When I use \usepackage[spanish]{babel} the following warning appears:

[img]http://img245.imageshack.us/img245/4766/sfkjgfjl3.png[/img]

How do I fix this?

It's telling you that it doesn't know how to hyphenate Spanish words and instead will use English words. You can accept that if the output is OK, otherwise, assuming you are using MiKTeX then
[quote]To select the new language, do:
Start-> Programs-> MiKTeX 2-> Settings, and select the Languages tab. Select spanish from the list, press the Apply button. Then select the General tab and press the Update Formats button. [/quote]
More at [url=http://www.tex.ac.uk/cgi-bin/texfaq2html?label=newlang]Using a new language with Babel[/url]


http://v.mercola.com/blogs/public_blog/Shift-Happens----Some-of-the-Most-Amazing-Statistics-You-Never-Knew-11829.aspx

Some of the most Shocking statistics.


My teacher told me to use calculator for this question, however i want to know how to do this algebraically, help me.

Q. Solve 2x=e^(-0.5x).

Thanks~

By calculus ($2x$ is strictly increasing, $\mathrm{e}^{-\frac12x}$ is strictly decreasing) we immediately get that there exists only one root and there must exist at least one root, since  $\lim_{x\to\pm\infty}\pm2x=\pm\infty$, $\lim_{x\to-\infty}\mathrm{e}^{-\frac12x}= \infty$ and $\lim_{x\to\infty}\mathrm{e}^{-\frac12x}= 0$.
The numerical approximation of the only root is $0.40777670940448035$. But I don't think it is an algebraic number or can be expressed by values of elementar functions, because it equals $2u$, where $u\mathrm{e}^{u}=\frac14$ and solutions int $w$ of the equation $w\mathrm{e}^{w}=z$ don't tend to be algebraic or elementar for rational $z$.

okai, so i have to use calculator for this!

anyway, thank you heaps!


[hide="Hint"] The polynomial $f(x)-f(1)$ has five roots. [/hide]

judging by there being 5 equal f's,
[hide]
am I wrong to say that f(1)=f(2)=f(3)=f(4)=f(5)=0?
If that is the case,
(x-1)(x-2)(x-3)(x-4)(x-5)=0
so a= -1+-2+-3+-4+-5=[b]-15[/b][/hide]

The official solution uses what t0rajir0u said, and then simply applies Vieta's sum.  I don't get why you can do that, do.

Sapphyre571, you got the correct answer, but I think your logic is a bit flawed.


function
integration
real analysis
real analysis unsolved

Is there a function,not necessarily continous, which graph is not symmetrical to the y-axis, that suffices following equation:

$ \int \limits_{\minus{}a}^a f(x) dx \equal{} 2 \int \limits_{\minus{}a}^0 f(x) dx$

Consider:

$ \int f(x)\,dx\equal{}F(x)$

then:

$ \int^{a}_{\minus{}a} f(x)\,dx\equal{}F(a)\minus{}F(\minus{}a)$

$ 2\int^{0}_{\minus{}a} f(x)\,dx\equal{}2F(0)\minus{}2F(\minus{}a)$

now:

$ \frac{d}{da}\left(F(a)\minus{}F(\minus{}a)\right)^{'}\equal{}f(a)\plus{}f(\minus{}a)$

$ \frac{d}{da}\left(2F(0)\minus{}2F(\minus{}a)\right)^{'}\equal{}2f(\minus{}a)$

so we have:

$ f(a)\plus{}f(\minus{}a)\equal{}2f(\minus{}a)$

$ f(a)\equal{}f(\minus{}a)$

Therefore, no..

milin did tacitly assume that $ f$ was continuous when he invoked the Fundamental Theorem of Calculus.

We could, of course, start with an even function $ f$ and then change its values in an asymmetric way on a set of points too small to change the integral. In that way, yes there is such a (discontinuous) function - not that that seems to be much of an answer.

I'm talking about something like this:

$ f(x)\equal{}\begin{cases}0,&x\equal{}\frac1n,n\in\mathbb{Z}\\1&\text{otherwise}\end{cases}.$

As I said, something of a cheap example.


Let $ f$ be a function defined on $ \text{N}_0 \equal{} \{ 0,1,2,3,...\}$ and with values in $ \text{N}_0$, such that for $ n,m \in \text{N}_0$ and $ m \leq 9, f(10n \plus{} m) \equal{} f(n) \plus{} 11m$ and $ f(0) \equal{} 0.$ How many solutions are there to the equation $ f(x) \equal{} 1995$?

A. None
B. 1
C. 2
D. 11
E. Infinitely many

[hide="Er..."] Unless I've misread the problem, $ 11 | f(n)$ for all $ n \in \mathbb{N}_0$ (in fact, $ f( \overline{d_n ... d_2 d_1 d_0}) \equal{} 11(d_n \plus{} ... \plus{} d_2 \plus{} d_1 \plus{} d_0)$), so there are no solutions.  Did you mean $ f(10n \plus{} m) \equal{} 11 f(n) \plus{} m$?  [/hide]

[hide="Solution"]

Just like previous poster said, all answers will be in the form of 11 x something so clearly, since 11 doesn't divide 1995, there is no such value or A. [/hide]


When I try to log in to AoPS Wiki, it says that

Login error:
You have not specified a valid user name.

What is the problem here?

I imagine it's because you have a slash in your name...

Probably; AoPS Wiki does a lot of strange stuff.

My contributions don't show (probably because of the underscores).

i_like_pie: I fixed the contribution thing for you...


If $x+y=4$ and $xy=2$ then find $x^6+y^6$.

Find two four digit numbers whose product is $4^8+6^8+9^8$.

Ok I got the first problem, and I kind of get the 2nd one. But can someone explain more on it?


Let $ n\in N^*$. Prove that
$ \sum^n_{j\equal{}1}\frac{1}{C^j_n}\equal{}\frac{n\plus{}1}{2^n}\sum^{n\minus{}1}_{j\equal{}0}\frac{2^j}{j\plus{}1}$


Prove that for all positive integers $ n$: $ \frac{2n}{3n\plus{}1} \le \displaystyle\sum_{k\equal{}n\plus{}1}^{2n}\frac{1}{k} \le \frac{3n\plus{}1}{4(n\plus{}1)}$.

[quote="moldovan"]Prove that for all positive integers $ n$: $ \frac {2n}{3n \plus{} 1} \le \displaystyle\sum_{k \equal{} n \plus{} 1}^{2n}\frac {1}{k} \le \frac {3n \plus{} 1}{4(n \plus{} 1)}$.[/quote]

We have, for $ 0 < a < x \le y$,  $ \frac {1}{x} \plus{} \frac {1}{y} < \frac {1}{x \minus{} a} \plus{} \frac {1}{y \plus{} a}$
$ \Leftrightarrow (x \minus{} a)(y \plus{} a) < xy \Leftrightarrow 0 < a(y \minus{} x \plus{} a)$, which is obviously true.

So, $ \frac {2n}{3n \plus{} 1} \equal{} \frac {n}{2}(\dfrac{1}{\dfrac{2n \plus{} (n \plus{} 1)}{2}} \plus{} \dfrac{1}{\dfrac{2n \plus{} (n \plus{} 1)}{2}})$
$ \le \sum_{k \equal{} n \plus{} 1}^{2n}\frac {1}{k} \le \frac {n}{2}(\frac {1}{n \plus{} 1} \plus{} \frac {1}{2n})$

Now $ \frac {n}{2}(\frac {1}{n \plus{} 1} \plus{} \frac {1}{2n}) \equal{} \frac {3n \plus{} 1}{4(n \plus{} 1)}$
$ \Leftrightarrow n(\frac {2(n \plus{} 1)}{n \plus{} 1} \plus{} \frac {2(n \plus{} 1)}{2n}) \equal{} 3n \plus{} 1$, which is obviously true, done.

Equality for left equality holds $ \Leftrightarrow n \equal{} 1$
Equality for right equality holds $ \Leftrightarrow n \equal{} 1,2$


Putnam
linear algebra
matrix
induction
college contests

Define a sequence $\{u_n\}_{n=0}^{\infty}$ by $u_0=u_1=u_2=1,$ and thereafter by the condition that
$\det\begin{vmatrix} u_n & u_{n+1} \\ u_{n+2} & u_{n+3} \end{vmatrix}=n!$
for all $n\ge 0.$ Show that $u_n$ is an integer for all $n.$ (By convention, $0!=1$.)

[u][b]The author of this posting is : blahblahblah[/b][/u]
____________________________________________________________________

We can show by induction that $u_{n}u_{n+1} = n!$.

After that, we just need to show that $u_n \vert n! + \left(n+1\right)!$, which isn't that difficult, using the above.

I calculated $u_0$ to $u_{11}$ and still didn't see the above pattern.

[quote="blahblahblah"]... and still didn't see the above pattern.[/quote]
In fairness to blahblahblah, "above" refers to another now-vanished post.

The pattern is this: $u_{n+2}=(n+1)u_n$, so that $u_{2k}=1\cdot3\cdot5\cdots(2k-1)$ and $u_{2k+1}=2\cdot4\cdot6\cdots(2k).$

Since the recursion uniquely defines the sequence (proof by induction), once you have shown that the above formulas satisfy the recursion, that suffices as a proof.

[quote="Kent Merryfield"][quote="blahblahblah"]... and still didn't see the above pattern.[/quote]
In fairness to blahblahblah, "above" refers to another now-vanished post.[/quote]

Strange things happen - in my file, blahblahblah's post was the very first reply to the original message!

  Darij

You're correct, darij, the pattern I was referring to was $u_nu_{n+1}=n!$

But the problem is that you does not prove that the above solution is the only one to the defined sequence. Probably there are other possible solutions that contain non-integer. Therefore, how could we prove this is the unique solution?

The recursion $ u_nu_{n\plus{}3}\minus{}u_{n\plus{}1}u_{n\plus{}2}\equal{}n!$ can be solved uniquely: $ u_{n\plus{}3}\equal{}\frac{u_{n\plus{}1}u_{n\plus{}2}\plus{}n!}{u_n}$.

A nice problem... We can either guess(?) or with some manipulation conclude that for odd $n$ $u_n$ is the product of all even positive numbers less than $n$ and for even $n$ it is the product of all odd positive numbers less or equal to $n$ . Proof then goes by induction.


geometry
geometric transformation
reflection
circumcircle
homothety
absolute value
geometry solved

In a triangle $ABC$ of the area $S$ , point $H$ is the orthocenter , $D,E,F$ are the feet of the altitudes from $A,B,C$ and $P,Q,R$ are the reflections of $A,B,C$ in $BC,CA,AB$, respectively. The triangles $DEF$ and $PQR$ have the same area $T$ . Given that $T>\frac 35 S$ , prove that $T=S$.

Are you sure the problem is correct? Geometer's Sketchpad does not show that those two triangles can have the same areas...

Shobber, it's well possible that the triangles DEF and PQR have the same area. Just check out some obtuse-angled triangles ABC. So the problem is indeed correct.

I will actually show a stronger assertion:

[color=blue][b]Problem.[/b] Let ABC be a triangle with area S. Let H be the orthocenter of triangle ABC, and D, E, F the feet of its altitudes from the vertices A, B, C, respectively. Furthermore, let P, Q, R be the reflections of the points A, B, C in the lines BC, CA, AB. Assume that the triangles DEF and PQR have the same (non-directed) area T. Prove that either $T=\frac35S$ or T = S.[/color]

[i]Solution.[/i] First some preliminaries:

Let O be the center and R the radius of the circumcircle of triangle ABC.

For an arbitrary point M in the plane of triangle ABC, the triangle formed by the orthogonal projections of the point M on the sides BC, CA, AB of triangle ABC is called the [i]pedal triangle[/i] of the point M with respect to triangle ABC.

According to the Gergonne theorem on pedal triangles (see http://www.mathlinks.ro/Forum/viewtopic.php?t=5658 posts #2, #12 and #13), the area of the pedal triangle of the point M with respect to triangle ABC is $\frac{S\cdot\left|OM^2-R^2\right|}{4R^2}$; hereby, the expression $\left|OM^2-R^2\right|$ is actually the absolute value of the power of the point M with respect to the circumcircle of triangle ABC.

The triangle DEF, formed by the feet of the altitudes of triangle ABC, is the pedal triangle of the orthocenter H with respect to triangle ABC; thus, its area equals $\frac{S\cdot\left|OH^2-R^2\right|}{4R^2}$.

Let N be the nine-point center of triangle ABC; then, the area of the pedal triangle of this point N with respect to triangle ABC equals $\frac{S\cdot\left|ON^2-R^2\right|}{4R^2}$.

The triangle PQR is the reflection triangle of triangle ABC; according to [url=http://forumgeom.fau.edu/FG2003volume3/FG200311index.html]Darij Grinberg, [i]On the Kosnita point and the reflection triangle[/i], Forum Geometricorum 3 (2003) pages 105-111[/url], Theorem 4, this triangle PQR is the image of the pedal triangle of the nine-point center N of triangle ABC in the homothety with center at the centroid of triangle ABC and factor 4. Hence, the area of triangle PQR is $4^2=16$ times as large as the area of the pedal triangle of the nine-point center N. Since the area of the pedal triangle of the nine-point center N is $\frac{S\cdot\left|ON^2-R^2\right|}{4R^2}$, we thus obtain that the area of triangle PQR is $16\cdot\frac{S\cdot\left|ON^2-R^2\right|}{4R^2}$.

Now, as we know that both the area of triangle DEF and the area of triangle PQR are equal to T, we get $T=\frac{S\cdot\left|OH^2-R^2\right|}{4R^2}=16\cdot\frac{S\cdot\left|ON^2-R^2\right|}{4R^2}$.

Hence, $\left|OH^2-R^2\right|=16\cdot\left|ON^2-R^2\right|$. But it is well-known that the nine-point center N of triangle ABC is the midpoint between the circumcenter O and the orthocenter H. Thus, $OH=2\cdot ON$, and we get

$\left|OH^2-R^2\right|=16\cdot\left|ON^2-R^2\right|=\left|16\cdot ON^2-16R^2\right|$
$=\left|4\cdot\left(2\cdot ON\right)^2-16R^2\right|=\left|4\cdot OH^2-16R^2\right|$.

Now, two cases are possible: Either $OH^2-R^2=4\cdot OH^2-16R^2$ or $OH^2-R^2=-\left(4\cdot OH^2-16R^2\right)$.

First we consider the case when $OH^2-R^2=4\cdot OH^2-16R^2$. Then, we quickly obtain $OH^2=5\cdot R^2$. Thus, $T=\frac{S\cdot\left|OH^2-R^2\right|}{4R^2}=\frac{S\cdot\left|5\cdot R^2-R^2\right|}{4R^2}=\frac{S\cdot 4R^2}{4R^2}=S$.

Now, consider the case when $OH^2-R^2=-\left(4\cdot OH^2-16R^2\right)$. Then, we get $OH^2=\frac{17}{5}\cdot R^2$. Thus, $T=\frac{S\cdot\left|OH^2-R^2\right|}{4R^2}=\frac{S\cdot\left|\frac{17}{5}\cdot R^2-R^2\right|}{4R^2}=\frac{S\cdot \frac{12}{5}R^2}{4R^2}=\frac35S$.

Hence, we get either T = S or $T=\frac35S$. And the problem is solved.

  darij


Of all the copy machines produced by CopyCo Inc., $ 80\%$ work correctly, and $ 20\%$ are broken.  A simple copying test is done on all machines before they are sent out.  A special paper is put in and copied.  When the copied sheet comes out, it either says ``Broken'' or ``Ok'' on it.  If the machine is really broken, the copied sheet will always say ``Broken''.  However, if the machine is not broken, $ 10\%$ of the time the test sheet will come out saying ``Broken'' anyway.  If a random machine is tested, and the sheet comes out saying, ``Broken'', what is the probability that the machine is really broken?

Relative to all possibilities and given that it comes out broken, there is a $ 0.8\cdot0.1\equal{}0.08$ probability that it is working, and a $ 0.2$ probability that it is broken. Of this $ 0.2\plus{}0.08\equal{}0.28$ probability that this happens, there is a $ \frac{0.2}{0.28}\equal{}\boxed{\frac57}$ probability that it is really broken.


If a five question True/False test is given and $ 60\%$ is a passing grade, what is your percent chance of passing if you guess on each question?

By symmetry (i.e. it is just as likely to get $ 3$ or more questions right as it is to get $ 3$ or more questions wrong), the answer is $ \boxed{\frac12}$.

You must get at least 3 of the questions correct to get $ 60\%$ of the questions correct.  Clearly the probability of this happening is the same as getting at least 3 of the questions wrong.  These three together encompass all possibilities, so the probability of getting at least 3 right is $ \boxed{50\% }$

math154, they're slightly different problems :wink:

[quote="5849206328x"]math154, they're slightly different problems :wink:[/quote]

Dang, somewhat late? :P  But yes, I do apologize for being dyslexic in both of my previous posts.


[color=darkred]Prove that for any point $M$ which belongs to the incircle of the triangle $ABC$ there is the following identity,

where $2p=a+b+c\ :$ $\boxed{\ a\cdot MA^{2}+b\cdot MB^{2}+c\cdot MC^{2}=abc+2(p-a)(p-b)(p-c)\ }\ \ (*)\ .$

See http://www.mathlinks.ro/Forum/viewtopic.php?t=113516 [/color]

[color=darkblue][b]Proof.[/b] I wanted to verify the [b]Gemath's identity[/b] through this well-known identity. Denote the points $D\in BC\ ,\ E\in CA\ ,\ F\in AB$ which belong to the incircle $w=C(I,r)$ of the triangle $ABC\ .$ Thus, $EF=2(p-a)\sin \frac{A}{2}$ and $m(\widehat{EDF})=\frac{B+C}{2}\ .$ The Gemath's identity from the topic mentioned above becomes $\sum 2(p-a)\sin^{2}\frac{A}{2}\cdot MD^{2}=8\prod \left[(p-a)\sin\frac{A}{2}\right]\ ,$ i.e. $\sum a\cdot MD^{2}=4(p-a)(p-b)(p-c)\ .$

Apply the Stewart's theorem to the cevian $[MD]$ in the triangle $BMC\ : \ a\cdot MD^{2}+a(p-b)(p-c)=(p-c)\cdot MB^{2}+(p-b)\cdot MC^{2}$ a.s.o.

Adding these three identities obtain $\sum a\cdot MD^{2}=\sum a\cdot MA^{2}-\sum [a(p-b)(p-c)]$ $\implies$ $\sum a\cdot MA^{2}=\sum a(p-b)(p-c)+4(p-a)(p-b)(p-c)\ .$

Prove easily that $(p-a)(p-b)(p-c)=pr^{2}$ and $\sum a(p-b)(p-c)=2pr(2R-r)\ .$ In conclusion, $\sum a\cdot MA^{2}=2pr(2R+r)=abc+2\prod (p-a)\ .$

[b]Remark.[/b] Can obtain directly this identity from the remarkable relation $a\cdot XA^{2}+b\cdot XB^{2}+c\cdot XC^{2}=2p\cdot XI^{2}+abc$ for $X\in C(I,r)\ .$[/color]

That is a nice problem! but we can prove simplely in your solution:
$\sum aMA^{2}=(a+b+c)MI^{2}+\sum aIA^{2}$
$\sum aIA^{2}=abc$ and when $M\in$ incircle $\Rightarrow MI^{2}=r^{2}$

You didn't read the remark from the my last reply.

oh sorry Virgil  :blush: , I have a nice problem as a present for you:
(maybe it is old but the solution quite new)
Given are equaliteral triangle $ABC, (O)$ is incircle tagent $BC,CA,AB$ at $A',B',C'$ with any $M$ lies on small arc $B'C'$ and $A'',B'',C''$ are projection of $M$ to the sides prove $\sqrt{MA''}=\sqrt{MB''}+\sqrt{MC''}$
A proof:
The equal is equaliteral
$MA''=MB''+MC''+2\sqrt{MB''}\sqrt{MC''}\Leftrightarrow (MA''-MB''-MC'')^{2}=4MB''MC''\Leftrightarrow MA''^{2}+MB''^{2}+MC''^{2}=2(MA''MB''+MB''MC''+MC''MA'')(*)$
we can prove $\overrightarrow{MA''}+\overrightarrow{MB''}+\overrightarrow{MC''}=\frac{3}{2}\overrightarrow{MO}\Rightarrow MA''^{2}+MB''^{2}+MC''^{2}+2\sum\overrightarrow{MB''}{\overrightarrow{MC''}=\frac{9}{4}r^{2}\Rightarrow}$ $MA''^{2}+MB''^{2}+MC''^{2}-\sum MB''MC''=\frac{9}{4}r^{2}(1)$(here we notice $\angle BMC=\angle CMA=\angle AMB=120^{0}$)
from $MA''+MB''+MC''=3r \Rightarrow MA''^{2}+MB''^{2}+MC''^{2}+2\sum MB''MC''=9r^{2}(2)$
Choose fit coefficiens and multiply $(1),(2)$ we will have $(*)$
The solution maybe quite long but the equal $(1),(2)$ have nice meaning, with it we can creat more nice other equal.


Προσπαθώ να λύσω αυτήν την άσκηση εδώ και μέρες  και δεν τα καταφέρνω. Μήπως μπορείτε να με βοηθήσετε;
Δίνεται η συνάρτηση $ f(x)$ με $ x{\in}R$ και $ f(nx \plus{} (1 \minus{} n)y) \equal{} nf(x) \plus{} (1 \minus{} n)f(y)$ και $ n{\in}[0,1]$. Να δείξετε ότι $ f(0) \equal{} f(1)$

Έστω η συνάρτηση $ f(x)\equal{}ax$ με $ a\neq 0$

Έχουμε:

$ f(nx \plus{} (1 \minus{} n)y) \equal{} nf(x) \plus{} (1 \minus{} n)f(y) \iff$

$ a[nx \plus{} (1 \minus{} n)y] \equal{} nax \plus{} (1 \minus{} n)ay \iff$

$ anx \plus{} (1 \minus{} n)ay \equal{} nax \plus{} (1 \minus{} n)ay$, που ισχύει

Ενώ έχουμε $ f(0)\equal{}0\neq a\equal{}f(1)$

Δεν ισχύει αυτό που λες γιατί σε επόμενο υποερώτημα μου ζητάει να αποδείξω ότι $ f(x) \equal{} f(0)$, δηλαδή ότι είναι σταθερή

gregg o pontios θέλει να πει ότι ενώ η $ f(x)\equal{}ax, \ a\neq 0$ ικανοποιεί την σχέση που έδωσες στην αρχή, δεν ικανοποιεί το συμπέρασμα $ f(0)\equal{}f(1)$. Μήπως λοιπόν η άσκηση δίνει και κάτι άλλο ως δεδομένο στην αρχή? Γιατί αν είναι έτσι φοβάμαι ότι ο pontios έχει δίκιο. Πολλές φορές τυχαίνει να βρίσκουμε κάποια άσκηση που να είναι λάθος. Άνθρωποι είναι και αυτοί που τις κατασκευάζουν.

Φιλικά,
Αλέξανδρος

Δεν νομίζω αλλά δεν είμαι και σίγουρος. Την άσκηση την βρήκα στο διαδίκτυο από το mathsforyou.
Απλά αυτό που λέω (και ίσως να μην στέκει αλλά έτσι το βλέπω :blush: ) είναι ότι αν η f είναι σταθερή τότε επαληθεύει την συνάρτηση.

gregg prospa8hsa na s apanthsw me pm alla den ta katafera. gia kapoio logo oti pm prospa8w na steilw paramenei sto outbox kai oxi sto sentbox.
Mporeis na m dwseis to link ap opou vrhkes thn askhsh?
Oso gia to allo me tous migadikous, |1/z_i|*|z_i|=|1/z_i*z_i|=|1|=1
ara 1>3b/a*|1/z_i|pou sunepagetai pollaplasiazontas kai ta 2 melh me a/3b oti a/3b>|1/z_i|
Kalutera twra?

Paidia xilia suggnwmh pou eprepe na to steilw edw pera alla opws proeipa den mporw na steilw pm!an kapoios gnwrizei pws mporw na to dior8wsw auto as m steilei.

Hlias


Let $ABC$ be a triangle and $h_a$ be the altitude through $A$. Prove that \[ (b+c)^2 \geq a^2 + 4h_a ^2 .  \]

Obviously, $ah_a = 2K$ where K is the area.

Then the ineq transforms into $a^2(b+c-a)(b+c+a) \ge 4K^2$.

which becomes after Heron and cancelling out, $4a^2 \ge (b-c+a)(a-b+c)$

which is equivalent to $3a^2 \ge 0$.

Well, isn't that quite brutal? Just try the following:

Let $H$ be the foot of the altitude $h_a$. Then $(b+c)^2 = b^2+2bc+c^2\geq BH^2 + h_a^2 + 2BH\cdot CH + CH^2 + h_a^2\geq$ $(BH+CH)^2 + h_a^2 = a^2 + h_a^2$.

So in fact, the inequality is very unsharp - we only have equality if $h_a=0$.

Peter

The actual official question of RMO-1996 was :

Let $ ABC$ be a triangle and $ h_a$ the altitude through $ A$. Prove that
                                \[ (b\plus{}c)^{2}\geq a^{2}\plus{}4h_{a}^{2}.\]

Let $D$ be the projection of $A$ onto $BC$. Let $DB=m$ and $DC=n$.

$1 \ge cos(B-C) \Rightarrow 1 \ge cosBcosC + sinBsinC \Rightarrow 1 \ge \frac{mn}{bc} + \frac{h^2_a}{bc} \Rightarrow bc \ge mn + h^2_a.$

$(b+c)^2 = b^2 + 2bc + c^2 \ge h^2_a + n^2 + 2(mn + h^2_a) + h^2_a + m^2 = n^2+2mn+m^2 + 4h^2_a = a^2+4h^2_a$

Equality holds when $cos(B-C) = 1$ or $\angle B = \angle C$

I have an easier version of duohead's proof.

Let $X$ be the foot of the altitude from $A$ to $BC$
$BX= c \cos B$
$CX= b \cos C$
$h_a = b \sin C = c\sin B$

$(b+c)^2 \ge BX^2 + CX^2 + 2 BX \cdot CX$
$\iff b^2 + c^2 + 2bc \ge b^2 + c^2 + 2 BX \cdot CX + 2 h_a^ 2$
$\iff bc \ge bc \cos B \cos C + bc \sin B \sin C$
Now the inequation  can be simplified as
$bc \ge  bc(\cos B \cos C + \sin B \sin C) = bc \cdot cos (B - C)$
So we get $1 \ge \cos (B - C)$
Since this is true we get the desired conclusion

Take the image of the A about BC and call it A'. Then, draw AC' parallel to BC and AC' = BC, that implies AC'CB is a parallelogram. Notice, angle(A'AC') = 90 also A'C = b and CC' = c. Now, apply triangle inequality in triangle(CA'C') and we are done!!


A circle has an area of 100pi, what is twice the circumference minus 8pi?

[url=http://lmgtfy.com/?q=circumference]Circumference[/url]

The radius is $ 10$, so the circumference is $ 20\pi$.

$ 2(20\pi) \minus{} 8\pi \equal{} \boxed{32\pi}$


trigonometry
geometry
trig identities
Law of Sines

The longest side will be the side opposite the largest angle.

[quote="davidlizeng"]you only give angle measures. that's not enough to determine the side lengths. you need to give at least one side length, then you can use the law of sines.[/quote]
He doesn't ask for the length of the longest side of the triangle, just [i]which[/i] side is the longest. If you actually compute the angles, you find that $\angle{F}=90^{\circ}$, so $\triangle{DEF}$ is a right triangle. Surely you agree that $\overline{DE}$, the hypotenuse, is the longest side.

I do not have to give the length of any side of the triangle to find the answer here. I'm not reviewing trigonometry right now.  So, the law of sines means absolutely nothing to me at this point.

[quote="Interval"]In triangle DEF, an exterior angle at D measures 170º, and m<E=80º.  What is the longest side of the triangle ?[/quote]
[hide]
D is 10 degrees, and E is 80 degrees. that means F > E > D.
F is the largest angle, so the side opposite F is ED.
[/hide]


Let be given a point $M$ inside a quadrilateral $ABCD$.Prove that:\[ min\{\widehat{MAB},\widehat{MBC},\widehat{MCD},\widehat{MDA}\}\le\frac{\pi}{4} \]

It's a generalization of Brocard Point (?)
It seems that it is really old enough. You can check this @ India MO 1998. (If I weren't false)


If anyone could help me with the following problem,I would appreciate it :)

 [b]Problem[/b]:
 If there is a switch from deposits into currency,what happens to the federal funds rate?
 Could you show to me a supply and demand analysis of the market for reserves?

  Thanks in advance.

The federal funds rate is set by the Federal Reserve to influence bank overnight rates. That rate wouldn't change just because consumers withdraw their funds. To reign in possible inflation that might arise from increased liquidity outside the banking sector, the FF rate may be raised. However if there is a massive bank run destabilizing the banking system the rate may sharply go down to meet the increased liquidity demands from banks.
If you meant if the [i]Federal Reserve[/i] switches its deposits into currency, I would think that they would raise the FF rate instead (if by deposits you mean deposits by banks being held by the Federal Reserve).

Yes, I'm necroposting but I thought this question deserved an answer, however late it may be.


induction
inequalities proposed
inequalities

Let $ a_i, b_i, c_i$ sequences of positive numbers for $ i \equal{} 1,2,\dots,n$.

(1) Prove that
\[ \sum \frac {a_i^2 \plus{} b_i^2 \plus{} c_i^2}{a_i \plus{} b_i \plus{} c_i} \geq \frac {(\sum a_i)^2 \plus{} ( \sum b_i )^2 \plus{} (\sum c_i)^2}{\sum (a_i \plus{} b_i \plus{} c_i)}\]
(2) Prove that
\[ \sum \frac {a_i^2 \plus{} b_i^2 \plus{} c_i^2 \plus{} k(a_ib_i \plus{} b_ic_i \plus{} c_ia_i)}{a_i \plus{} b_i \plus{} c_i} \leq \frac {(\sum a_i)^2 \plus{} ( \sum b_i )^2 \plus{} (\sum c_i)^2 \plus{} k(\sum a_{i}\sum b_{i} \plus{} \sum b_{i}\sum c_{i} \plus{} \sum c_{i}\sum a_{i} )}{\sum (a_i \plus{} b_i \plus{} c_i)}\]
holds for $ k \equal{} 8$.

(3)* Is $ k \equal{} 8$ smallest possible  :?:

(4)* Is $ k \equal{} 2$ biggest possible such that the [i]converse[/i] of (2) still holds (as in (1) for $ k\equal{}0$) :?:

The first one (for $ k\equal{}0$) is easy using Cauchy-Schwartz.

Concerning the general one: It is equivalent to
\[ (k\minus{}2)\cdot\sum\frac{a_{i}b_{i}\plus{}b_{i}c_{i}\plus{}c_{i}a_{i}}{a_{i}\plus{}b_{i}\plus{}c_{i}}\leq(k\minus{}2)\cdot\frac{\sum a_{i}\sum b_{i}\plus{}\sum b_{i}\sum c_{i}\plus{}\sum c_{i}\sum a_{i}}{\sum (a_{i}\plus{}b_{i}\plus{}c_{i})}\]

As you claim in (4) that $ k\equal{}2$ is the biggest possible value that the converse holds, $ k\equal{}2$ should be the smallest possible value such that the original one holds...

Sorry I didn't have the time to have a closer look at it...

[quote="Martin N."]The first one (for $ k \equal{} 0$)
Concerning the general one: It is equivalent to
\[ (k \minus{} 2)\cdot\sum\frac {a_{i}b_{i} \plus{} b_{i}c_{i} \plus{} c_{i}a_{i}}{a_{i} \plus{} b_{i} \plus{} c_{i}}\leq(k \minus{} 2)\cdot\frac {\sum a_{i}\sum b_{i} \plus{} \sum b_{i}\sum c_{i} \plus{} \sum c_{i}\sum a_{i}}{\sum (a_{i} \plus{} b_{i} \plus{} c_{i})}\]
[/quote]
 :blush: I saw of course that for $ k \equal{} 2$ we have an identity in (2), but it didn't occur to me to just subtract that identity from the general ineq :wallbash:

I'll prove it by induction, for $ i\equal{}1,2$, we have to prove (denoting $ m\equal{}a\plus{}b\plus{}c$, $ n\equal{}x\plus{}y\plus{}z$)
$ \frac{(a^2\plus{}b^2\plus{}c^2)m^2\plus{}(x^2\plus{}y^2\plus{}z^2)n^2}{2}\geq(ax\plus{}by\plus{}cz)mn\mbox{,}$
which is obvious by AM-GM and Cauchy-Schwartz.

Assuming the statement holds for some $ n$, we get
$ \sum_{i\equal{}1}^{n\plus{}1}\frac{a_{i}b_{i}\plus{}b_{i}c_{i}\plus{}c_{i}a_{i}}{a_{i}\plus{}b_{i}\plus{}c_{i}}\leq
\frac{\sum_{i\equal{}1}^{n} a_{i}\sum_{i\equal{}1}^{n} b_{i}\plus{}\sum_{i\equal{}1}^{n} b_{i}\sum_{i\equal{}1}^{n} c_{i}\plus{}\sum_{i\equal{}1}^{n} c_{i}\sum_{i\equal{}1}^{n} a_{i}}{\sum_{i\equal{}1}^{n} a_{i}\plus{}\sum_{i\equal{}1}^{n}b_{i}\plus{}\sum_{i\equal{}1}^{n}c_{i}} \plus{} \frac{a_{n\plus{}1}b_{n\plus{}1}\plus{}b_{n\plus{}1}c_{n\plus{}1}\plus{}c_{n\plus{}1}a_{n\plus{}1}}{a_{n\plus{}1}\plus{}b_{n\plus{}1}\plus{}c_{n\plus{}1}} \leq \frac{\sum_{i\equal{}1}^{n\plus{}1} a_{i}\sum_{i\equal{}1}^{n\plus{}1} b_{i}\plus{}\sum_{i\equal{}1}^{n\plus{}1} b_{i}\sum_{i\equal{}1}^{n\plus{}1} c_{i}\plus{}\sum_{i\equal{}1}^{n\plus{}1} c_{i}\sum_{i\equal{}1}^{n\plus{}1} a_{i}}{\sum_{i\equal{}1}^{n\plus{}1} a_{i}\plus{}\sum_{i\equal{}1}^{n\plus{}1}b_{i}\plus{}\sum_{i\equal{}1}^{n\plus{}1}c_{i}}\mbox{,}$
and the induction is complete.  :)

it doesn't look nice at all...sorry :blush:

@spanferkel: how did you prove this?

My proof of \[ \sum\frac{a_{i}b_{i}\plus{}b_{i}c_{i}\plus{}c_{i}a_{i}}{a_{i}\plus{}b_{i}\plus{}c_{i}}\le \frac{\sum a_{i}\sum b_{i}\plus{}\sum b_{i}\sum c_{i}\plus{}\sum c_{i}\sum a_{i}}{\sum (a_{i}\plus{}b_{i}\plus{}c_{i})}\]
is based on the same lemma $ \sum\frac{x_{i}y_{i}}{x_{i}\plus{}y_{i}}\leq\frac{\sum x_{i}\sum y_{i}}{\sum x_{i}\plus{}\sum y_{i}}$ that is used [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=1688547#1688547]here[/url] (which is a special case of Minkowski). Just put $ x_i: \equal{}a_i$ and $ y_i: \equal{}b_i\plus{}c_i$, then sum up cyclically. By the way, I had gotten the $ k\equal{}8$ case the same way via $ x_i: \equal{}2a_i\plus{}c_i$ and $ y_i: \equal{}2b_i\plus{}c_i$.  :)


A uniform spring whose unstressed length is $l$ has a force constant $k$ . The spring is cut into two pieces of unstressed length $l_1$ and $l_2$ where $l_1=nl_2$ and $n$ is an integer . What are the corresponding force constant $k_1$ and $k_2$ in terms of $n$ and $k$ ? Does your result seem reasonable for $n=1$ ?

I'd say $\frac{k}{n+1}$ and $\frac{nk}{n+1}$, don't you think so?

What concept do you use ? Is it Hooke's Law ?Cause I cant seem to figure out the right formula to do  :(

I think you got the answers backward.  The shorter spring would have the greater force for a given displacement withing the elastic zone of both springs.

Actually, now that I think about it they should be both $k$ no matter how long they are. Should have known this right from the beginning... :)

bus -- You're right.  However, the shorter spring piece will have the greater force at equal displacement inside the elastic zone of both spring pieces.

Actually, I derived the general case of the spring constant when a spring is cut into many pieces by just considering that the extension(tension etc.) was a function of the no. of coils....

[quote="wizard2378"]However, the shorter spring piece will have the greater force at equal displacement inside the elastic zone of both spring pieces.[/quote]And why is that? Shouldn't it be the same if they both have the same $k$?


vector
inequalities unsolved
inequalities

[quote="Axelshmmy"]$ X, Y ,Z$  are three vectors on  $ R^3$  prove that

$ llX \minus{} Zll \plus{} llZ \minus{} Yll \plus{} llX \minus{} Zll \plus{} llX \plus{} Y \plus{} Zll > \equal{} llXll \plus{} llYll \plus{} llZll$[/quote]
It's true in $ \mathbb R^n$ too.
A squaring kills it. :wink:


Mr. Harter drives from Chicago to Milwaukee on Sunday.  On Monday when he makes the return trip, he figures that if he increases his speed by one-fifth, he can make the trip in 80 minutes.  How many minutes did the first trip take?

Since the distances are equal, we can write \[ \left(\frac{6}{5}r\right)(80)\equal{}rt \Rightarrow t\equal{}\boxed{96}\]


It's been several weeks and I haven't seen a new Alcumulus video come out.  When will a new one come?

Firstly, it's Alcumus. For some reason, a lot people think it's Alcumulus. I don't know why. Just a small thing, unimportant.

Videos are given to you based on your progress. As you do more problems, the rate at which videos are given reduces dramatically. Thus, cherish the great videos while you are "Alcumus-young."  :)

Or, if you have nothing to do, you could hone your skills rewatching, for the cost of a little Karma. Most people get more the second time.  :)

[quote="r15s11z55y89w21"]Or, if you have nothing to do, you could hone your skills rewatching, for the cost of a little Karma. [/quote]
You don't lose (or gain) Karma as long as you comment on the video every time. You can rate it only the first time, and gain 5 Karma.

[quote="maybach"]When will a new one come?[/quote]
They made most of them late last year (October-November 2008). Intro to Algebra is probably coming out next year.

[quote="r15s11z55y89w21"]

Videos are given to you based on your progress. As you do more problems, the rate at which videos are given reduces dramatically. Thus, cherish the great videos while you are "Alcumus-young."  :)
  [/quote]
Is it possible to watch videos on the official Video List? (sometimes it works for me...)


function
limit
trigonometry
calculus
calculus computations

Let $a,b,c$ be non zero constant numbers.
Evaluate

\[\lim_{r\to\infty} \frac{\cos \frac{a}{r}-\cos \frac{b}{r}\cos \frac{c}{r}}{\sin \frac{b}{r}\sin \frac{c}{r}}\]


calculus
integration
algebra
function
domain
Ring Theory
superior algebra

Let R be a commutative ring and I is an ideal of R .Prove that :
(i) The quotient ring R/I is an integral domain if and only if I is a prime ideal.
(ii) The quotient ring R/I is a field if and only if I is a maximal ideal.
Can you prove this clause for me . Thanks a lot .

Review the definitions of an integral domain, a prime ideal, a field, and a maximal ideal.

(i)  What does $ ab \equal{} 0$ mean in $ R/I$?
(ii)  What does not having a multiplicative inverse mean in $ R/I$?


blogs trigonometry MATHCOUNTS algorithm AMC AIME function

This is where you can talk about solutions. Request them or answer them, it's up to you. I'll eventually post all the solutions but I'll answer the ones request in the order they are first.

Here are the answer keys.

[hide="Answer"]
1. A
2. A
3. B
4. B
5. E
6. E
7. D
8. A
9. B
10. D
11. B
12. B
13. IGNORE
14. E
15. D
16. A
17. A
18. C
19. D
20. C
21. B
22. C
23. C
24. E
25. B [/hide]

I put IGNORE for #13 because the correct answer is not one of them (I should be 24 but I put 42).

When we post our answer choices on the blog site, will everyone else see them?
Sorry, still kinda confused to what to do.

how do you solve the two trig ones? 11 and 13?

Yes.

People will be able to see your answers but it doesn't matter as long as you post your own (thus, more of AoPS Honor Code).

i remember a few of the problems from a mathcounts practice i had and from the mathcounts problem series. lol.
the one about the pattern in the square root could be solved in two ways. multiply the two outer numbers and add 1 or multiply the two inner numbers and subtract 1.

The trig ones were annyoing.I don't think I got them. Near the end (about 8 or so) my computer crashed, and it took me a while to get it on, so I didn't really answer the last few. Doesn't matter, I couldn't have solved them.

Could someone post a solution to #9? I still can't figure it out  :blush:

I'm going to grade the solutions that I received and send the score back to the owner first. Then, I'll post the solutions so it might be a while.

[hide="9"]$ gcd(n^2 \plus{} 7,n\plus{}4) \equal{} gcd(7\minus{}4n,n\plus{}4) \equal{} gcd(23,n\plus{}4)$ (by Euclid algorithm), and there are $ 86$ numbers divisible by $ 23$ in the range given.[/hide]

hm no time to take the test for me but after glancing over the square root pattern one, you could just note that the inside number has a units digit of 1, so the answer must be the answer with 1 as the units digit  :P

[quote="azjps"][hide="9"]$ gcd(n^2 \plus{} 7,n \plus{} 4) \equal{} gcd(7 \minus{} 4n,n \plus{} 4) \equal{} gcd(23,n \plus{} 4)$ (by Euclid algorithm), and there are $ 86$ numbers divisible by $ 23$ in the range given.[/hide][/quote]
Ugh. I somehow used the Euclidean Algorithm as well but wrote something down wrong so I got gcd(7,16)...so stupid of me.

I would have done this Mock AMC if I didn't have a Spanish project to work on...

Anyway, I just looked at number 14, and if you know something about linear recurrences...
[hide="Then..."]Maybe you can show that $ R_n\equal{}6R_{n\minus{}1}\minus{}R_{n\minus{}2}$[/hide]
I did this in my head, so I might be wrong, but if anyone can verify that what I am saying is true...

That's what I did. Did you get 9 as the answer?

Edit: Wait, I just read your post and realized that you didn't take it.

[quote="xpmath"]
Ugh. I somehow used the Euclidean Algorithm as well but wrote something down wrong so I got gcd(7,16)...so stupid of me.[/quote]

At first I had $ 7 \plus{} 16 \equal{} 25$, and it took me quite a bit of time to figure out what I did wrong.  :lol:

OMG I thought that exact same thing for about the whole test. I jus went with dividing the num by the denom, but it's the same thing really.

[quote="Silverfalcon"]I'm updating solutions to problems that were asked now (and soon too). So, keep asking for solutions for ones that you didn't get!  Difficult problems don't benefit you unless you learn how to do them!

[hide="Solution for Number 6"]
This is a Mathcounts problem from the book [i]The All-Time Greatest Mathcounts Problems[/i].  You can use log but there is much easier way.  It is from 1987 State Individual #9, which is current Target Round (name actually changed in 1988 so this is problem from last Individual Round!).

Note the pattern when the units digit was written:

$ 2^0 \equal{} 1$ (This is the ODD one that does not belong in pattern)
$ 2^1 \equal{} 2$
$ 2^2 \equal{} 4$
$ 2^3 \equal{} 8$
$ 2^4 \equal{} 6$
$ 2^5 \equal{} 2$
$ 2^6 \equal{} 4$

So, the pattern is 2,4,8,6....  Note that 11213 = 11212 x 4 + 1.  So, the units digit is same as the first term, which is 2. Now, we're going to subtract 1 to get the final answer $ \fbox{1}$. [/hide]

There is no need to log. Plus, you can't use it (unless you have $ \log 2$ and $ \log 3$ memorized) this year anyway.[/quote]
Yes I got the answer this morning using modular arithmetic(not all parts of my solution). Anyway, I don't see why you have to give the information about the number of digits: 3376! It makes me more curious to solve it by hand without using modular and the only method I know relates to number of digit is logarithm function! It's a nice problem :) .

i used mod 3 for the one about the jugs, and got two possible cases that i just had to test out.
im glad i saw the simple solution for the one about the units digit of that crazy exponent of 2. that's the only solution that came to my head .  :D

*btw, can anyone give a solution to #17. the first thing that came to my mind when i saw the equations was symmetry, and naturally i added them up. unfortunately i didn't know how to proceed from there as i couldn't find a slick way to factor or manipulate the resulting expression. :blush:

[hide="17"]
Summing all three equations gives $ x^2 \minus{} 2xy \plus{} 2y^2 \plus{} 6yz \plus{} 9yz^2 \equal{} (x \minus{} y)^2 \plus{} (y \plus{} 3z)^2 \equal{} 175$. $ x \minus{} y, y \plus{} 3z$ are both integers, and since for all integers $ a$, $ a^2 \equiv 0, 1 \pmod{4}$, it follows that the sum of two squares can only be $ \equiv 0, 1, 2 \pmod{4}$. However, $ 175 \equiv 3 \pmod{4}$, from which it follows that there are no solutions for $ x,y,z$.
[/hide]

I skipped 17 when I took it because I tend to have bad experiences with systems of equations on timed tests.

thanks azjps!
slick solution!

i wish i had seen that on the test as well.  :P

Apparently, I've reached maximum quota for uploading images.  So, I'm going to put up the images in my website and put the link here.

Since you guys are doing well on your own for easy to medium problems, I'm going to put up solutions to some of the later ones.  Let's start with the last one, which is absolutely beautiful problem!

Image:

http://hiddendiscoveries.com/resources/Mock+AMC+A+Number+25.JPG

[hide="Solution for Number 25"]
This is 1993 AIME Number 15. However, when you understand the solution at the end, you'll say, "Wow, that was pretty difficult but not bad at all!" Why? It doesn't require any knowledge beyond one year of school geometry. Trust me on that one.

First of all, description of the image.

$ r_1$ is the radius of the circle left of $ CH$ and $ r_2$ is the radius of the circle right of $ CH$.
$ P$ is the point that left circle is tangent to $ AC$ and $ Q$ is the point that left circle is tangent to $ AB$.
$ x$ is length of $ AH$ and $ y$ is the length of $ BH$. Thus, $ x \plus{} y \equal{} c$.
$ a,b,c$ are standard notation of triangle's sides.
$ h$ is the length of height.

Second, some key points.

This problem is difficult because it's hard to decide where to start. I'm going to start from $ HQ$ and focus on the left circle.  Note that $ HQ \equal{} r_1$ (if you don't get it, see that the segment connecting the center of left circle to $ R$ is a straight line parallel to $ HQ$). Then, $ AQ \equal{} x \minus{} r_1$. Since $ P$ and $ Q$ are both points of tangency, $ AQ \equal{} AP \equal{} x \minus{} r_1$.  So, $ CP \equal{} b \minus{} (x \minus{} r_1) \equal{} b \minus{} x \plus{} r_1$. Again, using same idea, we know that $ CR \equal{} h \minus{} (b \minus{} x \plus{} r_1) \equal{} h \minus{} b \plus{} x \minus{} r_1$. But, note that $ RH \equal{} r_1$ as well! (This is same way by comparing this segment to the radius of left circle going down and perpendicular to $ AB$).

So:

$ h \minus{} b \plus{} x \minus{} r_1 \equal{} r_1 \rightarrow r_1 \equal{} \frac {h \minus{} b \plus{} x}{2}$

Simiarly, $ r_2 \equal{} \frac {h \minus{} a \plus{} y}{2}$. Do NOT just pass by saying, "Oh okay." Work this out on your own so you completely get this!

Now, going back to our expression from the problem:

$ RS \equal{} |r_1 \minus{} r_2|$ (since we don't know $ r_1 < r_2$ or opposite; in the diagram, $ r_1 > r_2$ though)

$ \equal{} |\frac {1}{2} (h \minus{} b \plus{} x \minus{} h \plus{} a \minus{} y)| \equal{} \frac {1}{2} \cdot |(a \minus{} b) \plus{} (x \minus{} y)|$

Now, using Pythagorean Theorem:

$ h^2 \equal{} b^2 \minus{} x^2$
$ h^2 \equal{} a^2 \minus{} y^2$

Thus, $ b^2 \minus{} x^2 \equal{} a^2 \minus{} y^2$. Using difference of squares and rearranging, we get:

$ x \minus{} y \equal{} \frac {(a \plus{} b)(a \minus{} b)}{y \plus{} x}$

Now, substituting this back:

$ \equal{} \frac {1}{2} \cdot |(a \minus{} b) \plus{} \frac {(a \plus{} b)(a \minus{} b)}{x \plus{} y}|$

Recall that $ x \plus{} y \equal{} c$ so:

$ \equal{} \frac {1}{2} \cdot | \minus{} 1 \plus{} \frac {3987}{1995}| \equal{} \frac {996}{1995} \equal{} \frac {332}{665}$

Thus, $ m \plus{} n \equal{} 332 \plus{} 665 \equal{} \fbox{997}$. [/hide]

For number 25 you can just find angle A in terms of a cosine.

So just do the law of cosines backwords.  Youll get like 1993^2=1995^2+1994^2-2(1995)(1994)(cosA).

Youll get cosA=333/665 (make the calculation easier by using cool factoring tricks-also you dont have to solve it all the way because a factor of 1994 from the denominator will cancel out once you do the x/1994=cosA)

You can now find the lengths of the two segments broken up by the height

Then use that ratio to do HA/1994=cosA, so HA=1994(333/665)=998+332/665.

H is where the height touches AB (look at silverfalcons picture)

Then you will find like R+x where x (in silverfalcons picture its HA-R) is the segment that doesnt include the big radius but is still on one of the segments broken up by the height on the base (need a picture to explain this).  You find that R+x=998+332/665.  R is the bigger radius.

Then you will also find like r+x using the fact that tangents to a circle have equal lengths....just try to use some variables and youll find that r+x where r is the smaller radius.  r+x frmo some equation manipulation is 998 (try to set up some equations and subtract them)

Then do R+x-(r-x)=first answer-second answer which is 998+(332/665)-998= 332/665.
Calculation is a lot harder but since this was number 15 on AIME, 10-15 mintues is actually pretty quick (you can use alot fo (a-b)(a+b) and all that to find cosB which is the hard part).

How do you do that one with the basketball and free throws?

Please give your full solution, not just hint.  Moreover, none of the problems in this mock AMC requires knowledge of Ceva or Stewart's or anything like that. It is just elementary geometry.

I need a break from difficult problems so I'll put answers for easier problems first. Request any that you want to see the solutions!  :)

[hide="Solution for Number 1"]
This is 1993 National Sprint Round #23. It is pretty easy problem to person who's been to math competition for a while.  But for someone who's really new, this term is obscure --- telescoping.

Basically, if sequence telescopes, then the terms begin to cancel out.

First problem can be written as:

$ \frac{1}{1 \cdot 2} \plus{} \frac{1}{2 \cdot 3} \plus{} \frac{1}{3 \cdot 4} \plus{} \cdots \plus{} \frac{1}{99 \cdot 100}$

Now, here comes the trick for telescoping.

$ \frac{1}{n} \minus{} \frac{1}{n\plus{}1} \equal{} \frac{1}{n(n\plus{}1)}$

Using this idea, we can now note that:

$ \frac{1}{1} \plus{} \left(\minus{} \frac{1}{2} \plus{} \frac{1}{2} \minus{} \frac{1}{3} \plus{} \frac{1}{3} \cdots \minus{} \frac{1}{99} \plus{} \frac{1}{99}\right) \minus{} \frac{1}{100} \equal{} \frac{1}{1} \minus{} \frac{1}{100} \equal{} \frac{99}{100}$ [/hide]

[hide="Solution for Number 2"]
This is 1992 State Sprint #10. Honestly, to an average person, this problem is easier than the first one but since this is math competition and telescoping was pretty well-known, I put this one (even though it's state round) as number 2.

It is simple algebra..

$ 1000 \minus{} 2x \equal{} 100 \plus{} 3x \rightarrow 900 \equal{} 5x$

Now, we're trying to find what $ 1000\minus{}2x$ or $ 100\plus{}3x$ is.  So, instead of finding $ x$, I can multiply both sides by $ \frac{3}{5}$ to get this:

$ 540 \equal{} 3x$

Add 100 to that to get 640.

Of course, there is nothing wrong with finding $ x$. But this is one of those tricks that can save time (especially on the test like this). Because if you find $ x$, you'll have to find $ 3x$ again and well, why not just math in one step then?
[/hide]

[hide="Solution for Number 3"]
This is 1984 State Team #3. I've seen several problems similar to this and as someone already mentioned, there are two ways to approach this problem.

One way is pattern, and other one is algebra. I'll go to the algebra first.

We can write the inside as:

$ \sqrt{1 \plus{} n(n\plus{}1)(n\plus{}2)(n\plus{}3)}$

Group the product as $ n(n\plus{}3)$ and $ (n\plus{}1)(n\plus{}2)$ to get the $ n^2 \plus{} 3n$ and then just set $ y \equal{} n^2\plus{}3n$ and solve it.  Although this seems like an obvious way in this problem, there are AIME problems that can be solved using this manner.  So, if you didn't learn this method, learn it now! It's very effective.

Now, with pattern, note the final result.

5
11 = 5+6
19 = 5+6+8
29 = 5+6+8+10

This can be also written as $ 5 \plus{} 2(3 \plus{} 4 \plus{} 5 \plus{} \cdots)$. Now, since the inside product in the problem is $ 50 \cdot 51 \cdot 52 \cdot 53,$ we can write the result as:

$ 5 \plus{} 2(3 \plus{} 4 \plus{} 5 \cdots \plus{} 51) \equal{} 5 \plus{} \left(\frac{51 \cdot 52}{2} \minus{} 2 \minus{} 1\right) \cdot 2 \equal{} 2651$

Now, here is another important formula found by Gauss.

$ 1 \plus{} 2 \plus{} 3 \plus{} \cdots \plus{} n \equal{} \frac{n(n\plus{}1)}{2}$. The proof of this is as follows (using induction; now you don't need proof for AIME or AMC but anyway)

For $ n \equal{} 1$, this holds true (1*2/2 = 1).
Now, assume $ n \equal{} k$ is true.
Then, let's prove $ n \equal{} k\plus{}1$ is true.

$ 1 \plus{} 2 \plus{} 3 \plus{} \cdots \plus{} k \plus{} k \plus{} 1 \equal{} \frac{k(k\plus{}1)}{2} \plus{} k \plus{} 1 \equal{} \frac{k(k\plus{}1) \plus{} 2(k\plus{}1)}{2} \equal{} \frac{k^2\plus{}3k\plus{}2}{2} \equal{} \frac{(k\plus{}1)(k\plus{}2)}{2}$

which is same as we get on RHS if we set $ n \equal{} k\plus{}1$ so thus, this method holds true.
[/hide]

I apologize if I repeat any solutions from earlier in the thread. I'd like to help put answers to all of these problems up so that people may understand them.

[hide="Comment for #3"]
You can also notice that if the sequence is defined as $ \sqrt{1\plus{}(n)(n\plus{}1)(n\plus{}2)(n\plus{}3)}$, the final result is $ 1\plus{}(n\plus{}3)(n)$. Simple and quick, but useful.
[/hide]

[hide="Solution to 4"]
Fairly obvious. The maximum possible input into $ f(x)$ to use the $ f(x)\equal{}x\plus{}2$ definition is 9. Therefore the answer is $ 11$, $ B$.
[/hide]

[hide="Solution to 5"]
I found this somewhat confusingly worded, but it is straightforwards once you realize what exactly the problem is asking.
Let the arithmetic mean of the $ 12$ numbers be $ x$. Therefore, the sum of the $ 12$ numbers is $ 12x$. $ \frac{x}{12x} \cdot 100\% \equal{} \frac{25}{3}\%$, $ E$.
[/hide]

[hide="Solution to 6"]
This one relies on the pattern that occurs in the units digit of $ 2^n, n>0$. Observe:
$ 2^{4n\plus{}1} \equiv 2 \bmod{10}$
$ 2^{4n\plus{}2} \equiv 4 \bmod{10}$
$ 2^{4n\plus{}3} \equiv 8 \bmod{10}$
$ 2^{4n} \equiv 6 \bmod{10}$

Since $ 11213 \equiv 1 \bmod{4}$, the units digit of $ 2^11213\minus{}1$ is $ 1$.
[/hide]

[hide="Solution to 7"]
Since the vendor must split the jugs up in a $ 1: 2$ ratio, the sum of the jugs without the blue punch must be congruent to $ 0 \bmod{3}$. Since the sum of all the jugs fulfills this condition, only jugs congruent to $ 0 \bmod{3}$ may be kept for the vendor. This narrows the choices down to a 15 or 21 gallon jug. Seeing that it is impossible to choose jugs so that their capacities sum to $ \frac{(13\plus{}15\plus{}16\plus{}21\plus{}21\plus{}22)}{3}\minus{}15 \equal{} 31$, we can conclude that the vendor keeps a 21-gallon jug for himself, meaning the answer is $ D$.[/hide]

[hide="Solution to 8"]
This one is easily solved using proportionality. From the given capacity of the tank, the ratio of the radius of the base to the height of the cone is $ \frac{1}{6}$. If the tank is filled to 1/4 capacity, $ V_1\equal{}\frac{1}{4}V_0\equal{}\frac{\pi}{3}r^2h$. Substituting using the fact $ 6r\equal{}h$, we have $ V_1\equal{}\frac{1}{4}V_0\equal{}\frac{\pi}{3}\frac{1}{6^2}h^3$. Simplifying,
$ V_0\equal{}\frac{\pi}{3}\frac{1}{3^2}h^3$
$ \frac{\pi}{3}r_0^2h_0 \equal{}\frac{\pi}{3}\frac{1}{3^2}h^3$
$ r_0^2 h_0\equal{}\frac{1}{3^2}h^3$
Now, plug in the initial radius and height.
$ 16^2 \cdot 96 \cdot 3^2 \equal{} h^3$
Now, group the cubes on the LHS.
$ (16^3)(3^3)(2) \equal{} 2(48^3) \equal{} h^3$
and it is remarkably easy to see that h will take the form $ 48\sqrt[3]{2}$, and thus the answer is $ A$, $ 50$.
[/hide]

[quote="Silverfalcon"][hide="Correction for Number 11"]
I missed the error while I read the solution but there is an error in this problem. The correct problem should read:

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

Now, in the ARML book, the question has what I have.  But, in the solution, it started off with saying $ \sin(x \plus{} y) \minus{} \cos(x \minus{} y) \equal{} 0$... [/hide]

[hide="Solution for Number 11"]
In this solution, I'm going to solve the problem assuming that what I have (and the book) is wrong and go with what the book's solution says. There is an answer to $ \cos(x \plus{} y)$ but because of answer choice, we'll go with $ \cos(x \minus{} y)$.

I won't go into detail but essentially:

$ \sin(x \plus{} y) \minus{} \cos(x \minus{} y) \equal{} (\sin x \minus{} \cos x)(\cos y \minus{} \sin y) \equal{} 0$
after expanding using sine and cosine angle addition formula and factoring by terms.

Therefore:

$ \sin x \equal{} \cos x$ and $ \cos y \equal{} \sin y$

I think that this is the best part of this problem. Because $ \sin \frac {\pi}{4} \equal{} \cos \frac {\pi}{4}$, we can now say the following:

$ x \equal{} \frac {\pi}{4} \plus{} n \pi$
$ y \equal{} \frac {\pi}{4} \plus{} n \pi$

From here, we'll consider four cases.

Case 1: $ x \geq 0$
Solving $ \frac {\pi}{4} \plus{} n \pi \leq 1995$ yields 498 as the quotient with remainder.  The presence of remainder is important because this indicates that when $ n \equal{} 498,$ the resulted line is not a tangent to the circle.  So, counting $ n \equal{} 0$, there are 499 lines resulting in 998 intersections (above the x-axis and below the x-axis).

Case 2: $ y \geq 0$
Same as Case 1 so here are another 998 intersections.

Case 3: $ x \leq 0$
Do not assume that this is 998 as well! Note that the first line in $ x \leq 0$ starts off with $ \frac { \minus{} 3\pi}{4}$ and ends with $ \frac { \minus{} 1995 \pi}{4}$. So, this means that although there are 499 lines, the last line is TANGENT, thus only making one intersection.

Case 4: $ y \leq 0$
Same as Case 3 so total of $ 499 \times 2 \minus{} 1 \equal{} 997$.

Thus, the total number of intersections is $ 998 \plus{} 998 \plus{} 997 \plus{} 997 \equal{} 3990$ or $ \fbox{B}$. [/hide][/quote]
I don't get your solution in case 1? Why you can get 498 or 499 when solving that inequality? I got 0<=n<= 634.444 and I just can infer that n has 634 values :( . Someone please explains it to me!

There's a couple of typos:

[quote="Silverfalcon"]
In this solution, I'm going to solve the problem assuming that what I have (and the book) is wrong and go with what the book's solution says. There is an answer to $ \cos(x \plus{} y)$ but because of answer choice, we'll go with $ \cos(x \minus{} y)$.

I won't go into detail but essentially:

$ \sin(x \plus{} y) \minus{} \cos(x \minus{} y) \equal{} (\sin x \minus{} \cos x)(\cos y \minus{} \sin y) \equal{} 0$
after expanding using sine and cosine angle addition formula and factoring by terms.

Therefore:

$ \sin x \equal{} \cos x$ [i][u]or[/u][/i] $ \cos y \equal{} \sin y$

I think that this is the best part of this problem. Because $ \sin \frac {\pi}{4} \equal{} \cos \frac {\pi}{4}$, we can now say the following:

$ x \equal{} \frac {\pi}{4} \plus{} n \pi$
$ y \equal{} \frac {\pi}{4} \plus{} n \pi$

From here, we'll consider four cases.

Case 1: $ x \geq 0$
Solving $ \frac {\pi}{4} \plus{} n \pi \leq \boxed{\frac{1995  \pi}{4}}$ yields 498 as the quotient with remainder.  The presence of remainder is important because this indicates that when $ n \equal{} 498,$ the resulted line is not a tangent to the circle.  So, counting $ n \equal{} 0$, there are 499 lines resulting in 998 intersections (above the x-axis and below the x-axis).

Case 2: $ y \geq 0$
Same as Case 1 so here are another 998 intersections.

Case 3: $ x < 0$
Do not assume that this is 998 as well! Note that the first line in $ x \leq 0$ starts off with $ \frac { \minus{} 3\pi}{4}$ and ends with $ \frac { \minus{} 1995 \pi}{4}$. So, this means that although there are 499 lines, the last line is TANGENT, thus only making one intersection.

Case 4: $ y < 0$
Same as Case 3 so total of $ 499 \times 2 \minus{} 1 \equal{} 997$.

Thus, the total number of intersections is $ 998 \plus{} 998 \plus{} 997 \plus{} 997 \equal{} 3990$ or $ \fbox{B}$. [/quote]

The casework at the end isn't really necessary if you see that $ |x|\equal{}\left|\frac {\pi}{4} \plus{} n \pi\right| \le \frac{1995\pi}{4}$ gives us $ 998$ values of $ n$ and hence $ 998 \times 2 \minus{} 1 \equal{} 1995$ tangent lines, and then double for the $ y$ values.

Again, azips, how can you show that n has 998 values?? I just have the same result as I mentioned above :( .Can you show your solution to this inequality in detailed?
PS: Anybody has solution of problem 19 and 21? I really need to see the solution of these problems, especially to 19 when I can't find the number of ways to choose 3 segments form a triangle out of 45C3 sets of 4 segments.

I have been busy and forgot to keep my promise of providing solutions to ALL problems in this mock AMC. I'll do so as much as I get chance.

[hide="Solution to Number 24"]

I'll just type up the official solution from USAMTS website. This was USAMTS problem on Round 4 #5 in Year 10.

Note that this problem's answer holds true regardless of the shape of $ \triangle ABC$. The solution is nicely done with coordinates (somewhat brute forcing but pretty elegant).

$ \triangle ABC$ can be placed on a coordinate system with $ A$ on the origin, $ B$ at $ (4x,0)$, and $ C$ at $ (4y,4z)$. By the midpoint theorem, $ D \equal{} (2x\plus{}2y, 2z), E \equal{} (2y,2z), F \equal{} (2x,0), P \equal{} (x\plus{}y,z), Q \equal{} (2x\plus{}y,z),$ and $ R \equal{} (x\plus{}2y,2z)$.

Using the distance formula,
$ AQ^2 \equal{} (2x\plus{}y)^2 \plus{} z^2 \equal{} 4x^2 \plus{} 4xy \plus{} y^2 \plus{} z^2$
$ AR^2 \equal{} (x\plus{}2y)^2 \plus{} 4z^2 \equal{} x^2 \plus{} 4xy \plus{} 4y^2 \plus{} 4z^2$
$ BP^2 \equal{} (3x\minus{}y)^2 \plus{} z^2 \equal{} 9x^2 \minus{} 6xy \plus{} y^2 \plus{} z^2$
$ BR^2 \equal{} (3x\minus{}2y)^2 \plus{} 4z^2 \equal{} 9x^2\minus{}12xy\plus{}4y^2 \plus{} 4z^2$
$ CP^2 \equal{} (x\minus{}3y)^2 \plus{} 9z^2 \equal{} x^2 \minus{} 6xy \plus{} 9y^2 \plus{} 9z^2$
$ CQ^2 \equal{} (2x\minus{}3y)^2 \plus{} 9z^2 \equal{} 4x^2 \minus{} 12xy\plus{}9y^2 \plus{} 9z^2$
Adding these up gives $ 28(x^2 \minus{} xy \plus{} y^2 \plus{} z^2)$. Furthermore,

$ AB^2 \equal{} 16x^2$
$ BC^2 \equal{} 16x^2 \minus{} 32xy \plus{} 16y^2 \plus{} 16z^2$
$ CA^2 \equal{} 16y^2 \plus{} 16z^2$

Adding these up gives $ 32(x^2 \minus{} xy \plus{} y^2 \plus{} z^2)$.

That makes $ \displaystyle \frac{AQ^2 \plus{} AR^2 \plus{} BP^2 \plus{} BR^2 \plus{} CP^2 \plus{} CQ^2}{AB^2 \plus{} BC^2 \plus{} CA^2} \equal{} \frac{28(x^2 \minus{} xy \plus{} y^2 \plus{} z^2)}{32(x^2 \minus{} xy \plus{} y^2 \plus{} z^2)} \equal{} \frac{7}{8}$

Since this construction used an arbitrary triangle, the value is independent of the shape of $ \triangle ABC$. So, $ m\plus{}n \equal{} 15$ or E. [/hide]

[hide="Number 22 OFFICIAL SOLUTION"]
This is 2002 Putnam Problem so don't worry to see this problem because it won't happen. It's pretty nice problem though.

I'll first put the official solution first which is by Putnam committee itself.

QUOTED:

The probability is 1/99. In fact, we show by induction on $ n$ that after $ n$ shots, the probability of having made any number of shots from 1 to $ n\minus{}1$ is equal to $ \frac{1}{n\minus{}1}$. This is evident for $ n \equal{} 2$. Given the result for $ n$, we see that the probability of making $ i$ shots after $ n\plus{}1$ attempts is

$ \frac{i\minus{}1}{n} \cdot \frac{1}{n\minus{}1} \plus{} \left(1 \minus{} \frac{i}{n}\right) \cdot \frac{1}{n\minus{}1} \equal{} \frac{(i\minus{}1) \plus{} (n\minus{}i)}{n(n\minus{}1)} \equal{} \frac{1}{n}$

as claimed. [/hide]


${\sqrt[3]{2 x-1}=x \sqrt[3]{16}-\sqrt[3]{2 x+1}}$

${x=\frac{1}{2}\lor x=\sqrt{\frac{1}{4}+\frac{3 \sqrt{3}}{8}}}$

minsoens, Your solution is wrong. Try  ${ x = \frac{-1}{2}}$  and the others.
Before post a solution, try it on the problem, because of multiple valuable functions

My solution is the same, i just rewrite the fraction in other way

I think let $a=\sqrt[3]{2x-1}$and $b=\sqrt[3]{2x+1}$


George is making spaghetti for dinner.  He placed 4.01 kg of water in a pan and brings it to a boil.  Before adding the pasta, he adds 58 g of table salt to the water and brings it to a boil.  What is the temperature of the salty boiling water?  (Its a nice day at sea level so that the pressure is 1.00 atm.  Assume negligible evaporation of water.  Kb for water is 0.52 degress Celsius/molality.

First you have to find the molality of NaCl:
(58 g NaCl)(1 mole NaCl/58.4425 g NaCl) = .99243 mole NaCl
(.99243 mole NaCl)/(4.01 kg H2O) = .247488 m NaCl
So, the molality is .247488 m NaCl.

Then, I think that you have to multiply the molality by the Kb of water (0.52 degrees Celsius).
That would equal .12869 degrees Celsius.

Finally, you would add that number to 100 degrees, to get 100.13 degrees Celsius.

I did a search for this problem and found that the answer was 100.26 degrees Celius.  Did I do the problem right, or is the answer I found right?

I think I found where I messed up.  Since there are two ions in NaCl, I multiply (.247488 m NaCl) by (0.52 degress Celsius) by (2.0) to get .25739 degrees Celsius.

Then I add that to 100 and get the answer that I found, 100.26.  Is that answer right?

Please help.

i'd say that should be correct....the one with the vant hoff factor...100.26!!!

You're right it is the particle molarity.  If one had molecules with three ions, you would have to multiply it by 3.

it's molality that she used... colligative properties...


Find all pairs functions $f,g: \mathbb{N}_{0}\to\mathbb{N}_{0}$ such that \[f(m)-f(n)=(m-n)(g(m)+g(n))\quad \forall m,n\in\mathbb{N}_{0},\]
where $\mathbb{N}_{0}=\{0,1,2,...\}$.

Hello N.T.TUAN

 [quote="N.T.TUAN"]Find all pairs functions $f,g: \mathbb{N}_{0}\to\mathbb{N}_{0}$ such that
\[f(m)-f(n)=(m-n)(g(m)+g(n))\; \forall m,n\in\mathbb{N}_{0}, \]
where $\mathbb{N}_{0}=\{0,1,2,...\}$.[/quote]

$f(m)-f(n)=(m-n)(g(m)+g(n))$ $\Rightarrow $ $f(m)=m(g(m)+g(0))+f(0)$ $\Rightarrow $ $f(m)-f(n)=m(g(m)+g(0))-n(g(n)+g(0))$
And so : $m(g(m)+g(0))-n(g(n)+g(0))=(m-n)(g(m)+g(n))$, which implies $n(g(m)-g(0))=m(g(n)-g(0))$ and $g(n)=a*n+b$

Then $f(n)=n(g(n)+g(0))+f(0)=an^{2}+2bn+c$

And it is easy to verify that these necessary conditions work.

So the solution is :
$g(n)=a*n+b$
$f(n)=an^{2}+2bn+c$
$a,b,c\in\mathbb{N}_{0}$

--
Patrick


degf(X)= 2007
f(X) is an integer polynomial that have a2007=1
Prove that (f(X))^2 - 9 has no more than 2010 roots

And does $ a_{2007}$ refer to the coefficient of $ x^{2007}$ or $ x^0$?  (Conventions differ.)


inequalities
trigonometry
inequalities proposed

Let be given a triangle $ABC.$ Prove that: \[a^{2}\sin^{2}\frac{A}{2}+b^{2}\sin^{2}\frac{B}{2}+c^{2}\sin^{2}\frac{C}{2}\ge\frac{1}{12}\cdot (a+b+c)^{2}\]
It's really easy problem but do you have a solution without using [b]Chebyshev Inequality[/b]?

$a^{2}(1-cosA)+b^{2}(1-cosB)+c^{2}(1-cosC)\geq \frac{1}{6}(a+b+c)^{2}$
$\sum \frac{2a^{3}bc+a^{5}-a^{3}b^{2}-a^{3}c^{2}}{2abc}\geq \frac{1}{6}(a+b+c)^{2}$
$\sum 2a^{3}bc+a^{5}\geq \frac{abc}{3}(a+b+c)^{2}+\sum (a^{3}b^{2}+a^{3}c^{2})$

But $\sum a^{3}bc \geq \frac{abc}{3}(a+b+c)^{2}$ by AM-GM.
And I'm too lazy to do the other one, but hopefully it works by some kind of Schur-like thing.


linear algebra
matrix
linear algebra unsolved

Show that if dimV>1, then the set of normal operators on V is not a subspace of L(V).

P.S The problem doesn't specific whether V is on C or R. So I think it wants to show both.

Seems like all you have to do is find two $ 2 \times 2$ normal matrices whose sum is not normal. "Most" non-trivial choices work, for instance

$ A \equal{} \begin{pmatrix} 1 && 2 \\ 2 && 3 \end{pmatrix}$

$ B \equal{} \begin{pmatrix} 0 && 2 \\ \minus{}2 && 0 \end{pmatrix}$

which are obviously normal (symmetric and skew-symmetric, respectively) over either $ \mathbb{R}$ or $ \mathbb{C}$, and whose sum

$ A \plus{} B \equal{} \begin{pmatrix} 1 && 4 \\ 0 && 3 \end{pmatrix}$

is not normal, being triangular and not diagonal. For higher dimensions, just embed these matrices in larger ones in the obvious way.

[quote="Saradush"]Is A really normal? I think any normal operator has form  :blush:

a -b
b  a[/quote]

Nope. A [url=http://en.wikipedia.org/wiki/Normal_operator]normal operator[/url] $ A$ is one which commutes with its adjoint $ A^*$. In matrix terms, this means that $ A$ commutes with its transpose-conjugate, which is just transpose when $ A$ is real. A symmetric matrix $ A$ satisfies $ A\equal{}A^T$, so it obviously satisfies $ AA^T\equal{}A^TA$.


number theory
least common multiple
modular arithmetic
quadratics
induction
number theory proposed

Prove that this expression  $ a^2 \plus{} b^2 \plus{}c^2$ $ \equal{}$ $ 7$ does not have any answer in rational numbers.
(a,b,c are rational numbers).

First, reduce it to a problem in integers: assume that there's a solution, and let $ d$ be the LCM of the denominators of $ a,b,c$. After multiplying both sides by $ d^2$, the equation takes the form $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 7d^2$, $ a,b,c,d \in \mathbb{Z}, d \neq 0$.
It is enough to disprove the existence of solution to this equation. The squares mod $ 8$ are $ 0,1,4$.
If $ d$ is odd, then $ 7d^2 \equiv 7 \pmod 8$, but adding 3 (not necessarily distinct) numbers from the set $ \{ 0,1,4 \}$ never yields $ 7 \pmod 8$.
Thus, $ d$ is even. $ 7d^2 \equiv 0,4 \pmod 8$ and thus $ a,b,c$ are even (adding 3 (not necessarily distinct) numbers from the set $ \{ 0,1,4 \}$ yields 0 or 4 iff we take 0 and 4, which are quadratic residues that correspond to even numbers).

So for any existing solution $ (a,b,c)$ we can find another solution, $ (\frac {a}{2},\frac {b}{2},\frac {c}{2})$. By induction, $ 2^k |a$ for all $ k$'s, which implies $ a \equal{} 0$. Likewise, $ b \equal{} c \equal{} 0$, which implies $ d \equal{} 0$, a contradiction.
Hence, there are no rational solutions to $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 7$.


function
logarithms
algebra
floor function

Hi to all...

Again, if any of you have seen this topic posted in another "topic", please notify me and I'll go there (thanks).

The question is this: "[i]How many terminal zeros (consecutive zeros appearing at the end of a number) are there, in terms of n, for n![/i]"?

It's a nice easy problem that proved to be interesting. If you have a solution, please explain it as well.

 ;) Try not to make your posts too long, no one wants to read a post that spans for pages and pages...

[hide]in base ten, a zero is made by containing a factor of ten, or 5*2 as a product of primes. as every n! n>1 has more twos in its prime factors than fives, the number of fives determine the number of tens which are factors of n!, and therefore the number of zeroes on the end. if you factorise n! into a product of primes:

(2^a)(3^b)(5^c)(7^d)...(p^x) where p is the largest prime factor, then the number of zeroes after n! is c. c in terms of n is

c = [n/5] + [n/5^2] + [n/5^3] + ... to infinity, where [a] is the largest integer such that [a]<a (floor function brackets)
This is because we first write:

n! = 1*2*3*4*5*6*...*(n-1)*n
n! = 5(1*2*3*4*5*6*...)(1*2*3*4*6*...)
n! = 5(5(1*2*3...)(1*2*3*4*6*...)(1*2*3*4*6*...)

every fifth factor of n! can have a five taken out, every 5^2 th factor can have 5^2 taken out, every 5^3 th factor 5^3 taken out, etc.[/hide]

Very nice problem.

[hide="Solution"] I don't know if this has been posted or not, but it's a common concept, so it's good that it was posted again.

But you just have [n / 5] + [n / 25] + [n / 125] ... so on so forth, where that's the floor function of the expression.[/hide]

[hide]You are looking for the number of multiples of 10.  There will always be enough 2's to make a 10, so we're just counting the number of 5's in n!

$\displaystyle\sum_{i=1}^{\left[\log_5 n\right]} {\left[ \frac n{5^i} \right]}$
where [ ] is the greatest integer function[/hide]

This idea is easily extendable to a more general question, "How many zeroes are at the end of n! when it is written in base b, where b = pq for distinct primes p and q?"  However, there are lots of related questions that are sometimes different:

1) How many 0's are at the end of n! when it is written in base 4?
2) How about base 20?
3) How about base 12?


$n_{1},\ldots ,n_{k}, a$ are integers that satisfies the above conditions A)For every $i\neq j$,  $(n_{i}, n_{j})=1$ B)For every $i, a^{n_{i}}\equiv 1 (mod n_{i})$ C)For every $i, X^{a-1}\equiv 0(mod n_{i})$.
Prove that $a^{x}\equiv 1(mod x)$ congruence has at least $2^{k+1}-2$ solutions. ($x>1$)

Hehe, two problems from turkish MO of same "type".

i cant understand what is the question of the same type on the second round of 1993

http://www.mathlinks.ro/Forum/viewtopic.php?t=112708

not same year, still funny :)

The problem is not quite clear. Can some one from Turkey clarify this problem, especially condition C).

The condition $c)$ is problematic above. For the sake of convenience, I'll rewrite the problem.

Given positive integers $n_1,n_2,\dots,n_k,a$, satisfying:
$\bullet$ $(n_i,n_j)=1$ if $i\neq j$,
$\bullet$ $a^{n_i} \equiv 1 \pmod{n_i}$, for every $i$,
$\bullet$ $n_i \nmid a-1$,

show that the total number of integers $x\geq 2$, enjoying $a^x \equiv 1 \pmod{x}$ is at least $2^{k+1}-2$.

I will post my solution later, unless someone else already posts till then.

Enough waiting time ;) ... Here is the proof, that I've promised to post.

We begin by a few observations. First, notice that, $a$ is coprime with any $n_i$. Next,
[b]Claim[/b] None of the $n_i$'s are prime numbers.
[b]Proof[/b]
Assume that $n_i$ is prime for some $i$. By Fermat's little theorem, we have $a^{n_i-1}\equiv 1 \pmod{n_i}$. Combining this with the second item above, gives us $n_i \mid a-1$, contradicting with the preamble.$\square$

The next claim gives us a way to generate such $x$'s.
[b]Claim[/b] Let $I$ be a non-empty subset of $\{1,2,\dots,k\}$. Define,
$$
x_I \triangleq \prod_{i\in I}n_i.
$$
Then, $a^{x_I}\equiv 1 \pmod{x_I}$. Hence, we have shown the existence of at least $2^{k}-1$ such numbers.
[b]Proof[/b]
First, notice that $a^{n_i}\equiv 1 \pmod{n_i}$ gives us, $a^{x_I}\equiv 1 \pmod{n_i}$, for every $i\in I$. Since $n_i$'s are coprime, we also have,
$$
a^{x_I}\equiv 1 \pmod{x_I}.
$$

Next, we will show the existence of a second collection, in a way, analogous to $\{n_1,n_2,\dots,n_k\}$, and therefore will conclude the proof.
[b]Claim[/b] For every $i$, let $p_i$ be the smallest prime divisor of $n_i$. Then, the following holds:
$\bullet$ $n_i>p_i$ for every $i$.
$\bullet$ $\{p_i:i=1,2,\dots,k\}$ consists of $k$-distinct primes.
$\bullet$ $a \equiv 1 \pmod{p_i}$. Hence, $a^{p_i} \equiv 1\pmod{p_i}$.
Noticing that this collection satisfies $a^{p_i}\equiv  1 \pmod{p_i}$ for every $i$, we can generate $2^{k}-1$ other such numbers via first taking a non-empty subset $I$ of $\{1,2,\dots,k\}$, and then, letting $y_I =  \prod_{i\in I}p_i$. Since $n_i>p_i$ for every $i$, and $n_i$'s are coprime, this will generate $2^k-1$ other numbers, enjoying the desired property, and therefore establishing the proof.
[b]Proof[/b]
For each $n_i$, we have already shown that $n_i$ cannot be a prime number, and therefore, $n_i>p_i$. Also, since $n_i$'s are coprime, for any $i\neq j$, the set of prime divisors of $n_i$ and the set of prime divisors of $n_j$ are disjoint, hence, $p_i$'s are all distinct.

It remains to prove the last item. For this, note that by Fermat's theorem, we have $a^{p_i-1}\equiv 1 \pmod{p_i}$. Now, since $a^{n_i}\equiv 1 \pmod{p_i}$, we have,
$$
a^{(n_i,p_i-1)}\equiv 1 \pmod{p_i} \implies a\equiv 1\pmod{p_i},
$$
where the last assertion holds due to the fact that since $p_i$ is the smallest prime divisor of $n_i$, none of the prime divisors of $p_i-1$ can divide $n_i$, hence they are coprime.


trigonometry
inequalities unsolved
inequalities

For $ a;b \in R$.Find min and max of
$ F(a;b)=\frac{\sqrt{3}}{4} sin(a-b)+sin(\frac{\pi}{6}-a)+sin(b-\frac{\pi}{6})$

hello, to obtain all critical points you must solve the following equation system with respect to $ a$ and $ b$:
$ \frac{\sqrt{3}}{4}\cos(a\minus{}b)\minus{}\cos\left(\frac{\pi}{6}\minus{}a\right)\equal{}0 \newline$
and
$ \minus{}\frac{\sqrt{3}}{4}\cos(a\minus{}b)\plus{}\cos\left(b\minus{}\frac{\pi}{6}\right)\equal{}0$
Sonnhard.


integration trigonometry calculus algebra polynomial function quadratics

$ \int\frac{dx}{4\minus{}5sinx}$

i guess i'm supposed to use the weierstrauss method? but i have no idea what that is

so, help?

Make, as usual, $ \tan \frac {x}{2} \equal{} t$.

[hide="Then,"] $ \sin x \equal{} \frac{2t}{1\plus{}t^2}$, $ x' \equal{} \frac{2}{1\plus{}t^2}$, and so

$ P \equal{} \int \frac{1}{4\minus{}5 \sin x}\,dx \equal{} \int \frac{dt}{2t^2\minus{}5t\plus{}2}$.[/hide]

Expanding out the details of Carcul's post:

Since $ x\equal{}2\arctan t,$ we have $ dx\equal{}\frac{2}{1\plus{}t^2}\,dt.$

$ \cos\left(\frac{x}{2}\right)\equal{}\frac{1}{\sqrt{1\plus{}t^2}}$ and $ \sin\left(\frac{x}{2}\right)\equal{}\frac{t}{\sqrt{1\plus{}t^2}}.$

Hence, $ \cos x\equal{}\cos^2\left(\frac{x}{2}\right)\minus{}
\sin^2\left(\frac{x}{2}\right)\equal{}\frac{1\minus{}t^2}{1\plus{}t^2}$ and

$ \sin x\equal{}2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)\equal{}
\frac{2t}{1\plus{}t^2}.$

Make these substitutions in the original integral, and you will have the integral of a rational function. In this particular case:

$ \int\frac{dx}{4\minus{}5\sin x}\equal{}\int\frac{\frac{2}{1\plus{}t^2}\,dt}
{4\minus{}\frac{10t}{1\plus{}t^2}}$

$ \equal{}\int\frac{2\,dt}{4\plus{}4t^2\minus{}10t}\equal{}\int\frac{dt}{2t^2\minus{}5t\plus{}2}.$

I'll let you finish from there - the remaining steps include factoring the denominator and using partial fractions.

One note: that polynomial in the denominator does have real roots, which means that in any definite integral we must be careful about singularities. These singularities were not introduced by the substitution - they were present in the original integral. The original integral has singular points whenever $ \sin x\equal{}\frac45.$

This method is guaranteed to turn the integral of any rational function of the usual trigonometric function into a the integral of a rational function. If (and this is a very, very big if) you can factor the denominator of the resulting rational function into linear and quadratic factors (and do so in a closed form), then you can express the integral in elementary terms.

But as I said, that is a very big if. The Fundamental Theorem of Algebra guarantees that such a factorization exists, but not that you can find it in closed form.

wait, i am so confused

how do you make $ tan\frac{x}{2} \equal{} t$ ?

or even....

where did this come from?

$ x \equal{} 2arctan^{\minus{}1}t$ ?

i honestly have no idea what weierstrauss is so, please explain?

Karl Weierstrass (note spelling) was one of the mathematical giants of the 19th century. Of the many things he did, this little manipulation is surely one of the least important. But it does commonly have his name attached to it.

This - the Weierstrass method - is simply a certain specific [i]substitution[/i] in an integral. The substitution can either be written "forwards" as $ t \equal{} \tan\left(\frac {x}2\right)$ or "backwards" by solving that equation for $ x$ in terms of $ t.$ The result of that solution is $ x \equal{} 2\arctan x$ (or $ x \equal{} 2\tan^{ \minus{} 1}x$ if you prefer. Just don't say $ \arctan^{ \minus{} 1}x,$ which isn't what we mean.)

As for the explanation - the algebraic manipulations in my previous post are the explanation. There's really nothing deeper here - just those calculations, and the fact that they always "work" in a case like this. (If by "work" I mean they give you a rational function of $ t$ to integrate; finishing from there has its own complications.)

[quote="charlestreykang"]wait, i am so confused

how do you make $ tan\frac {x}{2} \equal{} t$ ?

or even....

where did this come from?

$ x \equal{} 2arctan^{ \minus{} 1}t$ ?

i honestly have no idea what weierstrauss is so, please explain?[/quote]

I've seen some "creative substitutions" in the forum as well.  I found this link that may provide some insight if you're really interested.

[url=http://pear.math.pitt.edu/Calculus2/week3/3_2.html#3_2li1.html]Good Short Article[/url].
[url=http://pear.math.pitt.edu/Calculus2/week3/3_2li4.html]and Here[/url].


Given a triangle $ABC$, let $K$ be the midpoint of $AB$ and $L$ a be point on
$AC$ such hat $AL = LC+CB$. Prove that $<KLB = 90$ if and only if
$AC = 3CB$.


NSEP stands for the National Standard Examination in Physics. It is the prelude to the Indian National Physics Olympiad...( INPhO)

The result is online at [url=http://www.hbcse.tifr.res.in/olympiads/nseres/phyres]www.hbcse.tifr.res.in/olympiads/nseres/phyres[/url]

I think letters should reach in a few days...

Please post if you got in! Mention your name, class and region.

I Got in...
Rushil Goel
Class XII
Chandigarh

I too got selected

Varun Agrawal
Class XII
Sonebhadra

(Congrats Rushil, you got selected in NSEC too. I didn't)

Rushil, is there a practical exam also in INPhO.
Please tell soon.

I really don't know!! Have been trying to figure that out for quite some time! But a friend of mine gave inpho last year but didn't talk about practical exam... he wasn't selected in theory! Maybe its done after the theory is cleared... please tell by email if you know!


[quote="kiler40"]Hi

I posted the same question on the introduction physics thread but i got no replies so i guess it doesn't belong there

I know that matter-antimatter collision makes alot of energy and that energy can turn into matter itself. So i was just wondering, why doesnt the energy surrounding spontaniously turn into matter (according to e=mc^2, a proton contains 3x10^-11 J so wouldnt it need just a bit of energy to be formed?)[/quote]

I am not sure about this, but I think that on the macro-scale entropy is what prevents energy turning into matter spontaneously, however on the fundamental particle scale particle-antiparticle pairs do form from the energy in the spacetime fabric, which is what gives rise to Hawking radiation. Correct me if I am wrong.. I have forgotten a lot of this stuff since a long time ago.

Turning a single gamma ray into a proton would violate several conservation laws: baryon number, charge, and momentum. Turning a single gamma ray into a proton-antiproton pair (or into an electron-positron pair) would violate the conservation of momentum. Hence it is possible only if something else carries away the required momentum, while carring away little energy, i.e., something heavy, like a nucleus. Therefore, a single photon has to interact electromagnetically with a nucleus to turn into proton-antiproton pair (or into an electron-positron pair). The gamma ray energy has to be at least 1 MeV for an electron-positron pair and at least 2000 MeV for a proton-antiproton pair. There are easier ways to generate antiprotons (high energy proton-proton collisions). Conversely, a proton-antiproton pair (or an electron-positron pair) do not anihilate into a single photon, because it would violate the conservation of momentum. First they have to come to a mutual rest and then they can anihilate into 2 photons of equal energy and opposite momentum. Collision of 2 photons of equal energy, just right to create a matter-antimatter pair, and with exactly opposite momentum, is very unlikely event, which is why it is not observed. For example, photons can have momentum in any direction with equal probability.

Hmm what about Hawking radiation then? It is quite a different situation from the one you described, because the energy is "borrowed" from the space-time continuum and is almost always returned nearly instantaneously, and the formation of the particle-antiparticle pair does not violate any conservation laws except on the Planck scale.

[quote="yetti"]I think the question was about real particles, not virtual ones. No energy has to be supplied to create a virtual particle-antiparticle pair, but they have to disappear within the time $\triangle t \approx \frac{h}{\triangle E}= \frac{h}{2mc^{2}}.$ For a proton-antiproton pair $m \approx 940\ \text{MeV/c}^{2},$ $h \approx 4 \times 10^{-21}\ \text{MeV sec},$ $\triangle t \approx 2 \times 10^{-24}\ \text{sec},$ which is short even on a nuclear scale. Unless the energy $2mc^{2}$ is supplied, the virtual particles become real and they go their separate ways ($\triangle x$ becomes large) without violating the energy conservation law. But a single photon with nothing real to interact cannot do this, because the energy-momentum relationship for photons is different from the energy-momentum relationship for protons/antiprotons, and the momentum uncertainty can be only $\triangle p \approx \frac{h}{\triangle x}.$[/quote]

Okay I see. But in the case of Hawking radiation the energy comes from the mass of the black hole itself, although the particle-antiparticle pair nearly always annihilate themselves and return the "borrowed" energy.


A one-player game is played as follows: There is a bowl at each integer on the $Ox$-axis. All the bowls are initially empty, except for that at the origin, which contains $n \geq 2$ stones. A move is either

(A) to remove two stones from a bowl and place one in each of the two adjacent bowls, or

(B) to remove a stone from each of two adjacent bowls and to add one stone to the bowl immediately to their left.

Show that only a finite number of moves can be made and that the final position (when no more moves are possible) is independent of the moves made (for a given $n$).

I don't think this is true. Take $ n\equal{}3$ and consider the following sequences of moves:

1. $ 0,0,3,0,0 \to 0,1,1,1,0 \to 0,2,0,0,0 \to 1,0,1,0,0$ (moves of type ABA)

2. $ 0,0,3,0,0 \to 0,1,1,1,0 \to 1,0,0,1,0$ (moves of type AB)

Both positions are final and they're not the same. Am I misunderstanding the moves?

No, the problem is correct (but I think the details can be explained in more detail)! First, it's impossible to make a moviment of the type $A$ in the beginning, because we need two bowls side by side with at least one stone. The figures represents the moviments according the official site of the olympiad.


find all function $f: \mathbb{N}\to \mathbb{N}$ such that

   \[f(m-n+f(n))=f(m)+f(n)\] for all $m,n \in \mathbb{N}$

[color=red][I LaTeXed it for you; please see [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=5261]this[/url] to learn it; also I moved it from "Algebra Open Questions" where it didn't belong; see also [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=144403]that thread[/url] before posting another problem with such a title/source. -mathmanman][/color]

[quote="Alisher"]find all function $f: \mathbb{N}\to \mathbb{N}$ such that
\[f(m-n+f(n))=f(m)+f(n) \]
for all $m,n \in \mathbb{N}$

[color=red][I LaTeXed it for you; please see [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=5261]this[/url] to learn it; also I moved it from "Algebra Open Questions" where it didn't belong; see also [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=144403]that thread[/url] before posting another problem with such a title/source. -mathmanman][/color][/quote]

I believe this is a problem from Italian TST,isn't it?

Looks like you're right [b]Tiks[/b], thank you.
( http://www.mathlinks.ro/Forum/viewtopic.php?t=89357 )

Topic locked.


algebra
polynomial
complex analysis
complex analysis unsolved

We are given the map $ w \equal{} P(z)$ of planes $ z$ and $ w$, where $ P(z)$ is polynomial with degree $ n$. Let $ A \equal{} \{z: P(z) \equal{} 0\}$ and $ B \equal{} \{z: P(z) \equal{} 1\}$. Prove that the number of points of the set $ A\cup B$ is not less than $ n \plus{} 1$.

Nice  approach (in my humble opinion) to this problem based on the beautiful Mason's theorem http://mathworld.wolfram.com/MasonsTheorem.html. (just write $ (1\minus{}P(z))\plus{}P(z)\equal{}1$) :)


geometry
analytic geometry
inequalities
parameterization

In the coordinate plane, the region $M$ consists of all the points $(x, y)$ satisfying the inequalities $y \ge 0$, $y \le x$ and $y \le 2 - x$ simultaneously.  The region $N$, which varies with the parameter $t$, consists of all the points $(x, y)$ satisfying the inequalities $t \le x \le t + 1$ and $0 \le t \le 1$ simultaneously.  What is the area of $M \cap N$?
(a) $-t^2+t+\frac{1}{2}$
(b) $-2t^2+2t$
(c) $1 - \frac{1}{2}t^2$
(d) $\frac{1}{2}(t-1)^2$

answer:
[hide]a[/hide]

explanation to come. (sorry, don't have time.)


arithmetic sequence
number theory proposed
number theory

Hello,

I suppose that many of you could enjoy this problem. It can turn to be messy if you are not carefully, but is not that hard.

Let $f$ by a bijection from $\mathbb{N}$ to $\mathbb{N}$. Show that there exist $a,b,c\in\mathbb{N}$ such that $a<b<c$ and $f(a)+f(c)=f(2b)$.

I have solved and the truth is that it really does not matter if you include or not 0 as a natural number.

Best regards,

Are you sure it's not $f(a)+f(c)=2f(b)$ instead of $f(2b)$? I seem to remember something like what I said being posted a very long time ago.

Anyway, looking at it from the point of view of $f^{-1}$, it's like this: we have to show that there are three numbers in arithmetic progresion $x-t,x,x+t$ s.t. $(f(x)-f(x-t))(f(x)-f(x+t))<0$, i.e. not both $f(x-t)$ and $f(x+t)$ are greater or smaller than $f(x)$ at the same time.

Assume the contrary. We have $f(n)>f(0)$ for all sufficiently large $n$, which means, by our assumption, that $f(n)>f(2n)$ for all sufficiently large $n$. This means that we can find $n$ s.t. $f(n)>f(2n)>f(4n)>\ldots$, so we get an infinite descending sequence of naturals, which is impossible.

Just an observation: I think orl posted, some time ago, a problem stating that for any $k$ the numbers $1,2,\ldots,2^k$ can be permuted by a permutation $\sigma$ so that there are no $a<b<c$ for which $\sigma(a)+\sigma(c)=2\sigma(b)$ . Of course, this means that there is such a permutation for $1,2,\ldots,n$ for any $n$. These two problems complete each other :). For the sake of completeness, let's prove the problem I mentioned:

We start with the sequence $1,2$ and when we have to pass from $2^k$ to $2^{k+1}$ we keep the old sequence as it was, and add $2^k+i$ right after $i$ for each $i\in\overline{1,2^k}$. For instance, our first sequence is $1,2$, the second sequence is $1,3,2,4$, the third sequence is $1,5,3,7,2,6,4,8$, and so on. This works because if the sequence containing $1,2,\ldots,2^k$ has no three properly ordered numbers in arithmetic progression, then the next one doesn't either, since modulo $2^k$ it looks just like the old one, except that it has two consecutive numbers for each residue modulo $2^k$.

Grobber,

You are right, the original statment was $f(a)+f(c)=2f(b)$ instead of $f(2b)$. But, would it be true with this modification??

This problem was also used at oral examination polytechnique


geometry
area of a triangle
Heron\u0027s formula

Circles $ A$, $ B$ and $ C$ are tangent as shown. The area of circle $ A$ is $ 16\pi$ square centimeters, the area of circle $ C$ is $ 16\pi$ square centimeters, and the area of circle $ B$ is $ \pi$ square centimeters. What is the number of square units in the area of $ \triangle ABC$?

[asy]unitsize(0.75cm);
pair A,B,C;
A=(-4,0);
C=(4,0);
B=(0,3);
draw(Circle(A,4),linewidth(0.7));
draw(Circle(C,4),linewidth(0.7));
draw(Circle(B,1),linewidth(0.7));
draw(A--B--C--cycle,linewidth(0.7));
label("$A$",A,SW);
label("$C$",C,SE);
draw(B--(1,4));
label("$B$",(1,4),NE);[/asy]

We find that the radius of circle $ A$ is $ 4$, the radius of circle $ C$ is $ 4$, and the radius of circle $ B$ is $ 1$. We now have an isoceles triangle with $ AC\equal{}8$, $ BC\equal{}5$, and $ AB\equal{}5$. Drawing the altitude from $ B$ to $ AC$ yields an altitude of $ x$ equal to $ 3$ (Because it is a $ 3$,$ 4$,$ 5$ right triangle). Thus the area is $ \frac{8\times{3}}{2}\equal{}\frac{24}{2}\equal{}\boxed{12}$. Alternatively, one could have used Heron's Formula to figure out the area, but I don't recommend it, at least not on this problem :D .


Can anybody post the problem of the 1st day of IMC 2007?

They must have a problem with the internet because I know Peter is there and he would have posted the questions by now. Either that, or they are having way too much fun :P

There is one small internet cafe and no other internet access! I'd post the problems but I don't have them with me, sorry! I'm sure someone will post the problems once we all go home. If no one else does it first then I will post them myself on Thursday evening (GMT -5).


algebra
polynomial
inequalities
algebra proposed

Find all polynomials $ P(x)\equal{}x^n \plus{} a_1 x^{n\minus{}1} \plus{} ....\plus{} a_0$

Where all coeficients are equal to $ \minus{}1$ or $ 1$

[quote="FOURRIER"]Find all polynomials $ P(x) \equal{} x^n \plus{} a_1 x^{n \minus{} 1} \plus{} .... \plus{} a_0$

Where all coeficients are equal to $ \minus{} 1$ or $ 1$[/quote]
I guess you mean to find all polynomials of a degree $ n$ that has $ n$ real roots and all its coefficients are equal to $ \minus{}1$ or $ 1$?Am i right?

[quote="FOURRIER"]yes i forgot to mention that all the roots are real and I can't edit it now.[/quote]
Let $ r_1,r_2,\dots,r_n$ be real roots of $ P$.
Then $ |r_1 \plus{} r_2 \plus{} \dots \plus{} r_n|^2 \equal{} 1$ and $ \left |\sum_{i<j} r_ir_j\right | \equal{} 1$,hence
$ \boxed{r_1^2 \plus{} r_2^2 \plus{} \dots \plus{} r_n^2\leq 3}$.
Also we know that
$ |r_1\cdot r_2\cdot\dots\cdot r_n| \equal{} 1$,therefore, $ \boxed{r_1^2\cdot r_2^2\cdot\dots\cdot r_n^2 \equal{} 1}$.
Using AM-GM inequality for boxed inequality and equation:
$ 3\geq r_1^2 \plus{} r_2^2 \plus{} \dots \plus{} r_n^2\geq n\sqrt [n]{r_1^2\cdot r_2^2\cdot\dots\cdot r_n^2} \equal{} n$,thus $ \boxed{n\leq 3}$.
If $ n \equal{} 1$,then
$ P(x) \equal{} x \plus{} 1$ or $ P(x) \equal{} x \minus{} 1$.
If $ n \equal{} 2$,then
$ P(x) \equal{} x^2 \plus{} x \minus{} 1$ or $ P(x) \equal{} x^2 \minus{} x \minus{} 1$.
If $ n \equal{} 3$,then obviously $ r_1^2 \equal{} r_2^2 \equal{} r_3^2 \equal{} 1$ and $ \sum_{i\neq j} r_ir_j \equal{} \minus{} 2$,hence
$ P(x) \equal{} x^3 \minus{} x^2 \minus{} x \plus{} 1$ and $ P(x) \equal{} x^3 \plus{} x^2 \minus{} x \minus{} 1$.