geometry
trigonometry
circumcircle
angle bisector

Question

Given $\Delta ABC$ with $AB>AC$.  The bisectors of the interior and exterior angles ar $A$ intersect line $BC$ at points $D$ and $E$, respectively.  Prove that

$\frac{\sqrt{AD^{2}+AE^{2}}}{CD}-\frac{\sqrt{AD^{2}+AE^{2}}}{BD}=2$

Please Explain All of Your Steps.

Thanks

Denote the sides of the triangle by $a=BC, b=CA, c=AB$ and the angles by $\alpha=\angle A,\beta=\angle B,\gamma=\angle C$

Since $AB>AC$, we have $\overline{BCE}$

\begin{eqnarray*}\angle DAB={\alpha\over 2}\land \angle EAB=\frac{\pi-\alpha}{2}&\implies& \angle DAE={\pi\over 2}\\ &\implies& \sqrt{AD^{2}+AE^{2}}=DE\end{eqnarray*}

Well-known (or easily proven by Sine Law): ${BD\over CD}={c\over b}\stackrel{BD+CD=a}{\implies}BD={ac\over{b+c}}\land CD={ab\over{b+c}}$

Apply Sine Law to $\triangle ACE$:

$\frac{EC}{\sin({\pi\over 2}-{\alpha\over 2})}=\frac{b}{\sin\angle BEA}\implies{EC\over b}=\frac{\cos{\alpha\over 2}}{\sin\angle BEA}$

Apply Sine Law to $\triangle ABE$:

$\frac{EB}{\sin({\pi\over 2}+{\alpha\over 2})}=\frac{c}{\sin\angle BEA}\implies{EB\over c}=\frac{\cos{\alpha\over 2}}{\sin\angle BEA}$

That gives ${EC\over b}={EB\over c}$, and substituting $EB=EC+a$ we get $EC=\frac{ab}{c-b}$

Then $DE=EC+CD=\frac{ab}{c-b}+{ab\over{b+c}}=\frac{2abc}{c^{2}-b^{2}}$

Then

\begin{eqnarray*}{DE\over CD}-{DE\over BD}&=& \frac{2abc}{c^{2}-b^{2}}\left(\frac{b+c}{ab}-\frac{b+c}{ac}\right)\\ &=& \frac{2abc}{c^{2}-b^{2}}\cdot\frac{b+c}{a}\cdot\frac{c-b}{bc}\\ &=& 2\end{eqnarray*}

NOTE: The external angle bisector also divides the opposite side in proportion with the two remaining sides, since we've got ${EC\over EB}={b\over c}$, except that in this case the difference (as opposed to the sum) of the segments is equal to $a$.

[color=darkblue][b]Lemma 1.[/b] If the points $B$, $C$, $D$, $E$ belong to the orientated line $d$, then $\boxed{\ \frac{\overline{DB}}{\overline{DC}}=-\frac{\overline{EB}}{\overline{EC}}\Longleftrightarrow\frac{2}{\overline{DE}}=\frac{1}{\overline{DB}}+\frac{1}{\overline{DC}}\ }\ .$

In this case we say that the points $D$, $E$ are [u]harmonical conjugate[/u] w.r.t. the points $B$, $C\ .$

A remarkable example : the points $B,C,D,E$ from the proposed problem.

[b]Proof.[/b] From the [u]Chasles' relation[/u], if $\{X,Y\}\subset d$, then $\overline{XY}=y-x\ .$ Therefore,
$\boxed{\ \frac{\overline{DB}}{\overline{DC}}=-\frac{\overline{EB}}{\overline{EC}}\ }$ $\Longleftrightarrow$ $\frac{b-d}{c-d}+\frac{b-e}{c-e}=0$ $\Longleftrightarrow$ $(b-d)(c-e)+(c-d)(b-e)=0$ $\Longleftrightarrow$ $\boxed{\ (b+c)(d+e)=2(bc+de)\ }\ .$
${\boxed{\ \frac{2}{\overline{DE}}=\frac{1}{\overline DB}}+\frac{1}{\overline{DC}}\ }$ $\Longleftrightarrow$ $\frac{2}{e-d}=\frac{1}{b-d}+\frac{1}{c-d}$ $\Longleftrightarrow$ $2(b-d)(c-d)=(e-d)(b+c-2d)$ $\Longleftrightarrow$ $\boxed{\ (b+c)(d+e)=2(bc+de)\ }\ .$

[b]The proof of the proposed problem.[/b] $AD^{2}+AE^{2}=DE^{2}$ $\Longrightarrow$ the requiered relation is equivalently with the relation
$2\cdot DB\cdot DC=DE\cdot (DB-DC)$ $\Longleftrightarrow$ $\frac{2}{DE}=\frac{1}{DC}-\frac{1}{DB}$ $\Longleftrightarrow$ $\frac{2}{\overline{DE}}=\frac{1}{\overline{DB}}+\frac{1}{\overline{DC}}$, what is trully using the lemma $1\ .$

I"ll present supplementary an important metrical property which is often useful in the triangle geometry.

[b]Lemma 2.[/b] The internal and external bisectors of the angle $A$ in the triangle $ABC$ cut the opposite side in the points $D$, $E$ respectively.
Then exist the relations  $AD^{2}=bc-DB\cdot DC$ and $AE^{2}=EB\cdot EC-bc\ .$
[b]Proof.[/b] The circumcircle $w$ of the triangle $ABC$ intersects again the lines $AD$, $AE$ in the points $D'$, $E'$ respectively.
Denote $p_{w}(X)$- the power of the point $X$ w.r.t. the circle $w\ .$
$1.\blacktriangleright$ $ABD'\sim ADC$ $\Longrightarrow$ $\frac{AB}{AD}=\frac{AD'}{AC}$ $\Longrightarrow$ $\boxed{\ AD\cdot AD'=bc\ }$ $\Longrightarrow$ $AD\cdot (AD+DD')=bc$ $\Longrightarrow$
$AD^{2}+DA\cdot DD'=bc$ $\Longrightarrow$ $AD^{2}+p_{w}(D)=bc$ $\Longrightarrow$ $\boxed{\ AD^{2}=bc-DB\cdot DC\ }\ .$
$1.\blacktriangleright$ $ABE'\sim AEC$ $\Longrightarrow$ $\frac{AB}{AE}=\frac{AE'}{AC}$ $\Longrightarrow$ $\boxed{\ AE\cdot AE'=bc\ }$ $\Longrightarrow$ $AE\cdot (EE'-EA)=bc$ $\Longrightarrow$
$EA\cdot EE'-AE^{2}=bc$ $\Longrightarrow$ $AE^{2}=p_{w}(E)-bc$ $\Longrightarrow$ $\boxed{\ AE^{2}=EB\cdot EC-bc\ }\ .$[/color]

Thank You very much for you proofs.

This was a 9th Grade Geometry proof before trig, so thanks.

You can totally skip the Sine Law in my proof and use the regular similarity (I just like to use Sine Law when posting since it's easier to follow).

OK, thanks.

I will try to understand the proofs better.

Thanks


function
trigonometry
quadratics
geometry
trig identities
Law of Sines

Given right triangle ABC with angle C being the right angle and AC=5 and BC=12, and that rays trisect angle C to intersect AB at points D and E (Point E is closer to B and D is closer to A). What is the length of CE?

[quote="TearOfMercury"]Hint: [hide]$\beta = \arctan(\frac{\bar{AC}}{\bar{BC}})$
and from there you can get the angle of the triangle CEB at E..[/hide][/quote]
what is arctan?

I thought you needed a right triangle for trigonometry but then you create a non right triangle once you have the trisectors...
I'm totally confused... :huh:

The angle $\beta$ is part of the right triangle ABC; its the angle at the point B; so you can use trigonometry. By trisecting the right angle you get the triangle BDE, where the angle at C is 30° and the angle at B is still the one from the right triangle.. the rest you can do with the Law of sines, which applies to any triangle.

In fact, we don't need trigonometric tables

[hide]
$\triangle CEB: \ \frac{\sin(\widehat{BEC})}{BC}= \frac{\sin 30^{o}}{BE}= \frac{\sin B}{CE}\Rightarrow $

$\frac{\sin(B+30^{o})}{BC}= \frac{\sin 30^{o}}{BE}= \frac{\sin B}{CE}\Rightarrow $

$\frac{BC}{CE}= \frac{\sin(B+30^{o})}{\sin B}= \frac{\sin B\cos 30^{o}+\cos B\sin 30^{o}}{\sin B}= \cos 30^{o}+\sin 30^{o}\cot B \Rightarrow $

$\frac{BC}{CE}= \frac{\sqrt{3}}{2}+\frac{1}{2}\cdot \frac{12}{5}= \frac{5\sqrt{3}+12}{10}\Rightarrow $

$CE = BC \cdot \frac{10}{5\sqrt{3}+12}= \frac{120}{5\sqrt{3}+12}=\cdots = \frac{40(12-5\sqrt{3})}{23}$[/hide]

let CE=x
WE HAVE           [color=darkred]BE={12/[5+x+12]}*13[/color]

                           BE/sin30=x/[5/13]

                 SUBSTITUTING FOR [i]BE[/i] WE HAVE A QUADRATIC

                             x^2+17x+20=0
                         hence  [i]CE[/i]=5.365424623862

[quote="TearOfMercury"]The inverse function of $tan$, sometimes also denoted as $tan^{-1}$. In other words $tan(\alpha) = \frac{a}{b}\Leftrightarrow arctan(\frac{a}{b}) = \alpha$.[/quote]

    it's not true in all the cases  :bruce:

I agree, that's not always the case if I may have my input.


Does anyone here play Reversi (othello)?
i love that game.  I play it all the time with myself.  Is there any strategy to win all the time?

I don't know if there's any known "winning strategy" yet, but I use a program called WZebra.  Its AI on the most difficult level, in my opinion, is close enough to a winning strategy. ;)


Calculate the molecular weight of the oxygen-carrier protein hemoglobin if this protein is 0.335% Fe by weight and each protein molecule contains four iron atoms.

1)between 1000 and 10,000 grams per mole

2)less than 1000 grams per mole

3)between 50,000 and 100,000 grams per mole

4)more than 100,000 grams per mole

5)between 10,000 and 50,000 grams per mole

Let's think what you want.  Grams per mole for hemoglobin (with Fe).

What do you know.  The hemoglobin is 0.335wt% Fe and 4 Fe atoms per hemoglobin molecule.

How much Fe would we have to have present to ensure that we had 1 gram-mole of hemoglobin (6.0234 X10^23 molecules)?  Why that would be 4 gram-moles of Fe.

What would that weigh?

How would you go from weight of Fe and wt% of Fe to total weight?

You should be able to see the answer and how to solve these types of problems from this.


Ex Pert calculated that if three times as many people worked twice as fast as the twelve people doing the work now, the job would be finished in eight days. Instead, Ex hired only four new people and all sixteen worked at the same rate as the original twelve. How many days did it take to finish the job?

With twelve people, it takes $ 8 \div \frac13 \div \frac12$, or $ 48$ days.

Thus, with sixteen people, we can form a proportion:

$ \frac{12}{48} \equal{} \frac{16}{x}$

$ \frac{1}{4} \equal{} \frac{16}{x}$

$ x$ is clearly $ \boxed{64}$.

The answer is 36 days.

If 36 workers, at double speed, take 8 days then,
36 workers at normal speed take twice as long, 16 days
12 workers take three times as long, 48 days

12 working 48 days each must yield the same product as 16 working x days each.
12*48 = 16*x
Or, the ratio of the numbers of workers is inversely proportional to the days taken to complete job.  {More workers must take less time.)
12/16 = x/48

Solving either equation gives the days count as x = 36


calculus
integration
calculus computations

Compute \[ I(a)=\int_0^a\frac{\arctan x}{x^2+ax+2}\,dx \]
Where $a>0$.

hello, Maple 14 can compute this integral, but the answer is terrible.
Sonnhard.


irrational number
algebra solved
algebra

Given a interger a>1
and a irrational number b,
and a big interger M
And a small positive real c

then (i think its true)
there exist a positive interger n>M
and a interger k ,such that

|(a^n)*b-k|<c

I think this is true
But I still cannot prove it by now. :(

You typed a "big" integer M, can I understand that M>0.

Yes.

And sorry for this.
I was trying to say
 a "positive interger
which can be very big".

Your statement is false.
Let $b=1/9-\sum_{i=2}^\infty 10^{-n!}$ (i.e. b is 0,1011101111... where 0's are placed on n!-positions), a=10, c=0.01.
Then for any integers n and k we have $|10^nb-k|>0.01=c$.


Hi everyone,

Could someone please tell me if this is correct?

I will do my homework and I will pass this class=I will not do my homework and I will not pass this class

Seven is an integer and seven is even=seven is not an integer and seven is not even

If T is continuous, then T is bounded=T is continuous and T is not bounded

I can eat dinner or go to the shows=I can eat dinner and I cannot go to teh show

If x is odd, then x is prime=x is odd and x is not prime

THe number x is prime only if x is odd=The number x is prime and x is odd

If I am not home, then Sam will answer the phone and he will tell you how to reach me=I am not at home Sam will not answer the phone, and he will not tell you how to reach me

If the stars are green or the white hourse is shinning, then the world is eleven feet wide=The stars are green, the white horse is shinning, and the world is not eleven feet wide.

Thank you very much

[color=red]EDIT by moderator: Moved from Geometry / Olympiad section.[/color]

Could someone please help me on this?

Thank you

If you expect a reasonable reply, post in the appropriate section and express yousrself clearly. Normally, relation = means equivalence, but you seem to use it as "is negation of." If this is the case, you are wrong in 6 cases out of 8. (The 3rd and the 5th negations are correct).

$ P, Q, R$ are statements, $ \neg$ is negation, $ \wedge$ is conjunction (and), $ \vee$ is disjunction (or), $ \Rightarrow$ is implication (if - then), $ \Leftrightarrow$ is equivalence (if and only if). You are supposed to use the following 3 propositions of logic:

$ \neg(P \vee Q)\ \Leftrightarrow\ \neg P \wedge \neg Q$
$ \neg(P \wedge Q)\ \Leftrightarrow\ \neg P \vee \neg Q$
$ P \Rightarrow Q\ \Leftrightarrow\ \neg P \vee Q$

For example, the 8th case:

$ P$: The stars are green.
$ Q$: The white horse is shining.
$ R$: The world is eleven feet wide.

Starting with $ (P \vee Q) \Rightarrow R,$ this is equivalent to $ \neg(P \vee Q) \vee R$ and the negation of this is $ (P \vee Q) \wedge \neg R.$ In words, "The stars are green [b]or[/b] the white horse is shining [b]and[/b] the world is [b]not[/b] eleven feet wide."

Thank you

[quote]If this is the case, you are wrong in 6 cases out of 8. (The 3rd and the 5th negations are correct).

 are statements,  is negation,  is conjunction (and),  is disjunction (or),  is implication (if - then),  is equivalence (if and only if). You are supposed to use the following 3 propositions of logic: [/quote]

Then wouldn't the statemente "I will do my homework and I will pass this class" be "I will not do my homework and I will not pass this class."?

The 1st case:

$ P$: I will do my homework.
$ Q$: I will pass this class.

"I will do my homework [b]and[/b] I will pass this class." in symbolic logic is $ P \wedge Q.$ Negation of this is $ \neg P \vee \neg Q.$ In words: "I will [b]not[/b] do my homework [b]or[/b] I will [b]not[/b] pass this class."

Thank you very much

What would the negation of "If and only if" be? -p V -q :maybe:

Is the negation of "If T is continuous, then T is bounded" T is continuous and T is not bounded
or T is not continuous and T is bounded. According to the last rule, wouldn't the second version be correct?

For d. I can eat dinner or go to the show, that would bee "I cannot eat dinner and I cannot go to the show,right?"

Thank you

Regards

1) Stop asking people on the internet to do your homework for you -- go ask you teacher for help instead.

2) Stop [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=39109070&t=187108]double-posting[/url].

3) Put your problems in an appropriate forum.

thanks everyone who helped. i am in college and am studying to be a teacher to help others. I would like to send my appreciation to all that assisted me. I do want to tell you all that I will be more than happy to  help you and your country when you need assistance. As all of us will age with time it will be me and my buds that will keep you and  yours alive with the coming years. I do hope you have enough health and insurance coverage to keep you alive!!!!!!!!

 :P

chocolatelover, would you mind if I asked you for a retraction of your last post ? (You have until 10:00 pm tonight to delete it.)

do we have A => not (B and C) <=> B and C => A

?


Putnam
floor function
function
modular arithmetic
algebra
college contests

What is the units (i.e., rightmost) digit of $\left\lfloor{\frac{10^{20000}}{10^{100}+3}\right\rfloor}$, where $\left\lfloor\cdot\right\rfloor$ represents the greatest integer function?


calculus
integration
probability
expected value
combinatorics unsolved
combinatorics

Let's imagine two sets A = {1,2,...,k} and B = {-1,-2,...,-k} for some natural k, then let's create n two-element sets X_1,X_2,...,X_n such that for each 0<i=<n X_i = {a,b} where a is from A and b is from B but |a|<>|b|. We know how do sets X_i look like and according to this we will choose the set C = {c_1,c_2,...,c_k} where |c_i|=i such that the number (denoted MAX) of sets X_1,X_2,...,X_n that have at least one common element with C is maximal. Determine the maximal constant 0<=c<=1 such that MAX>=[cn] for arbitrary n,k and sets X_1,X_2,...,X_n.
NOTE: [x] denotes the integral part of number x

Example:
k = 2, A = {1,2}, B = {-1,-2}
n = 4, X_1 = {-1,-2}, X_2 = {-1,2}, X_3 = {1,-2}, X_4 = {1,2}
we can choose C = {1,2} (in this case we have more possibilities)
the number X_i that have at least one common element with C is 3, X_1 and C have no common element.

From this example we can easily see, that c<=3/4, I think that c=3/4 is sufficient condition, but I can't prove it.
Could anybody help me with it? Thanks.

Probabilistic approach works perfect here! You are right $c=\frac{3}{4}$.

We will include $i$ in $C$ with probability $\frac{1}{2}$. Let's count expected number of sets $X_{k}$ s.t. $X_k\cap C\neq \varnothing$. Define aleatory variable $\xi_k$: $\xi_k=1$ iff $X_k\cap C\neq \varnothing$, and $\xi_k=0$ otherwise. Then we need to calculate $S=M(\xi_1+...+\xi_n)=M\xi_1+...+M\xi_k$. Since $X_k$ contain exactly two elements $a_k$ and $b_k$, $|a_k|\neq |b_k|$, then probability of $X_k\cap C=\varnothing$ is $\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}$, so $M\xi_k=\frac{3}{4}$. Finally, $S=\frac{3n}{4}$, hence result.

That's all!

Great method!  :D  :D  :D

Can someone tell me whether "aleatory variable" is right words here?  :blush:
My dictionary gives me too many variants and I don't know correct one.

Nice solution, Myth!

I haven't seen the phrase "aleatory variable" before.  (In fact, I had to look up the word "aleatory".)  I have seen the phrases "random variable" and "indicator variable" used instead.  (An indicator variable is a random variable that happens to be 0 or 1.)

Also, I have usually seen E() used for expected value, not M().

Oh, indeed, it is better to use "random variable". Thank you!
About M. It is standart notation here in Russia (I thought from "[b]M[/b]ean"). I didn't know that there is another notations :? Also we use D for dispersion.

That's interesting.  I am used to Var() for variance, which I assume is the same as dispersion.  Also Pr() for probability of an event.

[quote="Myth"]Yes, dispersion=variance, but my dictionary say that ди?пер?и?$\to$dispersion and in the same time variance$\to$ди?пер?и?. :lol:[/quote]

Unless you have a specific math dictionary you are better off with browsing the wolfgram math archive, e.g. [url=http://mathworld.wolfram.com/Variance.html]variance[/url].


function
trigonometry
calculus
derivative
algebra
polynomial
calculus computations

Generally, how do we extend the Taylor series definition to functions of several variable?
for example, how we find taylor series for $ f(x,y)\equal{}e^{xy}\cos{(x^2\plus{}y^2)}$? at (0,0)

Using [url=http://mathworld.wolfram.com/PartialDerivative.html]partial differentiation[/url], the definition is analogous.  That is, assume $ f(x, y)$ is a polynomial in $ x, y$ and take the partial derivatives with respect to $ x$ and $ y$ repeatedly.


algebra
polynomial
inequalities
calculus
derivative
calculus computations

How do you find what interval $x$ must be kept within in order to keep the difference between $\frac{1}{1-x}$ and its third order Taylor polynomial,
\[\frac{1}{2}+\frac{1}{4}(x+1)+\frac{1}{8}(x+1)^{2}+\frac{1}{16}(x+1)^{3}\], less than $0.05$?

My textbook has this,
\[R_{n}(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}\]
So in this problem, it becomes:
\[R_{n}(x) = \frac{\frac{24}{(1-x)^{5}}}{4!}(x+1)^{4}\]
So am I trying to maximise the $\frac{24}{(1-x)^{5}}$, and then substitute that value into the inequality and solve for $x$? I don't eve know what interval to maximise the derivative, and my textbook is confusing me  :oops:

The maximum value of the derivative depends on the choice of interval; trying to find the exact largest interval that works leads to a complicated algebraic equation. That's not worthwhile.

Instead, start by guessing an interval $I=[a-r,a+r]$ ($a=1$ for this problem). Find the maximum $M$ of the absolute derivative on this interval, then check whether the error is less than or equal to our tolerance $\epsilon$ ($.05$ in this case) on all of $I$. If it is, stop; I is the desired interval. If not, find $s$ such that $\frac{M}{(n+1)!}s^{n+1}=\epsilon$; $s$ will be less than $r$. In this case, the interval $[a-s,a+s]$ works, since $|f^{(n+1)}(x)|\le M$ for all $x\in [a-s,a+s]$ as well.

This procedure will usually find intervals that are smaller than the best possible interval. That's not a big deal; we could repeat with new guesses and get a closer approximation to the best possible interval.


[hide]$(x+y)^{2}=X^{2}+y^{2}+2xy=25+2xy\\25+2xy=1\Rightarrow 2xy=-24\Rightarrow xy=-12$ Now, we will have a few possibilities, but i am too lazy to post em  :D [/hide]

[quote="inom"]Solve the system
$x+y=1$
$x^{2}+y^{2}=25$[/quote]

[hide="A complete solution"]
Squaring the first equation, $x^{2}+2xy+y^{2}=1$
Subtracting from the second equation, $2xy=24$ and $xy=12$.

This is the point where alex__ib and anirudh left off. But continuing:

We solve the system
$x+y=1$
$xy=12$

Rewriting the first equation as $y=1-x$, we substitute:

$x(1-x)=12$
$x-x^{2}=12$
$x^{2}-x+12=0$
$(x-4)(x+3)=0$
So x can be $4$ or $-3$.

$(x,y)=(4,-3) \text{ or }(-3,4)$
[/hide]

Edit: Don't listen to me. I messed up pretty badly.

[quote="mathnerd314"][quote="inom"]Solve the system
$x+y=1$
$x^{2}+y^{2}=25$[/quote]

[hide="A complete solution"]
Squaring the first equation, $x^{2}+2xy+y^{2}=1$
Subtracting from the second equation, $2xy=24$ and $xy=12$.

This is the point where alex__ib and anirudh left off. But continuing:

We solve the system
$x+y=1$
$xy=12$

Rewriting the first equation as $y=1-x$, we substitute:

$x(1-x)=12$
$x-x^{2}=12$
$x^{2}-x+12=0$
$(x-4)(x+3)=0$
So x can be $4$ or $-3$.

$(x,y)=(4,-3) \text{ or }(-3,4)$
[/hide][/quote]
[hide]firstly, xy=-12
also (x-4)(x+3) doesn't equal x^2-x+12. It equals x^2-x-12, which is what your third equation should have been, if xy was 12, but it isn't 12 because, as other's have said, it's -12. so you should have x^2-x+12, but that doesn't factor to what you have. In fact it doesn't have real roots. So I'm now confused. somebody help me.[/hide]

[hide="Explanation"]Well, as shown before, we can square the first equation and subtract the second from it to obtain $xy=-12$. So, as said before, we just need to solve this system of equations: \[x+y=1\]  \[xy=-12\] We square the first one and subtract 4 times the second one from it to obtain
$x^{2}-2xy+y^{2}=(x-y)^{2}=49$.
Take the square root (assuming $x \geq y$), and we get
$x-y=7$.
Add this to $x+y=1$, and we can easily find that $x=4$, $y=-3$. However, we aren't done yet. Remember we assumed that $x \geq y$ earlier. If we go through the whole process again, this time assuming $y \geq x$, we'll find the solution $x=-3$, $y=4$, which is also valid, and those are our two only solutions.
(Alternatively, we could just notice that the two given equations were symmetric, so swapping the values of $x$ and $y$ will also lead to another solution.)[/hide]

[quote="anirudh"][quote="i_like_pie"][quote="inom"]Solve the system
$x+y=1$
$x^{2}+y^{2}=25$[/quote]
[hide]$x=1-y,\ x=\sqrt{25-y^{2}}$

$1-y=\sqrt{25-y^{2}}$

$y^{2}-2y+1=25-y^{2}$

$2y^{2}-2y-24=0$

$y^{2}-y-12=0$

$y=\frac{1\pm\sqrt{1+48}}{2}$

$y=\frac{1\pm7}{2}$

$y=\{-3,\ 4\}$[/hide][/quote]

you don't need the quadratic--just factor[/quote]
I know, but I like doing it the "general" way.

[quote="i_like_pie"][/quote][quote="anirudh"][quote="i_like_pie"][quote="inom"]Solve the system
$x+y=1$
$x^{2}+y^{2}=25$[/quote]
[hide]$x=1-y,\ x=\sqrt{25-y^{2}}$

$1-y=\sqrt{25-y^{2}}$

$y^{2}-2y+1=25-y^{2}$

$2y^{2}-2y-24=0$

$y^{2}-y-12=0$

$y=\frac{1\pm\sqrt{1+48}}{2}$

$y=\frac{1\pm7}{2}$

$y=\{-3,\ 4\}$[/hide][/quote]

you don't need the quadratic--just factor[/quote][quote="i_like_pie"]
I know, but I like doing it the "general" way.[/quote]

It'll probably take a long time :D


If the sides of a triangle are tripled, then the new area is what
percent of the original area?

Let's say the side lengths of a right triangle are 3, 4, and 5. If they're tripled, the new side lengths are 9, 12, and 15. The area of the original triangle is 6, and the area of the new triangle is 54, so we see that the areas increased $ \boxed{900\%}$.

Since the sides are tripled, the area must be 9x the original.  Thus, the answer is $ \boxed{900}$.

Quicker way. :P

Why must the area be 9 times the original? Please clarify.

isabella:Your use of an example is great, but how can you guarantee that it works for all triangles? (Besides the fact that it obviously does)

well you could go:
All triangles can be drawn so that they have base b and altitude a.
Then their area is ab/2
So if their side lengths are tripled, corresponding lengths of the triangle will also be tripled, so we have base 3b and altitude 3a
so the area is 9ab/2

Another way: let $ P_{ABC}$ be the area of $ \triangle ABC$. Then $ P_{ABC}\equal{}\frac{absin\gamma}{2}$. If we triple the sides, in formula for area $ ab$ is multiplied $ 9$ times and $ \frac{sin\gamma}{2}$ is unchanged. So area is multiplied $ 9$ times.

The ratio of the areas of similar figures is the square of the ratio of similitude.


abstract algebra
topology
geometry
college
linear algebra
group theory

Hey guys,

I was a non-math major as an undergraduate, and I am planning to attend grad school in pure math next year. I'm currently taking a theoretical computer science master's degree at Waterloo. I plan to do another master's degree at Waterloo in math, and then apply for various PhD programs. As an undergrad, I took the following math courses:

- analysis I-II (full year 2nd year, basic 1-variable upto uniform convergence)
- linear algebra (2nd year)
- group theory (3rd year)
- rings and modules (3rd year)
- metric space topology (3rd year)
- differential equations (3rd year)
- complex analyis (3rd year)

So I have not taken the standard 4th year courses that most undergraduates took at my university.

I had good grades in these courses, however I do not consider myself a particularily good problem solver (relative to the posters here  :) ). I struggle on problems that require one to go beyond the course material and come up with creative and insightful solutions. I also have never done contest level problem solving throughout high school or college. So, my question is: Should I spend the time between now and grad school

1) Learning more advanced math (self-study is the only option)
2) Improving my general problem solving abilities, and working through problem solving contests and books.

Will having very good problem solving skills be more beneficial than indepth knowledge of advanced topics for grad school? Thanks for any comments.

A both-and way to achieve your goals is to work through "transitions" books that have a high problem-solving emphasis, for example

[url=http://www.amazon.com/exec/obidos/tg/detail/-/0130144126/learninfreed]Mathematical Thinking: Problem-Solving and Proofs[/url] by John D'Angelo and Douglas West.

Be sure to go to the book's Web site to find the errata list. This book is a good overview of several areas of math, and it has a very strong problem-solving emphasis.

Hope this helps!

tokenadult: Thank you for the link. The book is available in the library.

Do you know of any slightly more advanced problem solving type books covering undergraduate mathematics. I found the book "Berkeley problems in mathematics" which seems to cover undergraduate material. This appears to be purely a collection of problems though. Will this book be too advanced for me?

You mentioned you were a non-math major, and then listed a pretty impressive set of math courses that you have already taken. I think you could get a lot out of the book of Berkeley problems, which is another good book. The Berkeley book has less exposition, but if I remember correctly it does show solutions for many (all?) of the problems included.

[quote="Isura"] So, my question is: Should I spend the time between now and grad school

1) Learning more advanced math (self-study is the only option)
2) Improving my general problem solving abilities, and working through problem solving contests and books.

Will having very good problem solving skills be more beneficial than indepth knowledge of advanced topics for grad school? Thanks for any comments.[/quote]

In graduate school, it will be much more beneficial to have in-depth knowledge of advanced topics.  Actually, even surface knowledge of advanced topics is more useful than no knowledge of them; it helps you choose which courses to take when you can really only take three and there are 34 interesting-looking classes being offered each term.  The specific skills that are honed through contest-style practice are not particularly useful for graduate-level mathematics.

In self-studying graduate-level mathematics, I strongly recommend trying to do ALL of the problems in whichever text you're reading.  As a graduate student, I spent part of each summer going over the details of texts I'd already studied, and the bulk of my understanding came while doing additional problems.

The specific tricks used to solve contest problems aren't a big deal, but problem solving in general is an essential skill- many problems won't broadcast what you need to do to solve them.
That list of courses is close enough to being a math degree that you should be sufficiently prepared for graduate school. If there are a few holes in what the university you go to expects, you can take one or two undergraduate courses to fill them.


function
limit
integration
calculus
real analysis
real analysis unsolved

Let $ u\colon\mathbf R^N\to\mathbf R$, $ N\ge1$, be a harmonic function such that
	\[ \lim_{n\to\infty} \int_{1 \le |x| \le 2} |u(nx)|\, dx \equal{} 0.
	\]
	Show that $ u \equiv 0$.

Note that the hypothesis doesn't directly say that $ u$ is bounded, but it comes very close to saying that.

Idea: Use the mean value property to represent $ u$ as the average over a ball. By taking the difference of two balls, show that $ u$ can be represented as its average over an annular region. Differentiate that to get a representation of $ \nabla u$ as an integral involving $ u$ over an annular region. Use the given condition to show that $ \nabla u\equal{}0,$ so $ u$ is constant. Once we have that, it won't be that hard to show that the constant is zero.

Not a fully elaborated proof, but it should be possible to fill in the details.


I don't consider this as "publicly" distributing the files because it is password protected, and I will only give the password to some people.

2000-2002 State MATHCOUNTS:

If you want the password, please private message me, and I will then decide if I should give it to you or not (most of the time I'll say yes though).

[size=67][color=red]AIME15: Removed files - use email/PM instead, please.[/color][/size]

When trading tests/giving tests, please use email or PM.


vector
group theory
abstract algebra
real analysis
real analysis unsolved

If A and B are convex sets then A intersect B is also convex?

I believe that I have proved this question by contradiction but my original train of thought was to try to prove by:

Proof: Let two points w, x be elements of A and the set A be convex. Likewise let there be two points y, z be elements of B and the set B be convex.

Case 1: WOLG Let w-x-y-z

The intersection of A and B is equal to the empty set.

Case 2: WOLG Let w-y-x-z

The intersection of A and B is equal to the line made by points x, y.
Therefor the intersection is convex

Case 3: WOLG Let w=y and x=z

The interxection of A and B is equal to A equal to B
Therefor the intersection is convex

My question is in Case 1 is the empty set considered convex allowing me to say the set is convex. And if it is convex does this prove the original question?

In case you are interested my proof by contradiction is:

If A and B are convex sets then A intersect B is also convex?

Proof: Let x, y be elements of A intersect B

Let there be a point c that lies on the line between x and y but not in the intersection of A and B. WOLG the point c is not an element of A since it is not in the intersection of A and B. But this is a contradiction to the set A being convex. Therefor A intersect B is convex.

I think that a simpler path to solve this problem is to consider the definition of convexity. A set $ A$ is convex if for every two points $ x,y$ of $ A$ the line segment $ \overline{xy}$ is contained in $ A$. If you, now, have two convex sets $ A$ and $ B$ and $ x,y$ are two points in the intersection $ A\cap B$, then we can deduce that $ \overline{xy}$
is contained in $ A$ (since $ x,y\in A$) and it is also contained in $ B$ (since $ x,y\in B$). Therefore $ \overline{xy}$ is contained in $ A\cap B$. Note that this proof is very similar to your proof by contradiction.

The truth is that the intersection of  [i]any[/i] family of convex sets (in any vector space) is convex. Properties
that pertain to a "belongs to" statement usually hold for arbitrary intersections. Examples:

The intersection of any family of subgroups of a given group is a subgroup.
The intersection of any family of subspaces of a given vector space is a subspace.
The intersection of any family of $ \sigma$-algebras over a set is $ \sigma$-algebra.
The interseciton of any family of topologies on a set is a topology.

and the list goes on and on...

I don't see how you can deduce like xy from the fact that both points are in both sets (I know it’s true but I don't see how you can just make that claim?). That’s why at first I used the 4 points and the different orders that they could be arranged in. Although I wasn’t sure if the empty set was considered convex allowing me to prove the convexity of the intersection. I guess my real question is can the empty set be considered convex?

Jondrette I was reading through your post again and I think I get what you are saying now. Because A is convex and B is convex and the points x and y lie in both sets the points between also lie in both sets (the intersection). Thats how you deduce that line xy is in the intersection.
-correct me if I’m wrong.

Answer to my own question if anyone is interested.

The empty set is considered convex but that still doesn’t allow me to prove that A intersect B is convex since there is no way to say for sure if the sets are linear.

My proof by contradiction is correct but I needed one more statement explaining why it was a contradiction. So the complete correct answer is (for the approach I took):

Proof: Let x, y be elements of A intersect B

Let there be a point c that lies on the line between x and y but not in the intersection of A and B. WOLG the point c is not an element of A since it is not in the intersection of A and B. But this is a contradiction to the set A being convex since c is a point between x and y not in A. And as the definition of convexity specifies that for any two random points picked in a set all points between also exist in the set. Therefore A intersect B is convex.

Yes, your proof by contradiction is correct, and if we "turn it around" to avoid the unnecessary contradiction what we get is exactly the argument Jondrette gave ;)

We could call the empty set "vacuously convex": to say that $ A$ is convex is to say that $ \forall x, y \in A \ldots$, and any statement that begins $ \forall x \in \emptyset$ is true.


A point $P$ is given inside an acute-angled triangle $ABC$. Let $A',B',C'$ be the orthogonal projections of $P$ on sides $BC, CA, AB$ respectively. Determine the locus of points $P$ for which $\angle BAC = \angle B'A'C'$ and $\angle CBA = \angle C'B'A'$

Let $ABC$ be inscribed in unit circle of complex plane. Then $a'=\frac12(p+b+c-\overline p bc)$, $b'=\frac12(p+c+a-\overline pca)$ and $c'=\frac12(p+a+b-\overline pab)$. Since triangles $ABC$ and $A'B'C'$ are simular and with same orientation we must have $\frac{a-b}{a'-b'}=\frac{b-c}{b'-c'}$, from which we easily get $\overline{p}(a-c)=0$. Therefore only such point is circumcenter of triangle $ABC$.

Can we not use complex numbers?  :maybe:

let $x,y,z$ be $AP,BP,CP$. from the condition we obtain that $\triangle ABC\sim \triangle A'B'C'$ and it's well known that the sides of the pedal triangle are
$ax/2R, by/2R, cz/2R$. then, we obtain that $x/2R=y/2R=z/2R$, ie. $x=y=z$ which implies that $P$ must be the circumcenter.


How can you show the product rule for two functions $ f,g$ in $ \mathbb{R}^n$?
that is: $ d(fg)_p \equal{} f(p)dg_p \plus{} g(p)df_p$, where $ p$ is a point in $ \mathbb{R}^n$

Of course this is easy for f,g in $ \mathbb{R}$, where you add and subtract $ f(x\plus{}h)g(x\plus{}h)$. But it is not clear to me how this would carry over for $ f,g$ in $ R^n$.

Are $ f,g$ from $ \mathbb R^n$ to $ \mathbb R$¿
But please ask this in the calculus forum: http://www.mathlinks.ro/index.php?f=296 .

ZetaX: Yeah, $ f: R^n\rightarrow R$, and $ g$ too, and each side of the eq. is a $ 1\times n$ matrix

If you try to echo the proof of the product rule for $ \mathbb{R}$...

$ D(fg)_p\equal{}lim_{h \rightarrow 0}\frac{\|f(x\plus{}h)g(x\plus{}h)\minus{}f(x)g(x)\|}{\|h\|} \equal{}
lim_{h \rightarrow 0}\frac{\| f(x\plus{}h)g(x\plus{}h)\plus{}f(x\plus{}h)g(x)\minus{}f(x\plus{}h)g(x)\minus{}f(x)g(x) \|}{\|h\|} \equal{}
lim_{h \rightarrow 0}\|f(x\plus{}h)\frac{g(x\plus{}h\minus{}g(x))}{h}\plus{}g(x)\frac{f(x\plus{}h)\minus{}f(x)}{h}\|$.

I am not sure how to proceed from here.. how does limit operate with $ \|\cdot\|$?.


integration
calculus
real analysis
real analysis unsolved

[quote="anik_andrew"][quote="aopsmember"]hello everyone

Is there a way i can write the following expression in terms of Q?

$\int\limits_{Q}^{\infty} f(x)\; dx = \frac{p}{r+b}$

thanks[/quote]

I can't understand what do you want... :blush:
Do you want to write $Q=\ldots?$[/quote]

yes, I want to write the expression in the form of Q = .

sorry for the confusion

[quote="anik_andrew"]And then what is it $p,r,b$ -constant? Or they are depend from $Q$?[/quote]

sorry about that.  p,r and b are constants.

:? Sorry. Then i don't knew...
I think that this is imposible in general case.... :(

Hmm, well if you use the fundamental theorem of calculus and assume that f has an inverse and that, say, f at infinity is 0. Although this leads to a trivial case.


am I right to say Potential energy is the energy stored within a system that can be transferred to the energy of some object in the system by a force in the system doing work on the object?

also am I right to say the Change in Potential Energy of a system is equal to the negative of the work done on an object in the system by a conservative force in the system?

and am I right to say a nonconservative force is one which when doing work to an object in a system causes the system to lose total mechanical energy.

[quote="Mango 070"]
am I right to say Potential energy is the energy stored within a system that can be transferred to the energy of some object in the system by a force in the system doing work on the object?
.[/quote]
Perfectly right
[quote="Mango 070"]
also am I right to say the Change in Potential Energy of a system is equal to the negative of the work done on an object in the system by a conservative force in the system?
.[/quote]
Of course you're right
[quote="Mango 070"]
and am I right to say a nonconservative force is one which when doing work to an object in a system causes the system to lose total mechanical energy
.[/quote]
Not completely right.
Mechanical Energy might increase, decrease or stay the same under the action of a non-conservative force.

ok, what exactly is a nonconservative force then?

Non-conservative forces come about when you ignore information. For example, if you don't account for heat generated due to friction as being part of total energy, then it would seem that the force of friction results in loss of energy. In reality, energy [i]is[/i] being conserved, but to keep track of that energy, you would have to model the scenario at a microscopic level. In many mechanical problems, you don't particularly care about how the dissipated heat is distributed among the many individual particles -- you can make a simpler and equally reliable model of how much relevant energy is being "lost" into an unusable form.


Help me solve this problem.
For given N and pair (A0, B0) find all pairs (A, B) such that for any integer X and Y if A0X+B0Y is divided by N then AX+BY is divided by N too (0<=A,B<N<=10000).

Suppose AX+BY is 0 mod N
If A and N are coprime, we can multiply by (A0/A) mod n to get A0X + B/A Y = 0 mod N, so (B-B/A)Y = 0 mod N. Thats easy to solve.

Theres a few cases to check here with coprime/not coprime numbers, but I'll leave that to you.


Prove that the maximum value of xy takes place when x+y=constant, if x=y

[hide]Expanding on Altheman's idea, by AM-GM:
$\frac{x+y}{2} \ge \sqrt{xy}$

We want equality (to maximize $xy$), or
$\frac{x+y}{2} = \sqrt{xy}$

Simplifying we have $(x-y)^2 = 0$
So $x = y$

There you have it.[/hide]

[quote="prototypex24"]Huh? I don't quite get you... if x=y, x+y=2x=2y so x=y is sufficient for xy to be maximum[/quote]

Yeah, that's what you have to prove :D


geometry
geometric transformation
abstract algebra
IMO
IMO 2007

Hi all,

Are the problems in imo given in english only?

The problems are translated into a number of different languages.  You can see all the different language versions of this year's IMO problems here:

http://www.imo2007.edu.vn/index.php?module=problem

It's kind of fun to try a language you don't know and see how much you can figure out from cognates and context.  (Obviously this works more easily with some languages than others!)


inequalities
geometry
circumcircle
vector
geometric transformation
reflection
rectangle

Let a, b, c and R be sides and circumradius of an arbitrary triangle ABC. Show that

$ a^2 + b^2 + R^2 \ge c^2$

when does the equality hold?

I have a trigonometric solution, but I want to see a pure geometric one.

Well,this one is not too difficult.Expand this one and we have the result:
[tex](OA^{->}+OB^{->}-OC^{->})^2 \geq 0 [/tex]
[tex]\Leftrightarrow R^2+a^2+b^2 \geq c^2 [/tex]
The equality holds when  $ A=B=30^0;C=120^0 $
 :)
{Can vector be called pure geometric? :?  :maybe: }

Very nice approach. Congratulations! But I'm still waiting to a pure geometric solution.

Here is a pure geometric proof:

Let O be the circumcenter of triangle ABC, and Z the reflection of O in the side AB. We will prove

$\text{\textbf{(1)}\ \ \ \ \ \ \ \ \ \ }CZ^2=R^2+\left( a^2+b^2-c^2\right) $.

Of course, then it will follow that $a^2+b^2+R^2\geq c^2$ with equality if and only if C = Z. One can easily show that C = Z is equivalent to A = B = 30° and C = 120°, so that the proof will be complete.

Remains to verify [b](1)[/b].

Since O lies on the perpendicular bisector of AB, the point Z is the reflection of O in the midpoint E of AB. In other words, E is the midpoint of OZ, too. Let D be the orthogonal projection of C on the line OZ. Then,

$CZ^2-R^2=CZ^2-CO^2=\left( CD^2+DZ^2\right) -\left(CD^2+DO^2\right)$
$ =DZ^2-DO^2=\left( DZ-DO\right) \cdot \left( DZ+DO\right) $.

Now,

$DZ-DO=OZ=2\cdot OE$

and

$DZ+DO=\left( DO+OZ\right) +DO=\left( DO+2\cdot OE\right) +DO=2\cdot DO+2\cdot OE=2\cdot DE$,

so that

$CZ^2-R^2=\left( 2\cdot OE\right) \cdot \left( 2\cdot DE\right) =4\cdot OE\cdot DE$.

Now, if F is the foot of the C-altitude of triangle ABC, then the quadrilateral CFED is a rectangle, and $DE=CF=h_{c}$, where $h_{a}$, $h_{b}$ and $h_{c}$ are the lengths of the three altitudes of triangle ABC.

If L is the foot of the A-altitude of triangle ABC, then < CAL = 90° - < ACL = 90° - C, while the circumcenter O of triangle ABC is the apex of the isosceles triangle AOB, thus yielding < OBE = < OBA = 90° - < AOB / 2 = 90° - 2C / 2 = 90° - C. Hence, < OBE = < CAL. Also, < OEB = < CLA = 90°. Hence, triangles OBE and CAL are similar, and

$\frac{OE}{EB}=\frac{CL}{LA}$.

Hence, we can continue our calculation above:

$CZ^2-R^2=4\cdot OE\cdot DE=4\cdot OE\cdot h_{c}=4\cdot OE\cdot \frac{h_{c}}{h_{a}}\cdot h_{a}$.

Now, it is well-known that

$\frac{h_{c}}{h_{a}}=\frac{a}{c}$;

thus,

$CZ^2-R^2=4\cdot OE\cdot \frac{a}{c}\cdot h_{a}=2\cdot OE\cdot \frac{a}{c/2}\cdot h_{a}=2ah_{a}\cdot \frac{OE}{c/2}$
$=2ah_{a}\cdot \frac{OE}{EB}=2ah_{a}\cdot \frac{CL}{LA}=2ah_{a}\cdot \frac{CL}{h_{a}}=2a\cdot CL$.

Finally, the generalized Pythagoras theorem states that

$CL=\frac{a^2+b^2-c^2}{2a}$,

so that

$CZ^2-R^2=2a\cdot CL=a^2+b^2-c^2$,

thus proving [b](1)[/b].

  Darij

[quote="shobber"][quote="Anh Cuong"]Well,this one is not too difficult.Expand this one and we have the result:
[tex](OA^{->}+OB^{->}-OC^{->})^2 \geq 0 [/tex]
[tex]\Leftrightarrow R^2+a^2+b^2 \geq c^2 [/tex]
The equality holds when  $ A=B=30^0;C=120^0 $[/quote]
I don't understand your vector solution... Can anyone please explain it?

And also, I'd like to ask Darij, how did you come up with this solution!? It is amazing![/quote]
Simply expanding it.
I don't know whether Darij found the solution from this, anyway, it's one of the possible way by expressing the vector OA+OB-OC geometrically.

[quote="mecrazywong"]I don't know whether Darij found the solution from this, anyway, it's one of the possible way by expressing the vector OA+OB-OC geometrically.[/quote]

Indeed, this is exactly how I did it: $\overrightarrow{OA}+\overrightarrow{OB}-\overrightarrow{OC}=\overrightarrow{CZ}$.

By the way, here is an alternative way to write up the vectorial solution: Applying [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15558]the fairly general geometric inequality[/url] to the point M = O and the reals x = 1, y = 1, z = -1, we get

$\left(1+1+\left(-1\right)\right)\cdot\left(1\cdot OA^2+1\cdot OB^2+\left(-1\right)\cdot OC^2\right)$
$\geq 1\cdot\left(-1\right)\cdot a^2+\left(-1\right)\cdot 1\cdot b^2+1\cdot 1\cdot c^2$.

This simplifies to

$OA^2+OB^2-OC^2\geq -a^2-b^2+c^2$.

But since O is the circumcenter and R is the circumradius of triangle ABC, we have OA = OB = OC = R, and thus this becomes $R^2+R^2-R^2\geq -a^2-b^2+c^2$, what simplifies to $R^2\geq -a^2-b^2+c^2$. Equivalently, $a^2+b^2+R^2\geq c^2$.

  Darij

I note the middlepoint M of the AB and symetrical point Z of the point O w.r.t. the point M. Then $4CM^2=2(CO^2+CZ^2)-OZ^2$
$\Longrightarrow 2(a^2+b^2)-c^2=2R^2+2CZ^2-4R^2 [1- \sin ^2 C]=2R^2+2CZ^2-4R^2+c^2\Longrightarrow$
$a^2+b^2-c^2=-R^2+CZ^2\Longrightarrow  CZ^2=R^2+a^2+b^2-c^2\ge 0 \Longrightarrow a^2+b^2+R^2\ge c^2.$

[quote="siuhochung"]what's Z?[/quote]
Darij wrote:
Let O be the circumcenter of triangle ABC, and Z the reflection of O in the side AB. We will prove

I am sorry, the point Z is the reflection of the point O by the line AB. Thank you, Dariij !

It is O.K. and will try a generalization of the form $a^2 +xb^2 +yc^2\ +mR^2\ge 0$, where $x,y\in R,\ m>0$. The wellknown inequality $a^2+b^2+c^2\le 9R^2$ isn't of this form !
I managed. See http://www.mathlinks.ro/Forum/viewtopic.php?t=46979

We may rewrite the inequality (2) in the form
$pa^2+qb^2+4R^2 \ge \frac{pq}{(\sqrt{1+p+q}-1)^2}c^2$.

Indeed Levi, your inequality is a nice particular case of (2) - case $p+q=8, p=4x, q=4y$:
$xa^2+yb^2+R^2 \ge xyc^2$, for $x+y=2$.

Another nice particular cases are the following:

$pa^2+qb^2+4R^2 \ge pqc^2$, for $p+q=3$;

$xa^2+yb^2+R^2 \ge \frac{xy}{4}c^2$, for $x+y=6$.

More general, for $x>0$ and $y>0$ such that $x+y=\frac{m^2-1}{4}$, $m>1$, we have
$xa^2+yb^2+R^2 \ge \frac{4xy}{(m-1)^2}c^2$.

[quote="levi"]What is your geometrical idea for your open generalization ? I didn't see. May be you present here or in a message. If not, I must discover it. Indeed, I like it.[/quote]

I think it's not open... usually if vasc does not say he conjectures something, he must have proved it before he posted it... and till now, you can see tons of unsolved problems on the forum proposed by vasc.... :P

Siuhochung, my the geometrical idea for the generalization proposed problem
(see http://www.mathlinks.ro/Forum/viewtopic.php?t=46979 ) is
selection of the point Z, i.e. the reflection of the point O by the any point which belongs to AB.
That's it ! You not to let go. I have not a place to twist because of you, ah me ! Savely, V.N. (lLevi).

I haven't a geometrical solution to the inequality
$\displaystyle pa^2+qb^2+4R^2 \ge \frac{pq}{(\sqrt{1+p+q}-1)^2}c^2$.
My solution is algebraic and a little trigonometrical ($4R^2=c/cosC, c^2=a^2+b^2-2abcosC)$.
Equality doesn't occur if $pq \le \sqrt{1+p+q}({\sqrt{1+p+q}-1)^2}$.


trigonometry
logarithms
LaTeX
function
trig identities
algebra
log identities

These are some problems from my honors trig class, I need help with them mostly because I'm not seeing the process on some of them and on others it has just been a while since I've worked with log and ln. Please use mostly trig identities and give fairly detailed solutions as i don't want to have to ask for help very often.

1.)$\log 10^{\tan{x}}=\tan{x}$

2.)$10^{\log{\mid\sin{t}\mid}}=\mid\sin{t}\mid$

3.)$-\ln \mid \sec{\theta}-\tan{\theta}\mid=\ln\mid\sec{\theta}+\tan{\theta}\mid$

4.)$\ln \mid \csc{x}-\cot{x}\mid=-\ln\mid\csc{x}+\cot{x}\mid$

5.)\[\frac{\sin{\alpha}\cos{\beta}+\cos{\alpha}\sin{\beta}}{\cos{\alpha}\cos{\beta}-\sin{\alpha}\sin{\beta}}=\frac{\tan{\alpha}+\tan{\beta}}{1-\tan{\alpha}\tan{\beta}}\]

Thanks for the log identities, I needed those but i feel really stupid now cuz i completely could have done #5 if I would have thought of using the trig sum and difference identities.

[hide="#5"]
We use the following identities:
$1.\;\sin(\alpha+\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta$
$2.\;\sin(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$
$3.\;\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$
$4.\;\tan\theta=\frac{\sin\theta}{\cos\theta}$

Using Identities $1$ and $2$, we have
$\frac{\sin(\alpha+\beta)}{\cos(\alpha+\beta)}$

Using Identity 4, we have
$=\tan(\alpha+\beta)$

Using Identity 3,
$\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$[/hide]

Thanks, after I used the trig sum identites for #5 it was a piece of cake but it's the first 4 I'm mainly having problems with cuz i haven't used logs in quite a while.

[quote="aahmadi"]Something that might help you is that you can choose a base, and use both sides of the equation as your exponent. For example:
ln(x)=5
e^(ln(x))=e^5
Now, since ln(x) is the same as log base e of x, then when e is your base and ln(x) is the exponent, then you are left with x.
x=e^5

So if we have:
log base 3 of x=5
3^(log base 3 of x)=3^5
x=243

Likewise, if he have:
3^x=pi
we could could take log base 3 of both sides of the equation which results in:
log base 3 of (3^x)=log base 3 of pi
x=log base 3 of pi

That should help take care of 1 and 2.

Btw, can anyone explain to me how to change my notation so that it is more mathy (like on ms equation editor).[/quote]

check out the LaTeX link on the left of the AoPS page

[quote="t0rajir0u"]A short list of the identities you might need. ;)

[b]Logarithmic Identities:[/b]

$\log x = \log_{10}x$
$\log_{b}a = c \Leftrightarrow a = b^{c}$

Corollary:  $\log_{b}(b^{a}) = a$
Corollary:  $b^{ \log_{b}a}= a$

$\log_{b}(a \cdot c) = \log_{b}a+\log_{b}c$
$b^{0}= 1 \Leftrightarrow \log_{b}1 = 0$
$\log_{b}(a^{c}) = c \log_{b}a$

Special case:  $\log_{b}(a^{-1}) =-\log_{b}a$

[b]Trigonometric Identities:[/b]

$\tan \theta = \frac{\sin \theta}{\cos \theta}$
$\sec \theta = \frac{1}{\cos \theta}$
$\csc \theta = \frac{1}{\sin \theta}$
$\cot \theta = \frac{\cos \theta}{\sin \theta}= \frac{\csc \theta}{\sec \theta}$

$\sin (\alpha+\beta) = \sin \alpha \cos \beta+\cos \alpha \sin \beta$
$\cos (\alpha+\beta) = \cos \alpha \cos \beta-\sin \alpha \sin \beta$
$\tan (\alpha+\beta) = \frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}$[/quote]
A couple things. First, here are some other important log identities that can be proven with the ones you provided but are still important.
$\frac{log_{a}x}{\log{a}y}=\log_{x}y$ (known as the change of base property)
$\log{a}-\log{b}=\log{\frac{a}{b}}$ (known as the division property)
$\log_{a}b=\frac{1}{\log_{b}a}$ (Known as the inverse property
$(\log_{a}b)(\log_{c}d)=(\log_{a}d)(\log_{c}b)$ (known as the lazy jim property)
Special case of lazy jim:
$(\log_{a}b)(\log_{b}c)=\log_{a}c$ (known as the chain rule)
Also, use the AoPS wiki for these identities as well, it is pretty good.

Oic how to do 3 and 4 now (no thanks to anyone else :mad: ...just kidding, lol), I must have been sleep typing that day. Just do the inverse function to both sides of equation ($e^{x}$) and then know your pythagorean identities. Btw bpms, is there another name name for that "chain rule" identity? Since it's kind of already used for calculus.....jw. Thanks for the identities anyways, I was unaware of your last one.

[quote="fireemblem13"][quote="bpms"]
$\frac{\log_{a}x}{\log_{a}y}=\log_{x}y$ (known as the change of base property)[/quote]
that should be $\frac{\log_{a}x}{\log_{a}y}=\log_{y}x$[/quote]
Typo, but I can't edit now :(
Can a mod please edit my post?


integration
calculus
calculus computations

Please help me ...i try several times..Please>.<

i have another question.
i have problem about finding the arc length..>.<

[quote="Thor"][quote="slammer"]i want to ask how to write some math's symbol on this forum?[/quote]
http://www.mathlinks.ro/Forum/viewtopic.php?t=5261[/quote]
thanks your help

how to solve this problem $\int\frac{x^2}{(1+x^2)^2}dx$ use integration by parts


A set $S$ consists of $1992$ positive integers among whose units digits all $10$ digits occur. Show that there is such a set $S$ having no nonempty subset $S_{1}$ whose sum of elements is divisible by $2000$.


inequalities
trigonometry
geometry unsolved
geometry

hello, we will prove this nice inequality and we use
$ \sin(A/2)=\sqrt{\frac{(s-b)(s-c)}{bc}}$
$ \sin(B/2)=\sqrt{\frac{(s-a)(s-c)}{ac}}$
$ \sin(C/2)=\sqrt{\frac{(s-a)(s-b)}{ab}}$
and the right-hande side is after some algebra
$ \frac{4R^2-8Rr+3r^2}{4R^2}=1-\frac{2r}{R}+\frac{3}{4}\left(\frac{r}{R}\right)^2=$
$ 1-\frac{16A^2}{abc(a+b+c)}+\frac{3}{4}\left(\frac{8A^2}{(a+b+c)(abc)}\right)^2$
after some ugly computation we get the equivalent inequality
$ {\frac{(a+b-c)(a+c-b)(b+c-a)(-b^3+2b^2c+2ab^2-9abc+2c^2b+2a^2b-a^3-c^3+2c^2a+2a^2c)}{a^2b^2c^2}\geq0}$.
Which is true because the second factor can be written as
$ a^2(c+b-a)+b^2(c+a-b)+c^2(a+b-c)+ab(a+b-c)+ca(c+a-b)+bc(b+c-a)\geq0$
Sonnhard.

[quote="Nebraska boy"]Show in any triangle the inequality

$ sin^4\frac {A}{2} \plus{} sin^4\frac {B}{2} \plus{} sin^4\frac {C}{2} \le \frac {4R^2 \minus{} 8Rr \plus{} 3r^2}{4R^2}$[/quote]

$ LHS\equal{}(\frac{1\minus{}\cos A}{2})^2\plus{}(\frac{1\minus{}\cos B}{2})^2\plus{}(\frac{1\minus{}\cos C}{2})^2\equal{} \frac{1}{4}(3\minus{}2\frac{R\plus{}r}{R}\plus{}$ $ 3\minus{}\frac{s^2\minus{}r^2\minus{}4rR}{2R^2})\equal{} 1\minus{}\frac{s^2\minus{}r^2}{8R^2} \le \frac {4R^2 \minus{} 8Rr \plus{} 3r^2}{4R^2}$

$ \Leftrightarrow s^2 \ge 16rR\minus{}5r^2$ - which is obviously true : $ IG^2 \equal{} \frac{1}{9}( s^2\plus{}5r^2\minus{}16rR)$

Show in any triangle the inequality
$$1- \frac{21r}{16R}+\frac{r^2}{8R^2}\geq \frac{a}{b+c} sin^2\frac {A}{2} + \frac{b}{c+a}  sin^2\frac {B}{2} + \frac{c}{a+b} sin^2\frac {C}{2} \geq 1- \frac{3r}{2R}+\frac{3r^2}{8R^2}+\frac{r^3}{4R^3}  \geq\frac{1}{2} (1-\frac{r^2}{R^2} ) \geq\frac{1}{2} (1-\frac{r}{2R} )\geq\frac{3}{8}$$
[size=75]SXTB (2)2020[/size]
[url=https://artofproblemsolving.com/community/c6h143963p815696]h[/url]


Ο παππούς είναι συνταξιούχος μαθηματικός, πολύ ιδιότροπος και ανυπόμονος.Θέλει τα αυγά του βρασμένα ακριβώς 15 λεπτά. Αλλιώς..........   Σας ζητά να του ετοιμάσετε ένα και το θέλει γρήγορα.Στο σπίτι όμως δεν υπάρχει κανένα ρολόι ή χρονόμετρο.  :(  Μόνο δύο κλεψύδρες. Μία που χρειάζεται 11 λεπτά για να αδειάσει και μία άλλη που χρειάζεται 7 λεπτά.Πώς θα μπορέσουμε να βράσουμε τα αυγά όσο τα θέλει ο παππούς; Ποιος είναι ο πιο γρήγορος τρόπος ;

[hide="ok then"]Γυρίζουμε ταυτόχρονα τις κλεψύδρες των 7 και 11.

Μετά από 7 λεπτά η 11άρα έχει ακόμη 4'
Γυρίζουμε την 7άρα.
Μετά από 4 λεπτά (δηλαδή μόλις αδειάσει η 11) στην 7άρα έχουν μείνει 3'.
Επομένως, αν αναποδογυρίσουμε την 7άρα θα μπορέσουμε να μετρήσουμε ακόμη 4'

$\begin{array}{c|ccc} 0\ min & 11 & 7\\ 7\ min & 4 & 0 \\ 7\ min & 4 & 7 \\ 11\ min & 0 & 3 \\ 11\ min & 0 & 4 \\ 15\ min & - & 0 \\ \end{array}$[/hide]

Αυτή είναι η πιο ορθή λύση. Υπάρχουν όμως και άλλες λύσεις [hide]Δεν χρησιμοποιούν όμως τον ελάχιστο χρόνο[/hide]. Μπορείτε να τις βρείτε ;;


Can any positive rational number can be written as the ratio of two triangular numbers? (i.e. of the form $ \frac {a(a \plus{} 1)}{b(b \plus{} 1)}$ for positive integers a and b).

Not sure if this is the right forum to ask, but I was thinking about this for the past few days and haven't made much progress, and I also haven't found any results on it through AoPS search or google.
Although, my math skills aren't very good, so I may be missing something obvious and it may even deserve to be in pre-olympiad. :P

Edit: Whoops, I meant to place this in the NT forum, could a mod move it there?

4 cannot be expressed as a ratio of two triangular numbers. the equation $ \frac{a(a\plus{}1)}{b(b\plus{}1)}\equal{}\frac{m}{n}$ is equivalent to $ m(2a\plus{}1)^2\minus{}n(2b\plus{}1)^2\equal{}m\minus{}n$. for $ m\equal{}4, n\equal{}1$, this equation becomes $ 4(2a\plus{}1)^2\minus{}(2b\plus{}1)^2\equal{}3$. letting $ 2a\plus{}1\equal{}x, 2b\plus{}1\equal{}y$, we have $ (2x\minus{}y)(2x\plus{}y)\equal{}3$, so we must have $ 2x\minus{}y\equal{}1, 2x\plus{}y\equal{}3$, i.e., $ x\equal{}y\equal{}1$. but then $ a\equal{}b\equal{}0$, which is impossible.

[quote="pluristiq"]4 cannot be expressed as a ratio of two triangular numbers.

>>> the equation $ \frac {a(a \plus{} 1)}{b(b \plus{} 1)} \equal{} \frac {m}{n}$ is equivalent to $ m(2a \plus{} 1)^2 \minus{} n(2b \plus{} 1)^2 \equal{} m \minus{} n$.  <<<

for $ m \equal{} 4, n \equal{} 1$, this equation becomes $ 4(2a \plus{} 1)^2 \minus{} (2b \plus{} 1)^2 \equal{} 3$. letting $ 2a \plus{} 1 \equal{} x, 2b \plus{} 1 \equal{} y$, we have $ (2x \minus{} y)(2x \plus{} y) \equal{} 3$, so we must have $ 2x \minus{} y \equal{} 1, 2x \plus{} y \equal{} 3$, i.e., $ x \equal{} y \equal{} 1$. but then $ a \equal{} b \equal{} 0$, which is impossible.[/quote]

No, the equation $ \frac {a(a \plus{} 1)}{b(b \plus{} 1)} \equal{} \frac mn$ is equivalent to $ m(2b \plus{} 1)^2 \minus{} n(2a \plus{} 1)^2 \equal{} m \minus{} n$.

true, 'twas a typo. $ a$ and $ b$ are just switched. nevertheless, you still get $ a\equal{}b\equal{}0$.


Ν.δ.ο.
$ [\sum_{i = 1}^n{x_i^{n - 1}]^\frac {1}{n - 1}\geq[\sum_{i = 1}^n{x_i^{n}]^\frac {1}{n}}}$  ια κάθε $ x_i\geq0$ και  $ n > 1$ φυσικος
Καποιος απ την παρεα σας στο παρελθον και εκτος forum μου ειχε δωσει μια ωραια λυση (ποιος?  :wink: )

Είναι η ανισότητα των βαρών (αν δεν κάνω λάθος). Βγαίνει εύκολα με Holder.

Η ερώτηση μου είναι γενικότερη και δεν αφορά μόνο την συγκεκριμένη άσκηση...Στην ανισότητα των βαρών δεν πρέπει πάντα να υφίσταται άθροισμα που να ισούτε με τη μονάδα από την υπόθεση????
Αν ναι (μπορεί να κάνω και λάθος γι΄αυτό ρωτάω) τότε στη συγκεκριμένη άσκηση πως εφαρμόζεται αν έχει δίκιο ο maniopas???? Μήπως εννοούσες των δυνάμεων ?

Παιδια βαλτε λυση για να σας απαντησω

Νδο $ \left[\sum_{i = 1}^n{x_i^{n - 1}\right]^\frac {1}{n - 1}\geq\left[\sum_{i = 1}^n{x_i^{n}\right]^\frac {1}{n}}}$,    $ \forall \ \ x_i \geq 0$ και $ n \in \mathbb{N}, n\geq1$
Για πιό ωραία, αισθητικά, αποτελέσματα βάζε πριν από αγκύλες ή παρενθέσεις ένα \left όταν ανοίγεις αγκύλη ή παρένθεση και ένα \right όταν κλείνεις.

πχ: \left[ ή \right].

[quote="maniopas"]Ναι, sorry  :blush: . Στα βάρη πρέπει οπωσδήποτε να έχουμε άθροισμα μονάδα. Α, επίσης είναι απλή εφαρμογή, όχι η γενική ανισότητα.[/quote]

Πάλι τόσο blind ρε παιδιά ... Μπορείς να μου ανοίξεις τα μάτια γιατί δεν βλέπω πως βγαίνει...
Αυτή την είχα λύσει πέρσυ αλλά τώρα που την ξαναβλέπω υπάρχει λύση σε μια γραμμή .... :wink:

Οντως σε (δυο..) σειρες αλλα και η αλλη εχει την αξια της γιατι εχει διαφορετικη διαδρομη απ τις συνηθισμενες..

Πολύ κομψό  :D

My solution : Θετω $ a_i \equal{} x_i^r$ τότε αρκεί νδο

$ \left( \sum_{i \equal{} 1}^n a_i\right)^{\frac {s}{r}}\geq\sum_{i \equal{} 1}^n a_i^\frac {s}{r}$  (1)

Θεωρούμε τη συνάρτηση $ f(x) \equal{} x^{\frac{s}{r}}$ αυτή είναι κυρτή προφανώς .
και $ \left(\sum_{i \equal{} 1}^n a_i,\ 0,\ 0,\ ...,\ 0\right)\succ \left(a_1,\ a_2,\ a_3,\ ...,\ a_n\right)$

οπότε από τη Majorasation Karamata inequalty παίρνουμε την (1).QED  :D

Επειδή δν ξέρω LaTeX την γράφω ειδικευμένα την ανισότητα για [size=150]τρεις όρους [/size]και μετά τη γενικεύω... Δλδ  √(a² + b² + c²) ≥ ³√(a³ + b³ + c³)

Έστω οι συναρτήσεις: y = x², y = x³, y =√x, y = ³√x
Έτσι παρατηρούμε ότι:
για 0 ≤ x ≤ 1 είναι x² ≥ x³    (1)
για x ≥ 1 είναι √x ≥ ³√x    (2)
Τώρα χωρίς βλάβη της γενικότητας υποθέτω 0 ≤ a ≤ b ≤ c , δλδ  c μεγαλύτερο απ'όλους. Η ανισότητα ισχύει για a = b = c = 0 άρα c > 0 και x₁ = a/c , x₂ = b/c .
Τώρα αρκεί ν.α.ο. :
√(x₁² + x₂² + 1) ≥ ³√(x₁³ + x₂³ + 1) ~(3)  για κάθε  0 ≤ x₁ ≤ x₂ ≤ 1, που είναι ισοδύναμη στην αρχική.
Αθροίζοντας   x₁² ≥ x₁³, x₂² ≥ x₂³ και 1>=1  ( από (1) ) :
S = x₁² + x₂² + 1 ≥ x₁³ + x₂³ + 1 = C ≥ 1
Από (2) τώρα √S ≥ √C ≥ ³√C , που αποδεικνύει την (3)

Με την ίδια διαδικασία καταλήγουμε στην ανισότητα:
(a^k + b^k + c^k + . . + z^k)^(1/k) ≥ (a^l + b^l + c^l + . . + z^l) ^(1/l)
που αληθεύει αν a, b, c, . . z ≥ 0 και k, l φυσικοί τέτοιοι ώστε  k < l ....

Δεν ξέρω κατά πόσο σωστή είναι και απόδεικνύει του κυρίου Πατήλα αλλά νομίζω ότι είναι στο στόχο......

Υ.Γ.  Την ψιλοέκανα και κάπως αλλιώς για τρεις αγνώστους πάλι αλλά ήθελε και λογαρίθμους, μερική παραγώγιση και Jensen με πολύ μεγάλη έκταση βέβαια....Θεώρησα δλδ τη συνάρτηση y=f(r,a,b,c)=(a^r + b^r + c^r)^(1/r) , τεσσάρων μεταβλητών, όπου r>1 και a,b,c>=0 .Αρκεί λοιπόν να δείξουμε ότι η
f(r,a,b,c) φθίνει σε σχέση με το r. Αυτό θα αποδεικνύει τη ζητούμενη ανισότητα f(2,a,b,c)>=f(3,a,b,c)  που θα αληθεύει καθώς 2<3..Αν αποδειχθεί αυτό στο τέλος προφανώς το αποτέλεσμα θα επιδέχεται και γενίκευση...Είναι δύσκολη όμως αυτή η απόδειξη και τεράστια......(μπορεί να κάνω και λάθος βέβαια οπότε σόρι προκαταβολικά)..........

Κομψες και συντομες οι λυσεις των Vaf  kai Sil
 Μπραβο Bil . Αυτο που εκανες για τρεις ορους μπορουσες με την ιδια διαδικασια να το κανεις και για $ n$ ορους και να τελειωνε.
[b]Παρατηρηση.[/b]
Δεν μπορει να εμαθα εγω ο 40αρης $ latex$ και να μην μπορεις εσυ ο 15αρης .Στο λεω αυτο γιατι δυσκολευτηκα να διαβασω την λυση σου. Επισης με αυτη την γραφη ελαχιστoι θα δινουν προσοχη στο post σου

Οκ κύριε Πατήλα, έχετε απόλυτο δίκιο και το καταλαβαίνω...Θα στρωθώ να μάθω.....! :oops:


If $ ab \equal{} k$, what is the maximum value of $ a^2 \plus{} b^2$?

This is maximized when $ a\plus{}b \equal{} ab$. Since $ a^2 \plus{} b^2 \equal{} (a \plus{} b)^2 \minus{} 2ab$, we have $ k^2 \minus{} 2k$.

hello, i think there is no maximum, because when we plugg $ b\equal{}\frac{k}{a}$ in the given term we get
$ f(a)\equal{}a^2\plus{}\frac{k^2}{a^2}$ and this function has only a local minimum.
Sonnhard.

Let me use my new found knowledge of Lagrange Multiplier:  :D
$ f(x)\equal{}a^2\plus{}b^2\minus{}t(ab\minus{}k)$
$ F_a\equal{}2a\minus{}tb\equal{}0$
$ F_b\equal{}2b\minus{}ta\equal{}0$
$ ab\equal{}k$

Thus the extremas are at $ a\equal{}b, a\equal{}\minus{}b$ from equation 1 and 2
 $ a\equal{}b$
$ 2a^2\equal{}2k$
or
$ a\equal{}\minus{}b$, which is $ \minus{}2k$
depending on values of k

How come we got different answers isabella and Sonnhard? Please clarify my mistake  :|

You can solve by following formula:
f(a)=a^2+\frac{k^2}{a^2} and this function has only a local minimum.

By
Abbeytallbert
[url=http://www.bestofcolleges.com]Bestofcolleges in US[/url]

$ a^{2}\plus{}b^{2}\equal{} (a\plus{}b)^{2}\minus{}2ab$
And given is $ ab \equal{} k$, a positive real number (I hope so).

So, $ a^{2} \plus{} b^{2}$ will be maximum when $ (a \plus{} b)$ is maximum as $ ab$ is constant.
Now $ a$ can take any large positive value which tends to infinity and correspondingly $ b$ will be just  $ \frac{k}{a}$ which is an infinitely smaller positive number depending on value of $ a$.
So all these argument make $ a^{2} \plus{} b^{2}$ an infinitely large number.

Conversely to find the minimum value of $ a^{2} \plus{} b^{2}$,  $ (a \plus{} b)$ should be minimum.
Using AM - GM inequality, we get that $ (a \plus{} b)$ is minimum when $ a \equal{} b$
So minimum value of $ a^{2} \plus{} b^{2}$ =  [b]2k.[/b]


Let $a,b,c > 0$ and $abc = 1$.  Show that
\[ \text{min}\{a^2b+b^2c+c^2a, ab^2+bc^2+ca^2\} \geq \text{max}\{a+b+c, ab+bc+ca \}. \]

$4$ ineqs can be solved in one :

$\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \ge a+b+c$ if $abc=1$.
It's well-known.


geometry
rectangle
perimeter
analytic geometry

I am given two points in a carteisan plane (x1,y1) (x2,y2) they are the co-ordinates of the diagonal of a rectangle. How can I find the length and width of the rectangle?

Nothing else is given, no perimeter, no area, and no theta

There are infinitely many rectangles, who has common diagonal

it is also given that x1 y1 x2 y2 will always be integers...any solutions?

It doesn't change anything. Even if $x_1, x_2, y_1, y_2$ are integers, then exist infinitely many different rectangles, with desired property.
To solve the problem we must know more

no there are not infinite solutions if you are given 2 opposite verticies and all the verticies are integers, and other coordinates of the rectangle must lie on the circle that has the diagonal as a diameter

[quote="dondigo"]Only [b]two[/b] vertices has integer coordinates. There is no information, that [b]all[/b]  must have integer coordinates[/quote]

i suppose that is true, but he probably meant the other 2 verticies, because it would be useless to say that the points that you are given must be integers


[quote="Hero Math"]$ (1)$ prove that if $ a,b$ and $ c$ are positive integers such that $ c^{2} \equal{} a^{2} \plus{} b^{2}$  then both $ c^{2} \plus{} ab$ and $ c^{2} \minus{} ab$ are also expressible as sum of square of two positive integers
$ (2)$ let $ a,b,c \in N$ such that $ (a \minus{} b)$ is a prime number and $ 3c^{2} \equal{} c(a \plus{} b) \plus{} ab$
Show that :$ 8c \plus{} 1$ is a perfect square[/quote]

[b](1)[/b] , Pythagoras Triples.
$ c^2\plus{}ab\equal{}t^2[(m^2\plus{}n^2)^2\plus{}(2mn)(m^2\minus{}n^2)]\equal{}[t(m^2\plus{}mn)]^2\plus{}[t(n^2\minus{}mn)]^2$,
Similarly $ c^2\minus{}ab\equal{}[t(m^2\minus{}mn)]^2\plus{}[t(n^2\plus{}mn)]^2$

[b](2)[/b] , let $ a\minus{}b\equal{}p$ and put $ a\equal{}p\plus{}b$.
The equation $ \Leftrightarrow b^2\plus{}(2c\plus{}p)b\plus{}(pc\minus{}3c^2)\equal{}0$
So the decriminant $ \equal{}(2c\plus{}p)^2\minus{}4(pc\minus{}3c^2)\equal{}(4c)^2\plus{}p^2\equal{}k^2$ is a perfect square

$ \Rightarrow p^2\equal{}(k\minus{}4c)(k\plus{}4c) \Rightarrow k\minus{}4c\equal{}1$ and $ 8c\plus{}1\equal{}k\plus{}4c\equal{}p^2$ is a perfect square.


We are given the convex quadrilateral ABCD. Let I be the intersection of the bisectors of angle BAD and BCD. Prove that we can inscribe a circle in ABCD iff S[IAB]+S[ICD]=S[IAD]+S[IBC]. This condition is equivalent to the fact that I lies in the line joining the midpoints of AC and BD, thus the problem will give a sufficent condition for a convex quadrilateral to have an inscribed circle. (Note that there is a standard result that if I is the center of the inscribed circle of ABCD the I lies in the line joining the two midpoints of AC and BD).


calculus
derivative
logarithms
calculus computations

Hey all, can someone check that I have this derivitive right? I'm really tired and I keep screwing it up. I need the particle derivitives of:

$ L \equal{} \frac { \minus{} 10}{\alpha}log{\frac {V_i^2}{V_o^2}}$ wrt $ V_i and V_o$

I get:

$ \frac {dL}{dV_i} \equal{} \frac { \minus{} 10}{\alpha ln{10}}\frac {2V_o^2}{V_i}$
$ \frac {dL}{dV_o} \equal{} \frac { \minus{} 10}{\alpha ln{10}}\frac { \minus{} 2}{V_i^2V_o}$

But these give crappy values, so I am almost sure I went wrong somewhere.

WHen you take the derivitive of $ \frac{V_{i}^{2}}{V_{o}^{2}}$ dont you get:

$ \frac{2V_i}{V_o^2}$?

hello, is $ \log$ the logarithm to the base $ 10$?
Sonnhard.

hello, i have got
$ L_{V_i}\equal{}\minus{}\frac{20}{\alpha V_i\ln(10)}$
$ L_{V_0}\equal{}\frac{20}{\alpha V_0\ln(10)}$
Sonnhard.


Let a,b and c be real numbers such that a>=b>=c>=0.Determine all pairs of non-zero real numbers (m,n) such that
     (a^m-b^m)*(b^n-c^n)>=(a^n-b^n)*(b^m-c^m)

a,b,c are fixed? Or in other words... We should find all m,n for each a,b,c? or we should find all m,n that works for all a,b,c for which a  >= b >= c>0?


function
trigonometry
algebra
functional equation
algebra unsolved

Find $ f,g: R\to R$ such that
1) $ (f(x)\plus{}g(x))^2\equal{}1\plus{}f(2x)$
2) $ f(x\plus{}y)\plus{}f(x\minus{}y)\equal{}2f(x)g(y)$

Are the two questions related or separate?

t0r, I assume the equations are simultaneous. It looks like the functional equation form of trigonometric functions definition-but are these two enough to guarantee uniqueness of the solution (i.e. are they only fulfilled for $ \sin$ nad $ \cos$), I'm not sure.

[quote="Algadin"]Find $ f,g: R\to R$ such that
1) $ (f(x) \plus{} g(x))^2 \equal{} 1 \plus{} f(2x)$
2) $ f(x \plus{} y) \plus{} f(x \minus{} y) \equal{} 2f(x)g(y)$[/quote]

Notice that $ f(x)\equal{}0$ and $ g(x)^2\equal{}1$ is a solution giving very nasty functions for $ g(x)$, any map of $ R \to \{\minus{}1,1\}$


trigonometry
integration
calculus
calculus computations

Let $ x \equal{} \cos^{3} t$ and $ y \equal{} \sin^{3}t$($ 0 \leq t \leq 2 \pi$). Also $ \rho(x,y) \equal{} k$.

Find $ I_0 \equal{} \int_{C} (x^{2} \plus{} y^{2}) \ dm$

So $ m \equal{} \int_{C} k \ ds \equal{} 3k \int_{0}^{2 \pi} \cos t \sin t \ dt$.

Then $ dm \equal{} 3k \cos t \sin t \ dt$

So does $ I_0 \equal{} 3k\int_{0}^{2 \pi} \left( \cos^{6} t \plus{} \sin^{6} t \right)(\cos t \sin t) \ dt \equal{} 0$.

Is this correct?

No. $ ds=\sqrt{(x'(t))^2+(y'(t))^2}dt=3\abs|\cos t\,\sin t|dt$. To get rid of $ |\cdot|$, integrate over $ 0\le t\le \pi/2$ and multiply the result by $ 4$.

They mean the same thing (integral with respect to arclength). I put $ ds$ because that's the notation I'm more used to.

Thanks

So why did they give the density $ \rho(x,y)$? I thought $ m$ was mass. I assume that you just plug in $ x$ and $ y$ and $ ds$ getting $ I_0 \equal{} 12 \int_{0}^{\pi/2} (\cos t^{6} \plus{} \sin t^{6}) \cos t \sin t \ dt$

Now I see: $ dm\equal{}\rho(x,y)ds\equal{}k\,ds$

So was my original post correct?

Thanks

No, see my formula for $ ds$ above. So much ado about nothing... but I got into top 45 posters, so I'm not complaining.  :)

$ dm \equal{} k \ ds$ (your formula)

$ ds \equal{} 3 \cos t \sin t \ dt$

So $ dm \equal{} 3k \cos t \sin t \ dt$ which is what I had.

And so $ I_0 \equal{} 3k \int_{0}^{2 \pi} (\cos^{6} t \plus{} \sin^{6} t)(\cos t \sin t) \ dt \equal{} 0$

Should I still integrate from $ 0$ to $ \pi/2$?

I'm tired of this discussion. All I wanted to say is that $ \sqrt{a^2}\equal{}|a|$, not $ a$.


[url]http://www.rethinkingphysics.com/preface.html[/url]
Someone led me to this site: It offers a solution to bridge the gap between classical and quantum physics.

Why hasn't anybody commented on this? I find the topic very interesting, but I lack the expertise to claim whether this series of papers have sense or are just a well-constructed attempt to simplify physics. Is there anybody here who can judge the reliability of these papers?

any project that claims to unify classical and quantum physics in five easy steps is probably drivel.

The steps are not so easy ;) But I share your opinion -- this is not likely to be correct. There are a few points which are questionable to me, but I just don't know whether it's my ignorance or the author is talking nonsense. That's just like the 3d time project you can find on the net.

A new web page is available which updates all previous sites and fills gaps in previous posts.

Visit: RethinkingPhysics-V3.net

This site shows how classical and quantum physics can be merged and Newton's gravity relation reconciled with general relativity, AND how gravity emerges at the particle scale. It also derives and expression for the proton/electron mass ratio.

I haven't studied the article in detail, at first glance, it appears, well, amatuerish. Firstly, the only Physics that can be done with high school Math was all done three hundred years ago, and the author is not even using single-variable calculus anywhere in the four papers.
Some problems that were obvious at first glance on the first page itself:

[quote]...but the well accepted equation of atomic physics $ m_e c^2 \equal{} \frac {e^2}{r_e}$...[/quote]
If by $ r_e$ he means the Classical radius of the electron, then that equation is by no means "well accepted" in atomic Physics, it is only a rough estimate, and if taken too seriously, immediately leads to big problems. Indeed, the mass and radius of electron present the biggest problem in Classical (and Quantum) Electrodynamics. All this has been explained very nicely in Feynman Lectures Volume 2 Ch. 28. Indeed, Feynman is most well known for solving this problem by his version of QED.

[quote]The strong force is given by $ F_s \equal{} \frac {\hbar c}{r^2}$...[/quote]
The [url=http://en.wikipedia.org/wiki/Quantum_chromodynamics]Strong Force[/url] is definitely [i]not[/i] an inverse square force. I don't know where the author got this idea.

[quote]Particles (e.g. the electron) are known to be rotating at high velocities,...[/quote]
Known to whom? I and authors of all QM books that I've read, didn't know this.

Indeed, I rate this article as little better than the Appolo Landing Hoax or the Bermuda Triangle Mysteries or [url=http://en.wikipedia.org/wiki/John_Hutchison]"The Hutchison Effect"[/url], i.e. Quackery.


floor function
combinatorics unsolved
combinatorics

a kangaroo is sitting at the point 1 (on the real line) and wants to visit (by jumping) $ 2,3,\dots,n$, all these points only once.
at the end, he will jump back to 1 (from the last visited point).

what is the max. possible sum of lengths of his jumps ?

This is too confusing, even for my own taste. But I do hope you get what I was trying to say...

[hide]
The maximum possible sum of the lengths of his jumps is $ (n\minus{}1) \plus{} (n\minus{}2) \plus{} (n\minus{}3) \plus{} ... \plus{} \left\lfloor\frac{n}2\right\rfloor \equal{} \frac{n(n\minus{}1)}2 \plus{} \left\lfloor\frac{n}2\right\rfloor$, which can be achieved by going in this path:

1 -> n -> 2 -> n-1 -> 3 -> etc.

Now, suppose there is a path with higher sum of lengths, and that the path is different from the one presented above.

At some point, the path must deviate from the path above. The last number that both paths visit at the same time is $ x$. The next number that our path would visit is $ y$. Again, at some point, the new path must visit $ y$. Then we can describe the new path as:

$ x$ -> $ i_1$ -> $ i_2$ -> ... -> $ y$ -> $ z$. (Sum: $ x \plus{} z \minus{} i_1 \minus{} y$)

Let's define an [i]iteration[/i] as a partial inversion of this path, so that it follows our path above by one more move ($ y$ gets visited after $ x$):

$ x$ -> $ y$ -> $ i_2$ -> $ i_1$ -> $ z$

We only changed two lengths -> $ |x\minus{}i_1|$ to $ |x\minus{}y|$ and $ |z\minus{}y|$ to $ |z \minus{} i_1|$.

Now, we have two cases:

Case 1($ i_1 > z$):

Then the sum of the new path is $ x \plus{} i_1 \minus{} y \minus{} z$. As $ i_1 > z$, this is greater than the sum above ($ x \plus{} z \minus{} i_1 \minus{} y$). So this path is an improvement.

Case 2($ z \ge i_1$):

Then the sum of the new path is $ x \plus{} z \plus{} y \minus{} i_1$. So this path is equivalent to the other one.

We get to the conclusion that iterations only raise the sum or leave it unchanged. After a finite number of iterations, we will reach our proposed path above, at which point we cannot iterate anymore. So the path proposed above is optimal.
[/hide]


linear algebra
matrix
vector
linear algebra unsolved

Suppose $J$ is a $n\times n$ matrix which all its elements are $1$. Suppose $A,B$ are 2 invertible matrices that $A+B=J$. Prove that \[ (1-s(A^{-1})(1-s(B^{-1}))=1  \], where $s(A)$ is the sum of the elements of $A$

Let $\alpha_i,\beta_i, i=\overline{1,n}$ be the eigenvalues of $A$ and $B$. So we can assume that $\alpha_1+\beta_1=n$ and $\alpha_i+\beta_i=0,i=\overline{2,n}$so knowing that $A$ and $B$ then $\alpha_i\neq 0, \beta_i\neq 0$. Making the computations the equality should verify.($Tr(A^{-1})=\frac{1}{\alpha_1}+\frac{1}{\alpha_2}+...+\frac{1}{\alpha_n}$)

i think there is a mistake in the exercice!
let A be the matrix with 1 above digonal and 2 on diagonal an let $B=J-A$

Omid could check the statement of this problem please ?

oh!it is not trace the statement is $(1-s(A^{-1})(1-s(B^{-1})=1$ where $s(A)$ means $\sum A_{ij}$.

Let $H$ be the row vector with all entries $1$ and $H^*$ be the column vector with all entries $1$. Then $H^*H=J=A+B$ but $HXH^*=s(X)$. So, our problem reduces to showing that $(1-HA^{-1}H^*)(1-HB^{-1}H^*)=1$, or, after opening the brackets, to $H(A^{-1}H^*HB^{-1}-A^{-1}-B^{-1})H^*$. As $H^*H=A+B$ we get $H(A^{-1}(A+B)B^{-1}-A^{-1}-B^{-1})H^*=0$, which is true since $A^{-1}(A+B)B^{-1}=A^{-1}+B^{-1}$

Excuse me I edited the post and Your solution is corretc iura


for one this it's not ISL 2003. For the second if it would be, I would ask you to remove it, and not post it again, ok? :)

[quote="A1lqdSchool"]Given four circles $k$, $k_{1}$, $k_{2}$, $k_{3}$ with centers $O$, $O_{1}$, $O_{2}$, $O_{3}$ and radii $R$, $R_{1}$, $R_{2}$, $R_{3}$, respectively, such that the circles $k_{1}$, $k_{2}$, $k_{3}$ lie inside the circle $k$.

Let $A$ and $B$ be two points on the circle $k$ such that the segment $AB$ is a diameter of $k$. Assume that the circles $k_{1}$ and $k_{2}$ externally touch each other, both touch the segment $AB$ (lying in one halfplane with respect to the line $AB$), internally touch the circle $k$ (lying inside the circle $k$) and externally touch the circle $k_{3}$. Also assume that the circle $k_{3}$ touches the segment $AB$.

Prove that $R_{3}\leq \frac{\sqrt 2-1}{4}\cdot R$.

(See the figure.)[/quote]

Hi!

I have something on it. I will assume that we have exactly the figure that is shown above.

Suppose WLOG $R=1$. Let $X$ be tangency point of $k_{1}$ and $AB$, $Y$ be the tangency point of $k_{2}$ and $AB$, $Z$ be the point of tangency of $k_{3}$ and $AB$.

We know that $XZ=2\sqrt{R_{1}R_{3}}$, $YZ=2\sqrt{R_{2}R_{3}}$, $XY=2\sqrt{R_{1}R_{2}}$. It follows that \[\sqrt{R_{3}}=\frac{\sqrt{R_{1}R_{2}}}{\sqrt{R_{1}}+\sqrt{R_{2}}}.\] On the other hand $OX=\sqrt{1-2R_{1}}$ and $OY=\sqrt{1-2R_{2}}$. So $2\sqrt{R_{1}R_{2}}=\sqrt{1-2R_{1}}+\sqrt{1-2R_{2}}$ :!:.

Therefore we are to maximize $R_{3}(R_{1},R_{2})$ assuming condition :!: . I have created separated topic for it. See http://www.mathlinks.ro/Forum/viewtopic.php?t=34099 .


function
trigonometry
ratio
algebra unsolved
algebra

Determine all real numbers $a$ such that the function \[ f(x) =\{ ax + \sin x\}\] is periodic. Here, $\{y\}$ is the fractional part of $y$.

[quote="Adriana N."]Determine all real numbers $ a$ such that the function $ f(x) \equal{}${$ ax \plus{} sinx$}  is periodic.Here {$ y$} is the fractional part of $ y$.
P.S. Please explain,don't write only the value/values of $ a$. Thank![/quote]

Let $ T$ be a period. Then $ f(x\plus{}T)\equal{}f(x)$ $ \forall x$

$ \iff$ $ a(x\plus{}T)\plus{}\sin(x\plus{}T)\minus{}ax\minus{}\sin(x)\in\mathbb Z$

$ \iff$ $ aT\plus{}2\sin(\frac T2)\sin(x\plus{}\frac T2)\in\mathbb Z$

Since $ aT\plus{}2\sin(\frac T2)\sin(x\plus{}\frac T2)$ is continuous, this means $ \sin(\frac T2)\equal{}0$ and $ aT\in\mathbb Z$

So $ T\equal{}2k\pi$ and $ a\equal{}\frac{n}{2k\pi}$

Hence the answer : $ \boxed{a\in\{\frac p{\pi},p\in\mathbb Q\}}$

[quote="Adriana N."]Determine all real numbers $ a$ such that the function $ f(x) \equal{}${$ ax \plus{} sinx$}  is periodic.Here {$ y$} is the fractional part of $ y$.
P.S. Please explain,don't write only the value/values of $ a$. Thank![/quote]

Using the definition of $ \{x\}$ we may write $ f(x) \equal{} ax \plus{} sinx \minus{} [ax \plus{} sinx]$
But since $ \minus{}1 \le \sin x \le 1$ we may finally write $ f(x) \equal{} \{ax\} \plus{}  \{\sin x\}$

Since $ \{\sin x\}$ is periodic with period $ 2\pi$ the function $ f(x)$ is periodic when $ a\equal{}0$

Consider now $ a \neq 0$. It is also $ \left\{a\left(x \plus{} \frac{k}{a}\right)\right\} \equal{} \{ax \plus{} k\} \equal{} \{ax\}\;\forall k \in \mathbb{Z}$
So the function $ \{ax\}$ is also periodic with period $ \;\frac1a$

Since $ f(x)$ is the sum of two periodic functions, in order to be periodic the ratio of two periods must be a rational number
So it means that   $ 2 \pi a$ must be a rational number

In conclusion, the infinite values of $ a$ are members of the real subset $ A\equal{}\{\frac{q}{\pi}, q \in \mathbb{Q} \}$

[quote="prester"] Using the definition of $ \{x\}$ we may write $ f(x) \equal{} ax \plus{} sinx \minus{} [ax \plus{} sinx]$
But since $ \minus{} 1 \le \sin x \le 1$ we may finally write $ f(x) \equal{} \{ax\} \plus{} \{\sin x\}$...[/quote]

Unfortunately, this is wrong. You cant write $ f(x) \equal{}\{ax \plus{} \sin x\}\equal{} \{ax\} \plus{} \{\sin x\}$

Choose for example $ a\equal{}\frac 3{\pi}$ and $ x\equal{} \frac{\pi}6$ : $ \{ax \plus{} \sin x\}\equal{} 0$ while $ \{ax\} \plus{} \{\sin x\}\equal{}1$

[quote="pco"][quote="prester"] Using the definition of $ \{x\}$ we may write $ f(x) \equal{} ax \plus{} sinx \minus{} [ax \plus{} sinx]$
But since $ \minus{} 1 \le \sin x \le 1$ we may finally write $ f(x) \equal{} \{ax\} \plus{} \{\sin x\}$...[/quote]

Unfortunately, this is wrong. You cant write $ f(x) \equal{} \{ax \plus{} \sin x\} \equal{} \{ax\} \plus{} \{\sin x\}$

Choose for example $ a \equal{} \frac 3{\pi}$ and $ x \equal{} \frac {\pi}6$ : $ \{ax \plus{} \sin x\} \equal{} 0$ while $ \{ax\} \plus{} \{\sin x\} \equal{} 1$[/quote]

Thank you  pco. You're obviosuly right. Sorry for mistake  :(

[quote="Adriana N."]Determine all real numbers $ a$ such that the function $ f(x) \equal{}${$ ax \plus{} sinx$}  is periodic.Here {$ y$} is the fractional part of $ y$.
P.S. Please explain,don't write only the value/values of $ a$. Thank![/quote]
If it is periodic then there exists a $ h \in \mathbb{R}$ s.t. $ f(x\plus{}h) \equal{} f(x) \iff$
$ \{ax\plus{}\sin x\} \equal{} \{ax\plus{}ha\plus{}\sin(h\plus{}x) \}$
$ \sin (h\plus{}x) \equal{} \cos x \sin h \plus{} \sin x \cos h$.
So: $ \{ax\plus{}\sin x\} \equal{} \{ax\plus{}ha\plus{}\sin(h\plus{}x) \} \iff ax\plus{}\sin x \minus{} (ax\plus{}ha\plus{}\sin(h\plus{}x)) \in \mathbb{Z} \iff$
$ p(x) \equal{} \sin x \minus{} \sin x \cos h \plus{} \cos x \sin h  \minus{} ha \in \mathbb{Z}$
Since $ p$ is continious we see that $ p$ is constant.
So $ p'(x) \equal{} 0 \iff \cos x ( 1 \minus{} \cos h) \minus{} \sin x \sin h \equal{} 0 \forall x$
Assume that $ \cos h \neq 1$. Then $ p'(x) \equal{} 0 \iff \cot x \equal{} \frac{\sin h}{1\minus{}\cos h} \forall x$ which gives a contradiction. So $ \cos h \equal{} 1 \iff h \equal{} 2n\pi, n \in \mathbb{Z}$.
Then $ p(x) \equal{} \minus{}ha \equal{} \minus{}2n\pi a \in \mathbb{Z}$ so $ a \equal{} \frac{t}{\pi}$ for some $ t \in \mathbb{Q}$.
If $ a \equal{} \frac{p}{q\pi}, p,q \in \mathbb{Z}$ then $ f(x) \equal{} f(x\plus{}2q\pi)$. So they fit. Hence the answer is $ a \equal{} \frac{t}{\pi},t \in \mathbb{Q}$


geometry
circumcircle
power of a point
radical axis
geometry unsolved

let $ BB_1$ and $ CC_1$ be the altitudes of triangle $ ABC$. and assume $ AB$ not equals to $ AC$. let $ M$ be the midpoint of $ BC$, $ H$ the orthocenter of $ ABC$ and $ D$ the intersection of $ BC$ and $ B_1C_1$.show that $ DH$ is perpendicular to $ AM$

I think it was posted before, but i didn't find it. So...
[hide="solution"]
Let $ w$ is circumcircle of quadrangle $ B_1CBC_1$ (so $ M$ is center of this circle)
Line $ AD$ - is polar of point $ H$ with respect to $ w$.
So we have $ MH$ is perpendicular to $ AD$
and
$ AH$ perpendicular to $ MD$.
So $ H$ is orthocencer of triangle $ ADM$.
and we have that $ DH$ is perpendicular to $ AM$(q.e.d.).[/hide]

[quote="SaYaT"]I think it was posted before, but i didn't find it. So...[/quote]
Yes, it has been posted before as [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=110887]Triangle, perpendicular lines[/url], by [b][size=100]Yimin Ge.[/size][/b]

Kostas Vittas

[quote="SaYaT"]
Line $ AD$ - is polar of point $ H$ with respect to $ w$.
[/quote]

Hello Sayat. Can you explain how you got this step. :blush:

Thanks in advance,
Befuddlers

[b]Theorem[/b] : [color=blue]If from a fixed point two secants are drawn to a circle and their extremities are connected in pairs , the opposite connectors intersect on the polar of the point.[/color]

As a direct consequence of this theorem we have $ H$ and $ D$ lie on the polar of $ A$ with respect to the circle centered at $ M$ passing through $ B,C,B_1, C_1$.

$ Q.E.D.$

Sayat could you please explain your method.

Here is my nice solution using [b]radical axis[/b].
X is the feet of altitude from A to BC.
[b]Circles[/b]
$AXM$,nine-point circle-XM                         (Radical axis)
$B_1C_1$,nine-point circle -$B_1C_1$       (Radical axis)
Hence D is a point on the radical axis of $BCB_1C_1$ and $AXM$.
Also note that power of H w.r.t $AXM=AH*HX=BH*HB_1$=power of H w.r.t $BCB_1C_1$
Hence H is also a point on the radical axis .
DH is the radical axis of circles $AXM$ and $BCB_1C_1$
Clearly the line joining the centres is AM and hence the proof.


hmm.

Suppose $g(x) = \arctan x - 3f(x)$
$g(0) = 0$, we want to prove that $g(x) > 0, x>0$, it is sufficient to prove that

\[g'(x) = \frac{(2+x^2)\sqrt{1+x^2}-2(1+x^2)}{(1+x^2)^{3/2}(1+2\sqrt{1+x^2})^2} > 0\]

or

\[2+x^2 > 2\sqrt{1+x^2} \Longleftrightarrow 1+\frac{x^2}{2} > (1+x^2)^\frac{1}{2}\]

the last inequality holds.


find four digit number such that 4*(multiply)ABCD=DCBA

Hmm ok, $2178\cdot4=8712$, now where's my money? :D

You should memorize this number; I've seen this problem in other competitions.


Determine the product of distinct positive integer divisors of $ n\equal{}420^{4}$.

This is a trivial computation once you have the result from [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=932844#932844]your other thread[/url].


geometry
trapezoid
circumcircle
power of a point
radical axis

[quote="Virgil Nicula"]Let $ ABCD$ be an isosceles trapezoid, where $ AB\parallel CD$ . For two points $ M$ , $ N$ which belongs to the sideline $ AB$ define the intersection $ P\in CM\cap DN$ . Prove that the point $ P$ belongs to the radical axis of the circumcircles of the triangles $ ADM$ , $ BCN$ .

[b][u]Remark.[/u][/b] This problem is an extension of

http://www.mathlinks.ro/viewtopic.php?t=187013 or

http://www.mathlinks.ro/viewtopic.php?p=818557#818557.[/quote]
[hide="Solution"]
Let $ \angle ADC = \angle BCD = \alpha$, $ \angle NDC = \beta$, $ \angle MCD = \gamma$ and allow the circumcircles of $ \triangle ADM$ and $ \triangle BCN$ to meet $ MC$ and $ ND$ at $ R$ and $ Q$ respectively.  Thus, $ \angle ADR = 180 - \angle AMC = \angle MCD = \gamma$.  Also, $ \angle ADN = \angle ADC - \angle NDC = \alpha - \beta$.  So, $ \angle QDR = \angle ADR - \angle ADN = \gamma - (\alpha - \beta) = \beta + \gamma - \alpha$.  Similarly, we can show that $ \angle QCR = \beta + \gamma - \alpha$ in the exact same way.  Therefore, $ QRCD$ is cyclic, so $ PQ\cdot PD = PR\cdot PC\implies \frac {PN}{PM} = \frac {PD}{PC} = \frac {PR}{PQ}\implies PN\cdot PQ = PR\cdot PM$, so the power of point $ P$ with respect to the circumcircles of $ \triangle ADM$ and $ \triangle BCN$ are equal.  Therefore, $ P$ lies on the radical axis of the circumcircles of $ \triangle AMD$ and $ \triangle BCN$. [/hide]


Prove that for any positive reals $ a,b,c,d$,

$ \frac{bcd}{a^3\plus{}3bcd}\plus{}\frac{cda}{b^3\plus{}3cda}\plus{}\frac{dab}{c^3\plus{}3dab}\plus{}\frac{abc}{d^3\plus{}3abc}\leq 1$.

[quote="massnet"]Prove that for any positive reals $ a,b,c,d$,

$ \frac {bcd}{a^3 \plus{} 3bcd} \plus{} \frac {cda}{b^3 \plus{} 3cda} \plus{} \frac {dab}{c^3 \plus{} 3dab} \plus{} \frac {abc}{d^3 \plus{} 3abc}\leq 1$.[/quote]
$ \sum_{cyc}\frac {bcd}{a^3 \plus{} 3bcd}\leq1\Leftrightarrow\sum_{cyc}\left(1 \minus{} \frac {3bcd}{a^3 \plus{} 3bcd}\right)\geq1\Leftrightarrow\sum_{cyc}\frac {a^3}{a^3 \plus{} 3bcd}\geq1.$
But $ \sum_{cyc}\frac {a^3}{a^3 \plus{} 3bcd} \equal{} \sum_{cyc}\frac {a^4}{a^4 \plus{} 3abcd}\geq\frac {(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2)^2}{a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4 \plus{} 12abcd}$
and $ \frac {(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2)^2}{a^4 \plus{} b^4 \plus{} c^4 \plus{}d^4\plus{} 12abcd}\geq1\Leftrightarrow \sum_{sym}(a^2b^2 \minus{} abcd)\geq0,$ which true by AM-GM.


Let $N=111...1222...2$, where there are 1999 digits of 1 followed by 1999 digits of 2. Express $N$ as the product of four integers, each of them greater than 1.

That way works.  Does anyone have another solution?

Sure.

[hide]We are looking at the number

$10^{1999} \times 11....1 + 2 \times 11....1$ =
$(10^{1999}+2)(111....1)$

That first factor is divisible by $2$ and $3$ since digits add to three, and thus we have four factors.[/hide]


http://www.mathlinks.ro/Forum/viewtopic.php?t=136494 ( In this box)  :wink: Locked!