Hmm, I think this is off topic here, but anyway...

I'm a Grade 12 high school student. I've never had a mark lower than 95 in math. Currently I'm taking an Algebra and Geometry course, and I'm starting to worry, because a good part of the curriculum is proofs, and I cannot do proofs.

I also can't solve problems.

I mean, I crumble under the most basic problems. I love mathematics and I love learning mathematics and I love doing mathematics, but I really have some sort of problem thinking logically. I don't understand. I always thought I was intelligent, but the fact that I can't solve simple problems or prove basic concepts makes me want to cry.

Is/has anyone in/been in the same boat as me? Is there any hope for me to be a good problem solver? I don't know what I'm supposed to do. "Practice," they say, but what am I to practice if even basic stuff gives me an aneurism?

To give you an idea of how much I suck, I can't solve this problem, which was from the middle school forum:

Take triangle ABC , such that angle C = 90 . Take two points D and E on line AB such that AD = DE = EB . Draw line segments CD and CE. That should divide ABC into three smaller triangles. If the area of CAD is 5, then what is the area of ABC?

I figure that AD = DE = EB = 1/3c, then 1/3c * the height of the triangle from c, h, = 5. ch then = 30, so the area of the triangle is 30, but that seems to apply to all triangles and I have a feeling it isn't right.

Can anyone give me a hint, a tip, anything, to get me started on this? I just can't do it and it's so frustrating!

Help! : :(

Here's a few hints.

Read one hint, and if you can't get it after trying for a while, then go on to the next hint. There are nine hints in total, and if you still can't get it after the eighth hint, don't take the ninth hint offensively :P . The hints are in sequential order, each one adding a little more knowledge.

[hide="Click to reveal hint 1"]
Draw a line through C, parallel to AB.

[hide="Click to reveal hint 2"]
Rotate the triangle a little bit.

[hide="Click to reveal hint 3"]
Rotate it so AB is on the bottom and C is on the top.

Draw the altitudes to the three mini-triangles.

Notice that the altitudes of the three triangles are the same.

Notice that the bases of the three triangles are the same.

Recall the formula for the area of a triangle.

If all of the triangles are the same area, and one triangle has an area of 5...

Get a tutor.

[/hide]
[/hide]
[/hide]

Dude Mosis,Don't take it too hard on yourself,that way u'll never progress...

Yah i was in the same situation in which u are...How do u think people get intelligent?They practice hell a lot and while practicing they try each question,however dumb for more than 15min and then see the solution,this way they develop a tendeny to have more endurance and they get used to many concepts and this develpos their logic....Though i'm still pretty dumb but i'm trying to improve :D ...the main thing is practice.

Just practice, do every problem u get ur hands on and read everything.  I think most people don't solve problems just because they have brilliant insight which they were born with; it's because they've seen a million problems like it and can apply a similar technique.

try to look at what other people do on that question, and if you still dont get how to do it, PM the user and ask him/her

GOOD LUCK :thumbup:  :bye:

[quote="shadysaysurspammed"]Dude Mosis,Don't take it too hard on yourself,that way u'll never progress...

Yah i was in the same situation in which u are...How do u think people get intelligent?They practice - a lot and while practicing they try each question,however dumb for more than 15min and then see the solution,this way they develop a tendeny to have more endurance and they get used to many concepts and this develpos their logic....Though i'm still pretty dumb but i'm trying to improve :D ...the main thing is practice.[/quote]
hmmm there are exceptions to REALLY bright students like eminem but I agree with you, mosis get practise alot and you will definately improve.

You probably need a good book with a lot of problems and good solutions. That way, you'll learn how to do the problems too. Textbooks only give you answers, and their problems only get so hard (but they have the basic concepts, which is essential). [url=http://www.artofproblemsolving.com/Books/AoPS_B_About.php]AoPS books[/url] have good problems and excellent solutions (if you get the solution part), but some of those problems are really hard.

how much has your teacher talked about proofs?

i also think you just need lots of practice and i also recomend the AoPS books

[quote="tianle"]is your text book called geometry and discrete math?? if so we are using the same text book and i can try to help u.[/quote]

Yes. It's red and it has some locus on the cover.

I was in the same boat as you last year in 7th grade. Then I started practicing a couple hours a day every day, and I got a few good problem solving book. And it worked. I skipped a year in math and I'm hoping to pass the AIME this year, so I would just recommend practice, practice, practice, ..., $\infty$

lol ok then pm me ur msn msger or AIM.. if u have... and we can do some problems together.. i'm also learning that book right now. u liv in canada? what school?

Have fun!
The inability to get questions you think are "easy" is not something to get upset about, just keep doing math problems for why you started looking at them in the first place: because you love the subject and you love trying problems out. I happen to be unable to solve lots of "easy" questions myself, but that's not something to get too upset about. What you learn from math and what you get out of trying to solve problems is what's most important, not whether you can solve every question you see or not. If you could solve every math question you looked at, math wouldn't be nearly as fun as you find it right now.
You ARE intelligent. My only tip: Keep up your love for mathematics  :) Keep going, and you'll be in great shape.

It's nice to know that the extraordinarily intelligent people around here have sympathy for others...like our teacher's say-"Moral values come before anything else ;) "


Are the angles in a 3-4-5 right triangle always the same? Or do they differ?

By SSS all 3-4-5 triangles are congruent and thus the angles are congruent.
If you are asking the angles within the triangle are equal to each other, the answer is no.

[quote="bpms"][quote="MichaelFaraday"]the angles can differ[/quote]
Um, actually the angles are always $\sin{\frac35}$ and $\sin{\frac45}$, hence the reason trig works...[/quote]

[cough]no trig in the classroom forum[/cough]

Yes, the corresponding angles in a 3-4-5 triangle are always equal. ragnarok23 basically said that all triangles with the same three side lengths are congruent. (Think of it this way: a triangular frame is rigid.) Therefore, since all such triangles are the same, the angles are also always the same.

By Using the SSS Theorem/Postulate, it is equal.

[quote="anirudh"]By Using the [b]SSS Theorem/Postulate,[/b] it is equal.[/quote]
SAS Similarity, SAS Inequality, SSS Similarity, and SSS Inequality are all [b]Theorems.[/b]

[quote="anirudh"]In my geometry book, it is listed as a Postulate.[/quote]
Well, say that to [b]my[/b] Geometry Book! :rotfl:

[quote="jackdillon"]I think the angles can differ![/quote]

If you think so, please give your reasoning :)

I can't say it loud enough. They are congruent by SSS and the angles are locked into place, virtually :arrow:


geometry
circumcircle
cyclic quadrilateral
geometry proposed

Let $H$ be the orthocenter of an acute triangle $\triangle{ABC}$. The tangents from $A$ to the circle of diameter $BC$ touch the circle at $P$ and $Q$. Prove that $P$, $Q$, and $H$ are collinear.

Actually, you just need to show that H lies on the polar of A with respect to the circle with diameter BC. This can be done in several ways. For instance, you may use the fact that a point P lies on the polar of a point Q with respect to a circle k if and only if the vectorial equation $\overrightarrow{MP} \cdot \overrightarrow{MQ} = r^2$ holds, where M is the center and r is the radius of k. Or use the fact that $AD \cdot HD = BD \cdot CD$, where D is the foot of the altitude from A to BC.

  Darij

Yes, Grobber, this is, of course, the simplest proof. How did I forget to note it!

  Darij

[quote="grobber"]It's just a particular case of the property: Let $ABCD$ be a cyclic quadrilateral s.t. $E=AB\cap CD$. Then $F=AC\cap BD$ lies on the polar of $E$ wrt the circumcircle of $ABCD$.[/quote]

Very nice, but how to prove this?

See Theorem 1 of Vladimir's first posting in

  http://www.cut-the-knot.org/cgi-bin/dcforum/ctk.cgi?az=read_count&om=390&forum=DCForumID6

  Darij

maybe someone has a solution without using polars ?

u can find polar free sol on http://www.kalva.demon.co.uk bad it's not nice sol.

[quote="surshu"]Can anybody post elementary solution?[/quote]

let M be midpoint of BC and D be feet of A-altitude.we have $ \angle APM = AQM =\angle D = 90^{\circ}$ hence 5 points A,P,Q,D and M are on a circle.
also we have $ AQ^{2}= AH.AD$ hence $ \angle APQ =\angle ADQ =\angle AQH$ also we have $ \angle APQ =\angle AQP$ hence P,Q,H are collinear.


group theory
abstract algebra
number theory

[quote="xxxyyyy"]Let $\phi(m)$ be the Euler's Phi function.

Let $e_{p}{(a)}=$the smallest exponent $e \geq 1$ so that $a^{e}\equiv 1 (mod p)$.

Now, prove that:
$e_{m}{(a)}$ always divides $\phi(m)$[/quote]

Hm..
[hide]$a^{\phi(m)}-1\equiv 0\mod m$
$a^{e}-1\equiv 0\mod m$
$m|a^{\phi(m)}-1$
$m|a^{e}-1$
$a^{\phi(m)}-1>a^{e}-1$ (?)
$a^{e}-1|a^{\phi(m)}-1$
$e|\phi(m)$[/hide]
not sure if any of the steps are valid  :|

[quote="t0rajir0u"][quote="ch1n353ch3s54a1l"]
$a^{\phi(m)}-1>a^{e}-1$ (?)
$a^{e}-1|a^{\phi(m)}-1$[/quote]

Not a valid step.  Just because one is bigger doesn't mean they divide.  For example, $a^{2}-1 \not | a^{3}-1$.  So you're basically assuming what you're trying to prove.

[hide="Discussion"] This is a necessary lemma in the proof of Euler's Totient Theorem.  A full treatment of it requires a little background in group theory.

$e_{m}(a)$ is known as the order of $a \bmod p$.  It's a powerful statement of group theory that the order of any element in a finite group divides the size of the group.  The group we are considering is the set of units $\bmod p$ under multiplication, which has size $\varphi(p)$, hence the statement.

Phrased in terms of group theory, a simple intuitive proof is easy (the rigorous one requires we be a bit more exact about a few things):

Consider the set $\{ 1, a, a^{2}, ... a^{e_{m}(a)-1 }\}$.  Now suppose we "multiply" the elements of this set by some $b$ that is not in the set, and add those elements to the set.  It's not hard to show that none of our new elements were in the set before, so the size of the set increases by $e_{m}(a)$.

We continue adding new elements in this way and we must eventually produce the entire group.  A set produced in this way has size a multiple of $e_{m}(a)$.  QED. [/hide][/quote]

Interesting... can you prove using number theory principles, not involve SETs in this?

I remember proving that in terms of group theory using a method with constructing right and left cosets (I think its called Lagrange's Theorem). It was at a summer camp a couple months ago, I probably still have the outline and lecture notes of that proof...

[hide="On second thought, I think I will post the rigorous proof (its pretty easy stuff)"]

We are going to prove a general result in group theory:
[b]Claim: Given a group $A, \forall g\in A, \ \ ord(g) \ | \ \ |A|$[/b]

[hide="Proof of Claim"]
[b]Definition:[/b] A [i]subgroup[/i] of a group $A$ is some subset which also satisfies closure under the same binary operation as $A$
[b]Definition:[/b] The [i]right coset[/i] of a set $H$ with respect to some element (not necessarily in the set) $g$ is denoted as $Hg$ and defined as such:
* $H=\{ g_{i}\}$
* $Hg=\{ g_{i}\times g\}$
The left coset is defined and denoted similarily (consider $g\times g_{i}$ as opposed to $g_{i}\times g$ ... remember we don't necessarily have commutativity)

Pick an arbitrary subgroup, $H$, of $A$
We break up this proof into three steps:
[hide="1) Show that any two right cosets of $H$ (w.r.t. some element of $A$) are either disjoint or equivalent"]This is the meat of the argument; pick two cosets $Hg$ and $Hg'$. Either
a) They are disjoint, as claimed
b) They are not disjoint; i. e. $\exists a \in Hg \& Hg'$
Let's explore this:
$a\in Hg=\{ h_{1}g, h_{2}g, ... \}$
$a\in Hg'=\{ h_{i}g', h_{2}g', ... \}$
Since $H$ is a group, we have closure (i.e. the "product" of any two elements is another element of the group)
So for any $h\in H, ha=h\times h_{k}\times g=h_{l}\times g$ ($h_{l,k}\in h$)
This means $ha\in Hg$, similarily $ha\in Hg'$. But we can do this for any arbitrary $h$; and since $a$ is just a member of $H$, we can generate any element of $H$ using the construction $ha$. Therefore (by exhaustion) $Hg=Hg'$, so two such cosets are either disjoint or congruent $\blacksquare$[/hide]
[hide="2) Show that the sum of all the right cosets of $H$ (w.r.t some element of $A$) equals $A$"]
This is obviously overkill; looking at the identity element alone (which must be present in each $H$ by definition of a subgroup) $g\in Hg \ \ \forall g$ (identity $e\times g=g$). So $\forall g\in A, g$ has been included in our construction, so $A$ is a subgroup of our construction. But our construction is a subgroup of $A$, so the two are equal $\blacksquare$[/hide]
[hide="3) Show that 1) and 2) imply the claim"]
Obviously, the order of each coset is equal to the order of $H$. Since $\sum_{i=1}^{m}Hg_{i}=A$ (for some $m$), we can compare orders:
$\sum_{i=1}^{m}|Hg_{i}|=|A|\implies m\times |H|=|A|$ using the fact that each coset has order $H$ and that each subset is disjoint or already listed. $m$ is an integer, implying that $ord(H) | ord(A)$
We are close to done; choose and arbitrary $g_{i}\in A$, and locate the smallest subgroup containing $g_{i}$. The order of $g_{i}$ is the same as the order of the subgroup (this is easy to show and has been mentioned earlier) so $\forall g\in A, \ \ ord(g) \ | \ \ |A| \blacksquare$[/hide][/hide]

The "claim" is known as Lagrange's Theorem, I believe.
[hide="This theorem implies the problem:"]
Let $A\equiv \{ a | gcd(a, m) = 1 \}\equiv \{a_{1}, a_{2}, ... a_{\varphi (m)}\}$
[b]* $A$ is a group under the binary operation of multiplication $\text{(mod m)}$[/b]
It's obvious that $|A|=\varphi(m)$, now it is easy to show that:[b]
*$e_{m}(a)=ord(a)$
By the claim we have just proven, $ord(a) | ord(A) \implies e_{m}(a) | \varphi (m) \blacksquare$
[/b]
[/hide]
[/hide]

[quote="t0rajir0u"][quote="xxxyyyy"]Interesting... can you prove using number theory principles, not involve SETs in this?[/quote]

But I am using number theory principles.  Number theory is very easy to understand from the perspective of group and ring theory  :D

And why can't I use the concept of a set?  It's a very useful concept!  You'll find it hard to prove much in number theory without the idea of a set.

[quote="ch1n353ch3s54a1l"][quote="t0rajir0u"]The group we are considering is the set of units $\bmod p$ under multiplication, which has size $\varphi(p)$, hence the statement.[/quote]

I'm not really sure what this means... could you explain?  :ninja:[/quote]

[b]Definition:[/b]  A group $G$ is a set and an operation $a*b$ that satisfies the following properties:

$\forall a, b \in G : a * b \in G$
$\exists I : a * I = a$
$\forall a \in G : \exists a^{-1}\in G : a * a^{-1}= I$
$(a * b) * c = a * (b * c)$
(Optional) $a * b = b * a$

[b]Definition:[/b]  Let $a$ be a member of some group $G$ and let $n$ be an integer.  Define

$a^{0}= 1$
$a^{n}= a * a^{n-1}$

Let's consider the set $\{ 0, 1, 2, 3, ... m-1 \}$ under the operation of multiplication $\mod m$.  Is this a group?

We find that the answer is no.  Many elements of this set do not have an inverse; $0$, or any nontrivial divisor of $m$.  If $m$ is prime, we can just use the set $\{ 1, 2, 3, ... m-1 \}$.

However, if $m$ is not a prime, we use the set $\{ a | gcd(a, m) = 1 \}$ (can you see why these are precisely the numbers with inverses?)  This is the "multiplicative group $\bmod m$," and it has size $\varphi(m)$, sometimes denoted $U_{m}$.  It's also the set of "units" (members with multiplicative inverses) in the [b]ring[/b] $\mathbb{Z}_{p}$, which is a related concept.

[b]Definition:[/b]  Let $a \in G$ for $G$ a finite group (a group with a finite number of elements).  The order of $a$ is the smallest positive integer $n$ such that $a^{n}= I$ (where $I$ is the identity element).  (The proof that $n$ exists is left as an exercise.)  We denote this $ord(a)$.

[b]Definition:[/b]  Let $|G|$ denote the cardinality of the set associated with the group $G$.

Then the original problem is a special case of the following statement.

[b]Lemma:[/b]  $\forall a \in G : ord(a) | |G|$[/quote]

[quote]And why can't I use the concept of a set? [/quote]
I never meant that you cannot use it... I just find it difficult to understand.. so thats all !!

Do not quote page-long posts for a single-sentence answer!

But there is a much more direct way to prove that if $k$ is minimal postive such that $a^{k}=e$, then $k$ divides all $m$ with $a^{m}= e$ (we work in groups).

me@home, you gave essentially the proof I gave - I didn't explicitly use the term "coset," though (we can use the unqualified word if we only want to talk about commutative groups).

ZetaX, I assume you mean $a^{k}= e, a^{m}= e \implies a^{(k, m)}= e$.  Yes, but then it remains to show the Totient Theorem, and I would rather prove it with this lemma instead.

[quote="me@home"]You mean proving that specifically in a more direct way than my proof? I'm pretty sure its close to the textbook way of proving it, its a nice proof.[/quote]
You proved Lagrange's theorem and not what I wrote, right¿
And I speak of a proof only with one third of the length ;)

@t0rajir0u: thats a closely related result and implies mine; and yes, the the totient theorem is needed then, but I would call it "known" (there is also a short proof for it) ;)

Yes, thats essentially what i did
how do yo get the upside down q mark

$a^{\phi(m)}\equiv 1\mod p\Rightarrow e_{m}(a)\leq\phi(m)$
assume that $e_{m}(a)$ does not divide $\phi(m)$, then $\phi(m)=q(e_{m}(a))+r$ with $1<r<e_{m}(a)$.
we have
$a^{\phi(m)}\equiv a^{q(e_{m}(a))+r}\equiv a^{q(e_{m}(a))}a^{r}\equiv 1a^{r}\equiv a^{r}\equiv 1\mod p$
since $1<r<e_{m}(a)$, we have a contradiction.

Jiang - Nice proof, of course.  You may   like to edit (2) typos.
[hide][quote]$\phi(m)=q(e_{m}(a))+r$ with $1<r<e_{m}(a)$.
[/quote]
Should be   $1\le r<e_{m}(a)$ [/hide]

It's not rigorous.  This is a lemma I would use to prove the Totient Theorem.

Can someone give a proof of the Totient Theorem without this lemma?

The proof of the totient theorem that I usually like goes something like this

Put $\varphi(n)=t$ to save me some typing.

Let $\{a_{1},a_{2},\ldots,a_{t}\}$ be the invertible elements of $\mathbb{Z}_{n}$.  It is straightforward to see that if $(a,n)=1$, then $\{a_{1},a_{2},\ldots,a_{t}\}=\{aa_{1},aa_{2},\ldots,aa_{t}\}$, whence \[a^{t}\prod^{t}_{i=1}a_{i}= \prod^{t}_{i=1}a_{i}\] Clearly $\prod^{t}_{i=1}a_{i}$ is invertible, whence $a^{t}=1$ in $\mathbb{Z}_{n}$.


ya seriously nat pwns u all. respect the swine D:

FEEL the Swine Flu!
FEEL the SARS!
lol

Wait since when does MO have a serious discussion???
But w/e.

I'll make up some frequently asked questions about Swine Flu on the spot.

* Do you have Swine Flu?
* Do I have Swine Flu?
* How do you get Swine Flu?
* How do you avoid Swine Flu?
* Should I tell people that you can't get Swine Flu from pork, or should I take advantage of low prices and spread the hype around?
* Is there going to be a pandemic?
* Who let the swine flu get here, anyways?

It is of my opinion that we should have had better border patrol.


group theory
abstract algebra
Galois Theory
superior algebra
superior algebra unsolved

[quote="JustSee12"]From Dummit and Foote:
Prove that if $ H$ is a nontrivial normal subgroup of the solvable group $ G$ then there is a nontrivial subgroup $ A$ of $ H$ with $ A \unlhd G$ and $ A$ abelian.

Bear in mind $ G$ is not necessarily finite; I think this would be much easier if it were.[/quote]

Let $ \{1\}\equal{}G_0 \subset G_1 \subset \cdots \subset G_n\equal{}G$ be a normal series of $ G$ and $ k\equal{}\min \{j: \ H \cap G_j \neq \{1 \} \}.$ Then $ A\equal{}H \cap G_k$ is obviously normal in $ G.$

Also, the map $ f: A \longrightarrow G_k/G_{k\minus{}1}$ defined by $ f(a)\equal{}aG_{k\minus{}1}$ is an injective group homomorphism because $ H \cap G_{k\minus{}1}\equal{}\{1 \}.$ Thus $ A$ is abelian.

[quote="JustSee12"]It's not obvious to me that $ H \cap G_k$ is normal in $ G$. I think you're using "normal series" in the sense that each $ G_i \unlhd G$, and as far as I know the existence of such a series (with abelian factors) is not guaranteed unless $ G$ is finite (in which case it's still quite difficult to prove).[/quote]

If we define that a group $ G$ is solvable if it has a [u]subnormal[/u] series, i.e. $ G_i \unlhd G_{i \plus{} 1},$ with all successive quotients $ G_{i \plus{} 1}/G_i$ abelian, then it's easy to see that $ G^{(n)} \equal{} \{1\},$ for some $ n$,

where $ G^{(n)}$ is the n-th derived subgroup of $ G.$ So we'll have this [u]normal[/u] series with successive abelian quotients: $ \{1\} \equal{} G^{(n)} \subset G^{(n \minus{} 1)} \subset \cdots \subset G.$ Conversely, every normal series

is obviously a subnormal series. So, in any group, the existence of a subnormal series (with successive abelian quotiens) is equivalent to the existence of a normal series (with successive

abelian quotients). I hope this helps.  :)


calculus
integration
algebra
function
domain
Ring Theory
polynomial

Helo. I stuck on a question, please help :)
 A finitely generated $ R$-algebra ($ R$ integral domain) is an integral domain $ A$, extending $ R$, with this proprty. There are $ a_1,...,a_l \in A$ such that there is no proper subring of $ A$ containing $ R$ and $ a_1,...,a_l$.

Suppose $ A = R[x_1,...,x_n]/P$ for some prime ideal with $ P \and R = {0}$ so that the quotient map restricts to an embedding of $ R$ into $ A$. Identifying $ R$ with its image in $ A$ prove that $ A$ is a finitely generated R-algebra.

By te way what does it mean "identifying R with its image in A"?

Thanks so much!!

So if the quotient map restricts to an embedding of $ R$ into $ A$, we can consider $ R$ as contained in $ A$ through this embedding. i.e. for any $ r\in R$ we look $ i(r)\in A$ where $ i: R\hookrightarrow A$ is the inclusion.

Alright now for the actual problem, it is a standard theorem that the quotient of a ring by a prime ideal is an integral domain. Basically, it is because prime ideals have the property that if $ ab\in P$ then either $ a$ or $ b$ is in $ P$. Thus after modding out by $ P$, there are no zero divisors because if $ ab$  is congruent $ 0$ in the quotient, then $ ab$ is in $ P$ thus $ a$ or $ b$ is in $ P$ thus in the quotient, $ a$ or $ b$ descends to $ 0$. For finite generation, notice that $ x_1,...x_n$ descend to a generating set of $ A \equal{} R[x_1,...x_n]/P$ by surjectivity of the quotient map so $ A$ is finitely generated as an $ R$ algebra. To elaborate on this, let $ f$ be an element in $ R[x_1,...x_n]/P$. Then it lifts to an element (more like an equivalence class of elements, but whatever, pick one of them) $ F\in R[x_1,...x_n]$ Now, we can write $ F$ as a polynomial (by definition of $ R[x_1,...x_n]$), say $ F(x_1,...x_n)$. Reducing this polynomial modulo $ P$ gives an expression for $ f$ in the quotient ring and thus $ x_1,...x_n$ descend to generators on the quotient.


geometry
function
analytic geometry
calculus
derivative

A spherical raindrop, starting from rest,falls under the influence of gravity.If it gathers in water vapor(assumed at rest) at a rate proportional to its surface, find its acceleration, assuming its initial radius to b negligible.

this is more of a maths prob than a phy 1, since its done once we get the reqd. DE, but im unable to get it.please help.
thanks...

Seems like the law is always the same: $ \frac d{dt}(mv)\equal{}F\equal{}mg$ where $ m$ is the mass, $ v$ is the velocity, and $ g$ is the acceleration of the free fall, isn't it? :). The only tricky thing is that $ m$ is not constant but changes according to the law $ \frac d{dt}m\equal{}\beta m^{2/3}$ where $ \beta$ includes the density of water and constants related to the geometry of the sphere. Can you continue from here?

So if I understand the problem correctly, then this is what I would do.

Let the z axis point downwards with the origin at the starting point of the water droplet.
Now you say that the droplet gains 'mass' proportional to its surface, then the $ dm$ mass gained by the droplet, while falling infinitesimal distance $ dz$ is:
\[ dm \equal{} \rho 4r^2\pi dz
\]
Where $ \rho$ is the mass density of the vapor and $ r$ is the radius of the droplet at a given moment.

Now lets assume that the 'vapor cloud' is homogenous. Then its density is constant so we also have fro the droplet:
\[ dm \equal{} \rho dV \equal{} \rho 4r^2\pi dr
\]
Now combining these two we get:
\[ \dfrac{dr}{dz} \equal{} 1
\]
That is:
\[ r(z) \equal{} z
\]
according to our choice of origin above.

So the mass of the droplet at a given moment is,(now I still write it as function of time, through the space coordinate):
\[ m(z) \equal{} \rho\dfrac43 r^3\pi \equal{} \dfrac{4\pi\rho}{3}z^3
\]
According to Newtons second law we have:
\[ mg \equal{} \dfrac{d(mv)}{dt} \equal{} m\dfrac{dv}{dt} \plus{} v\dfrac{dm}{dt} \equal{} m\dfrac{dv}{dz}v \plus{} v\dfrac{dm}{dz}v
\]
Where we used the fact that: $ v \equal{} \dot z$.

So our equation is:
\[ v\dfrac{dv}{dz} \plus{} \dfrac{v^2}{m}\dfrac{dm}{dz} \equal{} g
\]
plugging in the space derivative of the mass, we get:
\[ v\dfrac{dv}{dz} \plus{} \dfrac{3}{z}v^2 \equal{} g
\]
Now this is a straight forward differential equation for the $ v(z)$ function.
Notice that $ v\frac {dv}{dz} \equal{} \frac12\frac {d}{dz}(v^2)$. so lets use the $ u \equal{} v^2$ substitution:
\[ u' \plus{} \dfrac{6}{z}u \equal{} 2g
\]
AS we see, we got a simple nonhomogenous differential equation.

The $ U$ solution of the homogenous part:
\[ U' \plus{} \dfrac{6}{z}U \equal{} 0 \Longrightarrow U(z) \equal{} \dfrac{C}{z^6}
\]
Where $ C$ is some constant.

LEts look for the $ u_0$ particular solution in the : $ u_0(z) \equal{} \frac {C(z)}{z^6}$ that is $ C$ is now a function of $ z$. Plugging this back we have:
\[ \dfrac{C'(z)}{z^6} \equal{} 2g \Longrightarrow C(z) \equal{} \dfrac27gz^7 \Longrightarrow u_0(z)\equal{}\dfrac27gz
\]
So the general solution of the DE:
\[ u(z) \equal{} u_0(z) \plus{} U(z) \equal{} \dfrac{C}{z^6} \plus{} \dfrac27gz \equal{} v(z)^2
\]
Now we want that $ v(z \equal{} 0) \equal{} 0$ since it started from rest. As we see the above form of $ v(z)$ is not defined at $ z \equal{} 0$. The only way it could be defined is if when $ z\to 0$ then $ C\to 0$, so we could get a defined answer. But we definitavely want the previous condition to hold. So its lets choose $ C \equal{} 0$ that is $ C$ be constant zero.
So now we have for the velocity:
\[ v(z) \equal{} \sqrt {\dfrac27 gz}
\]
The acceleration is then:
\[ a \equal{} \dfrac{dv}{dt} \equal{} \dfrac{dv}{dz}v \equal{} \sqrt {\dfrac27 gz}\cdot \cfrac{\frac27 g}{2\sqrt {\frac27 gz}} \equal{} \dfrac{g}{7}
\]
So the acceleration of the water droplet is:
\[ a \equal{} \dfrac{g}{7}
\]
I hope I didnt make any errors since I latexed directly while solving.. :blush:


The positive integer $n$ has the property that, in any set of $n$ integers, chosen from the integers $1,2, \ldots, 1988,$ twenty-nine of them form an arithmetic progression. Prove that $n > 1788.$

Write $ 1,2,\ldots, 1988,$ in $29$-base expression. i.e. $1988=(2,10,16)_{29}$
Lemma: for every twenty-nine length of arithmetic progression. There are must be one term that contains '$28$' digit in its $29$-base expression.
Proof: we can let the ${n,n+d,n+2d,...n+28d}$ be the arithmetic progression. If $d$ is not divisible by $29$, one of term must contain '$28$' as the last digit. we can assume $29^{l}||d$, then one of the term must have '$28$' as last $(l+1)th$ digit.

By applying the lemma, we can choose those numbers, which doesn't contain '$28$' in their $29$-base expression. Then those numbers cannot form an 29 length of arithmetic progession. There are totally $1988-2.29-2.29-10+2=1864$ such that number. Therefore $n\geq1864>1788$.


geometry
conics
ellipse
geometry proposed

Given triangle $ ABC$ and an arbitray point $ P$ inside. Let $ l_1,l_2$ be two lines through $ P$ such that $ l_1\perp l_2$. Let $ X_1,X_2$ be two orthopoles of triangle $ ABC$ wrt $ l_1,l_2$,respectively. When $ l_1,l_2$ move, prove that $ X_1X_2$ passes through a fixed point.

Denote points $ M \equal{} AC\cap l_2 ,N \equal{} AB\cap l_2$.

$ A' ,C'$ be two points lie on $ l_1$ such that $ AA'\perp l_1,CC'\perp l_1$.

Let $ H,K$ be the orthocenters of $ ABC,AMN$,respectively.

Call $ S \equal{} AK\cap CH$.

We see that  $ A'X_1//AH,X_1C'//SH,A'C'//AS$ ,so triangles $ A'X_1C' ,AHS$ are homothetic.

On the other hand ,by [b]Thales' theorem [/b]we have:

$ \frac {PA'}{PC'} \equal{} \frac {MA}{MC} \equal{} \frac {KA}{KS}$.

So triangles $ A'X_1P,AHK$ are homothetic,we obtain $ PX_1//HK$( Because that both $ A'P//AK$ and $ A'X_1//AH$ ).

But from the [b]Goormaghtigh's theorem [/b],we have $ H,K,X_2$ are collinear.

Therefore , $ PX_1//HX_2 (1)$

Similarly, $ PX_2//HX_1 (2)$

Since $ (1)$ and $ (2)$ then $ X_1X_2$ passes through midpoint of $ PH$ which is a fixed point.

Thank to [b]Linh[/b] [hide="?"] than kinh :D [/hide] for your nice result.:)

Dear Mr.[b]Khanh[/b],
This result comes from two lemmas:
[b]Lemma 1:[/b] Denote $ X_1,X_2$ the orthopoles of two orthogonal lines through the orthocenter $ H$ wrt $ \Delta ABC$. Then $ H$ is the midpoint of $ X_1X_2$.
[b]Lemma 2:[/b] Denote $ X_1,X_2$  the orthopoles of two parallel lines $ l_1,l_2$ then $ X_1X_2\perp l_1$ and $ X_1X_2$ is equal to the distant of $ l_1,l_2$.
Thanks you for your nice solution.

Let $H$ be the orthocenter of $\triangle ABC$ and $A',B',C'$ the projections of $P$ on $BC,CA,AB.$ From the problem [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=498545]Orthopole lies on ellipse[/url] (post #2), we know that $X_1,X_2$ move on a ellipse $\mathcal{E}$ circumscribed to $\triangle A'B'C',$ further $A'X_1 \mapsto \ell_1$ and $A'X_2 \mapsto \ell_2$ are homographic. But since $\ell_1 \mapsto \ell_2$ is clearly an involution, then $X_1 \mapsto X_2$ is an involutive homography on $\mathcal{E}$  $\Longrightarrow$ $X_1X_2$ goes through a fixed point. Making $\ell_1 \equiv PA'$ and $\ell_1 \equiv PB',$ we figure out that the fixed point is none other than the center of $\mathcal{E},$ i.e. the midpoint of $\overline{PH}.$


Let $ a,b,c$ be nonnegative real numbers .

With : $ M\equal{}\frac{37(a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2}{2(ab\plus{}bc\plus{}ca)(a\plus{}b\plus{}c)^6}$

Prove that :

$ \frac{1}{a^2\plus{}2bc}\plus{}\frac{1}{b^2\plus{}2ca}\plus{}\frac{1}{c^2\plus{}2ab} \geq \frac{ab\plus{}bc\plus{}ca}{a^2b^2\plus{}b^2c^2\plus{}c^2a^2}\plus{}M$

[quote="yanglu"][quote="hedeng123"][quote="mai quoc thang"]Let $ a,\, b,\, c$ be nonnegative real numbers .

With : $ M \equal{} \frac {37(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2}{2(ab \plus{} bc \plus{} ca)(a \plus{} b \plus{} c)^6}.$

Prove that :

$ \frac {1}{a^2 \plus{} 2bc} \plus{} \frac {1}{b^2 \plus{} 2ca} \plus{} \frac {1}{c^2 \plus{} 2ab} \geq \frac {ab \plus{} bc \plus{} ca}{a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2} \plus{} M.$[/quote]
$ \frac {1}{a^2 \plus{} 2bc} \plus{} \frac {1}{b^2 \plus{} 2ca} \plus{} \frac {1}{c^2 \plus{} 2ab} \geq \frac {ab \plus{} bc \plus{} ca}{a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2} \plus{} \frac {747(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2}{2(ab \plus{} bc \plus{} ca)(a \plus{} b \plus{} c)^6}.$

The inequality  is holds.[/quote]
Your inequality is not true, hedeng, for counter-example, $ \,\,\, a \equal{} 200,\, b \equal{} 100,\, c \equal{} 1.$
 [/quote]
See http://www.irgoc.org/viewtopic.php?f=10&t=12&sid=17e26ff2c0593b9ed50b2951d78b031f&start=10 Example 121.

[quote="hedeng123"][quote="mai quoc thang"]Let $ a,b,c$ be nonnegative real numbers .

With : $ M \equal{} \frac {37(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2}{2(ab \plus{} bc \plus{} ca)(a \plus{} b \plus{} c)^6}$

Prove that :

$ \frac {1}{a^2 \plus{} 2bc} \plus{} \frac {1}{b^2 \plus{} 2ca} \plus{} \frac {1}{c^2 \plus{} 2ab} \geq \frac {ab \plus{} bc \plus{} ca}{a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2} \plus{} M$[/quote]

$ \frac {1}{a^2 \plus{} 2bc} \plus{} \frac {1}{b^2 \plus{} 2ca} \plus{} \frac {1}{c^2 \plus{} 2ab} \geq \frac {ab \plus{} bc \plus{} ca}{a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2}$
 $ \plus{} \frac {747(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2}{2(ab \plus{} bc \plus{} ca)(a \plus{} b \plus{} c)^6}$

The inequality  is holds[/quote]

sorry

$ \frac {1}{a^2 \plus{} bc} \plus{} \frac {1}{b^2 \plus{} ca} \plus{} \frac {1}{c^2 \plus{} ab} \geq \frac {ab \plus{} bc \plus{} ca}{a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2}$

$ \plus{} \frac {747(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2}{2(ab \plus{} bc \plus{} ca)(a \plus{} b \plus{} c)^6}$

is holds

and

$ \frac {1}{a^2 \plus{} 2bc} \plus{} \frac {1}{b^2 \plus{} 2ca} \plus{} \frac {1}{c^2 \plus{} 2ab} \geq \frac {ab \plus{} bc \plus{} ca}{a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2}$

$ \plus{} \frac {310(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2}{(ab \plus{} bc \plus{} ca)(a \plus{} b \plus{} c)^6}$

 is holds


function
real analysis
real analysis unsolved

I've come across this twice now, and haven't really seen how to deal with it properly. Say we take the subalgebra of functions of finite linear combinations of $ \{x^n : n\geq 2\}\subset C[0,a]$ for some $ a>0$. Is the closure of this (uniform) all the continuous functions on $ [0,a]$?

They way "little Rudin" states it is that we can say the closure is the whole thing if the algebra separates points (which it does) and vanishes nowhere. This set vanishes at $ 0$, though. Is the closure something more along the lines of $ \{f\in C[0,a] : f(0)\equal{}0\}$?

In general, is there a way to work around what to do if we know precisely how it doesn't separate points, or if we know where it vanishes? Another example would be that $ \{x^{2n} : n\geq 1\}$ vanishes at $ 0$ and doesn't separate $ x$ from $ \minus{}x$ if we are working on $ C[\minus{}1,1]$.

[quote="HilbertThm90"]Is the closure something more along the lines of $ \{f\in C[0,a] : f(0) \equal{} 0\}$?[/quote]

yes, c.f. [url=http://en.wikipedia.org/wiki/M%FCntz-Sz%E1sz_theorem]muntz-szasz[/url]

A minor observation (compared to blah's mighty theorem):

In "little Rudin" (I guess that's the Principles book) it is shown how to deal with your first case. $ \{ 1 \} \cup \{ x^n : n \geq 2 \}$ is dense in $ C([0,a])$, so if $ P_n \to f$ uniformly, then we can take $ Q_n(x) \equal{} P_n(x) \minus{} P_n(0)$, which implies $ Q_n \to f$ uniformly (provided that $ f(0) \equal{} 0$).

Oh, right. That makes sense. I knew there was some obvious way to rewrite them to work.

I'd heard of Muntz-Szasz, but just never looked it up.

I must say that I'm fascinated by one consequence of it: $ \{1\}\cup \{x^p : p \ \text{prime}\}$ works.  :)


[youtube]F-QA2rkpBSY[/youtube]

(If you want more, navigate to the youtube link and search for the next vid)

I personally agree with most of his statements.

"The Greatest Shortcoming of the Human Race is our Inability to Understand The Exponential Function"... what a statement.  :lol:


we have constant circle $C$ and line $\Delta$ .
A constant point $P$ lies on circle and $Q$ on line $\Delta$. circle  $C_1$ pass through $P$ & $Q$ and intersect $C$ in $R$ and $\Delta$ in $S$.
prove that $RS$ pass through a constant point.

Let $U$ be the second intersection of $PQ$ with the circle $\mathcal C$, and let $T$ be the second intersection of $RS$ with $\mathcal C$.

$\angle TUP=\pi-\angle TRP=\angle SRP=\pi-\angle SQP$, which shows that $\Delta\|UT$. Since $U$ is a fixed point, it means that $T$ is also fixed.


What is the units digit of $ 19^{19}\plus{}99^{99}$?

this is one of those types of problems with simple patterns.

first lets look at 19^19. it equals 19x19x19x19...
find the units digit of this product. the pattern for the units digit is 9-1-9-1-9... So the 19th units digit is 9.

Now look at 99^99. there is also a pattern here. it's 9-1-9-1-9-1...
The 99th units digit is 9. Now just add the two unit digits, and take the units digit of that sum. The sum is 18, and [b]8[/b] is the answer.


It is easy to note that (a^a)*(b*b)>=(a^b)*(b^a)(to prove it just take a>=b and devide both sides by (a^b)*(b^b)
it will reduce toa^(a-b)>=b^(a-b)By multiplying al inequalities from the above kind for a, b and c we will get the required inequality

$a^{b+c}b^{c+a}c^{a+b}\leq 1$
$\Leftrightarrow a^{a+b+c}b^{a+b+c}c^{a+b+c}\leq a^ab^bc^c$
$\Leftrightarrow a^ab^bc^c\geq 1$
$\Leftrightarrow a\ln a+b\ln b+c\ln c\geq 0$
Consider $f(x)=x\ln x$ where $x$ is positive. $f''(x)=\frac{1}{x}>0$, so $f(x)$ is a convex function. By Jensen's Inequality,
$a\ln a+b\ln b+c\ln c$
$\geq 3(\frac{a+b+c}{3})\ln(\frac{a+b+c}{3})$
$\geq 3\sqrt[3]{abc}\ln(\sqrt[3]{abc})$
$=0$
So the conclusion follows.

one more simple solution:

$a^{b+c}b^{c+a}c^{a+b}= (ab)^{c}(ac)^{b}(bc)^{a}\leq (\frac{a.bc+b.ac+c.ab}{a+b+c})^{a+b+c}\leq 1^{a+b+c}$

Can the Weighted AM GM inequality be used for positive relas?
It can only be used for positive rationals rite???
Please reply.

Thanks in advance.

No, it can be used for positive reals too. It would be very weird if it couldn't... In fact, the exponential function would be not continuous (meaning that small variations, like changing $ 3.1415$ (rational) by $ \pi$ (irrational) would be able to make the inequality becomes invalid).

Was "rite" an attempt to say "right"?

[quote="leepakhin"]$ a^{b \plus{} c}b^{c \plus{} a}c^{a \plus{} b}\leq 1$
$ \Leftrightarrow a^{a \plus{} b \plus{} c}b^{a \plus{} b \plus{} c}c^{a \plus{} b \plus{} c}\leq a^ab^bc^c$
$ \Leftrightarrow a^ab^bc^c\geq 1$
$ \Leftrightarrow a\ln a \plus{} b\ln b \plus{} c\ln c\geq 0$
Consider $ f(x) \equal{} x\ln x$ where $ x$ is positive. $ f''(x) \equal{} \frac {1}{x} > 0$, so $ f(x)$ is a convex function. By Jensen's Inequality,
$ a\ln a \plus{} b\ln b \plus{} c\ln c$
$ \geq 3(\frac {a \plus{} b \plus{} c}{3})\ln(\frac {a \plus{} b \plus{} c}{3})$
$ \geq 3\sqrt [3]{abc}\ln(\sqrt [3]{abc})$
$ \equal{} 0$
So the conclusion follows.[/quote]
$ a^{a} \geq a$, $ b^{b} \geq b$, $ c^{c} \geq c \Rightarrow  a^{a}b^{b}c^{c} \geq abc \equal{} 1$

[hide="Simple Solution"]
$ \& abc = 1  \\ \\
  \& {a^{b + c}}{b^{c + a}}{c^{a + b}} = {\left( {abc} \right)^{a + b + c}}\frac{1}
{{{a^a}{b^b}{c^c}}} = \frac{1}{{a^a}{b^b}{c^c}}$

now we know that if a quantity $ k\ge 1$. then obviously $ k^k \ge k$
and if $ k<1$. then we know that if we raise it to powers greater than 1, its value decreases. we can easily observe that when the exponent is less then 1, its value increases.

$ \therefore a^a\ge a, b^b\ge b, c^c\ge c \\
\implies a^ab^bc^c\ge abc = 1$

$ \implies {a^{b + c}}{b^{c + a}}{c^{a + b}} = {1 \over a^ab^bc^c} \le 1$
[/hide]

Applying the [b]GM-HM[/b] inequality to the positive reals [b]a,b,c [/b]with corresponding weights as [b]a,b,c[/b] respectively,we get
\[(a^ab^bc^c)^{1/(a+b+c)}>(a+b+c)/3>(abc)^{1/3}=1\],
where in the last step we have applied [b]AM-GM[/b] inequality.
Hence we get,
\[a^ab^bc^c>1\]
or,\[(1/bc)^a(1/ca)^b(1/ab)^c>1\]
or\[,a^{b+c}b^{c+a}c^{a+b}<1\]
as required.

[quote=Rushil]If $a,b,c$ are positive real numbers such that $abc= 1$, Prove that \[ a^{b+c} b^{c+a} c^{a+b} \leq 1 . \][/quote]
[rule]
[b]My Solution:[/b]
[rule]
[b][color=#00f]Weighted Power Mean[/color].[/b] Let $a_1, a_2, \cdots a_n$ and $w_1, w_2, \cdots, w_n$ be positive reals with $w_1 + w_2 + \cdots + w_n = 1.$ For any real number $r$, we define
$$
\mathcal{P}(r) =
\begin{cases}
(w_1a_1^r+w_2a_2^r+\cdots+w_na_n^r)^{\frac{1}{r}}&\text{ for }r\ne 0
\\(a_1^{w_1}a_2^{w_2}\cdots a_n^{w_n})&\text{ for }r=0.
\end{cases}
$$
If $r > s$, then $\mathcal{P}(r) \geqslant \mathcal{P}(s)$, equality occurs if and only if $a_1 = a_2 = \cdots = a_n.$
[rule]
By Power Mean with $r = 0, s = -1; w_1 = \tfrac{a}{a+b+c}, w_2 = \tfrac{b}{a+b+c}, w_3 = \tfrac{c}{a+b+c}$ applied to $a,b,c$ respectively, we get that
\begin{align*}
\left(a^ab^bc^c\right)^{\tfrac{1}{a+b+c}} &\geqslant \left[\sum_{\text{cyc}}\left(\frac{a}{a+b+c}\right)\frac 1a\right]^{-1}
\\&= \frac{a+b+c}{3}
\\&\geqslant \sqrt[3]{abc}
\\&=1
\end{align*}
whence $a^ab^bc^c \geqslant 1.$ So,
\begin{align*}
\implies& \left(\frac{1}{bc}\right)^a\left(\frac{1}{ca}\right)^b\left(\frac{1}{ab}\right)^c \geqslant 1
\\\implies& a^{b+c}b^{c+a}c^{a+b} \leqslant 1.\qquad\qquad\blacksquare
\end{align*}
[rule]

[hide=Solution]
We claim that $a^{a-1}\geq1$. If $a\geq1$, then $a-1\geq0$, so $a^{a-1}\geq1$. If $a\leq1$, then $a-1\leq0$, so $a^{a-1}=\left(\frac1a\right)^{1-a}\geq1$. Therefore, $a^{a-1}\geq1$, so $a^a\geq a$. Therefore, $a^ab^bc^c\geq abc=1$, so dividing both sides by $a^{a+b+c}b^{a+b+c}c^{a+b+c}=1$ gives $a^{-b-c}b^{-c-a}c^{-a-b}\geq1$, or $a^{b+c}b^{c+a}c^{a+b}\geq1$.

Wait did I screw up??

[hide=Solution]Multiplying both sides by $a^a \cdot b^b \cdot c^c$ shows that it's equivalent to proving $a^{a+b+c}b^{a+b+c}c^{a+b+c}=(abc)^{a+b+c}=1 \leq a^a b^b c^c$. WLOG $a \leq b \leq c$, so $b^a \leq b^b$ and $c^a \leq c^c$. Then $a^a b^b c^c \geq a^a b^a c^a = (abc)^a=1$, as desired. All steps are reversible, so we're done.[/hide]

[hide=@above]
$b^a\leq b^b$ is not true for $b<1$

Thanks, I’ll fix it when I get home from school.

It suffices to show that $a^ab^bc^c \geq 1$. Taking log of both sides, it suffices to show that $$x \cdot e^x + y \cdot e^y + z \cdot e^z \geq 0$$ where $x = \log a$ and cyclic permutations. But note that $$\sum x \cdot e^x \geq \sum x(x+1) \geq x+y+z = 0,$$ which is enough.

we do a tiny amount of trolling

Note that we have
\[ a^{b+c}b^{c+a}c^{a+b} \leq 1 \iff (abc)^{a+b+c} \leq a^ab^bc^c \iff a^ab^bc^c \ge 1 \]
but we have (by [url=https://aops.com/community/p1808247]USAMO 1974/2[/url]) that
\[ a^ab^bc^c \ge (abc)^{(a+b+c)/3} = 1. \]

Dividing by $(abc)^{a+b+c} = 1$, our inequality can be rewritten as
\[a^{-a} b^{-b} c^{-c} \leq 1 \iff a^a b^b c^c \ge 1 = abc.\]

This inequality follows from noting $a^a \ge a$, or $a^{a-1} \ge 1$, for all positive $a$. $\blacksquare$

WLOG $a\geq b \geq c$
take ln both sides
$(b+c)*\ln(a)+(c+a)*\ln(b) +(a+b)*\ln(c) \leq 0$
By Reversed Chebyshev
$(b+c)*\ln(a)+(c+a)*\ln(b) +(a+b)*\ln(c)\leq \frac{(2a+2b+2c)}{3}*(\ln(abc))\leq 0$

Rewrite this as,
\begin{align*}
a^{a+b+c}b^{a+b+c}c^{a+b+c} &\leq a^ab^bc^c\\
1 &\leq a^ab^bc^c\\
0 &\leq a\ln a + b\ln b + c\ln c
\end{align*}
Then as the function $f(x) = x \ln x$ is convex over $(0, 1)$ we find,
\begin{align*}
f(a) + f(b) + f(c) &\geq 3 f\left( \frac{a+b+c}{3} \right)\\
& \geq  (a+b+c) \ln \left( \frac{a+b+c}{3} \right)\\
&\geq 3\ln \left( \frac{a+b+c}{3} \right)
\end{align*}
Then noting that $\frac{a+b+c}{3} \geq 1$, and that $f$ is strictly increasing from $\left(\frac{1}{e}, \infty \right)$ we find,
\begin{align*}
3 \ln \left( \frac{a+b+c}{3} \right) \geq 3 \ln 1 = 0
\end{align*}

[b]Edit[/b] Oops, I have edited accordingly, simply suffices to note $a$, $b$ and $c$ are positive so we don't need convexity at $x = 0$.

@above.
[quote=Shreyasharma]Then as the function $f(x) = x \ln x$ is convex over $[0, 1]$ we find,
[/quote]
The function is convex over $(0,\infty)$ How one  can tell about the convexity at $x=0$

Rewrite the wanted inequality as $a^ab^bc^c\ge1$ and now we want $a \log a+b\log b+c\log c\ge 0$ for $\log a+\log b+\log c=0.$ Now since $a,b,c$ and $\log a,\log b,\log c$ are similarly sorted, by Chebyshev's inequality we have $a \log a+b\log b+c\log c\ge\frac{a+b+c}3(\log a+\log b+\log c)=0,$ done.

Funny, \[ a^{b+c}b^{c+a}c^{a+b} \leq 1 \Longleftrightarrow abc = (abc)^{a+b+c} \leq a^ab^bc^c \] We now claim that $x^x \geq x$ for all $x>0$. Indeed, note that if $f(x)=x^x$ then we have \[ f'(x) = (\ln(x)+1)x^x \implies f'(1)=1\] \[ f''(x) = (\ln(x)+1)^2x^x+x^{x-1} > 0 \] for all $x>0$. Thus, $f$ is convex up everywhere, implying that $f(x) \geq f'(1)(x-1)+1=x$. Thus, \[ a^ab^bc^c \geq abc \] as desired.

Multiplying by $a^ab^bc^c$ means we want to show $1\le a^ab^bc^c$. We take a logarithm to get that we want to show $0\le a\ln a+b\ln b+c\ln c$ where $\ln a+\ln b+\ln c=0$. Letting $a=e^r$, $b=e^s$, and $c=e^t$ means that we want to show $re^r+se^s+te^t\ge 0$ where $r+s+t=0$. Let $f(x)=xe^x$. Then $f'(x)=(1+x)e^x$. Therefore, $f(0)=0$ and $f'(0)=1$ so the tangent line to $f(x)$ at $x=0$ is $y=x$. Because $3f(0)=0$, we simply need to show that $xe^x\ge x$. This is true because $e^x< 1$ for $x< 0$ and $e^x> 1$ for $x> 0$ and $0\ge 0$, as desired.

My overly complicated solution:
Observe that $a^{b+c} b^{c+a} c^{a+b} = \prod_{cyc} \left(\frac{1}{a}\right)^a \le \left(\frac{3}{a+b+c}\right)^{a+b+c}$ (Weighted AM GM)
$\le \left(\frac{3}{a+b+c}\right)^{\frac{1}{3}}$ [AM-GM on the exponent and noticing that $a+b+c \ge 3$ from AM-GM]
$\le 1^{\frac{1}{3}}$ [AM-GM]
 $= 1$ $\square$


Use AP (addition principle) and MP (multiplication principle) to slove the following problem:
Let $x=${${1,2,3,...100}$} and let
$S=${$(a,b,c)|a,b,c\in X, a<b, a<c$}. (1)
Find $|S|$.
If someone has another way of doing this problem, please use it. Also, can someone please explain in detail what (1) means? Thanks in advance.

I believe the problem is asking for the number of ordered triples $(a,b,c)$ for which $a,b,c$ are all positive integers less than or equal to 100, and where $a<b$ and $a<c$.

[hide]For each value of $a$, there are $100-a$ choices for each of $b$ and $c$.  The answer is $\sum_{a=1}^{100}(100-a)^{2}=\sum_{k=0}^{99}k^{2}=\frac{99\cdot 100\cdot 199}{6}$.[/hide]

[hide="Solution"]For every chosen $a$, we can pick $b$ and $c$ among $100-a$ numbers which are greater than $a$. Since the ordered triples are asked for, and there's no condition on $b<c, b=c$ or $b>c$, we assume the most general case where any of the $100-a$ numbers can be put instead of $b$ and $c$, repeating allowed. That gives $(100-a)^{2}$ possibilities. Since $a$ can't exceed $99$, we get

$|S|=\sum_{a=1}^{99}(100-a)^{2}=\sum_{n=1}^{99}n^{2}=\frac{99\cdot 100\cdot 199}{6}=328350$[/hide]

EDIT: Beaten by a few seconds.


superior algebra
superior algebra solved

(Hunedoara is a Romanian county, or district, btw)

It's really easy, but it took me more than half an hour to solve it:

Let "*" be a binary operatiopn defined on (0,00), s.t. if x<y then x<x*y<y. Show that the operation isn't associative.

Suppose '*' associative . If we compare x and x*x , we obtain :
x < x*x ===> x < x*x*x < x*x  ===> x < x*x*x*x < x*x*x < x*x*x*x*x < x*x ===> x < x*x*x*x*x < x*x*x*x < x*x*x < x*x*x*x*x < x*x   (contradiction)
x*x < x ===> x*x < x*x*x < x  ===> x*x  < x*x*x*x*x < x*x*x < x*x*x*x < x===> x*x < x*x*x*x*x < x*x*x < x*x*x*x < x*x*x*x*x < x   (contradiction)
So x = x*x , for all x . And now : x < y ===> x < x*y < y  ===> x < x*x*y < x*y  < y ===> x < x*y < x*y < y (contradiction)
Manifestly '*' is a non associative operation.


I did some research. UT Dallas indeed has a bookstore. Its website can be found at [url]utdallas.bookstore.com[/url] Scroll down, on the left, there's a little blurb about HP7. They have a party starting at 10 PM on Friday, the day before the book is released, but because of the curfew and stuff I doubt we can get it then. But they do hold the books until 7/23 if you reserve them, so we can probably get them Saturday afternoon after class. However, according to the following campus map: [url]http://www.utdallas.edu/campusmap.html[/url], Waterview Park, where we are staying, is somewhat of a long walk to the bookstore (Waterview is in the top left corner if you can't find it), so we can probably all go as a big group.

So yeah I hope that was helpful.

Also the list price is 35 dollars, since we get 40 percent off by reserving, we get it for 21 dollars (20.99).

NOTE: Mi parentes have un-nixed the plan. CONTACT ME BEFORE SATURDAY and I'll have the book to you as soon as the parents deliver them - hopefully about 1 a.m.
Unless there's curfew problems, where I'll have them delivered at breakfast or something similar
Cost is 23 dollars to cover the tax.

so are you just going to sell them over there? because there might be other people not from aops who want it too.

unfortunately for them, I can't place advance orders and get this good a price if I wait until then.


Suppose that $ \frac{2}{3}$ of $ 10$ bananas are worth as much as $ 8$ oranges. How many oranges are worth as much is $ \frac{1}{2}$ of $ 5$ bananas?

$ \textbf{(A)}\ 2 \qquad
\textbf{(B)}\ \frac{5}{2} \qquad
\textbf{(C)}\ 3 \qquad
\textbf{(D)}\ \frac{7}{2} \qquad
\textbf{(E)}\ 4$

[quote="worthawholebean"]Suppose that $ \frac {2}{3}$ of $ 10$ bananas are worth as much as $ 8$ oranges. How many oranges are worth as much is $ \frac {1}{2}$ of $ 5$ bananas?

$ \textbf{(A)}\ 2 \qquad \textbf{(B)}\ \frac {5}{2} \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ \frac {7}{2} \qquad \textbf{(E)}\ 4$[/quote]

just set up as a fraction and you get 1 banana= 24/20 oranges then 5/2 bananas= 6/5 oranges times 5/2= 3 oranges

[hide="Solution"]Let $ b$ be the price of a banana, and $ o$ be the price of an orange. $ \frac{20}{3}b\equal{}8o$, or, multiplying each side by $ \frac{3}{8}$, $ \frac{5}{2}b\equal{}3o$, so the answer is $ \boxed{\textbf{C}}$.[/hide]


trigonometry
geometry
area of a triangle

Two intersecting circles C1 and C2 have a common tangent which touches C1 at P and C2 at Q. The two circles intersect at M and N, where N is nearer to PQ than M is. Prove that the triangles MNP and MNQ have equal areas.

hmm. extend $MN$ to $R$, where $R$ lies on $PQ$.
thus $PR=QR$ and $\sin PRN = \sin QRN$.

Denote the area of triangle $ABC$ by $ [ ABC ] $ .
$ [ PNM ] =  [PRM] - [PRN] = PR\cdot PM\cdot \sin PRN - PR\cdot PN\cdot \sin PRN = PQ\cdot PM\cdot \sin QRN - PQ\cdot PN\cdot \sin QRN = [QRM] -[QRN] = [MNQ]$.

Liyi, your solution can be simplified a little bit  :D

[quote="liyi"]hmm. extend $MN$ to $R$, where $R$ lies on $PQ$.
thus $PR=QR$[/quote]

(This is because $PR^2 = RN \cdot RM$ and $QR^2 = RN \cdot RM$, so $PR^2 = QR^2$ and PR = QR.)

Now, if P' and Q' are the orthogonal projections of the points P and Q on the line MN, then the triangles PRP' and QRQ' are congruent (because of PR = QR, < PRP' = < QRQ' and < PP'R = < QQ'R = 90), so PP' = QQ'. But the segments PP' and QQ' are the altitudes to the side MN in the triangles MNP and MNQ. Hence, in the triangles MNP and MNQ, the altitudes to the side MN are equal; and since the area of a triangle equals $\frac 1/2 \cdot$ side $\cdot$ corresponding altitude, it follows that the areas of triangles MNP and MNQ are equal.

  Darij

Well maybe its extremely easy but why PR 2 =RN*RM and QR 2 =RN*RM.It seems like there is one of the Euclids theorems applied but I dont see any right triangle and any altitude.

$\Gamma$ is a circle. $P$ lies outside it. $PT$ is a tangent of $\Gamma$ and the line $PMN$ intersects $\Gamma$ at $M$ and $N$.
[Theorem] $PT^2$ = $PM\cdot PN$.

Join $TM$ and $TN$. Suppose $M$ is nearer to $P$. we have $\angle PTM = \angle TNM$. and $\angle TPN$ is common, we have $\triangle PTM\sim \triangle PNT$, so $PT/PM = PN/PT$.


trigonometry
inequalities
inequalities proposed

Given a,b,c $ \ge 0$ and ab+bc+ca $ \equal{}1$ . Prove:
     $ \frac{{\left( {a^2  \plus{} bc \plus{} 1} \right)\left( {b^2  \plus{} ca \plus{} 1} \right)\left( {c^2  \plus{} ab \plus{} 1} \right)}}{{\left( {a^2  \plus{} 1} \right)\left( {b^2  \plus{} 1} \right)\left( {c^2  \plus{} 1} \right)}} \ge \frac{{125}}{{64}}$

[quote="leedt26"]Given a,b,c $ \ge 0$ and ab+bc+ca $ \equal{} 1$ . Prove:
     $ \frac {{\left( {a^2 \plus{} bc \plus{} 1} \right)\left( {b^2 \plus{} ca \plus{} 1} \right)\left( {c^2 \plus{} ab \plus{} 1} \right)}}{{\left( {a^2 \plus{} 1} \right)\left( {b^2 \plus{} 1} \right)\left( {c^2 \plus{} 1} \right)}} \ge \frac {{125}}{{64}}$[/quote]
I think your posting in this forum is [i]inappropriate[/i], please mods move this to the Inequalities [i]Unsolved [/i] Problems.... :oops:

[quote="leedt26"][quote="quykhtn-qa1"][quote="leedt26"]Given a,b,c $ \ge 0$ and ab+bc+ca $ \equal{} 1$ . Prove:
     $ \frac {{\left( {a^2 \plus{} bc \plus{} 1} \right)\left( {b^2 \plus{} ca \plus{} 1} \right)\left( {c^2 \plus{} ab \plus{} 1} \right)}}{{\left( {a^2 \plus{} 1} \right)\left( {b^2 \plus{} 1} \right)\left( {c^2 \plus{} 1} \right)}} \ge \frac {{125}}{{64}}$[/quote]
This inequality by my friend,vuonga2khtn :wink:
See it here: http://www.mathlinks.ro/Forum/viewtopic.php?t=266058[/quote]
can u post your solution by trig , quykhtn-qa1.[/quote]
If you want I can speak,
First,prove that:$ (1 \plus{} cos^2A)(1 \plus{} cos^2B)(1 \plus{} cos^2C) \ge \frac {125}{64}$
Second ,put $ \frac {bc}{(a \plus{} b)(a \plus{} c)} \equal{} x$,.....

[quote="quykhtn-qa1"][/quote][quote="leedt26"][quote="quykhtn-qa1"][quote="leedt26"]Given a,b,c $ \ge 0$ and ab+bc+ca $ = 1$ . Prove:
     $ \frac {{\left( {a^2 + bc + 1} \right)\left( {b^2 + ca + 1} \right)\left( {c^2 + ab + 1} \right)}}{{\left( {a^2 + 1} \right)\left( {b^2 + 1} \right)\left( {c^2 + 1} \right)}} \ge \frac {{125}}{{64}}$[/quote]
This inequality by my friend,vuonga2khtn :wink:
See it here: http://www.mathlinks.ro/Forum/viewtopic.php?t=266058[/quote]
can u post your solution by trig , quykhtn-qa1.[/quote][quote="quykhtn-qa1"]
If you want I can speak,
First,prove that:$ (1 + cos^2A)(1 + cos^2B)(1 + cos^2C) \ge \frac {125}{64}$
Second ,put $ \frac {bc}{(a + b)(a + c)} = x$,.....[/quote]
yes, it's nice proof . thanhks quykhtn-qa1


I dont know how to use code :S

Radical(6-x) -2 /Radical(3-x) -1

[quote="hunter34"]Is it:
$ \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$?[/quote]

yeah thats the question

Ok.

[hide="Hint"]
Multiply by the conjugate of each. So if you have $ \frac{\sqrt{a}-b}{\sqrt{c}-d}$ you multiply it by $ \frac{\sqrt{a}+b}{\sqrt{c}+d}$
[/hide]

[hide="Solution"]
\begin{eqnarray*}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}&=&\\ \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}\cdot\frac{\sqrt{6-x}+2}{\sqrt{3-x}+1}&=&\frac{x-2}{x-2}\\ \frac{x-2}{x-2}&=&\boxed{1}\end{eqnarray*}
[/hide]

hunter34 that is not called an inverse, it is a conjugate.

[quote="hunter34"]That is what I did.[/quote]

No, you multiplied by $ \frac{\sqrt{6-x}+2}{\sqrt{3-x}+1}$, which is not always 1, as stated by krsattack.

[quote="krsattack"]Your answer is wrong.  Plug in any value of x, because 1 is never a sol'n no matter what value you plug in.

New sol'n...[hide]To rationalize the denominator of $ \frac{\sqrt{a}-b}{\sqrt{c}-d}$, you DO NOT multiply by $ \frac{\sqrt{a}+b}{\sqrt{c}+d}$, which doesn't equal 1.  Instead, multiply by: $ \frac{\sqrt{c}+d}{\sqrt{c}+d}$, which does equal 1.

Now, we have:
$ \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}\cdot\frac{\sqrt{6-x}+2}{\sqrt{3-x}+1}$
$ =\frac{(\sqrt{6-x}-2)(\sqrt{3-x}+1)}{2-x}$

We can multiply out the top, but it will not help.  The fraction is irreducible and now has a rational denominator.[/hide][/quote]

If you plug in 2 you get 0/0 which shows that it can be reduced... right?

[quote="b-flat"][quote="brocool"]If you plug in 2 you get 0/0 which shows that it can be reduced... right?[/quote]

No, because $ \frac00$ is undefined.[/quote]

So how do you find the limit of the function at x=2?

Rerrange the expression so you do not get the indeterminate form, once you do that just plug in $ 2$.

Or L'Hopital's Rule. I think the answer is $ \frac{1}{2}$/


calculus
derivative
logarithms
calculus computations

I was wondering how you could find the critical point of $ f(x) = x^x $ I can estimate it by looking at the graph but is there a good way to calculate the critical point?

You have to calculate the derivative, of course.

$f(x)=x^x$

$\ln(f(x))=x\ln x$

$\frac{f'(x)}{f(x)}=1\cdot\ln x+x\cdot\frac1x=\ln x+1$

$f'(x)=x^x(\ln x+1).$

You can finish it from here.


How many 3-digit numbers are there such that the units digit is larger than the tens digit?

Units digit 1: 10 ways

Units digit 2: 20 ways

Continuing, we have $ 10\plus{}20\plus{}30\plus{}\ldots\plus{}90\equal{}\boxed{450}$.

This is incorrect. It doesn't take into account when they are equal.
Another way to think about this:
1. Both the tens and the ones digit can range from 0 to 9 (10 choices).
2. By symmetry, when the tens and ones digit are not equal, 1/2 of the arrangements will satisfy the given constraints.
3. 1 out of every 10 combinations has the same units and tens digit.
900 3 digit numbers. Answer is thus:
$\frac{900}{2} * \frac{9}{10} = 405$


calculus integration function inequalities trigonometry geometry limit

Let $ f: [0,\frac{\pi}{2}]\to R$ such that:
$ f(x)>0. \int_{0}^{\pi} 2f(x)dx <1$.
Prove that the equation $ f(x)\equal{}sinx$ has at least a solution in the inteval $ (0,\frac{\pi}{2})$

[quote="Moubinool"][quote="NguyenDungTN"] $ \int_{0}^{\pi} 2f(x)dx < 1$.
[/quote]

Could you check the interval of the integral ?[/quote]

Sorry, $ f(x)$ is a continuous  function in interval $ [0, \frac{\pi}{2}]$
and $ \int_{0}^{\frac{\pi}{2}} 2f(x)dx <1$

Can anyone help me?

The 2 in the integral is sort of confusing me, but here goes. The fact that $ f(x) > 0$ and the inequality is strict means that $ f(0) > \sin(0) \equal{} 0$. Now, suppose that the claim were not true, that there is no solution to $ f(x) \equal{} \sin(x)$ in $ (0, \pi/2)$. Since $ f$ is continuous, this must mean that $ f(x) > \sin(x)$ for all $ x$ in $ [0, \pi/2)$. But, \[ \int^{\pi/2}_0f(x)\mathrm{d}x < 1 \equal{} \int^{\pi/2}_0\sin(x)\mathrm{d}x\] from which it's easy to get a contradiction. I think this is right but that 2 in the statement of the problem is a nagging annoyance because it's never used in this solution, other than the fact that the integral of $ 2f(x)$ is greater than the integral of $ f(x)$.

[quote="pdt"]well we dont know if f(x) is continous or not[/quote]

In Nguyen's post before my last one he said that $ f(x)$ is continuous.

[quote="hadjimurad"]Since $ f$ is continuous, this must mean that $ f(x) > \sin(x)$ for all $ x$ in $ [0, \pi/2)$.[/quote]

Why? What about the case $ f(x) < \sin x, \forall x \in \left(0,\frac {\pi}{2}\right)$?  :(

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The way I see it, we have three cases:

1)If $ f(x) > \sin x, \forall x \in \left(0,\frac {\pi}{2}\right)$
Then  $ 1 \equal{} \int_{0}^{\frac {\pi}{2}} \sin xdx < \int_{0}^{\frac {\pi}{2}} f(x)dx < \frac12$, which clearly is a contradiction.

2)If $ f(x) < \sin x, \forall x \in \left(0,\frac {\pi}{2}\right)$
I cant prove this is not possible. :(
Intuitively I think this is possible, since the area must be less than $ \frac12$.There can be a curve which is entirely below $ \sin x$ and have area less than $ \frac12$. I could be wrong though :blush:

3) If neither of the above applies then $ \exists a,b \in \left(0,\frac {\pi}{2}\right): f(a) > \sin a$ and $ f(b) < \sin b$. Intermediate value theorem to $ g(x) \equal{} f(x) \minus{} \sin x$ wraps it up.

[quote="isomorphism"]Why? What about the case $ f(x) < \sin x, \forall x \in \left(0,\frac {\pi}{2}\right)$?  :( [/quote]

We know that $ f$ is continuous on $ [0, \pi/2]$, so if, for all $ x$ in $ (0, \pi/2)$, $ f(x) < \sin(x)$, then $ \lim_{x \to 0} f(x) \equal{} 0$. But, $ f(0) > 0$. So this case is impossible, because we're given that $ f(0) > 0$.


geometry
3D geometry
number theory
prime factorization

Ok so here is the problem:
find the smallest positive integer $ \ n$ where $ \ 2n$ is a perfect square and $ \ 3n$ is an perfect cube.
Thanks!

Well, for the 3n term, we HAVE to have two 3's at least. otherwise, it wont be a cube.

Similarly, for the 2n term, we HAVE to have one 2 at least, or else it wont be square.

But then if we have only one 2, it wont be a cube (3x3x3x2 is not a cube). So we need two more 2's

therefore the answer is 2x2x2x3x3=72, which is what athunder got i think.

Hey guys,
I'm kinda new to a lot of arbitrary mathematics not taught in schools, and even this confused me. Could you explain your reasoning for why a 2 and two 3s were required? Also, if there's a general forum page that explains the basics of these level problems, that would be great.

Thanks!

ok, uhhh i'll explain it.

$ 2n$ = perfect square
$ 3n$ = perfect cube

since we want the smallest number that fits this then the
prime factorization of the number will consist of only $ 2's$ and $ 3's$.

$ n$ has at;east one $ 2$ and one $ 3$.

so we multiply to the factorization three $ 2's$ and two $ 3's$.
The reason is that $ n \equal{} 2\cdot2\cdot2\cdot3\cdot3$
we need another $ 2$ to make it a perfect square, and guess what there is
a $ 2$ in $ 2n$ so its $ 2n \equal{} 2\cdot2\cdot2\cdot2\cdot3\cdot3 \equal{} \text{perfect square}$

same thing applies to $ 3n$.
 :lol:

2n=y^2
3n=z^3
so there should be one 2 and two 3's i think
and one 2 isnt a cube so you find the exponent
make the exponent of 2, x
make the exponent of 3, z
x/3=y and z/2=a
so the smallest that makes a integer is the ordered pair, 3,2
so you get 2*2*2*3*3 which is [b]72[/b]

Thanks guys, i learned, heh i cuda guessed and checked lol.


geometry
circumcircle
parallelogram
trigonometry
search
angle bisector
trig identities

If triangle ABC is a right triangle such that $C=90^{o}$, $BC=4$, and the length of the circumradius is 5, find the length of the bisector of B.

[quote="xenon_nightmare152"][hide="Dont trust this answer"]I got ~ 4.1833[/hide][/quote]
I m getting
[hide="answer"]
4.76
[/hide]
Is that right???

no need for that,see the diagram,
using trig you can find $angleB$ so then $1/2$ of it then you can find the lenght of the bisector.

[quote="WarpedKlown1335"]$\sqrt{37}$

[hide]Create a parallelogram ABCD and use 2(a^2 + b^2) = (d_1 + d_2)^2 where a and b are the lengths of two adjacent sides and d_1 and d_2 are the two diagonals.[/hide][/quote]

Are you sure you're not talking about the median? Anyway, $\sqrt{37}$ is not correct. 4.76 is also incorrect. The answer needs to be in terms of radicals (if it's not rational). This problem is from a book and I posted it because my answer did not match the one in the back, although I checked my solution several times and found no mistakes.

Very hard!!!
OK $CA^2=AB^2-BC^2=10^2-4^2=84$. Angle bisector means $\frac{BC}{CD}=\frac{BA}{DA}$, $DA=CD\frac{BA}{BC}$.

Put $CD+DA=CA$, $CD+CD\frac{BA}{BC}=CA$, $CD=\frac{CA}{1+\frac{BA}{BC}}=\frac{CA\times BC}{AB+BC}$

Put $BD^2=BC^2+CD^2=BC^2+\left(\frac{CA\times BC}{AB+BC}\right)^2$

Very beautiful $BD^2=\left(\frac{BC}{AB+BC}\right)^2(AB^2+BC^2+CA^2+2\cdot AB\times BC)$

$BD=\frac{BC}{AB+BC}\sqrt{AB^2+BC^2+CA^2+2\cdot AB\times BC}=\frac{BC}{AB+BC}\sqrt{2\cdot AB^2+2\cdot AB\times BC}$

Put in numbers and $BD=\frac{4}{14}\sqrt{280}=\frac{4}{7}\sqrt{70}$

Sorry I forget triangle  :blush:

[quote="deej21"]I also thought it meant side bisector and got $\sqrt{37}$, but did it for angle bisector also.

10000th user, that is a smart solution.  But there is a pretty easy one involving trig.

Hopefully we all notice that $c=10$.
If we draw the triangle and the angle bisector of B, we note that $\cos B = \frac 2 5$, and $\cos \frac B 2 = \frac 4 x$ where $x$ is the desired length.  We use the half-angle formula $\cos \frac \theta 2 = \sqrt \frac{1 + \cos \theta} 2$ (if you haven't seen this, this will be negative if theta is in quadrants 2 or 3, but it is in quadrant 1 here because it's a right triangle).

So, plugging in $\cos B$, we get $\frac 4 x = \frac {\sqrt{70}} {10}$ (simplifying the radical).  Solving, $x = \frac{4\sqrt{70}} 7$.  Nice problem.[/quote]
Sorry but I dont like trigonometer  :( . $\text{Trig}=\text{no fun}$

[quote="WarpedKlown1335"]OH my bad, you were looking for the angle bisector.

If $BD$ is the angle bisector of $\angle{B}$, $BD = \sqrt{(10)(4)(1 - \frac{84}{196})} = \frac{4\sqrt{70}}{7}$.  There's a theorem which gives you the length and it's derived from Stewart's and the angle bisector theorem.[/quote]

Yeh yeh, bisector formula say if triangle $ABC$ with sides $a,b,c$ and bisect $\angle C$ cutting side $c=p+q$, bisector length $t^2=ab-pq$  :trampoline:

The proof for Stewart's Theorem had trig in it (law of cosines if I remember correctly) and if anyone can prove it with no trig whatsoever, I'd be really happy.  But 10000th user, that solution's really really nice, I forgot to say that.

Thank you. I find by Search and is this proof good for you?
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=126152#p126152]http://www.mathlinks.ro/Forum/viewtopic.php?p=126152#p126152[/url]
Im sorry WarpedKlown1335, I have another URL you can read for proof
[url=http://www.cut-the-knot.com/pythagoras/corollary.shtml]http://www.cut-the-knot.com/pythagoras/corollary.shtml[/url]

Thanks for the solutions! I used trig (much easier that way, although the non-trig solution is quite impressive) and also got $\frac{4\sqrt{70}}{7}$, which means that the book was wrong.

Well that's the elegant proof for the angle bisector formula =), but I'm looking for a pure geometric proof of Stewart's (if there are any out there...).

*EDIT* NVM CUT-THE-KNOT PWNS *EDIT*

[quote="riddler"]But guys,
$2*5= Hypotenuse$ So that is that. Then the answer comes 4.76.[/quote]
even if rounded to 2 decimal places, the answer is still wrong

I think we all established the fact that the hypotenuse is $10$.  I don't get what you are saying, riddler.

If you get that $h=10$ then by trig you can get the angle between the $2$ sides given, after that you can find the bisector by trig as you know $1/2$ the angle you have just got.
Simple but then it comes as $4.76$. :?

All right.

You can use your calculator, using arccos and dividing and cos again, and get a decimal answer which is reasonably close to the real one, which is what you did.  But that is not how it works in math competitions and olympiads.  You must solve it without a calculator.  Doing trig without a calculator requires knowledge of lots of formulas (or at least how to derive formulas so you can make them up when necessary).  I gave a solution above involving the same concepts as yours, but using one formula that let me do it without a calculator, and get the exact, not close, answer.


I have worked on this problems for a while but haven't made any progress. So, I'll post to have some hints about its solution if anyone can solve it. By the way, this problem is from KOMAL Contest February, so it ended already!
Given $ n$ different subsets of the set of numbers $ 1,2,3,...,n$, prove that there exists a number, such that if it is left out of each subset, the remaining subsets will still be different.

[hide="solution"]Construct the graph with the subsets as vertices, and 2 connected with a numbered edge if the removal of that number would make those 2 subsets equal.

If the problem statement has a set of subsets which contradict it, then it's graph would have at least 1 edge labeld one, 1 edge labeld two, .... and 1 edge labeled n. Consider just these n edges. Since we have n edges on n vertices, we must have a cycle.

Let one of the numbers on an edge in this cycle be $ i$. Then the 2 subsets on either side of $ i$ on the cycle have one containing $ i$, and one not containing $ i$. Traversing the cycle the other way than crossing the edge $ i$ from the subset containing $ i$ to the one not containing $ i$, we see that we cannot lose the element $ i$ in our path of subsets, forcing the one not containing $ i$ to contain $ i$, contradiction![/hide]

[hide="High tech"]Let the $ n$ sets be $ S_1, S_2, \ldots S_n$. Let $ F^n_2$ be the field with $ 2$ elements, and consider the vector space $ F^n_2$, and let $ v_i$ be an element such that it's $ k$th component is $ 1$ iff $ k \in S_i$. Then $ v_i \plus{} v_j$ has $ 1$ in component $ k$ where exactly one of $ S_i, S_j$ contain $ k$. This means that if $ v_i \plus{} v_j$ has only one $ 1$ in it, deleting the element corresponding to where that $ 1$ is gives us a problem. Note that $ v_i \plus{} v_j$ is never the zero vector since no two of the sets are equal.

Consider the subspace $ V \equal{} \text{Span}(v_1 \plus{} v_2, v_2 \plus{} v_3, \ldots v_{n\minus{}1} \plus{} v_n)$. It's relatively easy to show this has all elements of the form $ v_i \plus{} v_j$ by taking partial sums. If $ (1,0,\ldots,0), (0,1,\ldots,0), \ldots (0,0,\ldots,1)$ are all in $ V$, then we have a problem since we can't delete any element to get distinct sets. But $ V$ is the span of $ n\minus{}1$ vectors and at most $ n\minus{}1$-dimensional, and if all those vectors were in $ V$, $ V$ would have $ n$ linearly independent vectors and have dimension at least $ n$. Contradiction.

Thus at least one of $ (1,0,\ldots,0), (0,1,\ldots,0), \ldots (0,0,\ldots,1)$ is not in $ V$. Delete whichever element corresponds to the vector that isn't.[/hide]


If   a>0 proved that the series (alternating series)
  a-a^1/2+a^1/3-a^1/4+...  devergent

but!!!   the series below converge!
     (a-a^1/2)+(a^1/3-a^1/4)+...

  a-(a^1/2-a^1/3)-(a^1/4-a^1/5)-...

PLEASE HELP ME!

A series $ s_n \equal{} \sum_{k \equal{} 1}^{n} a_n$ converges to a limit $ L$ when for any $ \epsilon > 0$ there exists an $ M$ such that for [b]any[/b] $ m > M$ we have

$ | s_m \minus{} L| < \epsilon$

The first series is divergent because $ \lim_{n \to \infty} |a_n| \equal{} 1$, so the series alternates between two different values and does not settle on a unique limit $ L$.

The other series converge because their terms are defined to converge to $ 0$ rather than to alternate between $ 1, \minus{} 1$; in particular,

$ a^{\frac{1}{2k \minus{} 1} } \minus{} a^{\frac{1}{2k} } \equal{} a^{\frac{1}{2k}} \left( a^{\frac{1}{2k(2k \minus{} 1)} } \minus{} 1 \right) \approx \frac{\ln a}{2k(2k \minus{} 1)}$

where we have used the approximation $ a^x \equal{} e^{x \ln a} \approx 1 \plus{} x \ln a$.  This is not entirely rigorous, but can be made so.

One way to think about it is that the sequence of partial sums $ s_n$ has the property that $ \lim_{n \to \infty} s_n$ does not exist, but

$ \lim_{n \to \infty} s_{2n} \equal{} L_1$
$ \lim_{n \to \infty} s_{2n \plus{} 1} \equal{} L_2$

both exist, and are [b]different.[/b]

Thank you  for your help  t0rajir0u,
I  think that I have to read more!
Thanks a lot again!


given quadrilateral ABCD inscribed in a circle, and point X which lies inside circle. Lines AX and DX intersects this circle at points M and N respectively. E and F are intersection points of lines AM and CD, DN and AB respectively. Prove that lines MN, EF, BC intersect at one point.

 :coool:

It is easy if we consider the non convex inscribed hexagon $ ABCDNM.$

So, by [b][size=100]Pascal theorem[/size][/b], we conclude that the points $ F\equiv AB\cap DN$ and $ Q\equiv BC\cap NM$ and $ E\equiv AM\cap CD,$ are collinear and the proof is completed.

Kostas Vittas.


please try to solve this good problem as I am not getting it.

$ I$ is a form I used to encounter for the set of integers.

[i][b]you should try this.
Its a very good problem given to me by my elder brother who is now taken up maths in college.[/b][/i]

Irrespective of how  good this problem is you still shouldn't be posting it on AoPS as it is a problem on the Indian postal problem sets. So please do not post solutions for this. sigh... I hoped we wouldn't have the same problem this year.. apparently not :)


Find all $ n$ such that $ 2n\plus{}5$ is a perfect square.

Is $ n \in \mathbb{Z}^{ \plus{} }$ or $ \mathbb{Z}$ or what?

[hide="Maybe"]

Note we can pick $ q \in \mathbb{Z}$ and generate $ n \equal{} 2(q(q \plus{} 1) \minus{} 1)$ and $ k \equal{} 2q \plus{} 1$ which indeed give infinite solutions. Where $ 2n\plus{}5 \equal{} k^{2}$

[/hide]


trigonometry MATHCOUNTS function calculus algebra polynomial search

I just bought a TI-84 plus graphing calc, so was it a good buy? Some of my friends are saying that the TI-89 is better, but I think it's way to expensive.

TI-89 is like a god. Imo, TI-89 would be the better buy because you can do so much more with it, but if you really can't put in the money required (even though you can use the TI-89 forever and ever and ever), then the TI-84 isn't so bad.

IMHO, I don't understand why one even needs the TI-84. Save yourself a 100 dollars and get a normal instead! If you need trig, get a scientific calculator! I mean, all they can do, is well, graph which isn't even that important on competitions.

Right now, I don't even have a basic calculator. And starting tomorrow, I need to start preparing for MATHCOUNTS and AMC.

[quote="pytheagle"]IMHO, I don't understand why one even needs the TI-84. Save yourself a 100 dollars and get a normal instead! If you need trig, get a scientific calculator! I mean, all they can do, is well, graph which isn't even that important on competitions.
[/quote]

Um, you're kidding right? Graphing calculators do so much more than that. You'll never survive in college and the real world without one.

Take for example that state target problem from this year...the one asking you to do 7^47-7^46 and find something...or something like that.

Just stick it into your TI-89 and it chugs the numbers for you.  With the solve-equation function, you can find the answer.

That's [i]nothing.[/i] You can do limits and all sorts of calculus stuff. Plug in any polynomial and it will spit out the roots. Write programs to do anything you need. 3D graphing? No problem! Factoring ugly polynomials? Easy as 3.14159....

[quote="AIME15"]Take for example that state target problem from this year...the one asking you to do 7^47-7^46 and find something...or something like that.
[quote]

That problem?  I just did that on my scientific calculator, wrote down the weird scientific notation thingy, guesstimated that the power of 7 would be more than 40, and found the answer(46) in about 30 seconds.  Unfortunately, I still only got a 5 on State target(of course, that would be a success if you heard my sprint score at state :blush: )

I actually used both the TI-83 my teacher provided and my own, and graphing calculators would help you a lot.  I don't have one yet though, but I guess my TI-30XA is ok for now.

[quote="pytheagle"] But anyway, it's still too expensive.[/quote]
[url]http://www.amazon.com/s/ref=nb_ss?url=search-alias%3Daps&field-keywords=TI-81[/url]

There you can get a 45 dollar TI-81.  I think it's pretty old though....

Or, for 135 bucks, here's the N-Spire......I bet it's good
[url]http://www.amazon.com/s/ref=nb_ss_0_4?url=search-alias%3Daps&field-keywords=ti+enspire&sprefix=TI+E[/url]

Here's the TI-30XA.  But you can get them for like 8.99 at Office Depot(I got one for science for my 7th grade year last week)
[url]http://www.amazon.com/s/ref=nb_ss?url=search-alias%3Daps&field-keywords=TI-30xa[/url]

Maybe I'll get a graphing calc next year............

[quote="pytheagle"]Well maybe the reason why I didn't see the importance of graphing calculators was because no one ever taught me how to use one. But anyway, it's still too expensive.[/quote]

If you've never used one, what basis do you have for determining whether it's a good deal or not?

Graphing calculators will serve you through university, especially if you major in math. Even in my high school math classes we need to use one.

Wow...another scientific vs graphing calc debate.
Okay, a TI-84 will be enough to get you through high school AP's but with a TI-89 you can do 3D graphing, evaluate differentials with just variables, no values, also, it has C programming capability instead of the uber fail BASIC interpreter. BUT, a scientific calc is really easy to use. I don't like the TI scientific calcs so I got the one that my brother has had since like 6 years, the Casio fx-115mx. Its really good, can solve equations, complex variables, pretty good integration. Basically everything but graphing.

[quote="pytheagle"]I HAVE used one. Several (like 100) times actually. It was the TI-84 Plus.

But the only thing I used it for was basic operations and it wasn't my own; it was my school's. And since I moved school's, I don't have access to graphing calculators anymore.[/quote]

Precisely my point.  :) (And the TI-84+ sucks, just saying.)

Or it could get even cheaper if you're willing to buy used.  Here's a TI-73 from 35 dollars(used) if you scroll down. [url]http://www.amazon.com/s/ref=nb_ss?url=search-alias%3Daps&field-keywords=TI-73[/url] Here are TI - 92's from 69.99
[url]ttp://www.amazon.com/s/ref=nb_ss?url=search-alias%3Daps&field-keywords=ti-93[/url]
And here you can download a FREE graphing calculator online if you have XP:
[url]malektips.com/xp_calculator_0008.html[/url]

Oh, and my science teacher has TI-84 basics.  They look WORSE than TI-83 Plus's(my math teacher has them)

Yeah i just bought this to get me through high school AP's that's all. For college, ima save up to get a TI-89.  :D

If you believe that TI-Basic is really weak at making programs, you can also make a C program on your computer and transfer it to the calculator in assembly format.

[quote="AceOfDiamonds"]dude the TI-84 plus is awesome... i don't know what you guys are talking about. It's programming is pretty awesome, even though TI-Basic is amazingly weak. You can program lots of really fun stuff :)[/quote]

Yes.  I bet the TI-84 PLUS is awesome.  But the basic just looks kinda old but its probably ok.........

i have the PLUS, it's awesome, so much that when ppls used our school's FAIL TI-15, they all lined up to use mine  :o

TI-84 Silver (+Games) FTW.

My school actually doesn't let us use the TI-89, because the interface is so much different from the 84 and 83.

Just asking, what does the TI-89 have that the TI-84 doesn't? :huh: (except for the better programming)

Look we don't actually NEED TI-84s or whatever right now, I have a TI-[b]34[/b] and I'm doing fine.

What's good on the TI-84s 89s whatever that can actually help a bunch on contest?

Actually 20 people in my school have TI 83 or higher, almost all are doing worse than me!

Why would you ever need TI-89?

If a competition has problems that get destroyed if you have a ti-89, then the competition probably isn't a very large or important competition....

Yeah, you can solve some mathcounts questions. That's pretty much the only other benefit. Laptop + Mathematica works better than ti-89 for almost everything except how easily they can be brought to tests.....

Yes, ti-89 can find limits.... thats not particularly helpful on anything other than tests, and you should know limits....

It's also much easier to use than the TI-84, and has pretty print, which makes viewing easier.

Well all I had was TI-34 until my teacher gave me a TI-84+ Silver at the end of the year (everyone present practically had a heart attack). I haven't learned how to really use the calculator, but it's already much better than TI-34 or even TI-83+ (which I used in math class a few years ago). Once I find out all it can do I'll like it even more...

[quote="AwesomeToad"]Actually 20 people in my school have TI 83 or higher, almost all are doing worse than me![/quote]

Congrats!

Are you talking about USAMO and later?
Or high school science/math?


Tree runners - I, II and III - are training in a straight street. The runners ocupied position measured from a fixed referential are described by $ S_{\text I}\equal{}5t\plus{}3$, $ S_{\text{II}}\equal{}2t\plus{}9$ and $ S_{\text{III}}\equal{}t^{2}\minus{}2t\plus{}9$
During the run,  what the number of times that the distance between runner I and II and the distance between runner II and III are equal?

$ (A)\ 1\ \ \ \ \ \ \ (B)\ 2\ \ \ \ \ \ \ (C)\ 3\ \ \ \ \ \ \ (D)\ 4\ \ \ \ \ \ \$

[hide="hint"]
For how many $ t$ does $ 5t\plus{}3\minus{}(2t\plus{}9)\equal{}2t\plus{}9\minus{}(t^{2}\minus{}2t\plus{}9)$ hold true?
[/hide]

[quote="math_music"]That lead to 2 solutions... but the answer actually is 3? What i've missing?[/quote]

[hide="Error correction"]The real equation is:

$ | 5t\plus{}3\minus{}(2t\plus{}9)| \equal{}| 2t\plus{}9\minus{}(t^{2}\minus{}2t\plus{}9)|$

$ |3t\minus{}6| \equal{} |t^{2}\minus{}4 t|$

We get two equations:

$ 3t\minus{}6 \equal{} t^{2}\minus{}4t$

$ 6\minus{}3 t \equal{} t^{2}\minus{}4t$

Solving separately, we get:

$ t^{2}\minus{}7t\plus{}6 \equal{} 0\implies (t\minus{}1)(t\minus{}6)\equal{}0$

$ t^{2}\minus{}t\minus{}6 \equal{} 0\implies (t\minus{}3)(t\plus{}2) \equal{} 0$

Since $ t>0$, this eliminates $ t \equal{}\minus{}2$. So, we do get $ 3$ solutions:

$ 1,3,6$[/hide]

Yeah, you need to restrict the interval so they have the same sign, alternatively make the absolute difference... Thanks vishalarul


Faith designs the following logo for a telephone company. The logo consists of a T divided into 5 regions. Faith has to colour each region with one of the three colours: red, green or gold. No two regions with a common edge can have the same colour; and each colour must be used at least once. How many possible colour schemes are available to Faith for her logo?

[img]http://img197.imageshack.us/img197/7374/logowft.jpg[/img]

[quote="KrazyFK"][hide="Solution"]Throughout the solution I will refer to the large trapezoid as the "middle"

For each possible choice for the colour in the middle, there are two cases for the left two triangles: They are the same colour, or they are a different colour. There are two possiblities for each of these cases. Suppose that the left two shapes are different colours. Then all 3 colours have been used already so that restriction can be lifted, giving 2 choices for each of the other 2 shapes, and 4 options in total for these 2 shapes. This means that if the two shapes on the left are a different colour there are $ 2 \cdot 4 \equal{} 8$ possible cases.

Suppose that the 2 shapes on the left are the same colour. Then one of the shapes on the right must use the missing colour. There are 4 possiblities for the right two shapes, and we remove the 1 possiblity which does not use the missing colour, giving 3 choices for the right two shapes. This means when the 2 shapes on the left are the same colour, there are
$ 2 \cdot 3 \equal{} 6$ possiblities.

This means there are $ 8 \plus{} 6 \equal{} 14$ possibilities for each possible choice of the middle colour, and since there are 3 choices for this colour, there are $ 3 \cdot 14 \equal{} 42$ colour schemes in total.

There are $ 42$ colour schemes available to choose from.[/hide][/quote]

Nice solution  :lol: But just a question, is there a reason why you chose to look at the middle and work around that?  :maybe:

[quote="KrazyFK"]Just the fact that the middle shares the most common edges, and therefore eliminates more possiblities than any other of the shapes.[/quote]

When i looked at this question, the two triangle to the left of the "middle" i thought they shared a common edge, do you know why i am wrong?

[quote="KrazyFK"]It is true that they lie on the same line, but when talking about edges of a shape, we generally refer to line segments, and in the two triangles none of the line segments which form them are shared.[/quote]

Thanks again  :lol:

Of course, if they do share a common edge, it does not take much to alter the solution. Simply remove the cases where the two triangles are the same colour to obtain $ 3 \cdot 8 \equal{} 24$

There are three possible colors for the large trapezoid. Then there are two possible colors for each of the three shapes surrounding the large trapezoid, and then there are two possible colors for the last shape so we have;

$ 3$ X $ 2^4$

$ 48$

However, we must subtract the possibilities that only use two of the colors, because this is the compliment. There are three colors to choose from that we won't use. Then there are two possibilities for the large trapezoid, and ten once we have chosen the large trapezoid's color, there is only one possibility for each of the other shapes. So we subtract
$ 3$ X $ 2$ or $ 6$ for a total of $ 42$.

I did it the same way as you but I don't know what you mean by "There are three colors to choose from that we won't use. Then there are two possibilities for the large trapezoid, and ten once we have chosen the large trapezoid's color".

Basically my logic to find the number of ways the T would be colored by only using two colors was we can choose any color out of the three to color the trapezoid, after this we can choose one of the triangles on the left to be one of the other two colors.  Then after knowing what color that triangle is, the other parts are fixed since we're only using two colors.  (E.g. the two triangles must be the same color, the parallelogram next to the trapezoid must also be the same color as the triangles, and the last triangle on the right has to be the same color as the trapezoid.)  So there are 3 * 2 = 6 ways to color the T using only two colors.


parameterization trigonometry vector conics ellipse function parabola

[color=blue][b](1)[/b][/color] A communication satelite with the mass m is on a geosynchronous circular orbit of radius $r_0$. Find the orbit radius $r_0$, the orbital velocity $v_0$ and expressions for the angular momentum $L_0$ and the mechanical energy $E_0$ of the satelite. Use Earth's radius $R_T \doteq 6.37 \times 10^6\ \text m$, the surface gravitational acceleration $g \doteq 9.81\ \text{m/sec}^2$ and the sidereal day $T_0 \doteq 24.0\ \text{hr}$.

On a circular orbit, the gravitational and centrifugal forces $\vec F_G, \vec F_C$ have the same magnitudes, but opposite directions. Let G be the gravitational constant, $M_T$ mass of the Earth, and $v_0, \omega_0 = \frac{v_0}{r_0}$ the satellite orbital and angular velocities.

$G \frac{m M_T}{r_0^2} = F_G = F_C = \frac{mv_0^2}{r_0} = m \omega_0^2 r_0 = \frac{4\pi^2 m r_0}{T_0^2}$

where we used the fact that the orbital period $T_0 = \frac{2 \pi}{\omega_0}$ on geosynchronous orbit is 1 sidereal day.

$\frac{r_0^3}{T_0^2} = \frac{G M_T}{4 \pi^2} = \text{const}$

which is Kepler's 3rd law. At Earth's surface, the gravitational force on a body with the mass m is

$F_G = G \frac{m M_T}{R_T^2} = mg,\ \ G M_T = g R_T^2$

Hence, Kepler's 3rd law becomes

$\frac{r_0^3}{T_0^2} = \frac{g R_T^2}{4 \pi^2}$

Expressing $r_0$ and substituting the values of $g,\ R_T,\ T_0 \doteq 24 \cdot 3600 = 86400\ \text{sec}$ and $\pi^2 \doteq 9.87$,

$r_0 = \sqrt[3]{\frac{g R_T^2 T_0^2}{4 \pi^2}} \doteq \sqrt[3]{\frac{9.81 \cdot 6.37^2 \times 10^{12} \cdot 86.4^2 \times 10^6}{4 \cdot 9.87}} \doteq$

$\doteq \sqrt[3]{75270 \times 10^{18}} \doteq 42.2 \times 10^6\ \text m = 42.200\ \text{km}$

The satelite orbital velocity $v_0$ is equal to

$v_0 = \omega_0 r_0 = \frac{2 \pi r_0}{T_0} \doteq \frac{6.28 \cdot 42.2 \times 10^6}{86400} \doteq 1535\ \text{m/sec} = 1.535\ \text{km/sec}$

The satellite angular momentum $\vec L_0$ is equal to

$\vec L_0 = \vec r_0 \times m \vec v_0$

Since $\vec v_0 \perp \vec r_0$,

$L_0 = m v_0 r_0 = m \omega_0 r_0^2 = \frac{2\pi m r_0^2}{T_0} = \text{const}$

which is Kepler's 2nd law. The satellite mechanical energy $E_0$ is the sum of its kinetic energy $K_0$ (cannot use $T_0$, already used for the sidereal day) and its potential energy $V_0$ in the gravitational field:

$E_0 = K_0 + V_0 = \frac{m v_0^2}{2} - G \frac{m M_T}{r_0} =$

$= \frac{m \omega_0^2 r_0^2}{2} - \frac{mg R_T^2}{r_0} = \frac{4 \pi^2 m r_0^2}{2T_0^2} - \frac{mg R_T^2}{r_0} =$

$= \frac{2 \pi^2 m}{r_0} \cdot \frac{r_0^3}{T_0^2} - \frac{mg R_T^2}{r_0} = \frac{2 \pi^2 m}{r_0} \cdot \frac{g R_T^2}{4\pi^2} - \frac{mg R_T^2}{r_0} =$

${= \frac{m g R_T^2}{r_0} (\frac 1 2 - 1}) = -\frac{m g R_T^2}{2r_0} < 0$

as required for a bound state of the satellite.

[color=blue][b](2)[/b][/color] By accidental firing the satellite engine, a velocity $\triangle \vec v$ toward Earth's center is added to the satellite orbital velocity $\vec v_0$, characterized by the boost parameter $\beta = \frac{\triangle v}{v_0}$.

[color=red][b](2a)[/b][/color] Assuming $\beta < 1$, calculate the parameters $l, \epsilon$ of the new elliptical orbit

$r = \frac{l}{1 - \epsilon \cos \theta}$

where $l = \frac{L^2}{GM_Tm^2}$ is called semi-latus-rectum :? and $\epsilon = \sqrt{1 + \frac{E L^2}{G^2 M_T^2 m^3}}$ is the orbit excentricity. L and E are the angular momentum and mechanical energy at this orbit, both conserved.

After the boost by $\triangle \vec v$, the new orbital momentum $\vec L$ is

$\vec L = \vec r_0 \times m \vec v = \vec r_0 \times m (\vec v_0 + \triangle \vec v) = \vec r_0 \times m \vec v_0 + \vec r_0 \times m \triangle \vec v = \vec L_0$

because the vectors $\triangle \vec v,\ \vec r_0$ are oppositely parallel, hence their vector product is zero. Thus the orbital momentum did not change and its magnitude is $L = L_0 = m v_0 r_0 = \frac{2\pi m r_0^2}{T_0}$ as before. Hence,

$l = \frac{L_0^2}{GM_Tm^2} = \frac{4 \pi^2 m^2 r_0^4}{g R_T^2 m^2 T_0^2} = \frac{4 \pi^2 r_0^4}{g R_T^2 T_0^2} = \frac{r_0^4}{r_0^3} = r_0 = \sqrt[3]{\frac{gR_T^2 T_0^2}{4 \pi^2}}$

At the circular orbit, the mechanical energy was $E_0 = -\frac{m g R_T^2}{2r_0} < 0$ and the excentricity

$\epsilon_0 = \sqrt{1 + \frac{2E_0 L_0^2}{G^2 M_T^2 m^3}} = \sqrt{1 - \frac{m g R_T^2}{2r_0} \cdot \frac{4 \pi^2 m^2 r_0^4}{T_0^2} \cdot \frac{2}{g^2 R_T^4 m^3}} =$

$= \sqrt{1 - \frac{4 \pi^2 r_0^3}{g R_T^2 T_0^2}} = \sqrt{1 - 1} = 0$

as expected (just checking the formulas). After the boost, the orbital velocity magnitude increased to

$|\vec v_0 + \triangle \vec v| = v_0 \sqrt{1 + \beta^2}$

the mechanical energy increased to

$E = E_0 + (K - K_0) = E_0 + \frac{mv_0^2(1 + \beta^2) - mv_0^2}{2} = E_0 + \frac{mv_0^2 \beta^2}{2}$

and the orbit excentricity became

$\epsilon = \sqrt{1 + \frac{2E L_0^2}{G^2 M_T^2 m^3}} = \sqrt{1 + \frac{2E_0 L_0^2}{G^2 M_T^2 m^3} + \frac{mv_0^2 \beta^2 L_0^2}{G^2 M_T^2 m^3}} =$

${{= \sqrt{\frac{v_0^2 \beta^2 L_0^2}{G^2 M_T^2 m^2}} = \frac{v_0 \beta L_0}{G M_T m}} = \frac{2 \pi r_0 \beta \cdot 2 \pi m r_0^2}{g R_T^2 m T_0^2}} = \frac{4\pi^2 \beta r_0^3}{g R_T^2 T_0^2} = \beta$

[color=red][b](2b)[/b][/color] Calculate the angle $\alpha$ between the major axis of the elliptical orbit and the radius vector of the engine misfire.

The elliptical orbit with the known parameters $l, \epsilon$ and the known focus (Earth's center) passes through the location of the engine misfire. The distance of the misfire from the focus is equal to the radius $r_0$ of the geosynchronous circular orbit. Substituting this into the ellipse equation yields the desired angle:

$r_0 = \frac{l}{1 - \epsilon \cos \alpha}$

$\cos \alpha = \frac{r_0 - l}{\epsilon r_0} = 0,\ \ \alpha = \frac \pi 2$

because earlier, we calculated $l = r_0$.

[color=red][b](2c)[/b][/color] Give the expressions for the perigee $r_{\text{min}}$ and apogee $r_{\text{max}}$ as a function of $r_0, \beta$ and calculate their values for $\beta = \frac 1 4$.

From the ellipse equation, the distance r from the focus is minimum for $\cos \theta = +1,\ \theta = 0$ and minimum for $\cos \theta = -1,\ \theta = \pi$. Therefore,

$r_{\text{min}} = \frac{r_0}{1 + \epsilon} = \frac{r_0}{1 + \beta},\ \ r_{\text{max}} = \frac{r_0}{1 - \epsilon} = \frac{r_0}{1 - \beta}$

because earlier, we calculated $\epsilon = \beta$. For $\beta = \frac 1 4$,

$r_{\text{min}} = \frac{4r_0}{5} \doteq 33,800\ \text{km},\ \ r_{\text{max}} = \frac{4r_0}{3} = 56,300\ \text{km}$

where we used the geosynchronous orbit radius $r_0 \doteq 42.2 \times 10^6\ \text m = 42,200\ \text{km}$ calculated previously.

[color=red][b](2d)[/b][/color] Determine the period T of the elliptical orbit as a function of $T_0$ and $\beta$ and calculate uts value for $\beta = \frac 1 4$.

Lenght of the ellipse major axis a is

$a = r_{\text{min}} + r_{\text{max}} = \frac{r_0}{1 + \beta} + \frac{r_0}{1 - \beta} = \frac{2r_0}{1 - \beta^2}$

By Kepler's 3rd law

$\frac{a^3}{T^2} = \frac{8r_0^3}{T_0^2}$

$T^2 = \frac{T_0^2}{(1 - \beta^2)^3},\ \ T = \frac{T_0}{(\sqrt{1 - \beta^2})^3}$

For $\beta = \frac 1 4$, we get

$(\sqrt{1 - \beta^2})^3 = \left(\sqrt{1 - \frac{1}{16}}\right)^3 = \frac{(\sqrt{15})^3}{64} = \frac{15 \sqrt{15}}{64} \doteq 0.9077$

$T \doteq \frac{T_0}{0.9077} \doteq 1.102\ T_0 \doteq 26.37\ \text{hr} \doteq 26\ \text{hr}\ 22\ \text{min}$

(could not resist to use the more accurate value of the sidereal day $T_0 \doteq 23.93\ \text{hr} \doteq 23\ \text{hr}\ 56\ \text{min}$).
[color=white].[/color]

[color=blue][b](3)[/b][/color] Calculate the boost parameter necessary to escape Earth's gravity. Determine the closest approach $r_{\text{min}}'$ to Earth's center as a function of $r_0$.

The escape path is a parabola with excentricity $\epsilon = 1$. Hence, the necessary boost parameter is $\beta = 1$. The closest approach is given (as before) by

$r_{\text{min}}' = \frac{r_0}{1 + \epsilon} = \frac{r_0}{1 + \beta} = \frac{r_0}{2}$

[color=blue][b](4)[/b][/color] Assuming $\beta > \beta_{\text{esc}}$:
(a) Determine the residual velocity $v_{\infty}$ as a function of $v_0$ and $\beta$.
(b) Determine the impact parameter b in terms of $v_0$ and $\beta$.
(c) Determine the angle $\phi$ of the asymptotic escape direction from the radius vector at the misfire location in terms of $\beta$ and calculate its value for $\beta = \frac 3 2\ \beta_{\text{esc}}$.

[color=red][b](4a)[/b][/color] At infinity, the potential energy of the satellite is zero and its mechanical energy E is equal to its kinetic energy $K_{\infty}$. Using the conservation of energy law,

$E = K_{\infty} = \frac{mv_{\infty}^2}{2}$

$E = E_0 + \frac{mv_0^2\beta^2}{2} = -\frac{mg R_T^2}{2r_0} + \frac{mv_0^2\beta^2}{2}$

$v_{\infty}^2 = -\frac{g R_T^2}{r_0} + v_0^2\beta^2$

Substituting Kepler's 3rd law [color=white].[/color]$\frac{r_0^3}{T_0^2} = \frac{g R_T^2}{4 \pi^2}$,

$v_{\infty}^2 = -\frac{4\pi^2 r_0^2}{T_0^2} + v_0^2\beta^2 = -v_0^2 + v_0^2 \beta^2 = v_0^2(\beta^2 - 1)$

$v_{\infty} = v_0 \sqrt{\beta^2 - 1}$

[color=red][b](4b)[/b][/color] Using the conservation of angular momentum $\vec L = \vec L_0$, at infinity, we have

$\vec L = \vec L_0 = \vec r_{\infty} \times m \vec v_{\infty} = \vec b \times m \vec v_{\infty}$

Since the vectors $v_{\infty} \perp \vec b$ are perpendicular, $L_0 = mv_{\infty}b$ and

$b = \frac{L_0}{mv_{\infty}} = \frac{mv_0r_0}{mv_0\sqrt{\beta^2 - 1}} = \frac{r_0}{\sqrt{\beta^2 - 1}}$

[color=red][b](4c)[/b][/color] For $\beta > \beta_{\text{esc}} = 1$, the satellite path is a hyperbola, but it has the same equation in polar coordinates as the ellipse, except now the excentricity is $\epsilon = \beta > 1$.

$r = \frac{l}{1 - \epsilon \cos \theta} = \frac{r_0}{1 - \beta \cos \theta}$

For $r \rightarrow \infty$, $\cos \theta \rightarrow \cos \theta_{\infty} = \frac 1 \beta$. This is an angle between the hyperbola asymptote and its major axis, which is perpendicular to the satellite radius vector at the misfire location. Hence,

$\phi = \theta_{\infty} + \frac \pi 2 = \arccos \frac 1 \beta + \frac \pi 2$

For $\beta = \frac 3 2\ \beta_{\text{esc}} = \frac 3 2$, we obtain

$\phi = \arccos \frac 2 3 + \frac \pi 2 \doteq 0.2677\ \pi + \frac \pi 2 = 0.7677\ \pi \doteq 138.2^\circ$
[color=white].[/color]


...then don't spam our forums.  We would like to keep this forum to CT math only, thank you very much.  We don't need to make this an outsider (i.e. Missouri) forum.

If you've got an issue with someone else, take it up with them privately.


inequalities AMC AIME USA(J)MO USAMO geometry MATHCOUNTS \/closed

I was wondering, what classes have been offered in the past?

Also, what classes will be offered in the future?

Also, (1) when is the expected date for AIME/Olympiad Inequalities and AIME/Olympiad Counting? I hope they begin sometime around this year's AIME and USAMO; (2) is it possible to obtain the transcripts and problem sets from past courses?

Good question, I was wondering the same thing.

For info on classes in the future, see this thread:

[url]http://artofproblemsolving.com/Forum/viewtopic.php?t=2025[/url]

It's still mostly accurate, and we may be adding a few other courses in addition to those listed (such as a complex numbers/trig class).

The classes we have offered or are running right now are:

Intro level: Number Theory, Counting
Intermediate/Advanced: Counting, Algebra, Geometry
Problem Series: AMC/AIME, MATHCOUNTS

As for 'buying' past classes, we'll think a bit about how we might best do so (i.e. whether we have the time to package them in a good format, etc.).  This may be something that has to wait a while.


function
number theory proposed
number theory

Is the set $\{n$ [b]s.t. base-10 expansion of [/b]$n^3$ [b]is free of zeros[/b]$\}$ infinite?

I am surprised : no answer !  :D  :D
Look at that: [url=http://mathworld.wolfram.com/Baxter-HickersonFunction.html]Baxter-Hickerson Function[/url]
It seems that for higher powers no formula is known and the question is surely still open if there are infinitely many zerofree powers!

:? I'm not surprised we don't find what has only been conjectured in 1999 after long research... ;)


1. a^2+b^2+(ab)^2 = 4567654.
Find all integer values of a and b so this equation is true.

2. A bit harder: a^2+b^2+c^2+(ab)^2+(bc)^2+(ac)^2+(abc^2) = 456787654.
Find all integer values of a, b, and c so this equation is true.

3. This may be a little bit too hard but:
(ab)^2 = a^2+b^2+c^2.
Find all integer values of a, b, and c so this equation is true.

a tad bit too difficult. if it isn't solved soon, i'll start posting hints, starting with:

[hide]remember simon's favorite factoring trick[/hide]

And what is simon's favorite factoring trick?

Try a search for it on this website.  Otherwise, just go and try the problems  :twisted:

Hm. Everyone will remember it as Simon's favorite factoring trick. I'm glad I have a bit of influence in the mathematics education of others.

In 20 years, when some of these folks are writing textbooks, you will be immortalized.  If they can spell you last name ;)

I'm stuck!

Searching through 70 pages to find, Simon's favorite factorin trick and I finally did. Here's what  have so far for number 1:

a^2+b^2+(ab)^2 = 4567654
a^2+b^2+(ab)^2 + 1 = 4567654 +1
(a^2 + 1)(b^2 + 1) = 4567655

4567655 = 5x281x3251
5x281x3251 = (a^2 + 1)(b^2 + 1)
Now I can't figure out any more.

You have this monster number, and you know its prime factors, and you know the forms of the factors you're looking for. Note that not every number is of the form, say, a2+1. We can't, for example, write 281 in that form. See if you can finish it off now.

Since 5 can be expressed as 2^2 + 1, then shouldn't 281x3251 be expressed as b^2 + 1, so b^2 = 913530 and then that's where I'm actually stuck at.