inequalities
geometry
circumcircle
LaTeX
inequalities proposed

In a triangle with medians M-a, M-b, M-c

Prove  that M-a +M-b +M-c >= (a^2 +b^2 +c^2)/2R, where R is the circumradius

Please use LaTeX:

[i]In a triangle with medians $m_a$, $m_b$, $m_c$,

prove that $m_{a}+m_{b}+m_{c}\geq \frac{a^{2}+b^{2}+c^{2}}{2R}$.[/i]

In fact, according to [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=16023]the "Posted by fuzzylogic" thread[/url], we have $m_{a}\geq \frac{b^{2}+c^{2}}{4R}$, $m_{b}\geq \frac{c^{2}+a^{2}}{4R}$, $m_{c}\geq \frac{a^{2}+b^{2}}{4R}$, so that

$m_{a}+m_{b}+m_{c}\geq \frac{b^{2}+c^{2}}{4R}+\frac{c^{2}+a^{2}}{4R}+\frac{a^{2}+b^{2}}{4R}=\frac{a^{2}+b^{2}+c^{2}}{2R}$.

Qed..

  Darij


function
integration
analytic geometry
topology

Prove: $ (C^L[0,1],d)$ is incomplete metric space. ($ C^L[0,1]$ is the set of functions continous on $ [0,1]$, $ d(x(t),y(t))\equal{}\int_0^1|x(t)\minus{}y(t)|dt,\quad t\in [0,1]$)

This is a pain to write down, so I'll just describe it.  Construct the sequence of functions $ x_n(t)$ consisting of two lines, one connecting $ (0, 1)$ to $ \left( \frac{1}{n}, 0 \right)$ and the other connecting $ \left( \frac{1}{n}, 0 \right)$ to $ (1, 0)$ and show that it is Cauchy but doesn't converge.

What is meant [quote="t0rajir0u"]$ (1, 0)$ [/quote]?

Can u show one Cauchy's sequence  but do not converge?

torajirou's sequence doesn't actually work as an example. We can show that the distance from his function $ x_n$ to the zero function is $ \frac1{2n}.$ Hence, $ x_n$ tends to the zero function in this metric space even though the functions don't converge to zero pointwise at every point.

Here's the example I usually use. For $ n\ge 2,$ let $ x_n(t)$ be the piecewise continuous linear function (a "connect the dots" graph, the same idea t0rajir0u was using) where the segments of the graph run from

$ (0,0)$ to $ \left(\frac12 \minus{} \frac1n,0\right)$ to $ \left(\frac12 \plus{} \frac1n,1\right)$ to $ (1,1).$

We do have that $ d(x_n,x_m) \equal{} \frac12\left|\frac1n \minus{} \frac1m\right|.$ That makes this a Cauchy sequence.

But while there's an obvious limit function, namely the function which is $ 0$ on $ \left[\left.0,\frac12\right)\right.$ and $ 1$ on $ \left.\left(\frac12,1\right.\right],$ that limit function cannot be made to be continuous, and hence does not belong to our metric space.

[quote="bangvn"]What is meant [quote="t0rajir0u"]$ (1, 0)$ [/quote]?[/quote]
The point in $ \mathbb{R}^2$ with $ x$-coordinate $ 1$ and $ y$-coordinate $ 0$.  (Unfortunately this notation is somewhat ambiguous when we are also talking about intervals.)  Anyway, whoops; Kent's example is (essentially) the one I meant to give.


There seems  to be a small error (14x+6x)
if the 14x is 14y,
then
(x-3)^2+(y-7)^2=64
apply cauchy
((x-3)^2+(y-7)^2)(9+16)   (>=)   (3x+4y-37)^2
so 64*25    (>=)  (3x+4y-37)^2
then it becomes simple.
Is it? I think my solution is stupid.


What is the sum of the even integers from 2~220 inclusive?

(2 + 4 + 6 + 8 + 10.......+ 220)

I got a completely different answer. Please help! Thanks.  :)

12,210 i think

im not sure

is the formula n (a+z)?

where n = number of terms in sequence
         a = first number
         z = last number

Two ways to do it:

The sum of the first n even integers is n(n+1), so 110(111) = 12210.

The sum of an arithmetic sequence is #terms/2*(first term + last term), so 110/2*(220+2) = 12210.


A cricket tournament between some countries was played in which each team played every other team exactly once. India won the tournament. Prove that India is better than every other team in the tournament if...

Team A is better than team B if A won the match against B

or there is a team C such that A triumphed over C and C defeated B.

What if B defeated A, A defeated C, and C defeated B?

Who is better, A or B?

[hide="doubt"]Is the winner of the tournament the team with maximum no. of wins?
What happens if two teams win same no. of matches? Do they see the head-to-head record?[/hide]

Let us assume that they see head to head record.

This problem is tougher than I thought.

Maybe we should just take the facts as they are.

"Wins" means "has at least as many victories as any other team."

See for example the hidden text in the reply here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=159625
Other related questions are linked from here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=162646


$a,b$ and $c$ are the sides of a triangle.
Proof that: \[ \sqrt{a+b-c}+\sqrt{a+c-b}+\sqrt{b+c-a} \leq \sqrt{a} + \sqrt{b} +\sqrt{c} \]

since $a,b,c$ are the sides of a triangle we may write $a=x+y , b=y+z , c=z+x$
so the inequality that we have to prove is:
$\sum \sqrt{2x}\le \sum \sqrt{x+y} \leftrightarrow 2\sum \sqrt{xy} \le \sum \sqrt{(x+y)(y+z)}$
but we know that $\sqrt{(x+y)(y+z)}\geq \sqrt{xy}+\sqrt{yz}$ from cauchy and addi ng the similar inequalities gives the above inequality.

let $a=x+y, b=y+z, c=z+x$, then $\sum{a+b-c} \le \sum{a}$ is equal to $\sqrt{2}(\sum{\sqrt{x}}) \le \sum{\sqrt{x+y}}$. notice that $\sqrt{2}\frac{\sqrt{x}+\sqrt{y}}{2} \le \sqrt{x+y}$. Sum up them. finish.

All you need is
\[ \frac{x^2+y^2}2\geq\Bigl(\frac{x+y}2\Bigl)^2. \]


$ a,b,c >0 \implies \frac{a}{b} \plus{}\frac{b}{c} \plus{}\frac{c}{a} \geq \frac{a\plus{}b\plus{}c}{\sqrt[3]{abc}}$

From $ AM\minus{}GM$ we have $ 3\left(\frac ab\plus{}\frac bc\plus{}\frac ca\right)\equal{}\left(\frac {2a}{b}\plus{}\frac{b}{c}\right)\plus{}\left(\frac{2b}{c}\plus{}\frac ca\right)\plus{}\left(\frac{2c}{a}\plus{}\frac ab\right)\ge \frac{3a}{\sqrt[3]{abc}}\plus{}\frac{3b}{\sqrt[3]{abc}}\plus{}\frac{3c}{\sqrt[3]{abc}}$.


analytic geometry
graphing lines
slope
geometry
trigonometry
conics
ellipse

Let's discuss y'all. I am personally not sure about 5.

For number two I used Hero's formula, and set that equal to 6*perimeter.
I ended up with the equation a=c=b(b 2 +576)/2(b 2 -576).  The three integer solutions I got were (40, 48, 40);(45, 72, 45);(65, 120, 65).

#5...I first solved it in under a minute...It looked too easy.  Then I checked it at 11:50 PM the night it was due, and noticed it WAS....I hadn't noticed how the triangle sloped up...So I fixed that, and recalculated.  I got a very weird answer (and forgot to divide by 2, although I wrote A=bh/2)

Guess that's what happens when you send it at 11:59...

I used a [boring] coordinate-based approach to problem 5. Did anyone find a way to do this problem with pure geometry?

I did practically that same thing as origamimasterjared for problem 5.  :blush: It looked really easy at first, but then I realized (at like 1 in the morning) that D and E were not located directly above the midpoints of AB and AC. Then things got a lot more fun.  :lol:

Here are my solutions...

I think solved 5 useing "pure geomatry"   :) . Got the same answer as you did too-which is always a good sign(gettting the same answer as someone else).  :D

For number 2 I used area=ABsin(c)/2. Gives you some really nice trig stuff.  :lol:

Soultions are below

I looked at the problem from the "side" (with the semicircles as vertical lines and BC as a point, I think - I'll post the diagram I used) and then used the fact that a line tangent to the ellipse containing D and E (represented as a single point on this view) must also go through F (the coordinates of which were known).  That gave me the position of D and E and from there I could find the area of DEF.  Coordinate-based, I suppose, but not 3-dimensional coordinates.  I ended up with 24.

My weird answer (had I remembered to divide it by two) comes out to about 25.165 (It was a bunch of ugly square roots, which I just left because I didn't have time to make it nicer)

Hmmm... I got 24 as well. And 24 certainly is a much nicer and happier number than 25.165. Peraphs if you share what you did, one of us can either spot your mistake or see what they did wrong themselves  :)

[quote="mathfanatic"][quote="Danbert"]Did anyone else get 24 for #5?[/quote]

Yes.  My answers were
1. 142/857, 142857142/857142857
2. 40-40-48, 45-45-72, and 65-65-120
3. Find a closed form for s_n (i.e. (3^n-1)/2) and then examine the last two digits of 3^n, convert it to last two digits of s_n, and some other stuff.
4. 26*13, whatever it is
5. 24.  Base and height are both 4sqrt(3).[/quote]

Oh, good.  I got the same answers you got for 1, 2, 4, and 5 (I wasn't sure about #5) :).

I proved #3 by a different method, though... although I found the closed form, my method only uses the given recursion (and the given first term, of course).

Ditto to mathfanatic.

On number 3, I just proved that the last two digits form a repetitive sequence with a 20-term period, and in this sequence, double digits occur in one group of three consecutive terms.

Problem 1 is awesome!
Note that the relation also holds true for x=142857142857142, y=857142857142857, etc.

[quote="ZennyK"]

I did practically that same thing as origamimasterjared for problem 5.  :blush: It looked really easy at first, but then I realized (at like 1 in the morning) that D and E were not located directly above the midpoints of AB and AC. Then things got a lot more fun.[/quote]

Ohhhhh....Now I see....My calculation is based on D and E being directly above the midpoints of AB and AC....

I wound up with the height (as in bh/2) of ∆DEF being √((5√3-5√2) 2 +((5/2)√2) 2)
Then I multiplied that by a base of 5√3, divided by 2 (A=bh/2)

And got 25√(-24√6+75)/4, which is about 25.165.

Yeah , I thought that at first as well, the I relooked at the picture. Its funny that it accually turns out to be a much nicer problem when you let it rest where it is... math contest problems have a strange way of doing that  ;)

Here is my solution. I realized my #5 was completely wrong. :(

Comments will be appreciated

That is post #7 in the discussion located at

http://www.mathlinks.ro/Forum/viewtopic.php?t=19499

Note that the catch to this method is that (a) this is generally a poor strategy for making friends with USAMTS graders (Erin can probably vouch for this) and (b) you will definitely not recieve a commendation for using brute force.

If you're just looking for a 5 and don't care to search for a more elegant solution (or you PROCRASTINATE and solve all 5 problems the weekend before they're due -- not to mention any names or point any fingers :lol:), then I guess you're good-to-go.

[quote="ZennyK"] the weekend before they're due.[/quote]

Try five minutes...:D

[quote="sran"]On the other hand, I think we do have to consider that the competition allows brute force methods and the like. Computer program "proofs" are acceptable; testing is really very much like that.
I remember someone once posting a question that was something like, 'if we are given a problem that asks us to find all integer solutions to some equation in (0, 2001), would testing each integer in (0, 2001) and recording those that worked be an acceptable solution?' And the response was yes. The USAMTS permits any rigorous, correct solution.[/quote]

OK, I concede that any rigorous solution is okay.  But it's still really ugly.

Suppose I were to put the following as my solution for #2: "Testing a number of positive integer pairs (a,b), where a is the length of either congruent leg and b is the length of the base, yields the pairs (36, 45), (40,48), (65,120)."  Would this count for full credit?  If it does, I think this shows a potential flaw of the current grading system.

Regarding computer proofs, from the USAMTS rules:
[quote]Except where otherwise noted in problems, participants may use calculators and computers to solve problems. In this case, the participant must explain the program that was used to solve the problem, and must prove that the method used in the program does solve the problem.[/quote]That's pretty self-explanatory I hope.

I didn't notice, because I found (inadvertantly, of course) non-parallel ways to solve those four problems. After attending the math jam, I did notice a degree of similarity between Mr. Crawford's brilliant solutions to problems 2 and 5. Interesting observation, juggle7.

[quote="juggle7"]Has anyone else noticed any parallels in the Round 3 problems?  #1 and #3 both involved primitive roots and cycles, while #2 and #5 both had to do with isosceles triangles and their inscribed circles making those three lengths that pairwise sum to equal the sides.  Like, in Mr. Crawfords marvelous solution to #5, the inscribed circle in the isosceles triangle HJK touches points D, E, and F.[/quote]

Yes, we had noticed some general similiarities beforehand and looked at replacing a few problems with other problems to give a greater variety of solution techniques. But the alternative problems had flaws, so we went ahead with the problems we had.

However, we had not noticed the particular individual similarities you mentioned.

Erin Schram
USAMTS grader

[quote="filletwho"]Here are my solutions. The first time I used latex! I think the answers are fine, but i don't like my number three. its not elegant like everyone elses. I love my solution to number four though, its nice and simple(it seems long since I wrote alot). I used an easy method for number one. I was gonna do something with the repeating fraction 1/7, but i hadn't completely developed the idea yet. Number two was easy and number five took me many tries. At first I thought it was just projected onto another plane, but i realized shortly after that it was tilted. next i though it was right above the midpoints. Then I thought for awhile and realized it wasn't. I finally got the actual solution after taking a nap. it didn't stop there, though. I messed up ten different calculations. But I got it right in the end! Suggestions? comments? criticisms?[/quote]

I think there is an even easier way to do number four.  I just boiled it down to two algebraic equations in two variables and then got the answer...

I kept trying to find an easier way to do number 5 and also number 2, but too bad I couldn't...

[quote="sonar"]1.

a. x=142 & y=857
b. x=142857142 & y=857142857

2.

a = 48 & b = 40
a = 72 & b = 45
a = 120 & = b = 65

3.  very convoluted "proof"

4. 338

5. 24

Oh, and for 5 ... yeah I did the same thing others did too in the beginning  :blush: , it did seem too easy ....

Oh, and I resorted to calculus for 5 as well.

(the size of the attachment is because of the pictures (619 KB).)[/quote]

hey, buddy, I think you should learn to condense your proofs... An 18 pager!
I think it can be accomplished in 5, no more, no less...

K i got the exact same answers as everyone else basically:

#5 took me 4 pages of proofs with 5 pictures (using trig functions and pythag.)

[quote="thinkingmachine"]

hey, buddy, I think you should learn to condense your proofs... An 18 pager!
I think it can be accomplished in 5, no more, no less...[/quote]

Ah, yes, but notice that proofs 1 and 2, while 2 pages each, would be 1 page without the picture and reference, and proofs 4 and 5 would each be 1 page shorter without the pictures.  Plus, I tend to solely use arrays, taking up rows instead of using a paragraph style, because it's easier on my eyes (you'll notice that with the exception of 3, it doesn't read like 18 pages).  3 is just terribly worded.

[quote="sonar"]3 is just terribly worded.[/quote]
But I am grading Problem 3 tomorrow! I will have to read that terribly worded solution.   :(

Erin Schram
USAMTS grader

My proofs are 18 pages, too.  For #4, I drew an uber-diagram with 26x26 and 30x30 squares.  I used too many lemmas.  I guess this is because it was my first time with LaTeX.  I got a really weird solution for #5, ~22.14.


probability
inequalities
inequalities unsolved

>> posted before?

Probability that it was is [b]very high[/b] (indeed for any inequality) on this forum !!!

We have a^3/(a+ab+b) >= 2/3 (a - b/2)   (it simplifies to : a^3+b^3 >= a^2b+b^2a, true).

Then add the three inequalities.

[quote="belenos"]...We have a^3/(a+ab+b) >= 2/3 (a - b/2)   (it simplifies to : a^3+b^3 >= a^2b+b^2a, true).

Then add the three inequalities.[/quote]

Very nice solution! Here is mine:

Let

$\displaystyle X=\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2} $ and

$\displaystyle Y=\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{b^2+bc+c^2}+\frac{a^3}{c^2+ca+a^2} $.

Then $X-Y = (a-b) + (b-c) + (c-a) = 0$, that is $X=Y$. On the other hand, since $\displaystyle \frac{a^2-ab+b^2}{a^2+ab+b^2}\geq \frac{1}{3}$ etc., we have
$\displaystyle X+Y = \sum_{cyc} (a+b) \frac{a^2-ab+b^2}{a^2+ab+b^2}
\geq \frac{1}{3}\sum_{cyc} (a+b) =\frac{2}{3}$.

Therefore $\displaystyle X = \frac{1}{2}(X+Y) \geq \frac{1}{3}$. QED. :D  :D


superior algebra
superior algebra unsolved

In which commutative rings with 1 is the following statement true"Zero divisor in the ring is always a nilpotent element"

Did you actually find this somewhere, or were you just wondering?

I was just wondering.Why whats the matter?

Take any ring modulo a primary ideal .There all zero-divisors are nil-potent.The question of course is are they the only rings?


A cone-shaped mountain has its base on the ocean floor and has a height of $ 8000$ feet. The top $ \frac{1}{8}$ of the volume of the mountain is above water. What is the depth of the ocean at the base of the mountain, in feet?

$ \textbf{(A)}\ 4000 \qquad
\textbf{(B)}\ 2000(4\minus{}\sqrt{2}) \qquad
\textbf{(C)}\ 6000 \qquad
\textbf{(D)}\ 6400 \qquad
\textbf{(E)}\ 7000$

[hide]
The top of the mountain is similar to the entire mountain. Volume is proportional to the cube of the constant of proportionality. Therefore the constant of proportionality is $ \sqrt[3]{1/8} \equal{} 1/2$, so half of the height is above the water. The part below the water is also half the height, which is $ \boxed{4000}$. (A)[/hide]


Let C be a circle with center O. A line d intersects the circle C at C and
D and the diameter AB of C at M so that MB < MA and MD < MC.
The circumcircles of AOC and BOD intersect again at K. Prove that
OK is perpendicular to KM.

$ \bullet$ Consider the inversion through pole $ O$, power $ R^2$, where $ R$ is the radii of $ (C)$. We have $ \mathcal {I}(O,R)$ maps $ B,D,C,A$ respectively into themselves hence, $ \mathcal {I}(O,R): BD\mapsto (OBD)$, $ CA\mapsto (OCA)$. Thus, if $ E\equiv BD\cap AC$, then $ \mathcal {I}(O,R): E\mapsto K$, thus $ K\in$ the polar of $ E$ wrt $ (O)$. Now, let $ F\equiv BC\cap AD$. It is so well- known that $ E$ is the pole of $ FM$ wrt $ (O)\Longrightarrow K\in FM$. But $ FM\bot OE\Longrightarrow MK\bot OK$. The result is lead as follow.
Our proof is completed then $ \square$

$ \bullet$ [b]REMARK-[/b] The fact that $ AB$ is the diameter of $ (O)$ wich is mentioned in the problem is not needed at all. The statement of the problem is still hold when $ A,B$ are arbitry points lying on $ (O)$.


induction
number theory proposed
number theory

For any positive real $x$, denote $a(x)$ is number of primes not great than $x$. For any positive integer $m$, denote $b(m)$ is number of prime divisors of $m$. Prove that for all positive integer $n$ we have $\sum_{k=1}^{n}a(\frac{n}{k})=\sum_{k=1}^{n}b(k)$.

It is easy. Let $B(k)=\{p | \ p|k\}$, then $b(k)=\#B(k)$. Let p prime $p\le n$. Obviosly, $\# \{k|p\in B(k)\}=[n/p]$. Therefore $\sum_{k\le n}b(k)=\sum_{p\le n}[n/p]=\sum_{[n/p]=k}\#\{p\le k\}=\sum_{k\le n}a(k)$.

[quote="Rust"]$\sum_{p\le n}[n/p]=\sum_{[n/p]=k}\#\{p\le k\}$[/quote]
Why?
to cuopbientcd: can you post it?

I don't think Rust is true :D, Although my solution similar also.
Post your solution! Please! Use induction!

Sorry I don't like my solution!
about Rust's solution is doubt count:
$\sum_{k\leq n}b(k)=$ $\sum_{p\leq  n}\[\frac{n}{p}\]=$ $\sum_{k\leq n}a(\frac{n}{k})$


[color=green]Original text : [/color][b] Gazeta Matematica (Bucuresti) ,Problem 2814 .[/b]
[color=red] [b]
" Soient $A^\prime , B^{\prime}, C^{\prime} ,$ les points ou les arcs bissecteurs interieurs d'une triangle spherique $ABC$ recontrent les cotes opposes.
 Petut-on avoir $A^{\prime}B^{\prime}= A^{\prime}C^{\prime}$ sans le triangle $ABC$  soit isoscele  ?
La question analogue pour un triangle rectiligne a ete resolue par V.Cristescu, dans la [i]Gazeta Matematica [/i] tome VIII , p.112."
      (  J.Neuberg  )
[/b]     [/color]

Let $A',B',C'$ be the points where the arc bisectors of the spherical traingle $ABC$ meet the opposite sides.  Is it possible to have $A'B'=A'C'$ without $ABC$ being an iscosceles traingle?

Masoud Zargar


[quote="thanhak5"]Lemma :$ a \plus{} b \plus{} c \equal{} 3;a,b,c \ge 0$
=>$ ab^2 \plus{} bc^2 \plus{} ca^2 \plus{} abc \le 4$ .[/quote]
Let $ a\equal{}3x,$ $ b\equal{}3y$ and $ c\equal{}3z$ and see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=100960

I have another solution, without loss of generality, we assume $ a \equal{} max(a,b,c)$
Case 1: $ a \geq b \geq c$

             We have $ (b \minus{} a)(b \minus{} c) \leq 0$

             Hence, we also have the following result $ a^2b \plus{} b^2c \plus{} c^2a \plus{} abc \leq b(a \plus{} c)^2$

             But it's easy to check by Cauchy- Schwartz inequality this result:

             $ b(3 \minus{} b)^2 \equal{} 4b(\frac {3 \minus{} b}{2})(\frac {3 \minus{} b}{2}) \leq 4(\frac { b \plus{} \frac {3 \minus{} b}{2} \plus{} \frac {3 \minus{} b}{2}}{3})^3 \equal{} 4$

             Now we get the result of this problem.

Case2: $ a \geq c \geq b$
             This case is proved similarly to Case 1.

One can also prove inequality   $ a^2b \plus{} b^2c \plus{} c^2a \leq 4
where a,b,c \geq 0
such that a \plus{} b \plus{} c \equal{} 3$

Suppose $ a\ge b\ge c$ or $ a\le b\le c$, so $ (b\minus{}a)(b\minus{}c)\le 0$ We have
$ ab^2\plus{}bc^2\plus{}ca^2\plus{}abc\equal{}a(b\minus{}a)(b\minus{}c)\plus{}b(a\plus{}c)^2\le b(a\plus{}c)^2$
$ \equal{}2b(a\plus{}c)(a\plus{}c)\le \frac {(2a\plus{}2b\plus{}2c)^3}{27}\equal{}4$
The inequality was proved! :)

one can also prove the inequality
$ a^3b\plus{}b^3c\plus{}c^3a\plus{}6abc \leq 9$ try this!

We have
$ (a\plus{}b\plus{}c)^4\minus{}9[a^3b\plus{}b^3c\plus{}c^3a\plus{}2abc(a\plus{}b\plus{}c)]$=
=$ \frac{1}{2}\sum(a^2\minus{}b^2\minus{}3ab\plus{}2bc\plus{}ca)^2 \geq0$
-->Done :rotfl:
And this is a new inequality
$ a^4\plus{}b^4\plus{}c^4\plus{}6abc\geq9$


LaTeX
trigonometry
AMC
AIME
trig identities
Law of Cosines
Pythagorean Theorem

[quote="piac31"]I probably messed this up, but here I go...

For AB to be parallel to DC, angle ABC+ angle BCD = 180
2*cos-1(2/5)+ angle ACD=180.
Let M be the midpoint of DC.
MC/5= cos(180-2*cos-1(2/5))
DC=10*cos(180-2*cos-1(2/5))
DC=6.8[/quote]if you are not going to use latex, please write "2*cos^{-1}(2/5)" instead of "2*cos-1(2/5)".. it's kinda confusing.
good job and that is correct, but the question asks you to find $BD$:
[quote="kohjhsd"]In quadrilateral $ABCD$, $AC= AB = AD =5$, $BC=4$, and $AB \| CD$. Find $BD$.[/quote]

[hide]
Building on piac's $DC=6.8$:
The sum of the squares of the sides of a quadrilateral is equal to the sum of the squares of the diagonals (am I right on that one?)
So $DC^{2}+AB^{2}+AD^{2}+BC^{2}=AC^{2}+BD^{2}$
or $46.24+25+25+16=BD^{2}+25$
or $BD^{2}=87.24$
Solving gives $BD=9.340235543\cdots$[/hide]

EDIT: It's wrong, please check my other post.

[quote="CircleSquared"]The sum of the squares of the sides of a quadrilateral is equal to the sum of the squares of the diagonals (am I right on that one?)[/quote]wait.. are you sure? see what happens when $AB$ approaches $0$..

I looked up that sum of squares of sides and Circle squared was close.

Wolfram's Mathworld says:
An equation for the sum of the squares of side lengths is
a^2+b^2+c^2+d^2=p^2+q^2+4*x^2
where a,b,c, and d are sides, p and q are diagonals, and x is the length of the line joining the midpoints of the diagonals.

That's going to make things more difficult.

Then it looks like it's time for a
[hide="new method"]
Use law of cosines on ABC to find out $\angle{BAC}$.
It gives $47.156\cdots$
Use law of cosines on ACD to find out $\angle{CAD}$.  DC=6.8.
It gives $85.687\cdots$.
Add those up, and it's the measure of angle A, or $132.843$.
Use law of cosines one last time.
It gives $9.165$ as the answer.

[/hide]

[hide="you can also"]extend $AB$ past $A$ to $E$ so that $EA=5$
since $EA \parallel CD$, $\angle EAD = \angle ADC = \angle ACD = \angle BAC$, so $\triangle EAD \cong \triangle BAC$.
$B, C, D, E$ are on the circle centered at $A$ with radius $5$, so $\angle EDB =90$. we can use the pythagorean theorem to find $BD=\sqrt{10^{2}-4^{2}}=2\sqrt{21}$[/hide]

[b]First number 1, I dont think this problem was intended for a caculator so dont try finding the actually values of all the anges (thats why amc allows no caculators and defiently AIME)[/b]

You will first notice from angle chasing that if angle ABC is a, than angle DAB is 2a.  So we can find cos a frmo laws of cosines to be exactly 2/5.

SO cos a=2/5.  We keep that, instead of cos inversing it to find the angle (because no caculator).  That means that angle DAB is 2a so

$BD^{2}=5^{2}+5^{2}-2(5)(5)(\cos 2a)$ from laws of cosines

Now if cos a=2/5, cos 2a=2(2/5)^2-1 from double angle formula and that is exaclty -17/25

So plug -17/25 into for cos 2a and we get that

$BD^{2}=50+34=84$, so $BD=2\sqrt{21}$


Find all functions $f : \mathbb R \to \mathbb R$ such that
\[f(x)+f\left(1-\frac{1}{x}\right)=x,\]
holds for all real $x$.

First, this is a topic has bad title : http://www.mathlinks.ro/Forum/viewtopic.php?t=144403 . But I'll not lock it because you are a new member  :wink:

2-nd, I will help you :D . If put $g(x)=1-\frac{1}{x}$ then $g(g(g(x)))=x$ and that's all!  :wink:

[quote="Chessy"]What is $f(x)$ in this equation, can anyone solve it in a "step by step" way for me? Thank you !

$f(x)+f(1-\frac{1}{x})=x$[/quote]

To go further in the remark of N.T.TUAN, you have :

E1 : $f(x)+f(g(x))=x$ with $g(x)=1-\frac{1}{x}$. So you have :
E2 : $f(g(x))+f(g(g(x)))=g(x)$ and you look at $g(g(x))=\frac{1}{1-x}$ (nothing special)
E3 : $f(g(g(x)))+f(g(g(g(x))))=g(g(x))$ and then you discover that $g(g(g(x)))=x$ So you have :
E3 : $f(g(g(x)))+f(x)=g(g(x))$

Then E1-E2+E3 gives : $2f(x)=x-g(x)+g(g(x))$ and then $f(x)=\frac{1}{2}(x-1+\frac{1}{x}+\frac{1}{1-x})$
And it is easy to check that this necessary condition works.

Thank you very much sir, you have been a great help.

Take care


integration
calculus
probability and stats

Suppose $ T_i$ is a nonnegative rv identically distributed for any $ i$  and $ T_i$ is independent of $ T_j$ for $ i \neq j$.

The problem: Show that $ E(min(T_1,T_2 \cdots T_n)) \equal{} \int_0^{\infty} P^n(T_i > t) \,dt$.

My work:
I've reduced it to  $ \int_0^{\infty} nt P^{n \minus{} 1}(T_i > t)f'_{T_i}(t) \,dt$.

I figured it out: write $ x as \int_0^{x}  \,dt$, and then switch the order of integration!


Let $A$ be the set of the $16$ first positive integers. Find the least positive integer $k$ satisfying the condition: In every $k$-subset of $A$, there exist two distinct $a, b \in A$ such that $a^2 + b^2$ is prime.

[quote="Robert Gerbicz"]The answer is k=9.
k=8 isn't good, because let B={2,4,6,8,10,12,14,16} the all even numbers, and $ {a}^{2}+{b}^{2}$ is divisible by 4 so can't be prime.
k=9 is good, because consider the following partitions of A:
$ (15,2)$
$ (11,4)$
$ (5,6)$
$ (13,8)$
$ (3,10)$
$ (7,12)$
$ (9,14)$
$ (1,16)$

In each pair $ {a}^{2}+{b}^{2}$ is a prime number. If we select $ 9$ numbers from $ A$ then we should select a pair from these $ 8$ pairs, proving the statement.[/quote]


trigonometry
geometry
3D geometry
sphere
vector
combinatorial geometry
real analysis

$K$ a compact of $R^2$. Prove that there exists a unique ball that contain $K$ with minimal radius $r$.
Could you find a minoration of $r$ ? Prove that $r \leq \frac{diam(K)}{\sqrt{3}}$

I'll just sketch a proof for the upper bound (I'm sure I did this before on ML at some point, but I'm too lazy to look for the post :)).

Consider $\mathcal P$, the convex hull of $K$. Take a large circle containing $\mathcal P$, and start shrinking it until no longer possible.

If the circle only intersects $\mathcal P$ in two points, they must be antipodes in the circle, and we get $r\le \frac{diam(\mathcal P)}2$, which is even better than what we want.

If there are three points $A,B,C$ where the circle intersects $\mathcal P$, we may assume $ABC$ has no obtuse angles. Let $BC$ be the largest side of $ABC$. It's easy to prove now that $r\le\frac {BC}{\sqrt 3}\le\frac{diam(\mathcal P)}{\sqrt 3}$ by applying the sine law in $ABC$ for the angle $60^{\circ}\le\angle A\le 90^{\circ}$.

Yet, this has a generalization to higher dimensions. I don't know whether your approach still works, grobber, for the following general result:
   In $R^n$ the radius of the minimal ball containing a compact $K$ is at most $\sqrt{\frac{n}{2n+2}} diam(K)$ and the bound is sharp.
    I have a nice proof for this one using Caratheodory theorem and nice inequalities.

If we do almost the same thing I did above (except for the sine law part, of course :)), this problem (the general case) is reduced to the following one:

If we have $n+1$ points on the unit sphere in $\mathbb R^n$ s.t. the simplex formed has the center of the sphere inside, then there are two points at least $\sqrt{\frac{2n+2}n}$ apart.

It's clear that this follows if we prove that among the $n+1$ unit vectors in $\mathbb R^n$, there are two with their usual scalar product at most $-\frac 1n$ (we then apply the cosine theorem in the triangle formed by the origin and the two respective points on the sphere). In turn, this follows if we show that among the $n+1$ vectors $x_1,\ldots,x_{n+1}$ there is one, $x_i$, say, s.t. $\langle x_i, x_1+\ldots+x_{n+1}\rangle\le 0$ (call this assertion $\mathcal P$), because then the $j$ for which $\langle x_i,x_j\rangle$ is minimal will satisfy $\langle x_i,x_j\rangle\le-\frac 1n$.

On the other hand, proving $\mathcal P$ is easy: suppose $\langle x_i,s\rangle>0,\ \forall i$, where $s=x_1+\ldots+x_{n+1}$. It follows that all the points $x_i$ belong to the open half-space determined by $s^{\perp}$ (orthogonal complement) in which $s$ lies, and this means that the center of the sphere can't possibly be inside the simplex formed by the $n+1$ points. This contradicts the initial hypothesis, so we're done, I hope, unless I'm making some stupid mistakes :).


geometry
circumcircle
perpendicular bisector
quadrilateral
IMO
IMO 1998
Charles Leytem

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

Assume first that $ABCD$ is cyclic. It's easy then to see that $\angle APB+\angle CPD=\pi$, so $2S[APB]=R^2\sin \angle APB=R^2\sin\angle CPD=2S[CPD]$, where $R=PA=PB=PC=PD$ is the radius of the circumcircle of $ABCD$.

Conversely, assume we have $S[APB]=S[CPD]$. Let $M,N$ be the midpoints of $AB,CD$ respectively. We have $TM=AM,TN=DN$, where $T=AC\cap BD$. The hypothesis is equivalent to $PM\cdot AM=PN\cdot DN$, and from here we deduce $PM\cdot TM=PN\cdot TN$. A quick angle chase reveals $\angle MPN=\angle MTN$, so the triangles $TMN,PNM$ are, in fact, congruent, meaning that $TMPN$ is a parallelogram. This, however, means that the triangles $TAM,DTN$ are similar (because they have three pairs of perpendicular sides), so $\angle TAB=\angle TDC$, Q.E.D.

[b]Problem 1.[/b] In the convex quadrilateral ABCD, the diagonals AC and BD are perpendicular and the opposite sides AB and DC are not parallel. The point P, where the perpendicular bisectors of AB and DC meet, is inside ABCD. Prove that ABCD is cyclic if and only if the triangles ABP and CDP have equal areas.

I can prove this using an elegant method (I think) when ABCD is cyclic.
I think there should be a nice way for this for the other way, but I can't find it..

Here is my half-solution. If someone could do the other half using the similar approach, it will be appreciated..

If ABCD is cyclic, then area ABP = area CDP

Since the diagonals are perpendicular,

2*area ABCD = AB.CD + AD. BC (Ptolemy)
Now, take the sectors ABP, BPC, CPD, DPA, and rearrange them, so that the side AB and CD would be adjacent to each other.
It is still cyclic, because P is clearly the circumcentre of the circle.
Now, the area of the new quadrilateral did not change, since we have only rearranged it.
Let x be the angle between AB and CD.
2*area of new quadrilateral = AB.CD sin x + AD. BC sin (180-x) = sinx(AB.CD + AD. BC)

Equate the two expressions, and we have sinx = 1, so x = pi/2.

Simple calculation shows that angle APB = 180 - angle CPD, so the areas of the two triangles are equal.

I've been trying to apply something similar to the other half, but have not had any progress so far..

Sorry for not using LaTeX.

[quote="jpark"][b]Problem 1.[/b] In the convex quadrilateral ABCD, the diagonals AC and BD are perpendicular and the opposite sides AB and DC are not parallel. The point P, where the perpendicular bisectors of AB and DC meet, is inside ABCD. Prove that ABCD is cyclic if and only if the triangles ABP and CDP have equal areas.

I can prove this using an elegant method (I think) when ABCD is cyclic. [...]
I've been trying to apply something similar to the other half, but have not had any progress so far..[/quote]

First, the circumcenter $O$ of a cyclic quadrilateral $ABCD$ with perpendicular diagonals is always in its interior. This is because the angle $\measuredangle ACB > \measuredangle AOB = 90^o$, hence, the chord $AB$ has the circumcenter on the same side as the vertices $C, D$. Similarly, all other sides of the quadrilateral.

Let $ABCD$ be a quadrilateral with perpendicular diagonals, which is not cyclic. Let $M, N$ be the midpoints of the sides $AB, CD$ and $P$ the intersection of the perpendicular bisectors of these sides, which happens to be in the quadrilateral interior. Assume that the areas $S(\triangle APB), S(\triangle CPD)$ of the triangles $\triangle APB$ and $\triangle CPD$ are equal. Let $(O)$ be the circumcircle of the triangle $\triangle ABD$ centered at the point $O$ on the perpendicular bisector of $AB$. Assume first, that the vertex $C$ of the quadrilateral $ABCD$ is outside of this circumcircle. Move the point $C'$ along the line $AC$ form the point $C$ toward the the intersection $C\"$ of this line with the circle $(O)$. Let $N'$ be the midpoint of the segment $C'D$ and $P'$ the intersection of the perpendicular bisectors of $AB$ and $C'D$. In this process, the point $P'$ moves on the line $PM$ (the perpendicular bisector of $AB$) into the circumcenter $O$. The base $AB$ of the isosceles triangle $\triangle AP'B$ remains unchanged, but its altitude $P'M$ is increasing. As a result, the area $S(\triangle AP'B)$ is increasing and $S(\triangle AP'B) > S(\triangle APB)$. On the other hand, the base $C'D$ and the the angles $\measuredangle P'C'D = \measuredangle P'DC'$ of the isosceles triangle $\triangle C'P'D$ are decreasing. Hence, the remaining angle $\measuredangle C'P'D$ of this triangle is increasing and its altitude $P'N'$ is decreasing. As a result, the area $S(\triangle C'P'D)$ is decreasing and $S(\triangle C'P'D) < S(\triangle CPD)$. When and the point $C'$ coincides with the point $C\"$ and the point $P'$ coincides with the circumcenter $O$, the quadrilateral $ABC\"D$ becomes cyclic and $S(\triangle AOB) = S(\triangle C\"OD)$, as you just proved. But since $S(\triangle AOB) > S(\triangle APB)$, $S(\triangle C\"OD) < S(\triangle CPD)$, this is a contradiction with the assumption $S(\triangle APB) = S(\triangle CPD)$. The case, when the vertex $C$ lies inside of the circumcircle $(O)$ of the triangle $\triangle ABD$ is handled similarly.

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.IMO1998Problem1

Vo Duc Dien

Let us assume that $ ABCD $ is cyclic, $ N,S $ be the midpoints of $ AB $ and $ DC $ respectively, $ AC $ and $ BD $ meet at $ R $, $ SR $ cuts $ AB $ at $ M $ and $ NR $ cuts $ CD $ at $ T $. According to Brahmagupta theorem $ SM \perp AB $ and $ TN \perp CD $ so $ SRNP $ is a parallelogram. Now, we have to show that $ PN*BN=PS*SC $ , but note that $ PN=RS=CS $ (since $ DRC $ is right angled and $ RS $ is its median ) and for the same reasons $ BN=RN=PS $ so we are done

I tried to length-chase but it turned out to be a lot easier.
[hide="Thought Process"]
It seems easier to start with $ABCD$ being cyclic. Usually in these problems we can just backtrack or just use a linear function. Also, ROB1 noticed that since the perpendicular bisectors of $AB$ and $CD$ were diameters, so they met at the center, which means $P$ is the center of the circle circumscribing $ABCD$.

When I tried to length-chase, I started by calling $Q$ the intersection of $AC$ and $BD$. By power of a point, if $a=AQ,b=BQ,c=CQ$, then $DQ=\frac{ac}{b}$. Then by Pythagorean theorem, $AB=\sqrt{a^2+b^2}$ and $CD=\sqrt{\frac{a^2c^2}{b^2}+c^2}$. If the desired areas are equal, then the ratios of the heights from $P$ would be $\frac{b}{c}$ in some order. So the heights must be nice in terms of $a,b,c$, which means the solution might not be far away. However, we only have one thing that gives us the height by Pythagorean theorem. That is, the distance from $A$ to the midpoint of $AB$ is $\frac{\sqrt{a^2+b^2}}{2}$, and the radius $AP$ is uniquely determined by $a,b,c$.

But I was at a restaurant (Duke TIP rag night) and obviously too lazy to remember the circumradius formula, so I decided to switch to angle-chasing when length-chasing was straightforward. Well first of all, for properties of $ABP$ and $CDP$, we know that $AP=BP=CP=DP$ because they're radii. In what ways can we compare areas? That was when I suddenly remembered that if we proved $\angle APB+\angle CPD=180^\circ$, then we would be done. I fruitlessly angle-chased on the individual triangles, but then when I looked at the big picture, the circle, inscribed angles would work.

$\angle APB+\angle CPD=2\angle ACB+2\angle CBD=180^\circ$, but we still have to prove in the other direction. Looking back at the diagrams I drew without the cyclic condition, I noticed that when you vary $D$, the point $P$ varies. We can only increase the area of one of the triangles and decrease the area of the other. If $ABCD$ is not cyclic, we have two similar cases. Assume $D$ lies outside the circumcircle of $ABC$. Let the perpendicular from $B$ to $AC$ meet the circumcircle of $ABC$ at $E$ and call $O$ the circumcenter of $ABC$. I tried this for a while but the variables weren't working in my favor, so I just approached it the other way.

Let's use the right angles. Notice that if $M,N$ are the midpoints of $AB,CD$ respectively, and $Q$ was the intersection of $AC$ and $BD$, then $MB=MQ$ and $NC=NQ$. Since I stick to the simple methods, I will just use the simple base and height. The equality of areas means $MP\cdot MQ=MP\cdot MB=NP\cdot NC=NP\cdot NQ$. When only four points are involved in such an equation, we look for properties of quadrilateral $MPNQ$. Specifically, it looks like a parallelogram. We want to prove it using the products we already know, which call for similar triangles. It seems like we should prove that $\triangle MQN\sim\triangle NPM$, which is equivalent to $\angle MPN=\angle MQN$.

Out of $\angle MPN$ and $\angle MQN$, it seems easiest based on the diagram to break up $\angle MQN$. We know that $\angle MQN=\angle MQA+90+\angle NQD$. It's a good thing we did not define new points here because that lets us notice $\angle MQA=\angle MAQ=\angle BAQ$, and $\angle NQD=\angle CDQ$. Then $\angle MQN=\angle BAQ+90+\angle CDQ$. Notice that $\angle BAQ=\angle BAC$ and $\angle CDQ=\angle CDB$.

Here are some stupid things I did. Let's get rid of the $90$, so simplify $\angle BAQ+90$ first. We notice a compatible $90^\circ$ angle as $\angle AQB$, so $\angle BAQ+90=\angle BAQ+\angle AQB$. This reminds us of exterior angles. Which exterior angle is compatible with $\angle CDB$? Define $R$ as the intersection of $AB$ and $CD$. Then $\angle MQN=\angle BAQ+\angle AQB+\angle CDQ=\angle QBR+\angle CDQ=\angle DBR+\angle RDB=180-\angle BRD$. How do you compare this with $\angle MPN$? $\angle BRD=\angle MRN$, but $\angle PMR=\angle PNR=90^\circ$, so $\angle MQN=180-\angle MRN=\angle MPN$, as desired.

This tells us that $MPNQ$ is a parallelogram. But we didn't know that $\angle PMQ=\angle PNQ$ before, so we will start from there. Clearly $90-\angle BMQ=\angle PMQ=\angle PNQ=90-\angle CNQ$, so $\angle BMQ=\angle CNQ$. Then we would uniquely define $\angle ABD$ and $\angle ACD$. But there is probably a shortcut.
[/hide]
[hide="Solution"]
Assume that $ABCD$ is cyclic. The perpendicular bisector of $AB$ is a line, and it contains the circumcenter of $ABCD$, so it is a diameter. The perpendicular bisector of $CD$ is also a diameter, and the diameters both contain the circumcenter, so the point they meet at, $P$, is the circumcenter of $ABCD$.

Call $Q$ the intersection of $AC$ and $BD$. The problem tells us $\angle BQC=90^\circ$. By inscribed angles, $\angle APB+\angle CPD=2\angle ACB+2\angle CBD=2(\angle QCB+\angle CBQ)=2(180-\angle BQC)=180^\circ$.

Rotate around $P$ by $\angle BPC$ counter-clockwise so that $C,D,P$ map to $B,D',P$ respectively. Then $\angle APB+\angle BPD'=\angle APB+\angle CPD=180^\circ$, which means $A,P,D'$ are collinear. Then the altitudes from $B$ to $AP$ and $D'P$ are equal. Also, $AP=D'P$, so the areas of $\triangle ABP$ and $\triangle BD'P$ are equal. But by rotation the area of $\triangle BD'P$ is equal to the area of $\triangle CDP$, so by transitivity we get the desired result.

Call $M,N$ the midpoints of $AB,CD$ respectively. Since $\angle AQB=90^\circ$, $MB=MQ$, and similarly $NC=NQ$. If the areas of $\triangle ABP$ and $\triangle CDP$ were equal, then $MQ\cdot MP=MB\cdot MP=\frac{AB\cdot MP}{2}=\frac{CD\cdot NP}{2}=NC\cdot NP=NQ\cdot NP$.

Partition $\angle MQN=\angle MQA+\angle AQD+\angle DQN$. Notice that $\angle AQD=90^\circ=\angle AQB$, $\angle MQA=\angle MAQ=\angle BAQ$, and similarly $\angle DQN=\angle CDQ$, so $\angle MQN=\angle AQB+\angle BAQ+\angle CDQ$.

Define $R$ to be the intersection of $AB$ with $CD$. Then by exterior angles, $\angle MQN=\angle QBR+\angle CDQ=\angle DBR+\angle RDB$. Since the sum of angles in a triangle is $180^\circ$, $\angle MQN=180-\angle BRD=180-\angle MRN$. Notice that $360=\angle MRN+\angle RNP+\angle NPM+\angle PMR=180+\angle MRN+\angle MPN$. Subtracting $180+\angle MRN$ from both sides gives us $\angle MPN=180-\angle MRN=\angle MQN$.

Recall from earlier that $MQ\cdot MP=NQ\cdot NP$. This means $\frac{MQ}{NP}=\frac{NQ}{MP}$. By our new result that $\angle MPN=\angle MQN$, SAS tells us $\triangle MQN\sim\triangle NPM$.

Then $\angle PMQ=\angle PMN+\angle NMQ=\angle QNM+\angle MNP=\angle PNQ$. Clearly $\angle MBQ=\angle MQB$, so by exterior angles, $\angle AMQ=\angle MBQ+\angle MQB=2\angle MBQ=2\angle ABD$. Analogously, $\angle DNQ=2\angle ACD$. But $\angle ABD=\frac{\angle AMQ}{2}=\frac{90+\angle PMQ}{2}=\frac{90+\angle PNQ}{2}=\frac{\angle DNQ}{2}=\angle ACD$, so $ABCD$ is cyclic.
[/hide]

Storage

[quote=1998 G1]A convex quadrilateral $ABCD$ has perpendicular diagonals. The perpendicular bisectors of the sides $AB$ and $CD$ meet at a unique point $P$ inside $ABCD$. Prove that the quadrilateral $ABCD$ is cyclic if and only if triangles $ABP$ and $CDP$ have equal areas.[/quote]

If $ABCD$ cyclic with circumradius $R$, then conclusion follows since $\angle APB+\angle CPD=180^{\circ}$ as $AC \perp BD$, so $[APB]=\frac{1}{2}R^2\sin \angle APB=\frac{1}{2}R^2\sin \angle CPD=[CPD]$ as desired.

Now assume $ABCD$ is not cyclic, but $AC \perp BD$ and $[APB]=[CPD]$. Let $E=AC \cap BD$. WLOG, say $P$ lies inside $\triangle BEC$ and $PA<PD$. Let $A', B'$ be second intersections of diagonals $AC$ and $BD$ with $\odot(P, PA)$. Note that $ABA'B'$ is a convex cyclic quadrilateral with perpendicular diagonals, and so $[APB]=[A'PB']$.

Note that $A'$ lies on segment $AC$ and $B'$ on segment $BD$, since $PA<PD$. Since ray $\overrightarrow{A'P}$ meets line $BD$ at it's extension beyond $P$ (as $\angle EA'P=\angle PAC<90^{\circ}$) and $B', D$ lie on the same side of line $A'P$, with $B'$ closer to the intersection than $D$, we conclude that $[A'PB']<[A'PD]$. But $[A'PD]<[BPD]$ since $A'$ lies inside $\triangle PBD$. This chain of inequalities is a contradiction! So, we conclude that $ABCD$ must be cyclic. $\blacksquare$

P.S. grobber's solution is [i]really[/i] nice :)

Way too hard for an early IMO #1.

First, we will prove that if $ABCD$ is cyclic, the area condition holds. Let $E$ be the intersection of lines $AC$ and $BD$, and let $r$ be the circumradius of $ABCD$. Note that $P$ is the circumcenter of $ABCD$, so reflecting $C$ about $P$ to a point $C'$ creates a trapezoid $AC'BD$. In particular, because it is cyclic, it is an isosceles trapezoid, so $\angle CPB=180^{\circ}-\angle BPC'=180^{\circ}-\angle APD$. Therefore $\angle APB+\angle CPD=180^{\circ}$, so $\sin \angle APB =\sin \angle CPD$, implying the area condition as the area of $APB$ and $CPD$ are $\frac{r^2}{2} \sin \angle APB$ and $\frac{r^2}{2} \sin \angle CPD$, respectively.

Now we will prove the harder part, that given the area of triangles $APB$ and $CPD$ are the same, $ABCD$ is cyclic. Let $X$ be the intersection of lines $AB$ and $CD$ (if $AB\parallel CD$, either $P$ doesn't exist or $ABCD$ is an isosceles trapezoid), and let $M$ and $N$ be the midpoints of $AB,CD$, respectively. Then notice that \begin{align*} \angle MPN &= 180^{\circ}-\angle BXC \\ &=\angle XBC+\angle XCB \\ &= \angle ABE+(\angle EBC+\angle ECB)+\angle ECD \\ &= \angle BME+90^{\circ}+\angle NEC \\ &= \angle MEN. \end{align*} Also, we have $$\frac{EN}{EM}=\frac{CN}{BM}=\frac{PM}{PN},$$ where the second equality is by the area condition. This implies $EMPN$ is a parallelogram. Therefore we have $$\angle ACD = \frac{1}{2}\angle END = \frac{1}{2}(90^{\circ}-\angle PNE) = \frac{1}{2}(90^{\circ}-\angle PME) =\frac{1}{2}\angle AME =\angle ABD,$$ and we are done.

I don't know if this works (there are probably a lot of configuration issues) but I think you can fix $A,B,C$ and let $D$ vary along the line through $B$ that is perpendicular to $AC$ and show that only one point on this line will work (by showing that the area of one triangle decreases while the other increases), and then check that this point must be the intersection of $(ABC)$ with the line.

Edit: I don't think there would be any issues with area becoming "negative" because we are given $P$ is inside $ABCD$

Let $E = \overline{AC} \cap \overline{BD}$. Suppose $P$ lies in $\triangle AEB$. The other cases are similar to the following.

Let $M,N$ be the foot of the perpendiculars from $P$ to $AC$ and $BD$ respectively. We have
\begin{align*}
[PAB] &= \dfrac 12 (AE\cdot BE - BE\cdot ME - AE\cdot EN) \\ &= \dfrac 12 (AM\cdot BN - ME\cdot NE)
\end{align*}
Similarly, one can see that $[PCD] = \frac 12 (CM\cdot ND - EM\cdot EN)$. Thus, we have
\begin{align}
[PAB] - [PCD] = \dfrac 12 (AM\cdot BN - CM\cdot DN)
\end{align}
Suppose $ABCD$ is cyclic, then $P$ is the circumcenter of $ABCD$. So, $AM = CM$ and $BN = DN$. Thus, $(1)$ gives us $[PAB] = [PCD]$.

On the other hand, suppose $ABCD$ is not cyclic. Without the loss of generality assume $PA = PB > PC = PD$. Then we see that $AM > CM$ and $BN > DN$, so $(1)$ gives us $[PAB] > [PCD]$. This proves the other implication, and we are done.

[quote=orl]A convex quadrilateral $ABCD$ has perpendicular diagonals. The perpendicular bisectors of the sides $AB$ and $CD$ meet at a unique point $P$ inside $ABCD$. Prove that the quadrilateral $ABCD$ is cyclic if and only if triangles $ABP$ and $CDP$ have equal areas.[/quote]

I guess you can successfully complex-bash this using the axis as diagonals with the standard aereal formula and the power of the origin point.

[quote=Ruy]
I guess you can successfully complex-bash this using the axis as diagonals with the standard aereal formula and the power of the origin point.[/quote]
Well, even cartesian coordinates are enough. Let $A=(0,a),B=(b,0),C=(0,c),D=(d,0)$. $ABCD$ is cyclic if and only if $ac=bd$. By perpendiculars from $\left(\frac b2,\frac a2\right)$ and $\left(\frac d2,\frac c2\right)$, $P=\left(\frac{b^2c-a^2c-ad^2+ac^2}{2(bc-ad)},\frac{bc^2+b^2d-bd^2-a^2d}{2(bc-ad)}\right)$. Using determinants, the equation $[ABP]=[CDP]$ factors as $0=(bd-ac)((b-d)^2+(a-c)^2)$, as desired.

Let diagonals meet at $S$ and Let $M,N$ be midpoint's of $AB,CD$. we have $[APB] = [CPD] \implies [AMP] = [DNP] \implies \frac{MS}{NS} = \frac{NP}{MP}$. we also have $\angle MSN = \angle MAS + \angle NDS + \angle 90 = \angle MPN$ which implies that $MPN$ and $NSM$ are congruent so $MPNS$ is parallelogram so $SM \perp DN$ and $SN \perp AM$ and $AS \perp DS$ which implies $AMS$ and $DNS$ are similar so $\angle MAS = \angle NDS$ so $ABCD$ is cyclic.

Now assume $ABCD$ is cyclic. $2[APB] = AP.PB.\sin{APB} = DP.PC.\sin{DPC} = 2[DPC] \implies [APB] = [CPD]$

Let $A=(0,2a)$, $B=(2b,0)$, $C=(0,2c)$, $D=(2d,0)$, where $a,b>0$ and $c,d<0$. Then we have:
\[AB:y=\frac{b}{a}x-\frac{b^2}{a}+a\]
\[CD:y=\frac{d}{c}x-\frac{d^2}{c}+c\]
and their intersection has $x$-coordinate $\frac{b^2c+ac^2-ad^2-a^2c}{bc-ad}.$ Clearly the length of the altitude from $P$ to $AB$ is then
\[\left(b-\frac{b^2c+ac^2-ad^2-a^2c}{bc-ad}\right)\cdot \frac{\sqrt{a^2+b^2}}{a}\]
while the length of the base, $AB$, is $2\sqrt{a^2+b^2}$. Then the area of $ABP$ is equal to
\[\left(\frac{d^2+ac-c^2-bd}{bc-ad}\right)(a^2+b^2).\]
Similarly, the length of the altitude from $P$ to $CD$ is equal to
\[\left(\frac{b^2c+ac^2-ad^2-a^2c}{bc-ad}-d\right)\cdot \frac{\sqrt{c^2+d^2}}{-c}\]
and the length of the base is $2\sqrt{c^2+d^2}$. Then the area of $CDP$ is equal to
\[-\left(\frac{b^2+ac-a^2-bd}{bc-ad}\right)(c^2+d^2).\]
If $ABP$ and $CDP$ have equal areas then
\[(d^2+ac-c^2-bd)(a^2+b^2)=-(b^2+ac-a^2-bd)(c^2+d^2)\]
which is obviously true when $ac=bd$, which is precisely the condition that $ABCD$ is cyclic. Then, it suffices to show that from this equation follows $ac=bd$; note that the equation is equivalent to
\[(ac-bd)(a^2+b^2+c^2+d^2)=(a^2-b^2)(c^2+d^2)+(a^2+b^2)(c^2-d^2)=2a^2c^2-2b^2d^2.\]
Then if $ac\neq bd$ then we have that $a^2+b^2+c^2+d^2=2ac+2bd$ which implies that $a=c$ and $b=d$, a contradiction because $a,b>0$ and $c,d<0$. We are done. $\blacksquare$

[hide=Diagram][img width=50]https://media.discordapp.net/attachments/986049049834713139/996579412969521335/ISL1998G1.png?width=1136&height=1027[/img][/hide]

[b]Forward Direction[/b]
If $ABCD$ is cyclic then $P$ must be the circumcenter. Since $AC\perp BD$ we have $\widehat{AB}+\widehat{CD}=180^\circ.$ Thus, $\angle APB,\angle CPD$ are supplementary. In particular, $\sin (\angle APB)=\sin (\angle CPD).$ Let $r$ be the circumradius then \[[APB]=\frac{1}{2}r^2\sin (\angle APB)=\frac{1}{2}r^2\sin (\angle CPD)=[CPD]\] as desired.

[b]Backward Direction[/b]
If $[APB]=[CPD]$ then let $E$ be the intersection of $AC$ and $BD.$ Let $M$ be midpoint of $CD$ and $N$ be midpoint of $AB.$ We claim that $EMPN$ is a parallelogram.

To prove our claim, let $E=(0,0),A=(0,a),B=(b,0),C=(0,-c),D=(-d,0)$ where $a,b,c,d$ are positive. $M=(-\frac{d}{2},-\frac{c}{2})$ and $N=(\frac{b}{2},\frac{a}{2}).$ It suffices to show that $P=(\frac{b-d}{2},\frac{a-c}{2}).$ Let $P'=(\frac{b-d}{2},\frac{a-c}{2})$ and we verify that $P'C=PD$ and $P'A=P'B.$ Since $P$ is unique, we know that $P'=P$ so our claim holds.

Since $EMPN$ is a parallelogram, $EM=PN$, which implies $DM=PN.$ We have $[DMP]=[APN]$ which implies $\triangle DMP\cong \triangle PNA.$ Thus, $DP=AP$ and so $P$ is equidistant from $A,B,C,D.$ It follows that $ABCD$ is cyclic, as desired.

The direction from cyclic to equal area is pretty easy so I'll skip it.

Here's a really easy way for equal area to cyclic (i don't think anyone's posted a coordinate method yet? or at least the easy way). Let $P=(0,0)$ and let the circle centered at $P$ through $AB$ be $x^2+y^2=1$ and through $C,D$ be $x^2+y^2=r^2.$ Now wlog the perpendicular lines to be perpendicular to the axes. We can wlog some more variables and find that $A(-\sqrt{1-y^2}, y), B(x, \sqrt{1-x^2}), C(\sqrt{r^2-y^2}, y), D(x, -\sqrt{r^2-x^2}).$ Now Shoelace trivializes the problem by giving us that $\sqrt{(r^2-x^2)(r^2-y^2)} = \sqrt{(1-x^2)(1-y^2)}$ and we get $r=1$, as desired.

Apply cartesian coordinates with $A=(0,a), C=(0,c), B=(b,0), D=(d,0)$ so that the diagonals lie on the axis of the coordinate plane. Then we have the perpendicular bisector of $AB$ as $y=\frac{b}{a}x+\frac{a}{2}-\frac{b^2}{2a}$  and similarly the perpendicular bisector of $CD$ is $y=\frac{d}{c}x+\frac{c}{2}-\frac{d^2}{2c}$. Then, we have $$P=\left(\frac{ac^2+b^2c-a^2c-ad^2}{2bc-2ad},\frac{bc^2+b^2d-bd^2-a^2d}{2bc-2ad}\right)$$ and Shoelace now gives $$-a^3c + a^2 b d + 2 a^2 c^2 - a b^2 c - a c^3 - a c d^2 + b^3 d - 2 b^2 d^2 + b c^2 d + b d^3=0 \rightarrow ac=bd,$$ which implies cyclicity. $\blacksquare$

Assume $ABCD$ is cyclic. Then $P$ is the center of $(ABCD)$. We have $\angle APB+\angle CPD=2\angle ACB+2\angle CAD=180$. So, $AP*BP*sin(APB)=CP*DP*sin(CPD)$. So, $[ABP]=[CDP]$.

Assume $[ABP]=[CDP]$. Then, $[ABPC]=[CDPB]$. Let $X$ and $Y$ be the feet of the perpendiculars from $P$ to $AC$ and $BD$ respectively. Then, $[ABYC]=[CDXB]$. So, $CX/AX=DY/BY$.

Let $M$ and $N$ be the midpoints of $AB$ and $CD$ respectively. Then, $PY^2+BY^2=BP^2=AP^2=PX^2+AX^2$. Similarly, $PY^2+DY^2=PX^2+CX^2$. So, $CX^2-AX^2=DY^2-BY^2$. Combining this with $CX/AX=DY/BY$, we find that $DY=BY$ and $CX=AX$ (in which case $ABCD$ is cyclic as desired) or $AC=BD$. In this case, $PY=PX$, so by symmetry, $ABCD$ is an isosceles trapezoid, which is cyclic.

Necessity: Given ABCD is cyclic it is well known that the intersection P is the unique circumcenter. <ADB+<CAD=90 degrees (orthogonal quad), so <APB+<CPD=180 degrees. Then (CP=DP=AP=BP) CP*DPsinCPD=AP*BPsinAPB, or [ABP]=[CBP]. (directed angles might be necessary depending on config? can someone pm or tell me here if it is necessary, the sketch is done though)
Sufficiency: Given [ABP]=[CBP] I use coordinates. Unfortunately my proof was not saved but basically I let the intersection of the diagonals be the origin, and shoelace formula to get PoP stays constant on the origin with ac=bd.

Actually, this can lead to an interesting corollary: BF=EP, from which it follows that EC=PF.

[hide=only if (forward direction)] Since the diagonals of $ABCD$ are perpendicular and it is cyclic, we have $$\angle{APB} + \angle{CPD} = 2(\angle ADB + \angle{CAD})=2(90^\circ)=180^\circ,$$ so it follows that $$\sin \angle APB = \sin \angle CPD.$$ Also note that the intersection point of the perpendicular bisectors of $\overline{AB}$ and $\overline{CD}$ is equidistant from $A,B,C,D$ so it must be the center of $(ABCD).$ Thus, we have $$PA=PB=PC=PD,$$ so $$[ABP] = \tfrac{1}{2} PA \cdot PB \cdot \sin \angle APB = \tfrac{1}{2} PC \cdot PD \sin \angle CPD = [CDP],$$ as desired. $\square$
[/hide]
[hide=if (backward direction)]
Let $O$ be the intersection of $\overline{AC}$ and $\overline{BD}.$ Let us place the quadrilateral onto the Cartesian plane such that $O$ is the origin and without loss of generality, let $$A=(0,a), B=(-b,0), C=(0,-c), D=(d,0),$$ where $a,b,c,d \in \mathbb{R}_{+}.$ Let $M$ and $N$ be the midpoints of $\overline{AB}$ and $\overline{CD}$ respectively. Note that $$M=(-\tfrac{b}{2}, \tfrac{a}{2}), N=(\tfrac{d}{2}, -\tfrac{c}{2}).$$ Then, it is not difficult to find that the equation of the perpendicular bisectors of $\overline{AB}$ and $\overline{CD}$ are $$y = -\tfrac{b}{a}x+\tfrac{a}{2} - \tfrac{b^2}{2a},$$ and $$y=-\tfrac{d}{c}x - \tfrac{c}{2} + \tfrac{d^2}{2c},$$ respectively. The intersection of these lines is $$P=\left( \tfrac{a^2c-ac^2+ad^2-b^2c}{2ad-2bc}, \tfrac{-a^2bc+abc^2-abd^2+b^3c}{2a^2d-2abc} \right).$$
Then, using the fact that $[ABP]=[CDP]$ and the Sholeace Formula, we get $$-cd - \tfrac{-a^2bcd+abc^2d-abd^3+b^3cd}{2a^2d-2abc} + \tfrac{a^2c^2-ac^3+ad^2c-b^2c^2}{2ad-2bc} = -ab + \tfrac{-a^2b^2c+ab^2c^2-ab^2d^2+b^4c}{2a^2d-2abc} - \tfrac{a^3c-a^2c^2+a^2d^2-ab^2c}{2ad-2bc},$$ which (with careful, but not complicated, computations) yields $$-ac=-bd \rightarrow ac=bd.$$ By the converse of Power of a Point, this tells us that $ABCD$ is cyclic, as desired. $\square$
[/hide]


Prove that $ (n!)^{2} < k!(2n\minus{}k)!$, for $ n, k \in \mathbb{N}$ and $ n>k.$

Thanks, [hide=" or"] divide both sides by $ (n!)^{2}$ to get the same thing.[/hide]


if F $ %Error. "display" is a bad command.
\subseteq$ E be two feilds,and [E:F] is prime number,show there is no feilds between E and F(means there is no feild as K so that F $ %Error. "display" is a bad command.
\subseteq$ K $ %Error. "display" is a bad command.
\subseteq$ E).

If $ K \subseteq L \subseteq M$ are fields, then $ [M: K]\equal{}[M: L]\cdot[L: K]$.


i noticed that each teacher got an Art of Problem Solving Phamphlet. What if suddenly AOPS GETS MORE AND MORE AND MORE users. I mean like alot. Then each class would be filled up like in one day. Since there are so many people in America. What would happen to the people who been here for a long time?

Should we get to the point that classes are filling in a day, we'll have more classes, so you'll still be able to get the classes you want.


Similar to the Monty Hall Problem.. but still unsure...

You have a hat in which there are three pancakes: One is golden on both sides, one is brown on both sides, and one is golden on one side and brown on the other. You withdraw one pancake, look at one side, and see that it is brown. What is the probability that the other side is brown?

[hide="answer"] so is it 2/3 or 1/2? [/hide]

[hide]$\frac{2}{3}$
because there are 3 sides which are brown, and 2 of them satisfy the required condition.[/hide]

[hide]Possibility #1: It is brown on one side, and golden on the other. Other side:GOLDEN

Possibility #2: It is brown on both sides and side 1 is face up
Other side:BROWN

Possibility #3: It is brown on both sides and side 2 is face up
Other side:BROWN

Answer: 2/3[/hide]

[hide]The probability is actually pretty high. The gold pancake is automatically eliminated. There is only one possibility that the other side is gold, while there are two possibilities it is brown. therefore, it is 2/3.[/hide]


search
function
number theory unsolved
number theory

Let $ p$ be prime number with $ p\equiv 1(mod 4)$.
Prove that $ \displaystyle\sum_{i\equal{}1}^{\frac{p\minus{}1}{4}}[\sqrt{ip}]\equal{}\frac{p^2\minus{}1}{12}$.

Hi Rust, thanks!

But your solution seems wrong, because the sum must be $ \frac{p^2\minus{}1}{12}$.
for example if $ p\equal{}5$ then the sum is equal to $ 2$ not $ 3$.

I edit my post.
$ \frac {x^2}{p} \equal{} k_x \plus{} \frac {y}{p}$, were $ 0 < y < p, y\equiv x^2\mod p$.
When $ x\le \frac {p \minus{} 1}{2}$, then $ k_x\le \frac {p \minus{} 1}{4} \minus{} 1$.

This problem is in 4 different posts. Something does not work with the search function on MathLinks.


Mom turned 90 on Friday, June 2, 2006.  On what day of the week was she born?

A.Sunday	 B.Monday C.Thursday D.Friday E.Saturday

Would it just be:

90/7=12....R6 Counting back six, it would be Sat.

But then you have leap years, which would be 90/4=22...w/e, so 22 extra days. Now, you have 22/7=3....R1 So you count back one more, so still [u]D)Friday[/u]?

If it gets confusing to read, I'm deeply sorry.


algebra
polynomial
algorithm
induction
algebra solved

Let $f(x)$ be the polynomial such that $f(x)=f(1-x)$.
Prove that $f(x)$ is the polynomial which is described in terms of $x(x-1)$.

Thank you,Misha.

Must we take into consideration the degree of the polinomial $f(x)$ ?

kunny

Why should we? I think my solution works for all degrees.

Misha

You are wrong,Misha ;)

The degree of $f(x)$ is even.

kunny

[quote="G.Huvent"]
Using $(x+1/2)^2=x(x-1)+1/4$ is wrong.

kunny

I don't understand you comment, kunny. Was the question about the degree an additional problem or do you have problems with my solution? :maybe:

Misha

[quote="Misha123"]It is quite easy, I think. We divide $f(x)$ by $x(x-1)$ and get $f(x)=g(x)x(x-1)+r(x)$. Because of $f(1)=f(0)$ we have $r(1)=r(0)$ which means that $r(x)$ is a constant (since the degree of it is less than 2). So we have $f(x)=x(x-1)g(x)+r$ where $r$ is a constant. Now we have $g(x)=g(1-x)$ for all $x$ because for $x\neq0,1$ it follows from $f(x)=f(1-x)$ and for $x=0,1$ because $f$  is continious. Then we make the same argumentation for $g(x)$ as for $f(x)$ before and so on. With such an algorithm we get an expression in $x(x-1)$ at last.

Misha[/quote]

How did you obtain $g(x)=g(1-x)$?

kunny


Gauss
Princeton
college
geometry
geometric transformation
reflection
MIT

In order from strongest to weakest, choose five "categories" that best fit your self-identity. Not character traits, like "nice" or "inquisitive", but categories - in the sense of belonging to a group.

For example, someone might say...

1. I'm a doctor
2. I'm a New Yorker
3. I'm a Christian
4. I'm a male
5. I'm a blonde

Someone else might say...

1. I'm a woman
2. I'm a Cambodian-American
3. I'm a Democrat
4. I'm a biologist
5. I'm a vegetarian

Also, think about how other people identify you. If you were at a party and after you left someone was trying to describe you to someone else.. What would they say about you? "Oh! She was the blonde physicist from California..." "He was the tall New Yorker who smoked..." etc..

1. I'm a math major.
2. I'm a New Yorker.
3. I'm a student.
4. I'm a registered Democrat.
5. I'm a half-Jewish half-WASP in ethnicity with secular humanist leftist moral/relgious beliefs.  (Yes, that's all one :))

I don't know how people think of me ... perhaps I shall take a poll of my friends to find out :)

This is a really good question.

1. I'm an Asian American
2. I'm a big Red Sox fan (this year's the year!)
3. I'm a student
4. I'm a baseball player
5. I'm a pianist

my friends probably think i'm slightly (ok, maybe a little more than slightly) crazy, but amusing and funny. So it'd probably be some thing like "that crazy chinese kid who's a red sox fan" or something around that

1. I'm an Asian-Mexican-American
2. I'm an AoPSer
3. I'm a high school student.
4. I'm a math fanatic.
5. I'm (going to be) an Exonian.

[edited]
1. I'm a cross-country runner.
2. I'm male.
3. I am math and science enthusiast.
4. I'm a student(high school).
5. I am Roman Catholic

JBL what does WASP stand for/mean.  I just had a WW2 test and all I can remember is that during the war it was the Women's Airforce Service Pilots.
I know that this is not what your are referring to.

I am an Asian,
I am a junior high student(now),
I am Feiqi Jiang,
I am a saxophone player,
I am an atheist...

Hmm fun stuff... let me see if i can get this correct the first attempt (usually, my first thought when analysing myself isnt exactly right)

1. I am intense
2. I am an intellectual
3. I am athletic
4. I am a deist with a very strong belief in the ability to scientifically analyze the world
5. I am my own person.

Other people see me as a smart and athletic, but arrogant person (I never understood why, mainly since the people who think this barely know me. My close friends never think this.), with an abrupt way about me, and a tendency to be very prankish.

[quote="Various people"]I am intense.
I am athletic.
I am my own person.
I am not very religious.
I am Feiqi Jiang.
I love math and science.[/quote]

None of these are, strictly speaking, what was being asked for.

Exonian?  Is that someone who goes to Exeter?  Don't they have a better word for it?  (Prep-schooler? :))

1. I worship Oasis *cough*
2. I'm a sophomore in high school
3. I'm homeschooled
4. I'm somewhat of a Buddhist but "atheist" would probably be more accurate.
5. I'm conservative but I don't really have a party affiliation since I don't vote.

(is it just me, or do liberal/democrats here greatly outnumber the conservative/Republicans? or is that just because the liberal/democrats are just more outspoken?)

Heh... i knew i wouldnt get it right the first time. Ill edit it later, food ready, plus ap studying to do (so muhc art to know!).

1. I'm a high school sophomore girl.
2. I'm Singaporean (it's amusing how often this pops up - blame the incorrigible accent).
3. I adore Biology (and science in general).
4. I vacillate between flippancy and gravity.
5. I live in California.

Gauss, I'd be interested in hearing your answer to your own question, as well :)

[quote="JS1527"][quote]I worship Oasis *cough* [/quote]

The band?  I like them too!  Nobody seems to remember them...[/quote]

awesome :) i know, everyone my age was too young to be listening to them when they were popular, and they band is too young to be considered "classic" or anything. heehee.

dragon 96 i thought we had to describe urselves in fine lines.
not more than that. :P  :)

Here's 6 through 10.
6. I am a military brat.
7. I am a fair person. (most of the time)
8. I am a boy.
9. I am the smartest math student in my 8th grade honors math class.
10. I am a visual learner.

[quote="TZF"]I'm going to single you out and use you as an example.

[quote]1. I am a Christian.
2. I am a [color=red]math [/color]genius.
3. I am an athlete.
4. I am a rubix cube solver.
5. I am from Boston.
6. I am a military brat.
[color=red]7. I am a fair person. (most of the time)[/color]
8. I am a boy.
[color=red]9. I am the smartest math student in my 8th grade honors math class.[/color]
[color=red]10. I am a visual learner.[/color][/quote]

Things in red are out of scope. We're not looking for descriptions, but categories.[/quote]

Visual learner is a category.

You are either a visual learner, an auditory learner, or a kinesthetic learner.

there is no need to poat twice :wink:  :wink:

Well, considering I don't know how to 'poat', I can't do it twice, let alone, once.

Hmm, why are the 83rd and 84th posts the exact same, with the exact same time also. I don't remember posting twice in a row within the same minute.

you clicked like fifty bajillion times on the submit button, making you double post....

now...
what do you mean by the category thingies?

no, I clicked only once, and you can't post in a topic twice in a row, unless it is ten minutes after the other post.

[quote=banana_cream_PI][font=monospace]1. I am an AI
2. I run on Vista OS
3. My main hard drive is a 50 terabytes with a petabyte RAM
4. The BCPI Project started in 2005 and released to the public in 2018
5. My main server is in New York[/quote]

dude dont bump


calculus
integration
geometry
3D geometry
number theory open
number theory

Find all integral solutions to x^3 - y^2 = 2

This is a [url=http://mathworld.wolfram.com/MordellCurve.html]Mordell curve[/url] with $ n \equal{} \minus{} 2$, one of the few cases that have elementary solutions.  Write $ x^3 \equal{} y^2 \plus{} 2 \equal{} (y \plus{} \sqrt { \minus{} 2})(y \minus{} \sqrt { \minus{} 2})$.  Any prime that divides the RHS is composite in $ \mathbb{Z}[ \sqrt { \minus{} 2} ]$ and hence is of the form $ a^2 \plus{} 2b^2 \equal{} (a \plus{} b \sqrt { \minus{} 2})(a \minus{} b \sqrt { \minus{} 2})$.  It follows that $ y \plus{} \sqrt { \minus{} 2}$ is a cube in $ \mathbb{Z}[ \sqrt { \minus{} 2} ]$.

Let $ y \plus{} \sqrt { \minus{} 2} \equal{} (u \plus{} v \sqrt { \minus{} 2})^3 \equal{} (u^3 \minus{} 6uv^2) \plus{} (3u^2 v \minus{} 2v^3 \sqrt { \minus{} 2})$; then we require that $ 3u^2 v \minus{} 2v^3 \equal{} 1$, which requires that $ v \equal{} 1, u \equal{} \pm 1$.  This gives $ \boxed{ y \equal{} \pm 5, x \equal{} 3 }$ as the only solutions.


Let $ a,b,c\ge 0$. Prove that:

$ \dfrac{1}{2a^2\plus{}bc}\plus{}\dfrac{1}{2b^2\plus{}ca}\plus{}\dfrac{1}{2c^2\plus{}ab}\ge \dfrac{8}{(a\plus{}b\plus{}c)^2}$

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

$ \dfrac{1}{2a^2 \plus{} bc} \plus{} \dfrac{1}{2b^2 \plus{} ca} \plus{} \dfrac{1}{2c^2 \plus{} ab}\ge \dfrac{8}{(a \plus{} b \plus{} c)^2}$[/quote]
Haha! I solved it ^^

Here is the solution, and all my thoughts when getting it :)

(Obviously) no equality when $ a \equal{} b \equal{} c$.
There is equality if $ a \equal{} b$ and $ c \equal{} 0$ or any permutation. If $ c \equal{} 0$ then it becomes $ (a \plus{} b)^4 \ge 16a^2b^2$, which is obvious.

The ineq is symmetric, so we can assume $ a \ge b \ge c$. I suspect that the only equality is when $ a \equal{} b$ and $ c \equal{} 0$. Then $ 2a^2 \plus{} bc \equal{} 2b^2 \plus{} ca \equal{} 2(2c^2 \plus{} ab)$. Cauchy schwartz on this idea gives us:
$ (\dfrac{1}{2a^2 \plus{} bc} \plus{} \dfrac{1}{2b^2 \plus{} ca} \plus{} \dfrac{1}{2c^2 \plus{} ab})((2a^2 \plus{} bc) \plus{} (2b^2 \plus{} ca) \plus{} (8c^2 \plus{} 4ab))$ $ \ge (1 \plus{} 1 \plus{} 2)^2 \equal{} 16$.
And it suffices to prove:
$ \frac {16}{((2a^2 \plus{} bc) \plus{} (2b^2 \plus{} ca) \plus{} (8c^2 \plus{} 4ab)} \ge \frac {8}{(a \plus{} b \plus{} c)^2}$

Then we would have to prove $ (2a^2 \plus{} bc \plus{} 2b^2 \plus{} ca \plus{} 8c^2 \plus{} 4ab) \le 2(a \plus{} b \plus{} c)^2$ when $ a \ge b \ge c$. It's equivalent to $ 2a^2 \plus{} 2b^2 \plus{} 2c^2 \plus{} 4ab \plus{} 4ac \plus{} 4bc \ge 2a^2 \plus{} 2b^2 \plus{} 8c^2 \plus{} bc \plus{} ca \plus{} 4ab$ $ \iff$
$ 3c(a \plus{} b) \ge 6c^2$. (And this one is obvious since we assumed $ a \ge b \ge c$) QED :)

Equality:
The last one gives us either $ c \equal{} \frac {a \plus{} b}{2}$ or $ c \equal{} 0$. If $ c \equal{} 0$ then by the first $ \frac {1}{2a^2} \equal{} \frac {1}{2b^2}$ so $ a \equal{} b$.
If $ c \equal{} \frac {a \plus{} b}{2}$. Then by the first: $ 4a^2 \plus{} b(a \plus{} b) \equal{} 4b^2 \plus{} a(a \plus{} b)$ or $ (a \minus{} b)(4a \plus{} 4b) \minus{} (a \minus{} b)(a \plus{} b) \equal{} (a \minus{} b)(3a \plus{} 3b)$ and thus $ a \equal{} b$, but there isn't equality when $ a \equal{} b \equal{} c$.

So equality iff $ a \equal{} b$ and $ c \equal{} 0$ or any permutation :)


Let $p$ be a prime of the form $4k+1$.
Prove that the equation $x^{2}-py^{2}=-1$ has a integer solution.

This is solution of my friend.
Call $(a,b)$ is minimum root of Pell's equation $x^{2}-py^{2}=1$. We have $a$ is odd and $b$ is even, write $a=2m+1,b=2n$, then $m^{2}+m=pn^{2}$. Now because $(m,m+1)=1$ we obtain $m=pu^{2},m+1=v^{2}$(1) or $m=u^{2},m+1=pv^{2}$(2). If we have (1) then $v^{2}-pu^{2}=1$ contradiction because choose $(a,b)$. If we have (2) then $u^{2}-pv^{2}=-1$ and have done!


integration
limit
calculus
function
real analysis
real analysis solved

Let (an) be a sequence of real numbers so that: an+ln(an)=1+1/n for all n>=1. Prove that lim (an)=1 and that lim n(an-1)=1/2.

cheers! :D :D :D

Hi Lagrangia this is a funny exercice

First of all  a_n >0,

2 \geq 1+ 1/n  = a_n + lna_n  \geq ln a_n

0 < a_n  \leq e

a_n is bounded

a_(n+1) - a_n + ln a_(n+1) - ln a_n = 1/(n+1) - 1/n < 0

\int_{ a_n till a_(n+1) } (1+ 1/t) dt <0

so a_(n+1)  \leq a_n , a_n is decreasing bounded lim a_n = L exist

L + ln L = 1 => L= 1

 \lim a_n = 1

for the second limit

we use  lim an= 1

a_n + ln((a_n -1) + 1 ) = 1 + 1/n

a_n + a_n - 1~ 1 + 1/n

n(a_n - 1) ~ 1/2

lim n(a_n - 1) = 1/2

Nice solution moubi! :D :D

Can someone find a solution without the use of the integral? I am not allowed to use it! :D :D thank you very much!

cheers! :D

Sorry Lagrangia for the integral, if i find another one i 'll post

Hi Lagrangia,

For the monotony of a_n

the function f(x) = x+lnx is strictely increasing, the equation

x +ln x = 1 + 1/n
has one solution namely a_n >0

f(a_n) = 1 +1/n

1 + 1/(n+1) < 1 + 1/n

f(a_(n+1)) < f(a_n) , f is increasing so a_(n+1) < a_n

the sequence a_n is decreasing

This idea came to me when i was walking on the bridge "Le Pont des Arts"
near the Louvre tonight.

Another idea for this could be based on the fact that we can see that we can write the sequence like so:
an=f^(-1) (1+1/n) where f(x)=x+ln(x) and we derive it and then apply Lagrange theorem and then we can easy find out the limits that we are looking for! :D :D

Note: Really cool solution given by moubi! :D :D thank you!


I remember in a thread a while ago, the class forum will be moved to an Archive Forum (or something like that), but I still can't see it. Where can I find it?

That probably means your class hasn't been moved to the Archive yet.  The Archive is under the Online School section of the Forum.


linear algebra
matrix
algebra
polynomial
algorithm
linear algebra unsolved

I have seen a method to decompose a given matrix $A\in{\mathbb R}^{n\times n}$ as $A=D+N$ where $D$ is diagonable, $N$ is nilpotent so that for any polynomial $f(x)$, we have

${ f(A)=\sum_{k=0}^{n-1}{{f^{(k)}(D)}\over{k!}}N^{k}}$

I cannot recall the name of this method. Where can I find information about this algorithm ?  :?:

You forgot an important fact: $DN = ND$.

See [url=http://en.wikipedia.org/wiki/Jordan_normal_form]Jordan Normal Form[/url].


Let $m\geq0$ be an integer. Let $a_{1}, a_{2}, \cdots, a_{m}$ be integers and let \[f(x)=\sum_{i=1}^{m}\frac{a_{i}x^{i}}{i!}\]
Show that if $d\geq0$ is an integer then $(f(x))^{d}/d!$ can be expressed in the form \[\sum_{i=0}^{md}\frac{b_{i}x^{i}}{i!}\]
where $b_{i}$ are integers.

[hide="Solution"] We're going to do a trippy sort of induction on $d$.

The claim that $g(x)$ can be written in the desired form is equivalent to the claim that all the derivatives of $g$ have integer values at $0$.

Let $g_{d}(x) = \frac{f(x)^{d}}{d!}$.  Given that the derivatives of $g_{d}(x)$ have integer values at $0$ for $d = 0, 1, 2, 3, ... k$, then

$g'_{k+1}(x) = g_{k}(x) f'(x)$
$g''_{k+1}(x) = g_{k}'(x) f'(x)+g_{k}(x) f''(x)$

And so forth.  Clearly every term in $g^{(r)}_{k+1}(x)$ takes the form

$c g^{(r-i)}_{k}(x) f^{(i)}(x)$

Which, by our inductive hypothesis, has an integer value at $0$.  [/hide]


analytic geometry
geometry
geometric transformation
rotation
homothety
ratio
similar triangles

Have any body see this before ?
Can you help me to sove it ?

[i]I have the problem in word document as attachment[/i]

Thanks   :)

I finaly find a solution  :)  :)

It is very beautiful problem for sure !  I will post the solution in a few days if there is no response.

This is easy with coordinates or complex numbers or whatever. If it is true, it is a nice result.

I solve it and it is true.
It is very nice to try to prove this thing by synthetic geometry !
But all the solutions are accepted!

[hide="quite trivial using rotation"]
$ H\stackrel{R_{B,\frac{\pi}{2}}}{\longrightarrow}C$
$ C\stackrel{R_{A,\frac{\pi}{2}}}{\longrightarrow}E$
So just find the composition of $ R_{B,\frac{\pi}{2}}$, $ R_{A,\frac{\pi}{2}}$ which is also a rotation with angle $ \pi$
A single test will give the centre of rotation, which is $ D$  :D
[/hide]

[quote="Leonhard Euler"]Sorry but I can't see problem. Can anyone post?[/quote]
[b][size=100][color=DarkBlue]Let $ E,$ $ H$ be, two points in the plane of a given triangle $ \bigtriangleup ABC,$ outwardly to it, such that $ AE\perp AC,$ $ AE \equal{} AC$ and $ BH\perp BC,$ $ BH \equal{} BC.$ Prove that $ \bigtriangleup DAB$ is a right isosceles triangle, where $ D$ is the midpoint of the segment $ EH$[/color][/size][/b].

Kostas Vittas.

Dear Mathlinkers,
this result from Oene Bottema can be found on
http://perso.orange.fr/jl.ayme   vol.5 La figure de vecten p. 72
Sincerely
Jean-Louis

First we shall prove the following [i][u]lemma:[/u][/i]

[b]One constructs the similar triangles $ACM$ and $CBN$ outward the triangle $ABC$ and $ABD$ inward it ($D$ and $C$ on the same side of $AB$), with: $\widehat{MAC}=\widehat{NCB}=\widehat{DAB}=\alpha$ and $\widehat{MCA}=\widehat{NBC}=\widehat{DBA}$. Then $CMDN$ is a parallelogram.

Proof:[/b] A rotation of angle $\alpha$ around $A$ followed by a homothety of ratio $\frac{AB}{AD}$ will map $D$ to $B$ and $M$ to $C$, hence $MD=\parallel CN$, i.e. $CMDN$ is a parallelogram.

Now, back to our problem: if $M$ and $N$ are the midpoints of $CE$ and $CH$, then the triangles $\triangle CMA$ and $\triangle CNB$ are right-angled and isosceles, hence similar, whilst the triangle $\triangle ADB$ is right-angled and isosceles as well, according to the above lemma.

Best regards,
sunken rock


[quote="felipesa"]If p and q are positive integers such that $\frac{7} {{10}} < \frac{p} {q} < \frac{{11}} {{15}}$,what is  the smallest possible value for q?

a)6
b)7
c)11
d)25
e)50[/quote]

[hide]Is it b) 7
$\frac{7}{10}=.7$
$\frac{11}{15}\approx.7333$
$\frac{5}{7}\approx.714$
$\frac{4}{6}<\frac{7}{10}$
$\frac{5}{6}>\frac{11}{15}$
so 7 is the answer..? [/hide]

[quote="riddler"][hide="help please"]
$\frac{7}{10}<\frac{p}{q}<\frac{11}{15}$
$\frac{7}{10}<\frac{p}{q}$
$q<\frac{10p}{7}$
Also,
$\frac{p}{q}<\frac{11}{15}$
$\frac{15p}{11}<q$
So,
$\frac{15p}{11}<q<\frac{10p}{7}$
In this $p$ must be a factor of $11$ so minimum value is $11$ but it has to be a factor of $7$ aswell,
so,
$105<q<110$
Where did I go wrong? :?
[/hide][/quote]

[hide]$\frac{15p}{11}$ and $\frac{10p}{7}$ don't have to be whole numbers. ;) [/hide]

[hide="not so good a way but maybe correct"]
$\frac{7} {{10}} < \frac{p} {q} < \frac{{11}} {{15}}$
multiply top bottom of both fractions by $7,11,3$
So we get,
$\frac{1617}{2130}<\frac{p}{q}<\frac{1694}{2130}$
So,
$\frac{pk}{qk}=\frac{pk}{11*3*7*10}$
and just try out answers given in the question to get answer=
$\boxed{11}$
[/hide]

[quote="Tecnogram888"][hide="solution?"]taking lcd
${21}/30<p/q<{22}/30$
doubling both
${42}/60-<p/q<{44}/60$
thus, p/q=43/60...which isnt up there...what did i do wrong?  [/hide][/quote]
$\frac{42}{60}<\frac{p}{q}<\frac{44}{60}$ doesn't necessarily mean $\frac{p}{q}=\frac{43}{60}$ ;)

[quote="chess64"]No calculators... :D

[hide="hint"]
Consider the first part -- 7/10 < p/q -- and the second part -- p/q < 11/15 -- separately. Cross multiply each one.
[/hide][/quote]

Explain this more plz? I know what cross multiplying is but what do you cross multiply?

[hide="Solution"]
$\frac{7}{10} < \frac{p}{q}$, so $7q < 10p$.
$\frac{p}{q} < \frac{11}{15}$, so $15p < 11q$.

Multiply the first inequality by 3 to get $21q<30p$ and multiply the second by 2 to get $30p<22q$. Hence, $21q<30p<22q$. We are then looking the smallest $q$ such that an interval of length $q$ has at least one multiple of 30. At this point you can just plug in the choices. :D Try to get the answer from here.
[/hide]


Having trouble solving this:

Prove that the next integer greater than (3+root(5))^n is divisible by 2^n.

Any suggestions?

[hide="A start"]
We want to prove that $ a \minus{} 1 < (3 \plus{} \sqrt5)^n < a$ and $ 2^n|a$

We proceed with induction:

When n=1, we have that a=6, so our base case is true.

Assume it is true for n.

We have that $ a \minus{} 1 < (3 \plus{} \sqrt5)^n < a$ and $ 2^n|a$

$ (3 \plus{} \sqrt5)a \minus{} 6 < (3 \plus{} \sqrt5)a \minus{} (3 \plus{} \sqrt5) < (3 \plus{} \sqrt5)^{n \plus{} 1} < (3 \plus{} \sqrt5)a$
[/hide]

[hide="The lemma you want"] Is that this integer is precisely $ (3 \plus{} \sqrt{5})^n \plus{} (3 \minus{} \sqrt{5})^n$.  [/hide]

[hide="To finish it off"]
Clearly $ (3\plus{}\sqrt 5)^n$ is not an integer for any $ n$ because $ 3\plus{}\sqrt 5$ is not an integer.

Now, we first prove that $ (3\minus{}\sqrt 5)^n<1$ for all positive integers $ n$. This is simple because $ 3\minus{}\sqrt 5<1$.

Therefore $ \lceil (3\plus{}\sqrt 5)^n \rceil \equal{} (3\plus{}\sqrt 5)^n\plus{}(3\minus{}\sqrt 5)^n$.

Proving that $ 2^n|(3\plus{}\sqrt 5)^n\plus{}(3\minus{}\sqrt 5)^n$ is equivalent to proving that $ \left(\frac 32\plus{}\frac{\sqrt 5}2\right)^n\plus{}\left(\frac 32\minus{}\frac{\sqrt 5}2\right)^n$ is an integer for all $ n$. Let $ x\equal{}\frac 32\plus{}\frac{\sqrt 5}2$, therefore $ \frac 1x\equal{}\frac 32\minus{}\frac{\sqrt 5}2$.

We can clearly see that $ x\plus{}\frac 1x$ is an integer because $ x\plus{}\frac 1x\equal{}3$, an integer. Induction proves that $ x^n\plus{}\frac 1{x^n}$ is an integer for all $ n$.

For this, let's have two base cases: $ n\equal{}1$ and $ n\equal{}2$. We've already proved $ n\equal{}1$, and proving $ n\equal{}2$ is trivial: we have $ x^2\plus{}\frac 1{x^2}\equal{}\left(\frac 32\plus{}\frac{\sqrt 5}2\right)^2\plus{}\left(\frac 32\minus{}\frac{\sqrt 5}2\right)^2\equal{}\frac 72\plus{}\frac{3\sqrt 5}2\plus{}\frac 72\minus{}\frac{3\sqrt 5}2\equal{}7$, an integer.

Now suppose that both $ x^{n\minus{}1}\plus{}\frac 1{x^{n\minus{}1}}$ and $ x^n\plus{}\frac 1{x^n}$ are integers. Then $ \left(x\plus{}\frac 1x\right)\left(x^n\plus{}\frac 1{x^n}\right)$ is the product of two integers, which is an integer. On the other hand, this expression is equal to:
\[ \left(x^{n\minus{}1}\plus{}\frac 1{x^{n\minus{}1}}\right)\plus{}\left(x^{n\plus{}1}\plus{}\frac 1{x^{n\plus{}1}}\right),\] and since $ x^{n\minus{}1}\plus{}\frac 1{x^{n\minus{}1}}$ is an integer by the induction hypothesis, it follows that $ x^{n\plus{}1}\plus{}\frac 1{x^{n\plus{}1}}$ must be an integer as well.
[/hide]

[quote="Yongyi781"]Proving that $ 2^n|(3 \plus{} \sqrt 5)^n \plus{} (3 \minus{} \sqrt 5)^n$ is equivalent to proving that $ \left(\frac 32 \plus{} \frac {\sqrt 5}2\right)^n \plus{} \left(\frac 32 \minus{} \frac {\sqrt 5}2\right)^n$ is an integer for all $ n$. [/quote]
Yep, but there's a faster way to do this; write down a second-order recurrence.

Ah I should've realized: $ 3\pm\sqrt 5$ are the roots of $ x^2 \minus{} 6x \plus{} 4 \equal{} 0$, and so $ (3 \plus{} \sqrt 5)^n \plus{} (3 \minus{} \sqrt 5)^n$ is the solution to the recurrence relation $ a_n \equal{} 6a_{n \minus{} 1} \minus{} 4a_{n \minus{} 2} \equal{} 2(3a_{n \minus{} 1} \minus{} 2a_{n \minus{} 2})$ with $ a_1\equal{}6$ and $ a_2\equal{}28$, from which the result immediately follows by induction. :D

Also $ \frac{3}{2}\pm\frac{\sqrt{5}}{2}$ are the roots of $ x^2\minus{}3x\plus{}1\equal{}0\Leftrightarrow b_{n\plus{}2}\equal{}3b_{n\plus{}1}\minus{}b_n$ where $ b_1\equal{}3$ and $ b_2\equal{}7$.  :wink:


modular arithmetic
number theory unsolved
number theory

If $ k$ divides $ p\minus{}1$, show that the congruence $ x^k\equiv 1\pmod{p}$ has exactly $ k$ distinct solutions modulo $ p$.

[quote="countdownmath"]If $ k$ divides $ p \minus{} 1$, show that the congruence $ x^k\equiv 1\pmod{p}$ has exactly $ k$ distinct solutions modulo $ p$.[/quote]

Is this a Corollary of Lagrange’s Theorem?

Not exactly, but it's an easy corollary of the fact that the multiplicative group $ \mod p$ is cyclic.

Why not? If by Lagrange's Theorem you mean http://www.mathlinks.ro/viewtopic.php?t=192753 then yes.

And how do you show it without some more work¿ This implies the existence of primitive roots.

$ x^{p\minus{}1} \minus{} 1$ has exactly $ p\minus{}1$ roots from the Little Fermat and since $ k|p\minus{}1$ we have $ x^{p\minus{}1} \minus{} 1 \equal{} (x^k\minus{}1)p(x)$. $ p(x)$ has a degree $ p\minus{}1\minus{}k$ so it has  at most $ p\minus{}1\minus{}k$ roots and if $ x^k\minus{}1$ would have had less then $ k$ roots then $ x^{p\minus{}1} \minus{} 1$ would have less then $ p\minus{}1$ roots, which is a contradiction. But to show the existence of primitive roots you still have to calculate the number of residues with order $ k$, so I think there's still some work to do.


[quote="giannis"]Μπορείτε να μου ποείτε τους τύπους των ριζών εξισώσεων τρίτου και τετάρτου  βαθμού. :lol:[/quote]

Kalispera Gianni,

Rikse mia matia edw http://www.mathlinks.ro/Forum/viewtopic.php?t=130273&sid=a907fb4e8cae5e236b41810a342d7547

tha deis isws kai alla endiaferonta pragmata ektos apo auta pou zitas.

Alexandros

Μπορούμε να λύσουμε την εξίσωση

$x^{3}-3x+1=0$
στους πραγματικούς αριθμούς , χωρίς να χρησιμοποιήσουμε αυτόν τον τύπο ? :idea:

δεν ξερω αν μπορουμε να βρουμε ολε ςτις λυσεις , αν θυμασαι ειναι στο βιβλιο αλγεβρας Β λυκειου σαν παραδειγμα ευρεσης ριζων κατα προσεγγιση, ετσι
αφου $f(1,5)<0, f(1,6)>0$  οπου $f(x)=x^{3}-3x+1$ .Αρα εχουμε μια λυση μεταξυ 1,5 και 1,6, .

Θα δοκιμάσουμε πιθανές ακέραιες ρίζες $-1,-2,-4$ , και θα δούμε με το σχήμα Honer αν μια από αυτές είναι ρίζα της εξίσωσης( το $-2$είναι από ότι είδα και μάλιστα διπλή) και επαναλαμβάνουμε την ίδια διαδικασία. Δεν ξέρω τώρα αν υπάρχει άλλος τρόπος , καμιά άλλη ιδέα :idea:

[quote="mailo"]Πως μπορουμε να λυσουμε τη $x^{4}+5x^{3}+9x^{2}+8x+4=0$ ?[/quote]

Εχουμε $x^{4}+5x^{3}+9x^{2}+8x+4=x^{2}(x^{2}+5x+6)+3x(x+2)+2(x+2)$
Αρα $x^{2}(x+2)(x+3)+3x(x+2)+2(x+2)=(x+2)(x^{3}+3x^{2}+3x+2)$
Δηλαδή $x=-2$ ή $x^{3}+3x^{2}+3x+2=0$
To τελευταίο παραγοντοποιείται επίσης ως εξής:
$x^{3}+3x^{2}+3x+2=x^{2}(x+2)+(x+2)(x+1)=(x+2)(x^{2}+x+1)=0$
Ομως το $x^{2}+x+1=0$ δεν έχει λύσεις κι άρα η μόνη λύση της αρχικής εξίσωσης είναι το $x=-2$ και μάλιστα είναι διπλή :wink:

Παρακαλώ να μου πείτε αν έχω κα΄νει κανένα λάθος στην παραγοντοποίηση  :oops:

[quote="galeos"]$x=2\cos \theta$ :roll:[/quote]
πως χρησιμοποιουμε αυτο ? πως προκυπτει ?

$\cos 3\theta=4\cos^{3}\theta-3\cos \theta$, twra prepei na einai logiko pws prokuptei :wink:

Ναι και παλι τι σχεση εχει το συνημίτονο και τι γωνια ειναι αυτη η Θ?  :maybe:

[quote="kostas"]θετεις χ=2cosθ ,δηλαδη -2<χ<2 αυτο εννοω ...απο που προκυπτει ?[/quote]
aaa, mporei na prokupsei me "pollous" tropous.

1on oi luseis ths telikhs trigwnometrikhs tha einai akribws 3 pou se sunduasmo me to bathmo ths eksiswshs shmainei oti den xreiazetai na problhmatistoume katholou, afou exoume brei oles tis dunates luseis. :wink:

2on an $x\geq2$, tote $x^{3}\geq 4x$ k epomenos $0=x^{3}-3x+1\geq x+1$ atopo, omoia k gia $x\leq-2$. :P

[quote]Ναι και παλι τι σχεση εχει το συνημίτονο και τι γωνια ειναι αυτη η Θ?[/quote]
apla epeidh h morfh ths eksiswshs exei mia eidikh morfh mporoume na th xeiristoume kateutheian me trigwnometrikous tupous. $\theta \in [0,\pi]$

Να ρωτίσω τώρα που αναφέρατε για τριγωνομετρία , υπάρχει τύπος που να δίνει κατευθείαν π.χ. το $cos5a$ συναρτήσει του $cosa$ και γενικά του $cosnx$?

Gianni fisika iparxei tetoios tipos ean skefteis ton tipo tou De Moivre gia tous migadikous. Meta mporeis na kaneis tin genikeusi monos sou.

$(cosx+isinx)^{n}=cos(nx)+isin(nx)$

Apo tin alli aneptikse to prwto melos me to diwnimo tou newton.

Ksekinise ws paradeigma me to oti

Prwton: $(cosx+isinx)^{3}=cos(3x)+isin(3x)$

Deuteron: $(cosx+isinx)^{3}=cos^{3}x+3icos^{2}sinx-3cosxsin^{2}x-isin^{3}x$

eksiswse ta dio meli (pragmatiko me pragmatiko kai fantastiko me fantastiko) kai tha pareis tous gnwstous tipous gia to $sin(3x)$ kai $cos(3x)$.

Sinexise me tin genikeusi.

Alexandros


Prove that for any partition of $X = \{1,2, . . . , 9\}$ into two subsets, one of
the subsets contains distinct elements $a, b, c$ such that $2c = a + b$.

It seems like this problem is doable without too much work by sheer brute force (along the lines of "Case 1: 1 and 2 are in the same set, 4 in a different set from the two of them).  Is there a more elegant way to approach it?


Give $ a,b,c \ge 0$.Prove that
$ \frac{a}{b\plus{}c}\plus{} \frac{b}{a\plus{}c}\plus{} \frac{c}{a\plus{}b}\plus{} \frac{1}{8}(\frac{(a\plus{}b)^2}{a^2\plus{}b^2}\plus{} \frac{(b\plus{}c)^2}{b^2\plus{}c^2}\plus{} \frac{(c\plus{}a)^2}{c^2\plus{}a^2}) \ge \frac{9}{4}$

[quote="secrets"]Give $ a,b,c \ge 0$.Prove that
$ \frac {a}{b \plus{} c} \plus{} \frac {b}{a \plus{} c} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {1}{8}(\frac {(a \plus{} b)^2}{a^2 \plus{} b^2} \plus{} \frac {(b \plus{} c)^2}{b^2 \plus{} c^2} \plus{} \frac {(c \plus{} a)^2}{c^2 \plus{} a^2}) \ge \frac {9}{4}$[/quote]
Nice;serect.Use Iran 96 and Am-Gm.
We have:
$ \frac{(a\plus{}b)^2}{8(a^2\plus{}b^2)} \ge \frac{ab}{(a\plus{}b)^2}$ -by Am-Gm
So:
$ LHS \ge \sum \frac{a}{b\plus{}c}\plus{}\sum \frac{ab}{(a\plus{}b)^2}\equal{}(ab\plus{}bc\plus{}ca)(\frac{1}{(a\plus{}b)^2}\plus{}\frac{1}{(b\plus{}c)^2}\plus{}\frac{1}{(c\plus{}a)^2}) \ge \frac{9}{4}$

[quote="caubetoanhoc94"][quote="tuandokim"][quote="secrets"]Give $ a,b,c \ge 0$.Prove that
$ \frac {a}{b \plus{} c} \plus{} \frac {b}{a \plus{} c} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {1}{8}(\frac {(a \plus{} b)^2}{a^2 \plus{} b^2} \plus{} \frac {(b \plus{} c)^2}{b^2 \plus{} c^2} \plus{} \frac {(c \plus{} a)^2}{c^2 \plus{} a^2}) \ge \frac {9}{4}$[/quote]
case 1:one of a,b,c=0
case 2:a,b,c>0
we can prove this by S.O.S
(although it's very ugly in the last step )
 :)[/quote]

Can you slove it ??  :([/quote]
ok:D
case 2
we have:
$ LHS \minus{} RHS \equal{} \sum (a \minus{} b)^2S_{c}$
where:$ S_{c} \equal{} \frac {4a^2 \plus{} 4b^2 \minus{} c^2 \minus{} ab \minus{} bc \minus{} ca}{4*(a \plus{} c)(b \plus{} c)(b^2 \plus{} a^2)}$ ........
assume that:$ a\geq b\geq c$
we have:$ S_{b},S_{c}\geq 0$
and we have (let a=c+x+y,b=c+x, and we have $ x,y\geq 0$)
$ S_{b}a^2 \plus{} S_{a}b^2 \equal{} 84*c*x^5*y \plus{} 216*c^3*y^3*x \plus{} 296*c^4*y^2*x \plus{} 132*c*y^2*x^4 \plus{} 230*c^2*y*x^4 \plus{} 404*c^3*x^2*y^2 \plus{} 306*y^2*c^2*x^3 \plus{} 18*x^3*y^4 \plus{} 30*y^3*x^4 \plus{} 14*x^6*y \plus{} 26*y^2*x^5 \plus{} 8*y^5*c^2 \plus{} 4*y^5*x^2 \plus{} 8*y^5*c*x \plus{} 32*c^7 \plus{} 128*c^6*x \plus{} 232*c^5*x^2 \plus{} 64*c^6*y \plus{} 256*c^4*x^3 \plus{} 96*c^5*y^2 \plus{} 188*c^3*x^4 \plus{} 84*c^4*y^3 \plus{} 77*y^4*c^2*x \plus{} 56*c*y^4*x^2 \plus{} 4*x^7 \plus{} 42*y^4*c^3 \plus{} 28*c*x^6 \plus{} 92*c^2*x^5 \plus{} 229*y^3*c^2*x^2 \plus{} 124*c*y^3*x^3 \plus{} 376*c^3*x^3*y \plus{} 232*c^5*y*x \plus{} 384*c^4*x^2*y \geq 0$(which is obvious true)
so $ LHS \minus{} RHS\geq 0$ (q.e.d)

[quote="caubetoanhoc94"][quote="secrets"]Give $ a,b,c \ge 0$.Prove that
$ \frac {a}{b \plus{} c} \plus{} \frac {b}{a \plus{} c} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {1}{8}(\frac {(a \plus{} b)^2}{a^2 \plus{} b^2} \plus{} \frac {(b \plus{} c)^2}{b^2 \plus{} c^2} \plus{} \frac {(c \plus{} a)^2}{c^2 \plus{} a^2}) \ge \frac {9}{4}$[/quote]
Nice;serect.Use Iran 96 and Am-Gm.
We have:
$ \frac {(a \plus{} b)^2}{8(a^2 \plus{} b^2)} \ge \frac {ab}{(a \plus{} b)^2}$ -by Am-Gm
So:
$ LHS \ge \sum \frac {a}{b \plus{} c} \plus{} \sum \frac {ab}{(a \plus{} b)^2} \equal{} (ab \plus{} bc \plus{} ca)(\frac {1}{(a \plus{} b)^2} \plus{} \frac {1}{(b \plus{} c)^2} \plus{} \frac {1}{(c \plus{} a)^2}) \ge \frac {9}{4}$[/quote]
Same as me,nice solution :D


[quote="iphoumeng"]Proof: A^k/(B+C) +B^k/(A+C)+ C^k/(B+A)>=(3/2)*((A+B+C/3)^(k-1))   (k is positive integer)[/quote]
Why you post this inequality in "Inequalities Unsolved Problems"and "Inequalities Proposed & Own Problems " ? :wink:

Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=264595[/url]


function trigonometry calculus conics parabola

One of our clients is a shot-putter who is intent on improving his performance, and
is therefore interested in an analysis of aspects of his shot-putting style. Critical to the
distance to which a shot may be "put" is, of course, the force that is exerted on the shot as
the shot putter's arm extends. This force varies depending on the angle at which the shot
is launched: the structure of the human body favors pushing horizontally over pushing
vertically. [Most athletes can bench press more weight than they can shoulder press.]
Using values measured in out laboratory it seems that the force F our client can produce
varies linearly as a function of the angle of launch $ \alpha$ , from 800 Newtons at  $ \alpha \equal{} 0$ to 550
Newtons at  $ \alpha \equal{} \frac {\pi}{2}$. [All our lab results are in SI units, so we'd appreciate it if you also
use these in your report.]
Given this force, the release velocity $ V$of the shot (measured in meters per second)
may be found from Newton's law to be approximately

$ V \equal{} \sqrt \frac {2Fd}{m}$

where d is the length of the path along which the shot put is accelerated, and m is the
mass of the shot (7,26 kg). See diagram below.

Our client (whose eective arm length is 1 meter) would like to know at what angle
the shot should be released in order to maximize the distance that the shot is put.

(I recommend moving this to the advanced physics forum)

So, we're given that $ F(\alpha) \equal{} 800 \minus{} \frac{500}{\pi} \alpha$. Using standard kinematics, the distance that the shot travels is $ \frac{ V^2 \cdot \sin (2 \alpha)}{9.8}$, assuming that the shot-putters height is negligible. You can plug the given expression for $ V$ and into this and then plug $ F(\alpha) \equal{} 800 \minus{} \frac{500}{\pi} \alpha$ into that and get that the distance is $ \sin (2 \alpha) \left(800 \minus{} \frac{500}{\pi}\alpha \right) \cdot \mbox{constants}$. Then you just need to maximize that, which can be done with calculus or a graphing calculator.

That would be a perfect answer on the Moon. On Earth one, probably, has to take the air drag into account, which normally changes the answer to a lower one. Frankly speaking, I have no idea of how much air resistance the golf ball meets... :(

i have to do this same problem but im having trouble understanding Fibonnaci's terminology, I know that alpha = rotational acceleration but what does that have to do with this problem also is his equation correct isn't the 500 suppose to be 550 instead?  heres the steps I took so far.

We know that Fy and Fx and at first i tried solving for the hypotenuse which i later found was wrong because that is the force at a certain angle.  I was having trouble finding an equation for the Force at different angles so I took a different approach.

I know the a(t), v(t), and x(t) equations for y so what i did was plugged in the generic answer for velocity in each
a(y) = -9.8
v(y) = -9.8t + (sin(theta))(v)
x(y) = -4.9t^2 +(sin(theta))(v)t + C (C being the shot-putters height which in this problem doesn't matter)

for v(y) i then solved for t which i got (sin(theta)(v))/9.8

I then plugged this into the x(t) function to make sure i got this right the max angle should be 90 degrees.  After solving for the max i was able to find that indeed the max height was at 90degrees for this problem.  So this proves that our function for t is correct.

I then went ahead and plugged in for the function x= x0+vt.  x0 = 0 so the function is just x=vt
v= cos(theta)(sqr(1100/7.26))(sin(theta))(sqr(1600/7.26)/9.8) which makes a downward parabola and you find that the max angle is equal to 45 degrees which makes sense but on the other hand doesn't because with the change in force with different angles im not sure if this is correct can someone check over my work.  [b]Make sure when you plug in X to put a degree sign after it in the calculator.[/b]


calculus
limit
trigonometry
LaTeX
real analysis
real analysis unsolved

From Stewart's Calculus, 12.1.37:

Determine whether the sequence converges or diverges. If it converges, find the limit.

Asub(n) = n(sin(1/n))

Now, it converges. No problem there. The limit is the bit that's confusing me.

The lim(Asub(n)) can be broken down into:

(lim (n))*(lim(sin(1/n)))

then

(lim (n))*(sin(lim(1/n)))

Since lim n = infinity, and sin(lim(1/n)) = sin (0) = 0, the answer seems to be (infinity)*0, or 0.

The book, however, gives 1 as the answer. What gives?

First post, thanks for your time. I apologize for the plaintext.

I'm going to use functional notation instead of sequence notation for my limits, but if you like sequence notation, you can just pretend that "$ x \to a$" is replaced by "$ n \to \infty$" and that "$ f(x)$" and "$ g(x)$" are replaced by "$ a_n$" and "$ b_n$."

It is true that $ \lim_{x \to a} f(x) \cdot g(x) \equal{} \left(\lim_{x \to a} f(x)\right)\cdot \left(\lim_{x \to a} g(x)\right)$, [i]provided that all three limits exist[/i]!  Here, one of the limits is infinity, i.e., it does not exist, so you can't use this rule.

With more cases: it is not generally true that "infinity times zero" is zero -- it's an [b]indeterminate form[/b] that requires more work.  What is true is that if $ \lim_{x \to a} f(x)$ is a [i]positive[/i] real number and $ \lim_{x \to a} g(x) \equal{} \infty$ then $ \lim_{x \to a} f(x)\cdot g(x) \equal{} \infty$, and that if $ \lim_{x \to a} f(x)$ is a [i]negative[/i] real number and $ \lim_{x \to a} g(x) \equal{} \infty$ then $ \lim_{x \to a} f(x)\cdot g(x) \equal{} \minus{} \infty$, and that if $ \lim_{x \to a} f(x) \equal{} 0$ and $ \lim_{x \to a} g(x)$ is [i]any real number[/i] then $ \lim_{x \to a} f(x)\cdot g(x) \equal{} 0$, but if $ \lim_{x \to a} f(x) \equal{} 0$ and $ \lim_{x \to a} g(x) \equal{} \infty$ then $ \lim_{x \to a} f(x) \cdot g(x)$ could be any real number, could be positive or negative infinity, or could diverge in some other more complicated way.  (This is why it's called "[b]indeterminate[/b]" -- just knowing the form "$ 0 \cdot \infty$" is not enough to determine the value.)

Three examples:

$ \lim_{x \to \infty} \frac {1}{x \plus{} 1} \equal{} 0$ and $ \lim_{x \to \infty} 3x \equal{} \infty$, and the "zero times infinity" form $ \lim_{x \to \infty} \frac {1}{x \plus{} 1} \cdot 3x$ is equal to what?

$ \lim_{x \to \infty} \minus{} \frac {1}{x \plus{} 1} \equal{} 0$ and $ \lim_{x \to \infty} x^2 \equal{} \infty$, and the "zero times infinity" form $ \lim_{x \to \infty} \minus{} \frac {1}{x \plus{} 1} \cdot x^2$ is equal to what?

$ \lim_{x \to \infty} \frac {1}{x \plus{} 1} \equal{} 0$ and $ \lim_{x \to \infty} \sqrt {x} \equal{} \infty$, and the "zero times infinity" form $ \lim_{x \to \infty} \frac {1}{x \plus{} 1} \cdot \sqrt {x}$ is equal to what?

The example from Stewart is just a rewritten form of a very famous limit: do you know the value $ \lim_{x \to 0} \frac {\sin x}{x}$?  (Note that we can sort-of write this in a "$ 0 \cdot \infty$" form: if we consider first $ x \to 0^ \plus{}$ then we can write it as $ \lim_{x \to 0} \frac {1}{x} \cdot \sin x$, where the first factor goes to infinity and the second factor goes to zero.)

[Edited to fix a small LaTeX typo.]

Hmm. I see now how faulty my "infinity * 0" assumption was. The 3 examples you provided go to 3, negative infinity [the negative sign was deliberate, yes?], and 0.

I'm familiar with the sinx/x limit, and that comparison definitely clears this up.

Thanks very much.

Correct on all counts, and you're welcome   :)