \rightarrow \infty}\frac{\ln\left(\frac{1}{a_{n}}\right)}{\ln n}$

exists

and they have the same limitation
 :D

If the limit of the first expression is $ L\neq 0$

then $ \frac{a(n)}{a(n\plus{}1)}\sim 1\plus{} L/n$

and

$ \ln(a(n))\minus{}ln(a(n\plus{}1))\sim  L/n$

and we have equivalence of partial sum

$ ln(a(1))\minus{}ln(a(n))\sim  L.ln(n)$ since $ \sum_{k\equal{}1}^{n}1/k \sim  ln(n)$


Glory to God,

I am online right now, please help me with these math problems:

Why does raising a number to the nonzero power gives you an answer of one?

[quote="kameela"]
Why does raising a number to the nonzero power gives you an answer of one?[/quote]

Do u mean raising to the zero power?

Is there a formula or is this just a rule?

Ex:  5 to  the zero power

I think you can think of it this way...
x^0=x^(-1)*x^1
(Because -1+1=0)
x^(-1) is 1/x, and x^1 is x. (1/x)*x is obviously 1. So x^0=1.
Eh, I can't remember how to reach the conclusion that a power with a negative exponent is the reciprocal of the base to the additive inverse of the exponent...

It it has been agreed upon that $0^0=1$ and my TI-89 would agree   ;) .  Here's a link that should help: [url]http://mathforum.org/dr.math/faq/[/url]

[quote="joml88"]It has been agreed upon that $0^0=1$[/quote]

I disagree.  This has not been "agreed upon".  The article explains this is a controversial definition.  Another  [url=http://mathforum.org/library/drmath/view/55764.html]Ask Dr. Math article[/url] refers to 0^0 as:  "This, too, is an indeterminate form.  Its logarithm is 0 * infinity."

Without knowing the context in which the expression is used, you cannot blindly assume that 0^0 is always 1.

On the other hand, you are always safe in treating expressions of the form $0^0$ as indeterminate.  [This merely means that you must take additional steps (such as application of L'Hospital's Rule) to justify the value of the expression.]

[quote="LynnelleYe"]Eh, I can't remember how to reach the conclusion that a power with a negative exponent is the reciprocal of the base to the additive inverse of the exponent...[/quote]

It's based on the rule that (a^b)/(a^c)=a^(b-c). So if we have a negative exponent, we can let b=0, and we get (1)/(a^c).

We can directly show that a^0=1 based on (a^b)/(a^c)=a^(b-c)

2e4=16
2e3=8
2e2=4
2e1=2
2e0=1
2e(1/2)=.5
2e(1/4)=.25

notice each power multiplies by 2 obviously so if 2e1=2 then 2e0=1

replace 2 with n and the answer with the nem
ans/n will always equal the power below it and if you keep it up
ne1=n
ne0=ne1/n=1

Hello, kameela!

[quote]Why does raising a number to the zero power gives you an answer of one?[/quote]

I've taught that it comes from the Division Rule for Exponents:  $\frac{a^m}{a^n} \;= \;a^{m-n} $

So, for example: $\frac{a^5}{a^3} \;= \;a^{5-3} \;= \;a^2$ . . . and: $\frac{a^2}{a^6} \;= \;a^{2-6} \;= \;a^{-4} \;= \;\frac{1}{a^4}$

Then someone asks, "What about $\frac{a^5}{a^5}$ ?" ... (What if the exponents are equal?)

According to the Division Rule, we get: $a^{5-5} \;= \;a^0$ ... but what is that?

Hey, we already know that $\frac{x^5}{x^5} = 1$, don't we? ... ("A thing divided by itself is one", right?)

So, although $a^0$ looks strange to us, we have to agree that it equals one (1).

. . . And [i]that[/i] is why $a^0 = 1.$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

You asked about $0^0$ . . . it is designated "indeterminate".

Where could such an expression come from?

One possibility is: $\frac{0^3}{0^3} \;= \;0^{3-3} \;= \;0^0$
But the problem is silly at the very beginning: $\frac{0^3}{0^3} \;= \;\frac{0}{0} \;= \;?$

The "number' $\frac{0}{0}$ is called "indeterminate".
It can be any number . . . It has [b]too many[/b] values.
So, "they" have desingated $0^0$ as another "useless number".

This is different from another "useless number": $\frac{2}{0}$
This is called "undefined" because there is [b]no[/b] value for it.

Both $\frac{2}{0}$ and $\frac{0}{0}$ are mathematical "garbage", but there are [i]two[/i] kinds of garbage, you see.
Very few teachers ever notice that distinction, let alone explain it.


Verify:

$ cos (2x) / (1 \plus{} sin (2x)) \equal{} (cot x \minus{} 1) / (cot x \plus{} 1)$

We have

$ \frac{\cot x\minus{}1}{\cot x\plus{}1}\equal{}\frac{\frac{\cos x}{\sin x}\minus{}1}{\frac{\cos x}{\sin x}\plus{}1}\equal{}\frac{\cos x\minus{}\sin x}{\cos x\plus{}\sin x}\equal{}\frac{\cos x\minus{}\sin x}{\cos x\plus{}\sin x}\cdot\frac{\cos x\plus{}\sin x}{\cos x\plus{}\sin x}$

$ \equal{}\frac{\cos^2x\minus{}\sin^2x}{\cos^2x\plus{}\sin^2x\plus{}2\sin x\cos x}\equal{}\frac{\cos(2x)}{1\plus{}\sin(2x)}$, as desired.


counting
derangement
probability
combinatorics unsolved
combinatorics

N people are sat in n seats, how many rearrangements of the people are there such that no one is sat in their original seat? Thanks

If $ N \equal{} n$, this is the problem of counting derangements, and the numbers are called by names such as rencontres numbers or subfactorials.  The associated sequence is [url=http://www.research.att.com/~njas/sequences/A000166]A000166[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url].  Searching for any of the keywords in the first sentence, or reading through the OEIS entry and following links to resources, will provide plenty information.

I've used this as an example early on in a probability course. One thing I do is invite speculation. As JBL did, I assumed that $ N \equal{} n$ -- that it was the same number of seats as people. I ask the question in this form: what is the probability that at least one person sits in his or her original seat? The speculation I invite is to ask what happens in the limit as $ n$ gets large. Of course, many students are too cowardly to guess, but among those that do guess there are always some who say that the probability tends to 1 as $ n\to\infty,$ while there are always others who say that the probability tends to $ 0$ as $ n\to \infty.$ I don't point any fingers of blame towards either faction here - this is a very hard problem to gain reasonable intuition over - but both factions are wrong.

The particular things I use this as en exercise in are the law of inclusion-exclusion and the uses of conditional probability. In particular, let $ E_i$ be the event that the $ i$th person returns to the $ i$th seat. Clearly $ P(E_i) \equal{} \frac1n.$ What we have to work for is $ P(E_i\cap E_j) \equal{} P(E_i)P(E_j\,|\,E_i),$ and then $ P(E_i\cap E_j\cap E_k)$ by a similar conditional chain, and so on.


[quote="mathnerd314"]mathnerd314's average on 10 ten-point quizzes is 7.8. He scored at least one 5, at least three 7's, at least two 9's, and at least one perfect score. What is the sum of all the distinct positive values for mathnerd314's median score?[/quote]
[hide]
Is the answer 22.5? I'll post my solution if it is, as I'm not so sure about this answer.
[/hide]

[quote="Ignite168"][quote="Captain Sugar"][hide]i found that he needed 24 more points to get his average of 7.8.  then, i took all the scores he gave me and put them in a line from smallest to largest. then i plugged a few sets of numbers in to get 22.5.[/hide][/quote]
There's a mistake with that though. What if that was his average on 15, 20, or even 100 10 point quizzes, it doesn't say it's his average on 10 10 point quizzes.
[/hide][/quote]

yes it does- it says "mathnerd314's average on 10 ten-point quizzes is 7.8."

if there was the numbers of 10-point quizzes was undefined, it would be unsolvable.

edit: it wouldn't be unsolvable, it would be anything between 4 and 10 that was an integer or an integer + 1/2.
4, 4.5, 5, 5.5, 6, 6.5, etc.

I stand corrected, this is one of the stupid mistakes that accounts for a huge chunk of my grade lost on practice Sprints and Targets  :blush:

Is there any way to either solve algebraically or systematically find all the possibilities? I'd hate to be in perpetual fear of missing a possibility.

You can systematically find all the possibilities...

the last 3 have to add up to 24

find the number of partitions

With a 10
4,10,10
5,9,10
6,8,10
7,7,10

With a 9 and no 10
6,9,9
7,8,9

With an 8 and no 9 or 10

8,8,8

But only paranoid people like me do that...I always make stupid mistakes and take the wrong precautions


probability
symmetry
combinatorics unsolved
combinatorics

2n tosses of a  fair coin,there will be exactly n  heads  and  n  tails.Find the exact probability.

It's a classic binomial, isn't it?
${{2n}\choose{n}}\frac{1}{2^{2n}}$.

Another classic problem:
Player $A$: $n+1$ tosses of a fair coin.
Player $B$: $n$ tosses of a fair coin.
What's the probability that player $A$ has more heads than $B$?

It'd be the closed form of $\sum_{k_{1}> k_{2}, \\ 1 \leq k_{1}\leq n}\binom{n+1}{k_{1}}\binom{n}{k_{2}}\frac{1}{2^{2n+1}}$.  But how would we simplify that?

The answer to the second problem is $\frac{1}{2}$.  One way to see why is by symmetry: If we switch the outcome of each coin, then we change a success event to a failure, and vice versa.

Here you have another way to prove the second problem:

Let $H_{A}+T_{A}=n+1$ be the heads and tails after the $n+1$ tosses of the fair coin by player $A$.
Let $H_{B}+T_{B}=n$ be the heads and tails after the $n$ tosses of the fair coin by player $B$.
By superposition and symmetry, we have that the probability of
$P(H_{A}+H_{B}=k)=P(T_{A}+T_{B}=k)=$
$=P(H_{A}+T_{B}=k)=P(T_{A}+H_{B}=k)=$
$={2n+1\choose k}\frac{1}{2^{2n+1}}$.
Now,
$P(H_{A}>H_{B})=P(H_{A}>n-T_{B})=P(H_{A}+T_{B}>n)=$
$=\sum_{k=n+1}^{2n+1}{2n+1\choose k}\frac{1}{2^{2n+1}}=\frac{1}{2}$.

And yet another one:

After $n$ tosses of each player, there are three cases:
Case 1) (with probability $P_{A}$), player $A$ has more heads.
Case 2) (with probability $P_{B}$), player $B$ has more heads.
Case 3) (with probability $P_{0}$), players $A$ and $B$ are tied.
Player $A$ wins automatically in case 1, and in case 2 only if he/she gets one more head after the $(n+1)$-th toss. So, if the coin is fair,
$P(H_{A}>H_{B})=P_{A}+\frac{1}{2}P_{0}$.
By symmetry, $P_{A}=P_{B}=p$. And $P_{0}=1-2p$, because $P_{A}+P_{B}+P_{0}=1$.
Finally,
$P(H_{A}>H_{B})=p+\frac{1}{2}(1-2p)=\frac{1}{2}$.

I like it because there are many ways to prove it!


geometry parallelogram rectangle rhombus

[quote="Quevvy"]Determine the area of the largest parallelogram that can be inscribed in a circle with a radius of $20\sqrt{6}$.[/quote]

[hide]
EDIT: [the explanation of my assumtion that it is a square that gives the greatest area] opposite angles in a parralleogram are equal, and since it is cyclic, the opposite angles add to 180.  so all the angles are 90, so it is a rectangle.  for the area of the rectangle to be maximized, it is a square...

diagnol is $40\sqrt{6}$.  So side is $40\sqrt{3}$.  $40\sqrt{3}^{2}=4800$[/hide]

[hide="solution"]If a quadrilateral is inscribed in a circle, then opposite angles must add up to 180 because of some random geo. theorem I forget the name of.  So it becomes clear that our parallelogram is a rectangle.  And with a little bit of visualizing in the brain, it also becomes clear that our rectangle must be a square to have superior area.  So, the formula for the area of a square(or any rhombus for that matter) is $\frac{d_{1}d_{2}}{2}$, where the $d$s are the diagonals.  The diagonal is $40\sqrt{6}$, so our area is $4800$.[/hide]

[quote="Quevvy"]Determine the area of the largest parallelogram that can be inscribed in a circle with a radius of $20\sqrt{6}$.[/quote]

[hide="solution"]The opposite angles must add up to 180 because it is inscribed in a circle.  Since the quadrilateral is a parallelagram, the opposite angles are equal which means all the angles are 90.

Draw one diagonal of the rectangle.  This is also the diameter because it is the hypotenuse of a right triangle.  We see that the base of the right trangle always stays the same, no matter where the vertices are moved.  To ensure the maximum height, the rectangle must b a square.

Since the radius is $20\sqrt{6}$, the side of the square is $40\sqrt{3}$.  Therefore, the area is $\boxed{4800}$[/hide]


There is an interesting article about the [url=http://www.zmag.org/content/print_article.cfm?itemID=6751&sectionID=33]elections 2004[/url] by [url=http://www.zmag.org/chomsky/index.cfm]Noam Chomsky[/url] in the [url=http://www.zmag.org/Third_.cfm]Z Magazine.[/url]

[quote="hello"]I quicky went over the article. Seems like Chomsky didn't like any of the candidates.[/quote]

That is not the point of the article.

Chomsky is good primarily for his sardonic one-liners.
I've enjoyed many of those, and the Z magazine piece seems to have its
share, but do understand Chomsky's schtick for what it is, i.e. public
performance art aimed at establishing (or posturing) Chomsky's
exceptionalism rather than saying anything in particular about the
elections, the tsunami, Middle East peace or the 2004 Red Sox.
(Well, there was TV footage of Chomsky in baseball cap during the
playoffs, so he may actually have something to say about the latter.)

Z magazine is largely an uploading ground / mutual adoration society
for a group of writers notably including Chomsky.  His pieces are,
I suspect, pretty much published hot-off-the-keyboard without tremendous
editorial scrutiny.  The Z website has many pieces that are, literally,
straight from his keyboard (or tape recorder).  In particular,
Chomsky's peculiar tendencies concerning facts and citations may be
more freely indulged in Z than elsewhere.

Some contra Chomsky sites:

http://www.mekong.net/cambodia/chomsky.htm      (analysis of Chomsky
       Khmer Rouge controversy)

http://www.wernercohn.com/Chomsky.html      (re: Chomsky and anti-Semitism)

http://jim.com/chomsdis.htm          (over-the-top Randroid antiChomsky site)

Personally I really like his articles. I will attend [url=http://www.ed.ac.uk/news/giffordlectures0405.html]his lecture soon.[/url]  :)

BTW it seems nobody actually wants to discuss about the content of the article. Maybe just few people read the article properly.

One explanation is that nobody wants to discuss the article,
but another is that the article doesn't say anything.

Chomsky's deep, insightful, penetrating analysis of the
2004 election appears to be limited to well-worn cliches
about low voter turnout, a small lead not equalling a
mandate, the prominent role of advertising and limited
voter knowledge, etc.    His article lacks references or
even proper editing that might have weeded out some
nonsensical remarks and increased the overall coherence.
As I mentioned, whatever Z magazine's overall editorial
standard, they don't hold Chomsky to any standard
whatsoever, and the result is sentences such as:

"The right of deceit must be guaranteed to the immensely
powerful and pathological immortal persons created by radical
judicial activism to run the society."

This is a pity, because as he gets older,
Chomsky tends even more to repeat himself and
reminisce (or self-congratulate) about having been
in the protest movement from the early days.
"Pauca sed matura" would do his writings a world of good.


AwesomeMath
summer program
MathPath
AMC
AIME
geometry
MATHCOUNTS

I spent a summer at both so I think I'm qualified to make a decent assessment.

AwesomeMath is a problem-solving based program. It focuses on problems that you would see on mathematical contests, notably the AIME and olympiads. In terms of classes, it's quite rigorous; tests are at the end of every week to make sure that you're understanding the material well. It's also somewhat confined in terms of classes - my year and I think last year as well you were in the same two classes for all three weeks.

MathPath, on the other hand, is much more free in terms of the classes; you pick a new set each week, and they're less formal, no tests or anything like that. They cover a broader range of topics, and you can pick what you're interested in. To name some off the top of my head, basic Linear Algebra, Combinatorial Proofs, Knot Theory, Hyperbolic Geometry, and various Problem Solving classes ranging from Chapter MathCounts to approaching Olympiad level. Since most of the classes are one week long, the focus is more on exposure than deep theory or whatever.

In general, I think it's fair to say that MathPath is to AwesomeMath as exposure is to depth. With the lectures in History of Math, guest lectures (Conway!), short classes (required) on proofs, and wide array of breakouts, you'll see alot of interesting new stuff at MathPath and may discover an interest that you never knew of before (for example, after taking the linear algebra crash course and also learning about axioimatic Euclidean geometry I became interested in how mathematics is constructed "from the bottom up"). At AwesomeMath you'll pick, with the help of Zuming (?), two areas of problem solving to focus on and delve into the depths of them, tackling hard problems and concepts within them.

As for environment, keep in mind the age difference. MathPath is for middle school students, while AwesomeMath is for high school students and a small percentage (maybe 20-25%) of middle school students. From my experience, MathPath is more of a tightly bound family, while AwesomeMath is more like high school, a set of cliques that of course interact with each other but much less so than at MathPath. There are a few reasons for this; again the age difference, also AwesomeMath is shorter, there are more people, and you only have one set of kids whom you take classes with (or maybe two? Again I don't know exactly what happened last year).

Another note on age difference: because the average age at MathPath is much younger, the staff will keep a much closer eye on you. For example, dorm meetings every night, you have to meet your counselor for every meal, you have to travel in groups, with an "adult" if you're going to a more obscure part of the campus, etc.

Fun: Well, they're both fun. You'd have fun at both. And as for selectiveness of admission, I feel like AwesomeMath is less discriminating, but I may be wrong. I don't know if there are reiable statistics for this.

CatalystOfNostalgia really covers it well...

i went to awesomemath last year. [didnt go to mathpath]
(so im probably going to be pretty biased)

like catalystofnostalgia said, the classes do go pretty in depth, and the problems u solve/or attempt to solve during the problem sessions with the TA's [after the lecture with the prof.] are quite difficult & challenging. but you learn alot.
in addition, the classes are pretty flexible. if u realize after a couple of days in your class that ur not in the right level of classes, ur very welcome to switch. furthermore, ur professors are all really really good.
oh and yea, last year it was 2 classes for 3 weeks. 3 hrs a day for each class ['cept for sun./weekends] 1.5 hrs = lecture 1.5 hrs=prob. session

awesomemath has around ~110-120 kids. idk how many mathpath has. but yea, the awesomemath does have a higher ave. age, whether we're "clique-y" or not, im not sure. but i do know that ppl did become really close friends with other. :P my 2 bffs were girls i met at camp XD maybe we are a little "clique-y" but no one is unfriendly or unaccepting. everyone was really nice.
umm, and yea, a lot more freedom at amsp XP

whether u go to mathpath or amsp, you should interact /open up to ur peers there. socialize & make friends in addition to learning math. make it a experience u'll nvr forget.

amsp = amazing fun! im sure mathpath is prolly as well ....

i highly suggest you go to awesomemath as some point in ur life, b/c it rly is an amazing experience, socially and mathematically.
for me, it was the best 3 weeks of my life & ik for others it was as well. really lives up to its name xP
[so yea, apply to amsp & go XD]

Thank you so much! I'll take those comments into consideration.

This thread is about a great question!

Let me put my cards on the table.  I am the Academic Director of MathPath and years ago I was faculty in the Hampshire Program.

My bottom line is that anyone who loves math should try more than one summer program, because there are many styles of mathematics and of learning, and one should expose oneself to several.  Similarly, one should go to a different graduate school than undergraduate school, develop some expertise in more than one math specialty, etc.

For young students who are eligible for both MathPath and AwesomeMath, I recommend they (apply to) go to MathPath first, and then later (perhaps after a second year at MathPath!) go to AwesomeMath and/or some of the older high school camps (Hampshire, the Ross Program, PROMYS, MathCamp; see [url]http://www.ams.org/employment/mathcamps.html[/url] for a good listing).

So, that’s my conclusion; what are my reasons?

CatalystOfNostalgia wrote very well. S/he summarizes the difference between MP and AM as exposure vs depth.  S/he calls MathPath exposure because there is so much choice and so little required work.  Both exposure and depth are important to get in the long run, but exposure may be better for middle school age kids to get first. Math summer camp may be the first time they are asked to do math all day all week, and you don’t know your stamina for it until you try it.  At MathPath you can pick a concentration level that fits the stamina you find you have. There are problem sets in many MathPath classes, but they are either done in class or as optional homework. There are also challenge problems (e.g., the Problem of the Day) and prizes for submitting outstanding solutions. There are students who do math 24/7 at MathPath and gain a lot of depth, by doing all of the above, by asking special questions of our distinguished visitors and our staff, by working on their own research ideas in their own time, by giving each other challenge problems.  And there are students who just come away with general ideas.  Because there is little pressure to perform, almost everyone has [i]math fun[/i] and well as general fun.

There is another advantage to exposure.  Mathematics is too big for anybody to ever learn all of it.  But you need to get an idea of its vastness and what parts might interest you.  As another student who attended both MP and AM wrote to me, “I learned about things I had never even thought about or heard of at MathPath. At Awesome Math the classes were more fundamental; one could have researched the topics on-line.” (I would have used the word “standard” instead of “fundamental”; no one claims that the classes at AM are elementary or easy.) Balance this exposure with doing depth on other occasions and you get an excellent mathematics education.

Another difference between our camps:  At AwesomeMath, competitions are never far out of mind. This is reasonable given who runs AM: oustanding problem writers and competition coaches.  And I am hardly against competitions; 25 years ago I was the head of all the AMC exams, something I was thrilled to lead because as a kid I had really liked competition problems. (My name is on the cover of Contest Book V.)  But competitions should always be a lead-in to mathematics per se, and I wonder if sometimes AM (or at least a lot of the students who attend) lose sight of that.  There are many excellent mathematicians (the majority, in fact) who either were not especially good at competitions or weren’t interested in them.  At MathPath we encompass all views.  If you want to practice for competitions there are competition courses every week at various levels.  But you never have to take a single one of them and many students don’t.  Some students are much more interested in the ideas of the distinguished research mathematicians who occasionally visit MathPath and give a glimpse of their current research.

Finally, here’s an area where I don’t know if MP and AM differ, and I would love feedback.  The kids at MathPath are zany and a lot of their fun comes from bouncing off each other, from singing bizarre songs to telling strange jokes, to yelling “I lose”, to writing parodies of bad mathematical writing, to the so-called talent show they put on at the end.  As one student put it mildly, the best thing about MP was “Hanging out and having fun with smart kids”; another said the best was “The dorms, friends, math! I also really enjoy Megan's salsa classes.”  Do kids at Awesome have fun the same zany way, or is it more of a grind, making it more of a school (or competition training program) than a camp?  From reading various posts in this forum, it does look like AM kids definitely have fun too, maybe the same way.

The lead staff of MathPath are happy to answer any questions you have.  You can contact us through [url]http://www.mathpath.org/contact.htm[/url]

I wouldn't go so far to call AwesomeMath a school (although I'm being a hypocrite since I sort of did in my first post); it definitely is, up to a point, a microculture of it's own. As I said, due to the age difference and size difference it's more clique-y, but there I know we had a similar type of thing at AwesomeMath - people playing the game (which I personally find kind of annoying) and a vast expanse of inside jokes from Track 3, most of which were probably kind of disturbing to me at my young and ripe age...maybe not. We also had fun things at AwesomeMath - in my year and I believe last year there was a Karaoke night, where people frankly made fools of themselves, a dance which I missed because I left early, and a scavenger hunt which turned into a game of "who can survive the rain".

But from my experience MathPath has a more clearly defined microculture, and I said earlier, it's more of a tightly bound family.

There was also the Six Flags trip which was really fun. && the talent show was quite interesting. :P
I don't believe AM should be described as 'school' but more like a 'really fun learning & bonding experience'.

But, all in all , I think [based on the MP account] we do similar "zany" things, if not more.

[quote="pinkmuskrat"]How do these two prominent camps compare with one another? I'm taking about the kids there, difficulty, fun, environment, and selectiveness of admission. Thanks![/quote]

i can speak about mathpath, but not awesomemath.

[NOTE: This post is extremely long]
[NOTE: From Mathpath 2008. Does not guarantee same options for Mathpath 09]
[NOTE: mathpath.org has a tab "Why Mathpath". You should also read that]
[EDIT: I realized Mr. M posted here!]

Mathpath is a summer camp for primarily middle schoolers, although under some special circumstances some freshman may be able to attend. The classes there have different levels, from chapter Mathcounts to infinite dimensional tic-tac-toe (my head is still spinning), to Lexicode (4*4=6? 1+1=0?), along with nice plenaries, ranging from the escher effect to projective geometry, there was pretty much something for everyone.

On an average weekday, you'd wake up aroudn 7:30, 7:45, and get into counsuler groups to go to breakfast. If anyone disagrees, that's fine, but it really woke you up. I hadn't had much experience walking a long way before, so every breakfast was really an eye-opener for me.

Heh, meals were fun. You really didn't want to get stuck in the back of the line (69 kids take an hour to get food), so you really wanted to get there first. The rule was if there is a group in front of you, you can't pass them. At a certain point, there were two different ways to go, and every time it would be a race to a certain intersection. Since there was no running, 'speedwalking' became an active sport.

Anyway, after breakfast you would go to history of math, which was pretty funny. The teacher would always have this window open of this sculpture, and any time he introduced a new mathematician, he would say "Would you like to see a picture of [insert]?", and bring up that window. History of math isn't that hard to explain, the name kinda says it all.

After that, your first "Breakout" would occur. There were many different choices. From Chapter Mathcounts to AMC 10 to AIME, or from basic Number Theory to a round of "Pwn the other team" (i.e. one team solves a problem and presents the solution, the other team attempts to pwn it, such as a flaw or an non-proven statement). These were fun, and lasted about an hour.

Then the pleneries would start. From Pick's theorem to Projective geometry, no one knew it all, and no one was completely lost. These each lasted an hour, and a visiting teacher would usually teach a certain topic or related topics for a week.

Lunch. Self-explanatory (for food options, see below)

After lunch you had the option of returning to the dorms, which contained pool tables, ping-pong, and a non functioning air-hockey table. Half the time you could hear an Ipod blaring, or the thud of a ball hitting a counsaler in the head. Your other option was to go to another building, which contained a convienance store, better pool tables, computers (no Runescape... cough), and a lounge where you could just pretty much sit around. (This lasted an hour)

After this leisurely break, Steven Maurer (Mr. M) would discuss well-written, and... not so well-written solutions to the qualifying test (It was kinda funny when the students started trying to talk about what was wrong iwth Mr. M's solution). This lasted about 20 minutes, and helped to improve general writing skills.

Your second breakout then began. See first breakout. However, this one lasted for 1.5 hours, instead of the hour above.

Dinner. Self-explanatory

After about 6:00 when dinner ended, you were pretty much free to do what you wanted. There were some evening activities, such as Soccer or Basketball, or simply some leisurely rest time to cool down, or some study time for those who needed it. The pool and ping-pong still applied, and computers were available for use for some time.

Mandatory bed was 10:00, with you having to be in rooms by 9:30.

*Food*

Breakfast had plenty of options. Eggs were a given, with another platter being ham or bacon. Waffles were always available, and sausage sometimes was. There were an assortment of juices, water, or soda for the non-morning people (see the poster)

Lunch always had assorted options. Pasta with/out sauce was always possible, along with pizza. Depending on the day, there might be lasanga, or other foods. If you were in the mood for something cold, sandwiches were always an option. Also, most lunches a grill would be open, with burgers, hot dogs, corn dogs, and the like.

Dinner again had assorted options, depending on the day. Pasta and Pizza were again a given, and sandwiches were always available. A "special item", which could be omelets, chicken, or w/e, was also included. Ice cream, cookies, and sometimes cake finished up the eating.

**

*Buildings*

The "Davis center" was the buliding I mentioned earlier that contained computers, a lounge, pool, convienence store, etc. We were able to go there form 1:00-2:00 and after the math day ended.

Chittenden & Buckham: Dorms. This contained a rec center, with pool and ping pong, as well as a lounge, where you could probably see towering zome buildings, and cards on the table, and a laundry center for... kinda self-explanatory  :P.

Library: While you can't check out the items (we aren't students at University of Vermont), there were plenty of computers open for public use, at least until closing time (8:00 PM).

I can't recall the names of the lecture buildings, but they don't exactly need description.

**

*Teachers*

George Thomas (Dr. T): Taught hyperbolic geometry, and an excellent soccer player.
Steven Maurer (Mr. M): taught the writing pleneries, discussing qualifying test solutions (written by actual students), and an occasional course.
Alan Lippert (Mr. L): taught the Mathcounts classes (Although I beat him on a sprint  :P ), and gave the daily wrap up problem of the days (yes, with prizes  :D ).
John Conway: Needs no description (if you really need one, google him)
Francis Su: Taught pleneries about cake-cutting
Jennifer Bieneke: Taught graph theory for 2 days
John Voight: Taught about the Escher effect, Pick's theorem.
Vladimir Chernov: Taught everyone weird proofs that 1=2
Jeff Dinitz
Thomas Drucker: Taught two weeks of history of math.
Robin Hartshorne: Taught Euclidean constructions
Susan Holtzapple (Mrs. H): Helped with the Mathcounts classes (managing to delay every class by about half an hour)
Glen Van Brummelyn: Taught two weeks of history of math (and plays soccer)
John Mccleary: Taught some geometry
Kip Sumner: didn't actually teach, but ran the camp.
Sam Vandervelde (Dr. V): Taught Farey sequences
**

*I don't know what else to put here*

All in all, to make a long post short, mathpath was awesome, and I'm planning and returning for '09 (if possible).

*BOGTRO*

AwesomeMath was rigorous. You'd need to take a lot of notes.

Or you can fall asleep during a problem session and get voted "most likely to fall asleep in class" like I did.

lol i remember that.
aka. get enough sleep. or get coffee in the morning :) [latter's a better choice so u wont miss out on the fun XD]

I think this has been an excellent thread, with really useful information (especially in the posts from kids who attended MathPath!). These posts will be really helpful to people trying to decide between camps, and I hope many of them see it.

CatalystofNostalgia and BOGTRO, I hope you will identify yourselves to me privately.  Since you both attended MathPath, you have my email address (or can find it again on the MathPath website).  And BOGTRO, since you say you want to come back, have you contacted me?  As I hope you know from the website, you have to contact me to find out what you must do to reapply - often nothing but send an updated application form.  Even if you are not sure you can come back (because of other family summer commitments, because of finances, whatever) it never hurts to find out sooner if you are readmitted.  It also helps us - the sooner we know our likely enrollment, the better.

So, all you people out there who think you [i]might[/i] want to attend MathPath, get your applications in.  Applications are coming in at a steady pace now - well ahead of last year's rate - and so the camp may fill up early.

My experience may or may not be outdated (I was MathPath 04 and AMP 06 and 07), but I thought MathPath was a lot less intense than AMP. MathPath seemed to care less about whether you learned the material or not, whereas Awesomemath had tests every week and occasional homework writeups, and the profs and assistants sort of went around and made sure you were following. While the tests and homework don't really count for anything, they're there, and it's Asian bad if you don't do them. On the other hand, the profs at MathPath just gave us problems on the material they covered in class and it was really up to us whether we wanted to do them or punt them.

This kind of sounds like MathPath is slackbad, which is obviously not the case. Actually, at MathPath I kind of liked doing math when I wanted to, where I wanted to, in a less competitive atmosphere than I'd ever been before. It's kind of a nice break from the school year when you're cramming for Mathcounts here and AMC there or whatever it is middle school students do.

But really, both camps are loads of fun. You'll learn a lot at either one. My opinion is that which one you go to this summer will matter less than where you go for undergrad. That is to say, not very much at all. (That was a joke, I think.)

[EDIT: Foodwise, I can only talk about the food at UTD. It was pretty bad in 06, but I understand it's gotten a lot better since then. In 07 they got Aramark to bring us food, meaning the selection is limited for a certain day, but at least it's not Chick Fil-A every day. Not sure what they did last year.]

Super revival...

Thanks everyone! I attended MathPath last year and have been accepted to AwesomeMath for the upsoming summer.

I've been to Awesomemath 3 times now;this uear will be my fourth
I love awesomemath...and ive heard great things about mathpath as well.
But, i went to awesomemath dallas and santa cruz last year, and at both camps i made friends who had gone to mathpath and were still eligible age to go but said they wanted to try something new; and they like awesomemath a lot(according to what ive talked with them about it) so yea...

:D:D yay awesomemath

I will say that in regards to food at the UTD AMSP, there is now a brand-new cafeteria on campus (I think they just finished construction) that should provide better food than what we've had for the past three years.

[quote="CatalystOfNostalgia"]there was a Karaoke night, where people frankly made fools of themselves.[/quote]
Hey, watch yo self.

Honestly, I can't say or judge math path as I know nothing about it, but AwesomeMath was simply the best experience of my life. In my area, no student has ever qualified for USA(J)MO until this year when I did. The accomplishment of course had a lot to do with the effort I had put in after AwesomeMath, but the program was undoubtedly the majority of the reason for the effort. AwesomeMath is a fun environment and many students probably thought I was an idiot and that I was just all fun/no work while I was there, those students would be partially correct. Though, I did pick up a lot of concepts and now, as I study much harder as I was inspired by my highly intelligent peers, I see a lot of the things I learned back in AMSP pop up again, and they are instantly revived from the back of my head.

AwesomeMath is just simply amazing and the results I and my local peers have witnessed in my personal development this year, yet I managed to still have the time of my life while I attended the camp.

I went to MathPath and not Awesome Math, and so I will just tell about MathPath. The month that I spent at MathPath was one of the best months of my life.

Things (from best to worst):

1. The other people there were very smart and many (about two thirds) were probably better than me at math. (This probably gave you nothing because you don't know how good I am at math...) I was amazed though, and I loved it how I fitted in so well. There is no place at all, where you can get that same experience. Going to MathPath is like being on AoPS for 4 full weeks. (And sleeping on AoPS!!!) :D

2. Some of the mathematics there was very challenging, although there were classes that seemed kind of useless. My favorite class, which was the one I thought was the most difficult was Inversion taught by Dr V in the third week. Complex numbers was probably the most useless class that I took.

3. The evening activities were free and not very organized, which was actually more good than bad, because at EPGY, for example, the activities were way too organized. I enjoyed playing soccer, table tennis (not ping-pong, :P) 4-square (or 5-square,) and many other FUN games.

4. The food was very good probably better than at the other camps that I have been to, (EPGY and Sitka Fine Arts Camp) although it was very repetitive.

5. It was hard to sleep: it was hot and humid in the rooms. The fact that we had our own rooms and no room mates didn't help much. I fell asleep in class many times, not because the plenaries or breakouts were boring, but because it was hard to sleep. (It applies to MP 2010, but I am not sure about other years.) :sleeping:

6. Compared to EPGY and SFAC, the counseling was poor and, I would say terrible. (Sorry counselors.) They left for meetings every night, and we were downstairs, in the lounge, unsupervised from 8 to 9. In Sitka FAC  we did all the night activities with the  counselors (much safer.) Weeks 3 and 4, the last two weeks, were starting to have inappropriate stuff going on such as a lot of cursing (even more than at Fine Arts Camp) and late in the third week, there was cross dressing day, which was when boys exchanged cloths with girls. (My opinion: very inappropriate...) On the other hand, the counselors did things that they totally didn't have to do. For example, we weren't aloud to cross Grand Ave without a counselor. (I crossed "illegally" three times.) My opinion: this was a useless rule, since we have about 99.9% the ability to cross a street of a counselor's ability. This rule led to the fact that we couldn't go to breakfast when we wanted to, and we had to go to breakfast with our counselor groups (about ten people.) At Sitka Fine Arts Camp, we got to go to breakfast when we wanted to, which I liked a lot better. (EPGY was even worse: the whole camp had to walk to breakfast together.)

Out of five stars, I would rate MathPath with 4.5 as a camp, and 5 as an experience.

Sorry about the fact that I didn't get to compare it with Awesome Math.

Your Soviet spy,
3333, the happiest number of all  :-D. (Can you prove it?)

[quote="stickfigure"][quote="mathluver2.718"]I'm not sure if this is true, but I get the idea that AM consists mostly of kids whose parents want them to improve at competition math, but maybe don't like math, while MP consists more of kids who like math in general. I'm not saying this is necessarily true; I have been to MP, but not AM, so I am also biased.[/quote]
While there definitely are some kids who attended AMSP JUST because their parents forced them to (unfortunately), I wouldn't be surprised if the same were true for MathPath. While most of the kids at AwesomeMath do want to improve their competition skills, there's no way you'd get sucked into that path if you lacked an initial love for math.
I went to AMSP, and a kid there solved a math problem while showering, got out of the shower to tell his roommate the solution, and then got back into the shower to resume his shower. If that doesn't show a love of math, I don't know what does..[/quote]

There was a kid at MathPath, who told me he hates math, and his mom just made him go there.


Without using Dirichlet's theorem, show that there are infinitely many primes ending in the digit $9$.

Assume there are only finitely many primes $\equiv -1 \mod 5$ and let $P$ be their product.
Let $q$ be any prime divisor of $(2P^2)^2-5$. Then $1= \left( \frac{5}{q} \right) = \left( \frac{q}{5} \right)$. Thus $ q \equiv \pm 1 \mod 5$.
Additionally $(2P^2)^2-5 \equiv -1 \mod 5$, so we can conclude that at least one times the $q \equiv -1 \mod 5$ case happens when $q$ runs through all prime divisors of $(2P^2)^2-5$.
But $q$ is trivially coprime to $(2P^2)^2-5$, a contradiction to the assumption.
And primes $\equiv -1 \mod 5$ always end in $9$.


Vasc, where are you?  :D Can you show me your proof?


geometry
3D geometry
sphere
angle bisector
geometry proposed

Given two planes $\pi _1,\ \pi _2$ intersecting orthogonally in space. Let $A,B$ be two distinct points on the line of intersection of $\pi _1$ and $\pi _2,$ and $C$ be the point which is on  $\pi _2$ but not on $\pi_1.$ Denote by $P$ the intersection point of the bisector of $\angle {BCA}$ and $AB$, and denote $S$ by the circumference on $\pi _1$ with a diameter $AB.$ For an arbiterary plane $\pi _3$ which contains $CP,$ if $D,E$ are the intersection points of $\pi_3$ and $S,$ then prove that $CP$ is the bisector of $\angle {DCE}.$

[hide]Consider the sphere through the circle $S$ and containing point $C$,and let $Q$ be the intersection of $CP$ with the sphere. Since $CP$ is the angle bisector of $\angle {ACB}$ and $\pi_2$ intersects $\pi_1$ perpendicularly, $Q$ is the center of the section of the sphere defined by circle $S$. Thus for any two points $D, E$ on $S$, $Q$ is the midpoint of arc $DQE$, and thus, $CP$ is the angle bisector of $\angle {DCE}$.[/hide]


(Nguyen Huu Minh Tuan)
Given $ a , b , c > 0$ and $ a \plus{} b \plus{} c\leq \frac {3\sqrt {3}}{2}$ .
Prove that: $ \sqrt {a \plus{} {b}^{3}} \plus{} \sqrt {b \plus{} {c}^{3}} \plus{} \sqrt {c \plus{} {a}^{3}}\leq \frac {3\sqrt {7\sqrt {3}}}{2\sqrt{2}}$

Try $ a \equal{} b \equal{} c\equal{}\frac {\sqrt {3}}{2}$

Sorry, I have edited it. I had a mistake in type :oops:


1) Show how each of the following conversions could be made by making use of a rearrangement reaction:

(a) Benzofenone -----> Benzanilide

(b) 2-Methylcyclohexanone -----> 3-Methylcycloheptanone

(c) Cyclohexanone -----> $ \epsilon$-caprolactone

(d) Toluene -----> Aniline

[hide="For third one"]Baeyer villager rearrangement!!
reagent- $ CF_3CO_3H$ or any other per-acid[/hide]
 :maybe:  :maybe:

[/hide]

That's right. What about the other ones?

has the second prob got anything to do with this?
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=213229

Yes, it does. What about parts (a) and (d)?

[hide="(a)"]

1. React with $ \ce{NH_{2}OH}$
2. Beckmann rearrangement
[/hide]

[hide="(d)"]

1. Benzylic bromination using NBS to benzyl bromide
2. Sommelet reaction to benzaldehyde
3. react it with hydroxyl amine to give benzaldoxime
4. beckmann rearrangement to give N-phenyl formamide
5. Hydrolysis to give aniline

Can also go via Hoffman bromamide.
[/hide]

Chemrock, I made this post specifically for Da Vinci to answer. If you are going to post here, then at least HIDE your answers. Also, your answer for (d) is ridiculously long.

[hide]
Instead can we perform benzylic bromination thrice using NBS and then use aq. KOH to get benzoic acid and then perform Schmidt reaction?
[/hide]

[hide="alternate"]
This is low yielding:
1. oxidise to benzoic acid
2. smidt reaction(involves rearrangement to isocyanate)
[/hide]

Oxidation of side chains with permanganate is an high yielding process.


Who are the students who got 40 and more points at IMO ?  :)

42 got Bela Andras Racz (Hungary) Jacob Tsimemramn(Canada)
Andrey Badzyan and Mykhail Dubashinsky(Russia)
41 I think got a Chinese and a Taiwanese student(but I don't know their names)
 40 got Andrei Negut (Romania) and Oleg Golberg(USA :)).

Maybe I missed someone..

[quote="iura"] and Oleg Golberg(USA :)).

Maybe I missed someone..[/quote]

 :maybe:

[quote="iura"]41 I think got a Chinese and a Taiwanese student(but I don't know their names)
 maybe I know the Chinese one who got 41,his name is HuangZhiyi

[quote="iura"]41 I think got a Chinese and a Taiwanese student(but I don't know their names)
..[/quote]

Tsai  Cheng  Chiang ,the boy from Taiwan.

What forum member has got most points btw? :)

I'm surprised how canadian had done full on IMO and how Gollberg coudn't :maybe:

[quote="Fedor Petrov"]Maybe, this Canadian (Tsimerman?) is a smart guy? I guess, he's not much more Canadian, than Golberg is American.[/quote]

Your guess is right.  :D

  Darij


A function $ f(x)$ is defined by $ f(x) \equal{} \frac{1}{2}[10^x\plus{}10^{\minus{}x}]$, for x in $ R$. SHow that

$ 2f(x)f(y)\equal{}f(x\plus{}y)\plus{}f(x\minus{}y)$

[hide="nevermind this solution"]Taking inspiration from the proof of the product-sum formula:

$ 2 \cos(x)\cos(y) \equal{} \cos(x \plus{} y) \plus{} \cos(x \minus{} y)$

and the similarity of f(x) and the formula $ \cos(x) \equal{} \frac {e^{ix} \plus{} e^{ \minus{} ix}}{2}$,

define the function $ g(x) \equal{} \frac {10^x \minus{} 10^{ \minus{} x}}{2i}$.

Lemma1 : $ f(x \plus{} y) \equal{} f(x)f(y) \minus{} g(x)g(y)$
and Lemma2 :$ f(x \minus{} y) \equal{} f(x)f(y) \plus{} g(x)g(y)$.

which follows from just multiplying those expressions out, $ \frac {10^{x \plus{} y} \plus{} 10^{ \minus{} x \minus{} y}}{2} \equal{} \frac {(10^x \plus{} 10^{ \minus{} x})(10^y \plus{} 10^{ \minus{} y})}{4} \minus{} \frac {(10^x \minus{} 10^{ \minus{} x})(10^y \minus{} 10^{ \minus{} y})}{ \minus{} 2}$ (don't feel like typing all the work), and similarly for the other.

Adding the two lemmas, we get our result.[/hide]

[hide]
Meh, let $ k \equal{} \minus{} ln(10)i$, then $ f(x) \equal{} \frac {10^x \plus{} 10^{ \minus{} x}}{2} \equal{} \frac {(e^{ik})^x \plus{} (e^{ \minus{} ik})^x}{2} \equal{} \cos (kx)$.

Since $ 2\cos(kx)\cos(ky) \equal{} \cos(kx \plus{} ky) \plus{} \cos(kx \minus{} ky)$, we have result.[/hide]


Seems to me like these AIM Math Jams are becoming popular.

I'll post the Math Jam schedule here and you can just reply to this thread if you want your's to be on here.

By the way, the rules will be at the discretion of the host.

Users:

kyyuanmathcount - poopyloopy2003  (Yes, I will get a new one soon)
DarkKnight - Kaoticity
ch1n353ch3s54a1l - ch1n353ch3s54a1l
ccy - theccy
LadyKn1ght - LadyKn1ghtz
Fanatic - Warcraft1fanatic
goldendomer - Ddavidw111
biffanddoc - biffanddoc
easyas3.14159... - Happynuthead
pianoforte - pianofortegauss
Klebian - Ibepulak
inyoung - azngirl1292
ignite168 - Jathan84

Thank you everyone, we had fun!

Ok people please don't post people's screennames without thier permission because i'm getting alot of annoying spammers.

Yeah, I only posted those who posted theirs.

[quote="easyas3.14159..."]Some more ppl:
aznmath1337-dont remember
cheesybread92-mysmartmouth
Ddavidw111-goldendomer
dunkruler23-slamchess
insanemorgoth-xxm0rg07hxx
luv er of ra in-Lucy4eva
mathlete1106-don't remember
mthxpnguinsxlife-don't remember
swkcorean-shinwoo
Zenith Xan-Zenith Xan

These are all members who've come before.[/quote]
OK. A few things. aznmath1337 is noneoftheabove.
Mathlete1106 and mthx pgnuinsxlife are either not AOPS users or not Mathcounts forum users.

Also, a few words of advice:
Don't make more than 1 math jam a week. If you want to host one, put it in the schedule enough ahead of time.
Last week, we had, what, 4, 5, 6 math jams? The first one was like the guinea pig. The second one, many people came. Third one wasn't bad. After that, it was basically repetition, with few people participating, and very boring.
I'd say one every 2 weeks or so, because that gives people enough time to see if it'll work in their schedules, and that'll ensure as much participation as we can get. A math jam with 4 or 5 participants is not really that effective. A math jam with 10 participants, all that are at a high level, gives people the opportunity to really practice and see how they compare with other AOPSers.

I'd be interested in joining one, although my chances of making nats are pretty low. my AIM sign name is azngirl1292

I could host a math jam on Thursday from 8 to 9 EST. I'll do it if enough people reply. (If you don't want to post your screen name, PM me)

I'm interested even though I just discovered AoPS VERY recently, and the fact Zenith Xan is an AoPSer amuses me as I knew him a long time before I discovered AoPS

AIM SN: Jathan84

I'm interested. You know my AIM. It's up there.^^^^^^

i could do one next week sometime, probably wednesday

I'll join.  My sn is cheesybread92 (it should be listed above).

I should be there too.

Oh, a note, I'll probably be there for any of your mathjams  :D .

wait.. 8-9 EST is eastern standard time right?

I don't want to host one; I'd be interested in attending, I think...
my sn is muffinluvr493
kk
GO MN!!!

Ok, I [i]might[/i] be able to host a math jam tonight or tomorrow, 8:00-9 EST. Post if you're interested.

Questions will be from past countdown and sprint tests...

I am very interested, but for the sake of me and everyone that needs to go to bed earlier on a weeknight than on a friday night, and you because you need time to prepare the problems, have the jam tomorrow instead.

Cool.  I should be there.  Your aim name is pianofortegauss, right?

[quote="Klebian"]I HIGHLY disrecommend you doing it tomorrow. No one will come, as you're giving almost no prior notice. But to each his own.[/quote]

I HIGHLY recommend you do it tommorow :P

I'm gonna try to do another mathjam on tuesday, 7-8 EST  :) .

EDIT: cancelled do to internet problems

Tuesday I can be there, my AIM is Atheist Temple.

Tuesday I can't be there, MathCounts practice :(

eh, I'm having internet problems at home (at school right now) so I wouldn't be able to host one anyways...

I am willing to host a last-minute AIM Math Jam on Wednesday at 6:00 PM, EST. If my idea gets off the ground, it will consist of more difficult questions, mainly Sprint round questions.

Again, my AIM is Atheist Temple.

its almost 6...i'll IM you to see if its on.

I would truly appreciate joining an AIM math jam. My screen name is DirtBob88.


[url]http://news.bbc.co.uk/2/hi/uk_news/education/6589301.stm[/url]

I haven't really taken a look at it yet but supposedly it's pretty hard. I figured if somebody was going to solve it it would be somebody here  :P, have fun!

PS: mods feel free to move this thread wherever, I don't really have a feeling for how hard this problem is yet.

Here are my thoughts.

1. Dot product = 0.
2. Find normals to the planes using cross products then use the dot product.
3. Dot product to get angle.


advanced fields
advanced fields unsolved

Can someone give me some advice on this problem?

Let A(n, r) be the number of ways to partition n into the sum of no more than r terms and B(n, r) be the number of ways to partition n into the sum of any number of terms each at most r (a term must be a positive integer). Prove that A(n, r) = B(n, r) for all 1 <=  r <= n

To each partition, associate its [url=http://mathworld.wolfram.com/FerrersDiagram.html]Ferrers diagram[/url]. To each Ferrers diagram which has all its rows of length $\le r$, associate the Ferrers diagram obtained from it by considering the rows to be columns and vice-versa. Ypu get a bijection from the diagrams with rows of length at most $r$ to those with at most $r$ rows.

Edit:

I didn't know there was a name for that: what I said can be put like this: to each partition, associate its [url=http://mathworld.wolfram.com/ConjugatePartition.html]conjugate partition[/url]. The partition has parts of size $\le r$ iff its conjugate partition has at most $r$ parts.


There are 2 different questions that you might be asking:

(1) What does that mean?

(2) How do I prove it?

Which one are you asking?

I think he's asking how to prove it.

You can prove it using induction - I think that's a easy way. We also learned another way to prove this during Intermediate Algebra Class manipulation sums, but I can't remember it.  :blush:

However, if we are trying to prove it, I think this would belong in the Intermediate Forum.

okay...

I will post it on the intermidate section

Or you can watch the proof without words animation, which is basically same as 3X.lich's proof.


geometry
rectangle
probability
ratio
number theory
prime numbers
area of a triangle

I will be holding a contest that will change your life. (its just a regular contest :D ) if you want to participate then sign up here.  There will be 5 rounds of problems.  in each round there will be 5 problems.  You will get 3 points answer for a correct answer and 5 points for a correct solution.  So sign up if you want to participate.

I'll sign up, but how can it change your life?

[quote="236factorial"]but how can it change your life?[/quote]
i just wrote that so people would sign up. :D

I'll sign up... what difficulty are the problems and how do you determine the winner in the end?

the difficulty is classroom math.  The winnner will be determined by getting 3 points for answer and 5 for solution.  There will be about 5 rounds.  There will be 5 problems in every round.

i think we'll start when we get 8-10 people.

Great.

How many rounds?

Do we send solutions?

Do we PM you?

How many questions per round???

 :D  :D

 :?  :?

[quote="Drunken_Math"]I'll join!  :)[/quote]
you already joined :roll:

Sounds fun, even though I'll get owned.  :P Sign me up

[quote="pianoforte"]for number 1, is his uncle giving him double what he gave last time, or giving so that the total is double what he already had?[/quote]
you got all the questions correct but you didn't have a good solution. :)

When is round two due? Can you please make it Saturday again. I probably won't have time to write out the solutions until Friday.

[quote="Geogr-ath"]I want to join... I'll probably stink but oh well! :oops:  :)  :D[/quote]
sorry, but its a little late to join. :(

[size=200]STANDINGS[/size]
1st:  nonie 78 pts
2nd: pianoforte 68 pts
3rd: biffandoc 35 pts

[size=200]ROUND 3[/size]
1) The digits 2, 3, 4, 7 and 8 are each used once to form a five-digit number. What is the probability that the tens digit is odd and the number is divisible by 4? Express your answer as a common fraction.

2) There are 8 red, 8 blue and 8 yellow marbles in a jar. What is the fewest marbles you can remove from the jar so that the ratio of red to non-red marbles is 3 to 7 and so that the ratio of yellow to non-yellow marbles is also 3 to 7?

3) The area of triangle A is 4. What is the area of a triangle whose sides are each 3 times as long as those of triangle A?

4) What is the units digit of $3^{33}$?

5) The sum of two prime numbers is 20. What is the greatest possible product of these two primes?

PM me the answers by next saturday....
and good job so far to all of you. :)

[quote="jensen"]i think these marathons are so nice.i hope that anyone join these.. :D[/quote]
well, everyone can join before round 1 starts and round 1 already started.

Is this over or something. ( or is everybody just really busy)

oh, i'm so sorry...ive been so busy with other things that i have forgotten about this.   pianoforte never submitted answers so you win, nonie! :D  :D
this thing will only allow me two smilies

so is this over or should I post another one

this contest is over so feel free to make your own. :)


I am trying to figure out the probability of getting exactly two pairs from a deck of 52 cards. I see a solution on awiki page, but I am looking for a different way.

I've solved a different problem where one needs to draw 3 of a kind and a pair. "full house"

so i am looking at the probabilities of choosing each card individually.

drawing first card is 100% or simply 1
second card must be 3 out of 51 since i have to match the 1st card and only have 3 of those left in deck
third card must be 2 out of 50 for the same reason as the 2nd. i keep choosing out of less cards too hence 51..50..so on
fourth card is 48 out of 49 since i can't have 4 of a kind, i only have 48 cards to choose from 49
fifth card is 3 out of 48 which is the pair for the 4th card.

multiplying all these probabilities out gives me .00144

using the same technique i am trying to solve the 2 pair problem

1st card is 100% as above
2nd card is 3 out of 51 same as above
3rd card is 48 out of 50 since i can't get any more of the 1st 2 cards
4th card is 3 out of 49 to match the 3rd card
5th card is 44 out of 48 since i can't get any of the 1st and 2nd pair. so 52 - 8 = 48 cards to choose from

however, multiplying this all out doesn't give me the right answer .0475

this is driving my crazy.  :oops: someone. please explain to me my mistake. if you are going to talk about a different solution, please do not waste your time or mine. Thanks in advance.

It's quite simple: you don't have to draw the cards in that order. The second card probably isn't the match for the first.

then why was i able to solver the 1st problem (full house) with my method ?! those pairs and 3 of a kind also don't have to be dealt in order. yet the method gave the right answer.

[quote="toyrunr"]yet the method gave the right answer.[/quote]check my name!


algebra polynomial calculus derivative quadratics induction

I didn't understand your question as stated.  But if you are looking for an application of finite differences, here is one:
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=10955]www.artofproblemsolving.com/Forum/viewtopic.php?t=10955[/url].

Basically, if you have a few terms of a sequence, you can find a polynomial that agrees with the sequence by using finite differences.

Can you give an example problem?  I don't understand what you mean when you say "finite difference equation".

When you say "finite difference sequence", I think you just mean "sequence".  Doesn't the problem have to specify that the sequence comes from a polynomial?  What is the degree of the polynomial? (If you don't specify, there are an infinite number of solutions.)

by going upward using the existing terms in our differencing?

Right, I let the first numbers in the rows (11, 5, -12, -6) be the coefficients in the formula.

Uhhh, I think he wants a derivation of the formula, not an explanation of how to use it.
[quote="Ravi B"]Okay, now you have a well-defined problem.  Let's call the first term "term 0", the second term "term 1", etc.  So we want to find term 29.  To do so, let's compute finite differences:
[code]
11    16     9   -16   -65  -144
    5    -7   -25   -49   -79
     -12   -18   -24   -30
         -6    -6    -6
[/code]
We then extract the first number from each row to get a formula.  The formula for term n is
\[11 + 5n - 12 {n \choose 2} - 6 {n \choose 3}.\]
Now plug in n = 29.[/quote]

I do mine with calculus.   Every time you create another difference sequence, it's estimating the next derivative with $dx=1$.  Look at these sequences:

4,9,14,19,...
5,5,5,...
It's the line $y=5x-1$, with $y\prime = 5$, the difference.

5,13,23,35,...
8,10,12,...
2,2,...
It happens to be the quadratic $y=x^2+5x-1$, with $y\prime\prime = 2$, the second difference.

4,35,116,271,...
31,81,155,...
50,74,...
24,...
This is the cubic $y=4x^3+x^2-1$, with $y\prime\prime\prime = 24$, the third difference.

11    16     9   -16   -65  -144
    5    -7   -25   -49   -79
     -12   -18   -24   -30
         -6    -6    -6
Now, we have to work backwards.  Your example has a constant third difference of -6.  Thus, the sequence can be described by a cubic polynomial, $y=a_0+a_1x+a_2x^2+a_3x^3$.  We know -6 is the third derivative, $-6=y\prime\prime\prime = 3!\times a_3\rightarrow a_3 = -1$.

We now want to find $y-a_3x^3=a_0+a_1x+a_2x^2$, so you must find the quadratic sequence by subtracting out the cubic term.

11, 16, 9, -16, -65, -144
-                           (-x^3)
______________________
12,24,36,48,60,72,...
which is 12x.

the xth term is $12x-x^3$

I forgot there is no calculus allowed, but that is the easiest way I could find.  I think you got far enough into calculus to know the power rule.

edit: no, I don't actually do that method.  I just thought it would be easier to grasp.

It might be possible to derive using $\displaystyle \sum_{i=1}^n i^k$ for various k.. but that looks messy.  Calculus is the easiest way.

Sorry, Rep123max, I didn't understand what you were asking.  The formula I used is called [url=http://mathworld.wolfram.com/FiniteDifference.html]Newton's difference formula[/url].  It can be proved by induction on the degree of the polynomial.  No calculus is required.  The key fact is that the first difference of say the sequence $n \choose 3$ is the sequence $n \choose 2$.

BTW - Just before the age of calculators (Not that long ago, btw, just say 30-40 years ago) almost all experimental scientiests (particulary those who used "mathematical tables" ) knew these techniques by heart.. If you look at old mathematical tables (Like astornomical data etc) these tables used to have 'first' and 'second' differences printed in the table to make the calculaitons easy.. :)


modular arithmetic
algebra
binomial theorem

Find the last two digits of $\sum_{x=1}^{2006}x^{5}$.

Suppose for example you have something like $\sum_{x=1}^{100}x^{3}$ that you want to compute mod 10.  Notice that

$11^{3}+12^{3}+\cdots = \left(10+1\right)^{3}+\left(10+2\right)^{3}+\cdots$

When you compute this mod 10, you get

$11^{3}+12^{3}+\cdots \equiv 1^{3}+2^{3}+\cdots\pmod{10}$

This follows from the binomial theorem since every term will have a factor of 10 except for the last.  You can apply this to the original question by noticing that the numbers 1 - 99 repeat twenty times in the series, and then you are left over with 6 more.

Oh yes, by the way, the answer to the example I just gave would be

$\sum_{x=1}^{100}x^{3}\equiv 10\sum_{x=1}^{9}x^{3}\equiv 0\pmod{10},$

I believe.


3. How many times does the graph of the function f(x) = 2*x^8-3*x^5-2*x^3-2 cross the x-axis? (without the calculator)

If you can locate the zeroes, then you get your answer.

That polynomial can't be factored :wink: It also has a degree of 8, for which there is no straight formula for finding the roots.

[hide="Hint"]$ f(x) \equal{} (2x^3 \minus{} 3)(x^5\minus{}1) \minus{}5$. Now you can probably use a non-calculus increasing/decreasing argument, and you can obviously use a calculus increasing/decreasing argument too.[/hide]

[quote="kunny"]You can use Decarte's rule of sign.[/quote]
Then if question like this:
How many real solutions does the equation x^4-4*x^3+x^2-2*x-1=0 have?
Which I cannot  use Decartte's rule? How do you do it?

For that problem, we can use a trick  :)

[b]Solution:[/b] We consider $ P(x\plus{}1)\equal{}(x \plus{} 1)^4 \minus{} 4(x \plus{} 1)^3 \plus{} (x \plus{} 1)^2 \minus{} 2(x \plus{} 1) \minus{} 1 \equal{} x^4 \minus{} 5x^2 \minus{} 8x \minus{} 5$

By Descartes' Rule of Signs, the above has one positive and one negative root, for a total of $ \boxed{2}$ roots.

[b]Motivation:[/b] Our problem was with the coefficients, so we sought another polynomial with different coefficients that we knew would have the desired property. Here, by depressing the quartic, we remove one sign comparison, which seemed potent.