geometry
rectangle
ratio
trigonometry
induction
CEMC
arithmetic sequence

It's already Friday evening, and it's been over two days since the contest, so I will post the problems now.

[b]Part A[/b]
[b]1[/b] If $ a=15$ and $ b=-9$, what is the value of $ a^2+2ab+b^2$?
[b]2[/b] A circular wind generator turns at a rate of $ 30$ complete revolutions per  minute. Through how many degrees will it turn in one second?
[b]3[/b] $ ABCD$ is a rectangle with $ A$ on the line $ y=x+10$, $ B$ is on the line $ y=-2x+10$ and points $ C$ and $ D$ are on the $ x$-axis. If $ AD=4$, what is the are of rectangle $ ABCD$?
[b]4[/b] In June, the ratio of boys to girls in a school was $ 3: 2$. In September, there were $ 80$ fewer boys and $ 20$ fewer girls, and the ratio of boys to girls was $ 7: 5$. What was the total number of students at the school in June?
[b]5[/b] The numbers $ 1,2,3,...,9$ are placed in a $ 3$ by $ 3$ square so each number can be used only once. The sum of the three rows, the three columns, and the two long diagonals are added together to get a "grand sum" $ S$. What is the maximum possible value of the grand sum $ S$?
[b]6[/b] $ O$ is the centre of the circle, $ AN$ is tangent to the circle at $ A$, $ P$ lies on the circle, and $ PN$ is perpendicular to $ AN$ (so that $ PN<AO$). If $ AN=15,PN=9$, find the radius of the circle.
[b]7[/b] Determine all ordered triplets of real numbers $ (x,y,z)$ which satisfy teh system of equations
$ xy=z^2$
$ x+y+z=7$
$ x^2+y^2+z^2=133$
[b]8[/b] A $ 1$ by $ 9$ rectangle made up of unit squares has $ A$ as its top left corner and $ B$ as its bottom right corner $ B$. There are various routes from $ A$ to $ B$ traveling along the segments so that no segement is visited more than once. Out of all these possible routes, determine:
[b]i[/b] the length of the routes which occurs most often;
[b]ii[/b] the number of different routes of this length.

[b]Part B[/b]
[b]1[/b] (a) If $ x-1,2x+2,7x+1$ are the first three terms of an arithmetic sequence, determine the value of $ x$.
             (b) For the value of $ x$ in (a), what is the middle term of the sequence
$ x-1,2x+2,7x+1,...,72$?
             (c) If $ y-1,2y+2,7y+1$ are the first three terms of a geometric sequence, determine all possible values of $ y$.
             (d) For each of the values of $ y$ in (c) determine the $ 6$th term of the geometric sequence $ y-1,2y+2,7y+1,...$.
[b]2[/b] Given a convex quadrilateral $ ABCD$. $ \angle{ABC}=\angle{BCD}=906o$, $ AB=9,BC=24,CD=18$. The diagonals $ AC$ and $ BD$ meet at $ E$>
(a) Find the area of the quadrilateral $ ABCD$.
(b) Show that the ratio $ DE: EB=2: 1$.
(c) Determine the area of triangle $ DEC$.
(d) Determine the area of triangle $ DAE$.
[b]3[/b] Alphonse and Beryl are back! :) They are playing a two person game with the following rules:
- Initially there is a pile of $ N$ stones, $ N\ge 2$/
- The players alternate turns, with Alphonse going first. On his first turn, Alphonse must remove at least $ 1$ and at most $ N-1$ stones form the pile.
- If a player removes $ k$ stones from the pile on their turn, the other player must remove at least $ 1$ and at most $ 2k-1$ stones on their next turn.
- The player who removes the last stone wins the game.
(a) Determine who will win if $ N=7$.
(B) Determine who will win when $ N=8$.
(c) Determine all values of $ N$ for which Beryl has the winning strategy and explain this strategy.
 A cat is located at $ C$, $ 60$ meters directly west of a mouse located at $ M$. The mouse is trying to escape by running at $ 7$ m/s in the direction $ 30^o$ east of north. The cat runs at $ 13$ m/s in a straight path that will intercept the mouse as quickly as possible.
(a) If $ t$ is the time in seconds that it takes for the cat to catch the mouse, determine the value of $ t$.
(b) Suppose that the mouse instead chooses a different direction to escape form the cat. Show that no matter in which direction it runs, all points of interception lie on a circle.
(c) Suppose that the mouse ins intercepted after running a distance of $ d_1$ meters in a particular direction. If the mouse would have been intercepted after running a distance $ d_2$ meters in the opposite direction, prove that $ d_1+d_2\ge 14\sqrt{10}$.

For A6, I messed up and got 18 :wallbash_red:
I only put (9,4,-6) because I didn't bother to permute the x and y, for A7. My others match.
My Bs match yours, except 3c and I didn't do 4.
So around 55 or so. Not bad in 7th grade.
EDIT-I thought in 3, it asked what is the best possiblee strategy for winning and why? or something like that.

Um, For question 4c, I believe that it was to prove that $ d_1 \plus{} d_2 <\equal{} 14\sqrt{30}$ :P

Also, how did you solve 3c???

No. It was $ d_1\plus{}d_2\ge 14\sqrt{30}$ :wink:
I will post the solution for 3c today or tomorrow.

4b i believe there is a 91/2, i can't remember what it is

4c after u done the circle, try to find the shortest chord

I got everything that rem got.

Now, I just hope my proof for 3c, merits some credit

[quote]I conjecture that Problem 7 will be too easy, 3 of the problems will have bad puns, all written questions will have multiple parts, 6 people will solve the last question, and Alphonse and Beryl will play a game.[/quote]

It looks like my first two predictions were wrong, and probably the fourth also.

[quote][b]3[/b] $ ABCD$ is a rectangle with $ A$ on the line $ y \equal{} x \plus{} 10$, $ B$ is on the line $ y \equal{} \minus{} 2x \plus{} 10$ and points $ C$ and $ D$ are on the $ x$-axis. If $ AD \equal{} 4$, what is the are of rectangle $ ABCD$?[/quote]Is that the exact wording of the question? $ A$ could lie above or below the axis, giving two different areas.

I agree that 4 is far too easy. It may have an elegant solution, but the trigonometry is not too complicated. I know that 4c has an elegant solution (unfortunately the algebraic solution is not much more work):
[hide="A solution"]It is not difficult to see that if the cat runs toward the mouse, then it gets caught after $ 21$m, while if it runs away from the mouse it gets caught after $ 70$m. We can apply AM-GM, and then we can apply Power of a Point, since the points where the mouse gets caught are on a circle. We get
\[ d_1 \plus{} d_2 \ge 2\sqrt {d_1d_2} \equal{} 2\sqrt {21\cdot 70} \equal{} 14\sqrt {30}.
\]
[/hide]

I would be interested to see if anyone has a geometric solution for #4b.

Fortunately, (for the case where the question number is 3), part c is quite challenging. It is not "difficult" in the end, but it takes a long time to see how to solve it, and even then writing a rigorous proof is a whole other task in itself. Hopefully the markers will allot a large portion of the marks for #3 to part c, and mark it strictly. This is probably the best way to avoid having a whole bunch of people within a handful of 80 marks, since I think only a few people could have written up #3c rigorously.

[hide="3c"]Lemma 1: Suppose that $ M$ satisfies the conditions $ a_1, a_2, \ldots a_n,$ where condition $ a_i$ states that
\[ M \not\equiv 2^{i \minus{} 1} \textnormal{ (mod }2^i).
\]
Then $ M \equiv 0 \textnormal{ (mod }2^n).$

Proof (by induction). The result is clear for $ n \equal{} 1.$ Now suppose that the result holds for $ n \equal{} k.$ Then $ M \equiv 0 \textnormal{ (mod }2^k),$ so $ M \equiv 0 \textnormal{ (mod }2^{k \plus{} 1})$ or $ M \equiv 2^n \textnormal{ (mod }2^{k \plus{} 1}).$ The second of these possibilities is ruled out by condition $ a_{k \plus{} 1},$ so we have $ M \equiv 0 \textnormal{ (mod }2^{k \plus{} 1}).$ This completes the induction.

Lemma 2: Suppose that player X receives $ T$ stones on their turn, $ n \in \textbf{N},$ $ T \equiv 2^{n \minus{} 1} \textnormal{ (mod }2^n),$ and $ T \neq 2^{n \minus{} 1}.$ Also suppose that X starts the game on this move, or that the other player took more than $ 2^{n \minus{} 2}$ stones on the turn before. Then X wins.

Proof (by induction): We see that if a player ever takes one stone on their turn, then each player must always take one stone in the future. Thus if $ T$ is odd, then X will win if he takes one stone. Hence the result holds for $ n \equal{} 1.$

We now suppose that the result holds for all $ n\le k$ and show that it then holds for $ n \equal{} k \plus{} 1.$ We proceed by contradiction and assume that there exists some $ T$ for which all conditions of the Lemma hold, but for which X loses. Then since all possible values of $ T$ are bounded below, there exists a minimal value for which X loses. We consider this value of $ T.$ By assumption, X starts the game or the other player took more than $ 2^{k \minus{} 1}$ stones on the turn before. Thus X can, and does, take $ 2^{k}$ stones on his turn.

If the other player takes $ 2^k$ stones on the next turn, then there are two possibilities. The first possibility is that there are $ 2^k$ stones remaining, in which case X takes them all and wins, a contradiction. The second possibility is that all the conditions of the Lemma are still satisfied, but X receives a lesser number of stones. This is a contradiction because we assumed $ T$ to be minimal (It is for this case only that a contradiction proof is useful. If not, we would have to use something like "infinite descent" which is less rigorous.).

If the other player does not take $ 2^k$ stones on his turn, then he must take some other number of stones which is at least $ 1$ and at most $ 2^{k \plus{} 1} \minus{} 1.$ We can write this number of stones as $ 2^{r \minus{} 1} l$ for positive integers $ r,l,$ where $ 2^{r \minus{} 1}$ is the greatest power of 2 dividing this number of stones, so that $ l$ is odd. We have $ r \minus{} 1 < k,$ and by assumption $ T \equiv 2^k \textnormal{ (mod }2^{k \plus{} 1}).$ Hence $ T \equiv 0 \textnormal{ (mod }2^r).$ Hence X now receives a pile with $ T'$ stones in it, where $ T' \equal{} T \minus{} 2^{r \minus{} 1} l \equiv 2^{r \minus{} 1} \textnormal{ (mod }2^r).$ If $ T' \equal{} 2^{r \minus{} 1},$ then X takes all the stones and wins, a contradiction. Otherwise, since the other player took more than $ 2^{r \minus{} 2}$ stones on the turn before, we can apply the Lemma for $ T'$ with $ n \equal{} r.$ It follows that X wins, again a contradiction.

In all cases we have a contradiction. This completes the induction and the proof of this Lemma.

By Lemmas 1 and 2, we now see that Beryl can win only if $ N$ is a power of $ 2$. On the other hand, suppose that $ N$ is a power of $ 2$. Then if Alphonse gives Beryl a number of stones which is a power of $ 2$, then he has taken at least half the stones, and Beryl can take the remainder and win. If Alphonse does not give Beryl a power of $ 2$, then she wins by Lemmas 1 and 2.[/hide]

How much do you think that the second part of A8 and 3c will be worth???

Hmm, will I get any points for just proving that Beryl can't win when there are an odd number of stones?

Can someone post a solution for A8?

I ran out of time to do 4b) and 4c) but these are my solutions for them:

My strategy for Beryl was: if Alphonse removes a*2^b (where a is odd) at some turn, then Beryl responds by taking 2^b on her next turn.  This is essentially the same as everyone else's approach I think.   Somehow, I failed to mention the word induction in my solution though :(

[quote="rem"]My solution for 3c is a bit different from Steven's although it's basically the same idea.

[b]Lemma 1[/b]: If $ N$ is odd and it is the turn of player $ X$ then $ X$ wins.
[b]Proof[/b]: Player $ X$ takes $ 1$ stone on each of his turns, then his opponent is forced to take $ 1$ stone (as $ 2*1 \minus{} 1 \equal{} 1$) and so after a pair of consecutive turns, the number of stones in the pile will always decrease by $ 2$, and so before player $ X$'s turn, the number of stones will always be odd, and eventually it will be $ 1$ and player $ X$ will take the last stone and win.
[b]Lemma 2[/b]: If $ N$ is not a power of $ 2$ and it is player $ X$'s turn, then player $ X$ wins.
[b]Proof[/b] Define $ 2^k$ to be the smallest power of $ 2$ such that $ 2^k$ does not divide $ N$. We will prove lemma 2 by induction on $ k$, where we define $ f(N) \equal{} 2^k$ to be the smallest power of $ 2$ such that $ 2^k$ does not divide $ N$. Then we need to prove if $ f(N) \equal{} 2^k$ and $ N$ is not a power of $ 2$ then player $ X$ wins.
[b]Base Step[/b]: $ k \equal{} 1$. This is true by lemma 1.
[b]Induction Step[/b]: Assume the result holds for $ 1,2,...,k$. We need to prove it for $ k \plus{} 1$. Then for $ k \plus{} 1$ have $ 2^{k \plus{} 1}$ to be the smallest power of $ 2$ such that $ 2^{k \plus{} 1}$ does not divide $ N$ and $ N$ is not a power of $ 2$. Then we can write $ N$ as $ m*2^{k \plus{} 1} \plus{} 2^k$ (where $ m$ is a positive integer), because $ 2^k|N$ but $ 2^{k \plus{} 1}$ does not and $ N$ is not a power of $ 2$.
Then, on his first turn, $ X$ takes $ 2^k$ stones. Then, his opponent can take any number from $ 1$ to $ 2^{k \plus{} 1} \minus{} 1$ stones. If he takes a number of stones different from $ 2^k$ then if the number of stones in the pile after his turn is $ M$ then $ 2^k$ does not divide $ M$ hence $ f(M)\le 2^k$ and by the induction assumption it follows that player $ X$ has a winning strategy form that point. So to survive, player $ X$'s opponent must take $ 2^k$ stones. ([b]That is the power of mathematical induction[/b] :D ) Then after the consecutive turns of player $ X$ and player $ X$'s opponent, the number of stones in pile becomes $ N' \equal{} (x \minus{} 1)2^{k \plus{} 1} \plus{} 2^k$. Then player $ X$ will just keep taking $ 2^k$ stones form the pile, his opponent is forced to take $ 2^k$ tones too, so after their two consecutive turns, the number of stones in pile will decrease by $ 2^{k \plus{} 1}$. Then, eventually before player $ X$'s turn there will be $ 2^{k}$ stones in the pile and he will take all of those and win. So the result is true for $ k \plus{} 1$ and lemma 2 follows.

Then if $ N$ is not a power of $ 2$ Alphonse wins by lemma 2. Otherwise if $ N$ is a power of $ 2$ in order to survive, Alphonse must take the number of stones so before Beryl's turn, the number of stones in the pile is a power of $ 2$. But then Alphonse must take at least $ \frac {N}{2}$ stones so Beryl can take the rest in the next turn as $ \frac {N}{2} < 2*\frac {N}{2} \minus{} 1$ as $ N\ge 2$. So Beryl wins only when $ N$ is a power of $ 2$.[/quote]

beautiful solution

Number 8:

A bijection between the vertical segments chosen and the path travelled.

Also, the number chosen has to be odd.

Thanks, longwan! :)
For question $ 8$ note that if the number of vertical segments is $ k$ then the path consists of $ 9$ horizontal and $ k$ vertical segments. We can "stick" any of the vertical segments into any of the $ 10$ gaps -- before all the horizontal segments, in between any two consecutive horizontal segments, or after all horizontal segments, if we fix all the horizontal segments and want to count the number of ways in which we can place $ k$ vertical segments. Then the number of routes of length $ 9\plus{}k$ is $ \binom{10}{k}$ where $ k$ is odd and $ 1\le k\le 9$. So the maximum number of routes of a fixed length $ 9\plus{}k$ is $ \binom{10}{5}\equal{}252$ which occurs when $ k\equal{}5$, i.e. length of the route is $ 14$.

For #4, it follows easily by the circle of apollonius.

My solution was very similar to rem's except I failed to deal with the trivial case when Alphonse first takes exactly half the number of stones  :oops:

Official Solutions, and averages (but not results) are now posted online at the CEMC website: https://cemc.math.uwaterloo.ca/english/contests/open.shtml

Sometime during my Christmas vacation last year, although that's not too specific since it was 4 weeks long  :D

Around December 20 would be my guess.

Hmm... well, I heard that the top 200 people after the initial 40-50 are chosen are sent a problem set to work on, so does this mean that the problem set will be sent out over the winter break? Also, does anyone know any details of this problem set? like, the difficulty? the time period allowed to work on them? Is it sent through school?... etc.

Thanks
 :D

According to Dr. Larry Rice at Math Circles, 60 people will be directly selected to write CMO, and 200 more will have to write a second contest which will be held some time in January, where the problems will be posted online, and people will have a week to submit their solutions. I am not sure how many of those will be selected to write CMO though.

are you sure? cuz vincent_five just said that the cutoff for CMO was 72 :'( i got 71 if ur sure sweet i have hope again


function
integration
calculus
derivative
real analysis
real analysis solved

$f_1,f_2 > 0$ Riemann integrable functions defined on $[0,1]$
Let $a,b >0; a+b=1$

$f_{n+2}=\sqrt {\int_{0}^{x}f_{n+1}^a(x)f_{n}^b(x)dx}$

Study the convergence of $f_n$.

Dear,Moubinool nice problem,but I can't come close to solution.The only things I got(obviuos of course) is that $f_{n}$ are differentiable with continious
derivatives($n{\geq}4$).Can you give a hint?Thank you.

It is slightly generalized problem, which I solved some time ago. Am I right, Moubinool?
I don't see any difficulties in repeating my previous proof. :?

Yes Myth, do you know where is the problem you solved ?

Initial problem http://www.mathlinks.ro/Forum/viewtopic.php?t=20994 with my solution.

Your first generalization http://www.mathlinks.ro/Forum/viewtopic.php?t=21560


I'm currently doing a project on chat rooms at school and we're going to present this project in Microsoft Powerpoint form.  However, one of the topics that I need to do research for is the [b]history of chat rooms[/b] in which I couldn't find information on.

I've checked World Book, Encarta, Grolier, etc...
I've checked through Google, Altavista, etc...

Yet I couldn't find any information.

This has never happened to me before and I need your help to help me find some information on the history of chat rooms(or how chat rooms were invented).

Thanks.

[url=http://en.wikipedia.org/wiki/Chat_room]Wikipedia is a wonderful thing.[/url]

Thanks for the site but it doesn't contain any information concerning the history of chat rooms...

I believe that the first widespread chat service was IRC, born sometime in 1988, try clicking on the IRC link and reading up on its history.

try http://en.wikipedia.org/wiki/Bitnet_Relay_Chat that is the precursor to IRC and probably the first chat system.

Thx everyone for the very very quick replies.

I'm going to try bitnet and see what happens.  I'll ask my teacher tomorrow.

Once again, thanks.  If you do find more information, please tell me.

Thanks again.

[quote="BHorseMath"]Chat rooms were first created by the military to communicate I believe.[/quote]
Isn't that the Internet?

The Internet was developed by ARPA (now DARPA) and called ARPAnet some decades ago (around 2)...


[quote="Jakey"]What are the answers for MOEMS 2007 Contesys 1-4? I just wanna check mine, just to make sure they are right. Or, you could share your answers too..., afterwhich i'l post mine.:D[/quote]
I'm pretty sure you can't discuss the test until a certain date because not everyone has taken it?


Barry wrote 6 different numbers, one on each side of 3 cards, and
laid the cards on a table, as shown. The sums of the two numbers on
each of the three cards are equal. The three numbers on the hidden
sides are prime numbers. What is the average of the hidden prime
numbers?
[asy]path box=(0,0)--(1,0)--(1,1.5)--(0,1.5)--cycle;
draw(box);
draw(shift(1.5,0)*box);
draw(shift(3,0)*box);
label("44", (0.5, .75));
label("59", (2, .75));
label("38", (3.5, .75));[/asy]

[quote="FantasyLover"]Every prime except 2 is odd number.

also, 44+a=59+b=38+c are also primes.

59+b is prime, and b should be odd.

but if you add odd and odd, it becomes even.

b shouldn't be odd.

there's only 1 prime number that's not odd;2.

so the sum of back and front of a card is 59+2=61.

61-44=17.
61-38=23.

the prime numbers are 17, 2, 23.

the mean of three numbers are 42/3=14.

Answer : 14.[/quote]

Or, you could just do these problems,

$59-44$

$59-38$

and $44-38$ and get

$15, 21,$ and $6.$

Then you add them together which gives you $42$, but it asked for the mean, not the sum.

So you do this: $42/3$ to get $14$ as the answer.

Hope I helped:)

I am having trouble wrapping my brain around this. The average of the differences between the cards we can see is also the average of the cards we cannot see. Could you help me understand why this works?


Prove that the sequence $(a_{n})_{n\geq 0}$,$a_{n}=[n\sqrt{2}]$, contains an infinte number of perfect square

Suppose that $x_{n},y_{n}$ has the propety$2x_{n}^{2}=y_{n}^{2}+1$
then $[\sqrt{2}x_{n}y_{n}]=y_{n}^{2}$
I left the detail to you. :)


1) Show that n^4 + 4 is composite for all n > 1.
2) Show that n^4+n^2+1 is composite for n > 1.

I'm guessing the solutions are similar. Is there a really clever factorization I am missing?

The last step is difference of squares. Thanks!

A more challenging one in similar style is $ 4^n\plus{}n^4$.

[quote="randomgraph"]A more challenging one in similar style is $ 4^n \plus{} n^4$.[/quote]
$ N \equal{} 4^n \plus{} n^4$. Assume $ N$ is prime $ \Rightarrow$ $ n$ is odd. So $ n \equal{} 2q \plus{} 1, q \in \mathbb{N}_0$.
So $ N \equal{} 4(2^q)^4 \plus{} n^4$ which is the Sophie Germain identity.


geometry
circumcircle
geometric transformation
reflection
trapezoid
symmetry
Euler

The medians $ AI , BJ $ of the triangle $ ABC $ intersect the circumcircle of $ ABC $ at $ K , M $ . If $ AK = BM $ ==> $ AC=BC $ $ ?????? $ . If not , find the condition such that $ AK = BM $ ==> $ AC=BC $ ?

$AK=BM$ is equivalent to ($MK\|AB$ OR $AM\|BK$). In the first case ($MK\|AB$), it shouldn't be too hard to deduce $CA=CB$. The question here is finding whether the second case can ever occur without $CA=CB$.

It's possible for it to be so, for continuity reasons. Do you have any dynamic geometry software? A dynamic sketch would convince you right away. Having a dynamic sketch, you could easily move the point $C$ so that $CA<CB$ at all times, for instance, and $AM,BK$ meet on one side of $AB$ in one position and on the other side in the other position.

I'm sure that : $ AK = BM $ doesn't infer $ CB=CA $ . We need to find a condition for triangle $ ABC$ such that : $ AK = BM  ===>  CB=CA $

I don't understand the kind of condition you want but, obviously, if G lies on the circle OAB, the reflection through the line OG maps A to B and K to M thus AK = BM.
Kind regards

I mean we can erect a non-isosceles triangle ABC satisfying : $ AK = BM $.
 This is the condition such that : $ AK=BM \Longleftrightarrow CA=CB $ :
   $ \alpha \leq 60 $ or $ \alpha \geq 90 $ ( $ \alpha = \angle ACB $ )

  Mr Le had two nice solutions for this problem ! ;)

Dear Tu!
Do you really think the solutions of Mr Le is nice?
 :D
can you post it?

dear Minh Khoa , I'm very lazy to post his solution (I think you know it  ) . This is a problem on M&Y . From me , its solution is nice ! :)

This one had a long stay in the open questions section, too... Let's finally solve the problem:

We will show the following theorem, which will yield the solution of the problem:

[color=blue][b]Theorem 1.[/b] Let ABC be a triangle, let B' and C' be the midpoints of its sides CA and AB, and let the medians BB' and CC' of triangle ABC intersect the circumcircle of triangle ABC at the points Y and Z (apart from the points B and C, of course). Then:

[b](a)[/b] We have BY = CZ if and only if either YZ || BC or BZ || CY. Hereby, the "either or" includes the possibility that both relations YZ || BC or BZ || CY hold, of course.
[b](b)[/b] We have YZ || BC if and only if b = c.
[b](c)[/b] We have BZ || CY if and only if the point G lies on the circle OBC. Hereby, G is the centroid and O is the circumcenter of triangle ABC, and the abbreviation "circle $P_1P_2P_3$" means the circle through three given points $P_1$, $P_2$, $P_3$.
[b](d)[/b] We have BY = CZ if and only if either b = c or the point G lies on the circle OBC.
[b](e)[/b] If BZ || CY, then the angle A of triangle ABC must satisfy $60^{\circ}\leq A<90^{\circ}$.
[b](f)[/b] If we are given that BY = CZ, and if we have $A\leq 60^{\circ}$ or $A\geq 90^{\circ}$, then b = c.[/color]

[i]Proof of Theorem 1.[/i] In the proofs of Theorem 1 parts [b](a)-(d)[/b], we will use directed angles modulo 180°.

[b](a)[/b] It is a well-known fact that two chords in a circle have equal lengths if and only if their non-directed chordal angles are equal. With directed angles, this fact becomes a bit more complicated: Two chords in a circle have equal lengths if and only if their directed chordal angles are equal or oppositely equal. Thus, the chords BY and CZ in the circumcircle of triangle ABC have equal lengths if and only if their chordal angles < BCY and < CYZ are equal or oppositely equal. In other words, BY = CZ if and only if < BCY = < CYZ or < BCY = - < CYZ. Now, < BCY = < CYZ is equivalent to < BCY = < CBZ (since < CYZ = < CBZ, as the points C, Z, Y and B lie on the same circle), and the equation < BCY = < CBZ is clearly equivalent to BZ || CY. Furthermore, < BCY = - < CYZ is equivalent to < BCY = < ZYC, and this is clearly equivalent to YZ || BC. Thus, we have BY = CZ if and only if BZ || CY or YZ || BC. This proves Theorem 1 [b](a)[/b].

[b](b)[/b] We have to show that YZ || BC if and only if b = c. Well, proving that b = c implies YZ || BC is trivial: If b = c, then the triangle ABC is isosceles, and YZ || BC follows from symmetry. So it remains only to show that YZ || BC implies b = c.

In fact, if YZ || BC, then the quadrilateral BZYC is a trapezoid. Since this trapezoid is inscribed, it must be an isosceles trapezoid (in fact, any inscribed trapezoid is isosceles). Thus, this trapezoid has an axis of symmetry, which must be simultaneously the perpendicular bisector of the bases BC and YZ of the trapezoid. Of course, this axis of symmetry must pass through the point of intersection G of the diagonals BY and CZ of the trapezoid. So, in other words, we have seen now that the perpendicular bisector of the segment BC passes through the point G. Thus, the perpendicular bisector of the segment BC has two points in common with the A-median of triangle ABC: namely, the centroid G of triangle ABC and the midpoint of the segment BC. Hence, the perpendicular bisector of the segment BC must coincide with the A-median of triangle ABC; thus, the point A lies on the perpendicular bisector of the segment BC, what yields AC = AB. Thus, b = c. Hence, the conclusion from YZ || BC to b = c is established, and Theorem 1 [b](b)[/b] is proven.

[b](c)[/b] Since the point O is the circumcenter of triangle ABC, and thus the center of the circle passing through the points A, B, C, the central angle theorem yields < BOC = 2 < BAC. But since the points B, C, A, Y and Z lie on one circle, < BAC = < BYC and < BAC = < BZC, so that 2 < BAC = < BAC + < BAC = < BYC + < BZC. Hence, < BOC = < BYC + < BZC, and thus

  < BGC - < BOC = < BGC - (< BYC + < BZC) = < BGC - < BYC - < BZC
                = < (BY; CZ) - < (BY; CY) - < (BZ; CZ) = < (CY; BZ).

Thus, < BGC - < BOC = 0° if and only if < (CY; BZ) = 0°. But the equation < BGC - < BOC = 0° is equivalent to < BGC = < BOC, what simply means that the point G lies on the circle OBC. The equation < (CY; BZ) = 0° is equivalent to BZ || CY. Thus, we see that the point G lies on the circle OBC if and only if BZ || CY. This proves Theorem 1 [b](c)[/b].

[b](d)[/b] Theorem 1 [b](d)[/b] follows trivially from Theorem 1 [b](a)[/b] after applying [b](b)[/b] and [b](c)[/b].

Now, we will prove Theorem 1 parts [b](e)-(f)[/b]. Hereby, we will use non-directed angles again.

[b](e)[/b] Assume that BZ || CY. First we will show that A < 90°.

In fact, BZ || CY yields < ZBC = 180° - < BCY. Hence, at least one of the angles < ZBC and < BCY must be $\geq 90^{\circ}$. We can WLOG assume that $\measuredangle ZBC\geq 90^{\circ}$. Since the quadrilateral AZBC is cyclic, we then have < ZAC = 180° - < ZBC, and thus $\measuredangle ZAC\leq 180^{\circ}-90^{\circ}=90^{\circ}$. Hence, $A=\measuredangle BAC<\measuredangle ZAC\leq 90^{\circ}$.

Hence, we have showed that A < 90°. Remains to prove that $60^{\circ}\leq A$.

In fact, assume, for the sake of contradiction, that 60° > A.

After Theorem 1 [b](c)[/b], since BZ || CY, the point G must lie on the circle OBC.

First we consider the case when the triangle ABC is acute-angled.

By the Euler line theorem, the orthocenter H, the centroid G and the circumcenter O of triangle ABC lie on one line; hereby, the point G lies on the segment HO and divides this segment in the ratio $\frac{HG}{GO}=2$.

Now, since the point O is the circumcenter of triangle ABC, by the central angle theorem, < BOC = 2 < BAC = 2A. On the other hand, for the orthocenter H of the acute-angled triangle ABC, we have 180° - < BHC = A (this is a well-known and almost trivial fact; see http://www.mathlinks.ro/Forum/viewtopic.php?t=660 post #6 for a proof), so that < BHC = 180° - A. Since 60° > A, we have $180^{\circ}=3\cdot 60^{\circ}>3A$, so that 180° - A > 3A - A = 2A. In other words, $\measuredangle BHC>\measuredangle BOC$. Since the points H and O lie in one and the same halfplane with respect to the line BC (namely, in the halfplane containing the point A - this is all because the triangle ABC is acute-angled), this yields that the point H lies inside the circle OBC. Also, it is clear that the point O lies on the circle OBC. Hence, since circles are convex, the point G, lying on the segment HO, must also lie inside the circle OBC. This contradicts with the result that the point G lies on the circle OBC. Hence, we have a contradiction.

So we have settled the case when the triangle ABC is acute-angled. Now remains to consider the case when the triangle ABC has a right or obtuse angle. Since the angle A is surely acute (since 60° > A), the right or obtuse angle must be one of the two angles B and C. We can WLOG assume that the right or obtuse angle is C. Then, the angles A and B are acute (since a triangle cannot have more than one right or obtuse angle).

The point G is the centroid of triangle ABC, and the segment CC' is the median from the vertex C. Thus, the point G lies on the segment CC'.

Since the point O is the circumcenter of triangle ABC, it lies on the perpendicular bisector of the side AB; in other words, $OC^{\prime}\perp AB$ for the midpoint C' of the side AB. Thus, < OC'B = 90°. On the other hand, since O is the circumcenter of triangle ABC, we have OB = OC, so that the triangle BOC is isosceles; since a base angle of an isosceles triangle is always acute, it follows that the angle < OCB is acute. In other words, $\measuredangle OCB<90^{\circ}$. Since < OC'B = 90°, this becomes $\measuredangle OCB<\measuredangle OC^{\prime}B$. Now, the points A and C lie in the same halfplane with respect to the line BO. This halfplane therefore also contains the midpoint C' of the segment AB. Hence, the relation $\measuredangle OCB<\measuredangle OC^{\prime}B$ shows that the point C' lies inside the circle OBC. The point C trivially lies on this circle. Since circles are convex, the point G, lying on the segment CC', must lie inside the circle OBC. This contradicts with the result that the point G lies on the circle OBC.

Hence, we have got a contradiction in every case. Thus, our assumption 60° > A cannot be true, and we must therefore have $60^{\circ}\leq A$. Together with A < 90°, this yields $60^{\circ}\leq A<90^{\circ}$, what proves Theorem 1 [b](e)[/b].

[b](f)[/b] Assume the contrary, i. e. assume that BY = CZ, and that we have $A\leq 60^{\circ}$ or $A\geq 90^{\circ}$, but not b = c. By Theorem 1 [b](a)[/b], from BY = CZ one concludes that either YZ || BC or BZ || CY. But YZ || BC yields b = c by Theorem 1 [b](b)[/b], and this is impossible by our assumption. Thus, we must have BZ || CY. Then, by Theorem 1 [b](e)[/b], we have $60^{\circ}\leq A<90^{\circ}$. On the other hand, we are given that $A\leq 60^{\circ}$ or $A\geq 90^{\circ}$. Hence, we must have A = 60°.

Now, we try to proceed like in the proof of Theorem 1 [b](e)[/b] above. In the case when the triangle ABC has a right or obtuse angle, we proceed in exactly the same way and get a contradiction. In the case when the triangle ABC is acute-angled, we can also proceed analogously, with the only difference that the point H lies not inside the circle OBC, but on this circle, and thus, since the point O also lies on this circle, the point G, lying on the segment HO, must lie inside this circle, contradicting the assumption that the point G lies on the circle OBC. Well, more precisely, almost contradicts. The only possible case when we don't get a contradiction is the case when the points H and O coincide (in this case, the point G, lying on the segment HO, must also coincide with the points H and O, and thus lies on the circle OBC). But H is the orthocenter and O is the circumcenter of triangle ABC, and the orthocenter and the circumcenter of a triangle coincide only if the triangle is equilateral. Thus, the triangle ABC must be equilateral. But this yields b = c, contradicting our assumption above that we don't have b = c. Hence, we get a contradiction in all cases, so that our assumption is wrong. Hence, Theorem 1 [b](f)[/b] is proven.

[i]Note.[/i] By some algebra, one can show that BZ || CY holds if and only if $a^4=b^4+c^4-b^2c^2$. This yields another proof of Theorem 1 [b](e)[/b].

  Darij


invariant
group theory
abstract algebra
superior algebra
superior algebra unsolved

a.  Show that $ |N_G(H): H|$ is equal to the number of right cosets of $ H$ in $ G$ that are invariant under the right multiplication by $ H$.
b.  Suppose that $ |H|$ is a power of the prime $ p$ and that $ |G: H|$ is divisible by $ p$. Show that $ |N_G(H): H|$ is divisible by $ p$.

Thks in advance.

[quote="Hidden Scofield"]
depending on the following lemma :

If $ G$ is a $ p \minus{} group$ , then $ |X|\equiv |X_o| \ mod \ p$, where $ X$ is a $ G \minus{} set$,

Where $ X_o \equal{} \{x\in X: g.x \equal{} x , \forall g\in G\}$

[/quote]

Note :

In the 1st question they are equal $ mod \ p$

 :wink:

Could u be more specific, please?

THank you.

Note that , $ S\equal{}X$ in the lemma above  :)

Hidden, you have to actually read his problem...

a) $ Hxh \equal{} Hx$ if and only if $ xhx^{\minus{}1}\in H$, so we are counting the elements of $ N_G(H)$ under some equivalence relation.  What is this relation?  Well, $ Hx\equal{}Hy$ if and only if $ xy^{\minus{}1}\in H$, which is the same as the equivalence relation for cosets of $ H$.

For this sort of problem, brave manipulations of the definitions will get you far.  But you have to get your hands dirty!  It's not really possible to "see" the answer without going through the details first.

Oh yes, we can alter the proof for right multiplication of cosets , I always use the left multiplication  :)

I take it back, you can extract the proof from what you posted, but the statement that it depends on some lemma about congruence mod p is not true...

Oh  :idea: , The subgroup should be $ p\minus{}group$  .

Thx LydianRain

THks everyone so much. I was so busy, so I could not go back sooner.

@Hidden : Could you tell me which book did u use here? Thks.

Thomas W. Hugerford , Algebra [Graduate Text]


This is from a Brasilian olympiad. I think that 'my' solution is a little  long.If you find something short and synthetic , please write.Thnks.

[quote="Leonhard Euler"]We have $ AE\parallel CG,AH\parallel CF$ and $ AC,EG,HF$ are concurrence(Brianchon theorem). Hence $ \triangle AEH$ and $ \triangle CGF$ are homothety. Hence $ EH\parallel FG$.[/quote]

 I put the figure and I read your solution. It seems nice and easy  :lol: .Thnks


inequalities
function
algebra unsolved
algebra

Let $ f: \mathbb{R}\rightarrow\mathbb{R}$ be a function which satisfies $ |f(x\plus{}y)\minus{}f(x)\minus{}f(y)|<1$ for all real numbers $ x,y$. Prove that $ |f(\frac{x}{2008})\minus{}\frac{f(x)}{2008}|<1$ for all real number $ x$.


Are you sure about the 5th one?
I got 45.
I got the radius to be 38/7

It is 032.
The answers are posted on the AMC website.
http://www.unl.edu/amc/e-exams/e7-aime/e7-1-aimearchive/2009-aa/09-2-AIME-Answers.shtml

I believe all of worthawholebean's on the first post were correct.


Does anyone know of a good book on sequences and series or a good website with lots of information and problems on sequences and series?

Have you checked some (standard) calculus/analysis books already, or are you not satisfied by them? [url=http://tutorial.math.lamar.edu/Classes/CalcII/SeriesIntro.aspx]Here[/url] the basics are explained.


vector
topology
calculus
derivative
advanced fields
advanced fields theorems

Given two differentiable manifolds $M_{1}$ and $M_{2}$, two vectors fields $X_{1}$ and $Y$ on $M_{2}$, and a diffeomorphism $f: M_{1}\rightarrow M_{2}$, is it true that $f^{*}[X,Y] = [f^{*}X,f^{*}Y]$, where $[,]$ denotes the Lie bracket and $f^{*}X$ the pullback vector field of $X$ (thus defined on $M_{1}$).

Note that the assertion is true if $X$ is assumed to be $C^{2}$ (for, in this case, $X$ generates a flow, and computations can be easily carried out). What if $X$ is only differentiable ?

That should be a pushforward, not a pullback. Pullbacks of vector fields don't work.

I think you need at least a $C^{2}$ environment; the Lie bracket isn't even a vector field if we don't apply it to $C^{2}$ functions.

This doesn't mean you need any extra derivatives of $X$ and $Y$ for this- the proof is a straightforward manipulation of notation:

$([f_{*}X,f_{*}Y]u)\circ f=f_{*}X(f_{*}Y(u))\circ f-f_{*}Y(f_{*}X(u))\circ f$
$=X(f_{*}Y(u)\circ f)-Y(f_{*}X(u)\circ f)=X(Y(u\circ f))-Y(X(u\circ f))=[X,Y](u\circ f)$ for any $u$, so $[f_{*}X.f_{*}Y]=f_{*}[X,Y]$


Let p^3-1 divisible p and p-1 divisible. Prove that p = q^2+q+1


limit
trigonometry
algebra unsolved
algebra

lim (x -sin x) /(x - tan x)
when   x---->0

lim (x -sin x) /(x - tan x)
$=lim{(1-cos x)/(1-(1/cosx)^{2})}$
$=lim(-cosx^{2}/(1+cosx))$
$=-0.5$
here I used the rule of L`Hospital :wink:


calculus
integration
geometry
geometric transformation
rotation
trigonometry
calculus computations

Let the curve $ C: y\equal{}x\sqrt{9\minus{}x^2}\ (x\geq 0)$.

(1) Find the maximum value of $ y$.

(2) Find the area of the figure bounded by the curve $ C$ and the $ x$ axis.

(3) Find the volume of the solid generated by rotation of the figure about the $ y$ axis.

[b](1) [/b]

$ y^2 \equal{} x^2(9\minus{}x^2)$

$ y^2 \equal{} \frac{81}{4} \minus{} (x^2\minus{}\frac{9}{2})^2 \ \leq \ \frac{81}{4}$

so, $ y_{max} \equal{} \frac{9}{2}$

[b](2)[/b]

$ C$ intersects the $ x$ axis at $ 0$ and $ 3$.

Area = $ \int_0^3 x \sqrt{9\minus{}x^2} \ dx \equal{} 9 \ sq. \ units$

hope i haven't made a calculation mistake there

[b](3)[/b]

i don't remember the formula exactly for rotation ...

is it  $ \pi  \int  f(y)^2 \ dy$ ?

If that's right, you get $ \pi \int_0^{4.5} (\frac{9}{2} \plus{} \sqrt{\frac{81}{4} \minus{} y^2}) \ dy$

The answers to (1), (2) are correct, but that of (3) is incorrect. (3) is not easy.

for (3), we can use the cylindrical shells formula

$ V = \int^{a}_{b} 2\pi x f(x) dx$

in our case,

$ V = \int^{3}_{0} 2\pi x^2\sqrt{9-x^2} dx
= 2\pi \int^{3}_{0} x^2 \sqrt{9-x^2} dx$

which we can evaluate by the substitution $ x = 3\sin\theta$

[quote="Attila the Hun"]for (3), we can use the cylindrical shells formula

$ V \equal{} \int^{a}_{b} 2\pi x f(x) dx$[/quote]

There is no problem in your solution, regrettably Japanese high school students don't study the formula.

student? University student or High school student?

Well, it applies to this test. As it is an entrance exam, I guess post-high school?

[quote="Tyl"]I guess post-high school?[/quote]

Yes. :)

So can these Japanese post-high school students use the formula (on this test, this exercise) presented by Timestopper_STG? If so I don't understand your comment really.

hello, you can use the formula $ V_y\equal{}\pi \cdot \int_{a}^bx^2\cdot|f'(x)|\,dx$.
Sonnhard.

[quote="Attila the Hun"]for (3), we can use the cylindrical shells formula

$ V \equal{} \int^{a}_{b} 2\pi x f(x) dx$

in our case,

$ V \equal{} \int^{3}_{0} 2\pi x^2\sqrt {9 \minus{} x^2} dx \equal{} 2\pi \int^{3}_{0} x^2 \sqrt {9 \minus{} x^2} dx$

which we can evaluate by the substitution $ x \equal{} 3\sin\theta$[/quote]

[quote="Timestopper_STG"]kunny, is it possible for any japanese student to use that formula on the test?
I mean what if they have learnt some theories by themselves.  :)[/quote]

Direct using the shell method is not allowed, if we use it, we should prove the formula except just answering type test, or not the type "justify your answer''

its very easy to prove. I'd rather prove it than do it using cross sections perpendicular to the y axis and drop down dead.


geometry
geometric transformation
homothety
projective geometry
power of a point
radical axis
perpendicular bisector

[quote="hophinhan"]The incircle of $ \triangle{ABC}$ touches $ BC,CA,AB$ at $ D,E, F$, respectively.
Let $ X$ be a point inside $ \triangle{ABC}$ such that the incircle of $ \triangle{XBC}$ touches $ BC$ at $ D$ also, and touches $ CX$ and $ XB$ at $ Y$ and $ Z$.
Prove that : $ FX\ ;\ EY$ and $ AX$ are concurrent at a point lies on a line perpendicular $ BC$ at $ D$.[/quote]
Let (I) and (I') be the incircles of triangles ABC and XBC, respectively.
Let FZ cut AX at K, EY cut AX at K'. We will prove $ K\equiv K'.$
Applying Menelaus's theorem for the triangle ABX we have $ \frac{FA}{FB}.\frac{ZB}{ZX}.\frac{KX}{KA}\equal{}1$ or $ \frac{KX}{KA}\equal{}\frac{ZX}{FA} (2)$
Similarly, $ \frac{K'X}{K'A}\equal{}\frac{XY}{EA} (2)$
From $ (1)(2)$ we get $ \frac{KX}{KA}\equal{}\frac{K'X}{K'A}$ so $ K'\equiv K$
Let $ FZ$ cut $ II'$ at $ K_1, EY$ cut $ II'$ at $ K_2$, we will prove $ K_1\equiv K_2$
Let $ I'Z\cap IF\equal{}\{M\}, I'Y\cap IE\equal{}\{L\}$
Since $ BF\equal{}BZ$ and $ \angle BFM\equal{}\angle BZM\equal{}90^o$ we obtain $ MF\equal{}MZ$, similarly $ LE\equal{}LY$
Applying Menelaus's theorem for triangle MII' we have:
$ \frac{FM}{FI}.\frac{K_1I}{K_1I'}.\frac{ZI'}{ZM}$ then $ \frac{K_1I}{K_1I'}\equal{}\frac{FI}{ZI'}\equal{}\frac{r_{ABC}}{r_{XBC}}$
Similarly, $ \frac{K_2I}{K_2I'}\equal{}\frac{r_{ABC}}{r_{XBC}}$ thus $ K_1\equiv K_2$
->QED

[b]Another approach:[/b]
First, let me restate the problem so that it can fits my solution:

[b]PROBLEM- [color=darkblue]Let $ \triangle ABC$ with $ (I)$ its incircle. $ (I)$ tangents to $ BC,CA,AB$ respectively at $ A_0,B_0,C_0$. Let $ X$ be the point inside $ \triangle ABC$ such that the incircle $ I_a$ of $ \triangle XBC$ tangents $ BC$ at $ A_0$ also. Let $ C_1,B_1$ be the tangency point of $ (I_a)$ with $ XB,XC$, respectively. Prove that: $ C_0C_1,B_0B_1, AX$ are concurrent at $ D$ and $ D\in II_a$.[/b][/color]

[i]Proof[/i]

Let $ N\equiv C_0B_0\cap BC$, $ N'\equiv B_1C_1\cap BC$.Then, it is very well- known that $ (NA_0BC) \equal{} (N'A_0BC) \equal{} \minus{} 1\Longrightarrow N \equiv N'$. Thus $ B_0C_0,B_1C_1, BC$ are concurrent. Hence by [b]Desgaurte theorem[/b], we conclude that $ \triangle B_0B_1B,\triangle C_0C_1C$ are $ 2$ perspective triangles. But $ A\equiv BB_0\cap CC_0$, $ X\equiv BB_1\cap CC_1$, $ D\equiv B_1B_0\cap C_1C_0\Longrightarrow A,X,D$ are collinear, which implies that $ AX,C_0C_1,B_0B_1$ are concurrent at $ D$.
Now, let $ B_2\equiv I_aB_1\cap IB_0$, $ C_2\equiv I_aC_1\cap IC_0$. It is very easy to note that $ \{B_2,B_1,B_0,B\}$ and $ \{C_2,C_1,C_0,C\}$ are sets of concyclic points. But $ BB_0 \equal{} BB_1$, $ CC_0 \equal{} CC_1\Longrightarrow B_2B_0 \equal{} B_2B_1$, $ C_2C_0 \equal{} C_2C_1$. Denote $ (C_2),(B_2)$ be the circle with radii $ C_2C_0,B_2B_0$, respectively. Then, $ (C_2)$ internally tangents to $ (I)$ at $ C_0$ and externally tangents to $ (I_a)$ at $ C_1$. Or in other word, $ C_0,C_1$ are the external center and internal center of homothety that turns $ (C_2)\mapsto (I)$, $ (I_a)\mapsto (C_2)$, respectively. Hence the [b]Monge-d'Alembert circle theorem[/b] now says that $ C_1C_0$ passes the internal center of homothety wich turns $ (I)\mapsto (I_a)$, let call that center $ D'$. With the same argument, we also have $ B_0B_1$ passes through $ D'$.Hence $ D'\equiv B_0B_1\cap C_0C_1$, which implies $ D'\equiv D$. Therefore $ D,I,I_a$ are collinear. The result is lead as follow.
Our proof is completed then.

[b]REMARK-[/b]
[hide]I am very happy when I am able to find out $ D$ is the internal center of the homothety wich turns $ (I_a)\mapsto (I)$. What a surprise for me, I have never thought of this property before. Nice problem my dear friend, [b]Hophinhan[/b][/hide]

$ FE$ and $ ZY$ go through the harmonic conjugate $ P$ of $ D$ with respect to $ (B,C).$ Then $ \triangle AFE$ and $ \triangle ZXY$ are perspective with perpectrix $ BC$ $\Longrightarrow$ $AX,FZ$ and $ EY$ concur at their perspector $ J.$ Note that $ PE \cdot PF \equal{}PY \cdot PZ \equal{} PD^2$ implies that $ FEYZ$ is a cyclic quadrangle $ \Longrightarrow$ $ JY \cdot JE \equal{} JZ \cdot JF.$  Also,  tangents $ (AC,XC)$ and $ (AB, XB)$ through the pairs of points $ (E,Y)$ and  $ ( F,Z)$ meet at the radical axis $BC$ of both circles $\Longrightarrow$ $ J$ is the center of the inversion taking one circle into another $\Longrightarrow$  $ J$ lies on $ ID.$

Thanks Linh ; Minh and Luis .... All solutions are very nice :D

Nice prolem,Nhạn!
Denote the points like my figure.
Let $ J$ be the intersection of $ IF$ and perpendicular line of $ FM$ from $ B$,$ K$ be the intersection of $ IE$ and perpendicular line of $ EN$ from $ C$.Easy to see that $ (J)$ and $ (K)$ are tangent to $ (I)$ and $ (I')$.
It's easy to that $ FE,MN,BC$ are concurrent at a point $ G$.Then $ GD^2 \equal{} \overline{GF}.\overline{GE} \equal{} \overline{GM}.\overline{GN}$ then $ F,E,M,N$ are concyclic then L lie on the radical axis of $ (J)$ and $ (K)$.
$ AE^2 \equal{} AF^2$ and $ XM^2 \equal{} XN^2$ so $ A$ and $ X$ also lie on the radical axis of $ (J)$ and $ (K)$.Hence $ A,X,L$ are collinear.
I'' lie on the perpendicular bisector of MN so $ I''M^2 \equal{} I''N^2$.
We have $ I''N^2 \minus{} NC^2 \plus{} CD^2 \minus{} BD^2 \plus{} BM^2 \minus{} I''M^2 \equal{} CD^2 \minus{} BD^2 \minus{} CD^2 \plus{} BD^2 \equal{} 0$ then by Carnot's theorem,L lie on the line perpendicular to BC from D.

[img]http://img29.imageshack.us/img29/8049/hahavnw.jpg[/img]


Solve for all values of x:

$\sin{x}+\cos{x}=\sqrt{2}$

kunny gave a hint to a really cool solution  :D

We have $2 = \left(\sin x+\cos x\right)^{2}+\left(\sin x-\cos x\right)^{2}= 2+\left(\sin x-\cos x\right)^{2}$. Therefore, $\sin x = \cos x$, so $\sin x = \cos x = \frac{\sqrt{2}}{2}$, i.e. $x = \frac{\pi}{4}+2k\pi$.

That's almost what I did, here's my solution:

[hide="A solution"]I squared:

$\sin^{2}{x}+\cos^{2}{x}+\sin{2x}=2$

$\sin{2x}=1$

The only places where sine theta is equal to 1 are 90+360k(degrees). So 45+180k are the solutions...[/hide]

Maybe I'm not getting something?

You squared. That makes that you include the solutions to $\sin x+\cos x =-\sqrt{2}$  :wink:

[hide="you could also do"]
\[\sin x+\cos x=\sqrt{2}\sin (x+45^\circ)\]
[/hide]


AMC
USA(J)MO
USAMO
search
geometry proposed
geometry

A circle (I) is inscribed in a quadrilateral $ ABCD$. Prove that $ (AI\plus{}DI)^2\plus{}(BI\plus{}CI)^2\equal{}(AB\plus{}CD)^2$

Please search on [b]Resources[/b] section when you've already known the source of problems. http://www.mathlinks.ro/viewtopic.php?p=17437#17437


Here's the problem I have no idea how to do! :(
Prove that with every c>8/3, there exist a real number k in order to have [k^c^n] is a prime number with every positive integers n([] is the greatest integer function).
Any hints for this problem will be very appreciated!

Nobody wants to help me or at least, gives me some ideas? :(


calculus
integration
logarithms
derivative
trigonometry
function
complex analysis

$ \int_{0}^{\infty}\frac{ln(x)^2dx}{1\plus{}x^2}$

Are there any ways to do this without contour integration? I did it with contour integration, but the problem is that I have no idea how to prove the residue theorem and am confused by the whole thing, so I don't feel like I can justify my answer

I think that such problem had better clear it up commonly in a range of [b]complex analysis[/b].

One small step: by making the change of variables $ x\mapsto \frac1x,$ we can show that

$ \int_0^1\frac{\ln^2x}{1\plus{}x^2}\,dx\equal{} \int_1^{\infty}\frac{\ln^2x}{1\plus{}x^2}\,dx$ and hence write the original problem as

$ 2\int_0^1\frac{\ln^2x}{1\plus{}x^2}\,dx.$

Now, expand $ \frac1{1\plus{}x^2}$ as a power series to get

$ 2\sum_{k\equal{}0}^{\infty}(\minus{}1)^k\int_0^1x^{2k}\ln^2x\,dx.$

Leaving to the reader the details of $ \int_0^1x^{2k}\ln^2x\,dx\equal{}\frac{2}{(2k\plus{}1)^3},$ we get

$ \sum_{k\equal{}0}^{\infty}\frac{4(\minus{}1)^k}{(2k\plus{}1)^3}$ as our answer.

Finding a closed for for that series isn't all that obvious, but writing it that way does open up several other possible techniques, including Fourier series methods.

Within the last couple of weeks, someone posted here a summation that is at least related to this. I haven't been able to find the topic (since I don't know what the title was).

The sum can also be attacked by numerical means. Using a modest number of terms and a Richardson extrapolation table, I get that

$ \sum_{k\equal{}0}^{\infty}\frac{4(\minus{}1)^k}{(2k\plus{}1)^3}\approxeq 3.87578458503748$

- which happens to agree, to 13 or 14 digits, with $ \frac{\pi^3}{8}.$ That latter coincidence isn't by itself a proof but should tell you whether your other methods are correct or not.

If we make the substitution $ x \equal{} \exp y$ to convert (twice) the integral into $ \int_{ \minus{} \infty}^{\infty} \frac {y^2}{\cosh y} dy$, which is essentially the second derivative of the Fourier transform of sech at zero, but the only evaluation of $ \widehat{\text{sech}}$ I've seen uses complex analysis (which could alternatively/more directly be used to sum the series by contour integrating $ \frac {\csc \pi z}{(z \plus{} 1/2)^3}$ or by initially contour integrating, probably) ...

edit -- Professor Merryfield, you were probably referring to http://www.artofproblemsolving.com/Forum/viewtopic.php?t=222690

The answer is $ \frac {\pi^3}{8}$. Here's one proof:

The results $ \ln \left(2 \sin \frac {x}{2}\right) \equal{} \minus{} \sum_{n \equal{} 1}^{\infty} \frac {\cos(nx)}{n}$ and $ \ln \left(2 \cos \frac {x}{2}\right) \equal{} \sum_{n \equal{} 1}^{\infty} ( \minus{} 1)^{n \minus{} 1} \frac {\cos(nx)}{n}$ can be easily derived. Therefore,

$ \ln \left(2 \sin \frac {x}{2}\right) \minus{} \ln \left(2 \cos \frac {x}{2}\right) \equal{} \ln \left(\tan \frac {x}{2}\right) \equal{} \sum_{n \equal{} 1}^{\infty} \frac { \minus{} \cos(nx) \minus{} ( \minus{} 1)^{n \minus{} 1} \cos(nx)}{n}$

$ \therefore \ln \left(\tan \frac {x}{2}\right) \equal{} \minus{} 2 \sum_{n \equal{} 1}^{\infty} \frac {\cos[(2n \minus{} 1)x]}{2n \minus{} 1}$. Now let's apply Parseval's identity:

$ \frac {1}{\pi} \int_{ \minus{} \pi}^{\pi} \left[\ln \left(\tan \frac {x}{2}\right)\right]^2\,dx \equal{} \sum_{n \equal{} 1}^{\infty} \left(\frac { \minus{} 2}{2n \minus{} 1}\right)^2$

$ < \equal{} > \frac {2}{\pi} \int_{0}^{\pi} \left[\ln \left(\tan \frac {x}{2}\right)\right]^2\,dx \equal{} 4 \sum_{n \equal{} 1}^{\infty} \frac {1}{(2n \minus{} 1)^2} \equal{} 4 \cdot \frac {\pi^2}{8} \equal{} \frac {\pi^2}{2}$

$ < \equal{} > \int_{0}^{\pi} \left[\ln \left(\tan \frac {x}{2}\right)\right]^2\,dx \equal{} \frac {\pi^3}{4}$.

Next, considering, on this last integral, the change of variables $ x \equal{} \pi \minus{} 2t$, then after a simple manipulation we arrive at

$ \int_{0}^{\pi /2} [\ln (\tan x)]^2\,dx \equal{} \frac {\pi^3}{8}$. To complete the "proof", we make the substitution $ \tan x \equal{} u$, and so

$ \boxed{\int_0^{\infty} \frac {(\ln x)^2}{1 \plus{} x^2}\,dx \equal{} \frac {\pi^3}{8}}$.

The series which Professor Merryfield arrived to has the value

$ \sum_{n\equal{}0}^{\infty}\frac{(\minus{}1)^n}{(2n\plus{}1)^3} \equal{} \frac{\pi^3}{32}$.

This can also be derived through Fourier series (for example, by considering the sinus Fourier series of the function $ f(x) \equal{} x(\pi \minus{} x)$).


[quote="othman"]proof that :
$\forall x\in\mathbb{R}_+^{\large*} . \exists n\in\mathbb{N}.\tan(\frac{x}{n} )\not\in\mathbb{Q}$ :D[/quote]

What about $x=\pi , n=4$  ??!!
:cool:

[quote="Diogene"]
What about $x=\pi , n=4$  ??!!
:cool:[/quote]

That's "for all $x$, there exists an $n$," not for all $x$ and $n$.

Let $x$ be a positive real number, and let $m_0$ be a natural number such that $\frac x{2^{m_0}} <\frac{\pi} 2$
Let $y_n$ be the decreasing sequence defined by $y_n = \tan (\frac x{2^{m_0+n}})$ .
Remember : $\tan (t/2) = \frac{\sqrt{1+\tan^2(t)}-1}{\tan (t)}$
If $y_n=\frac{p_n}{q_n}, \ y_{n+1}=\frac{p_{n+1}}{q_{n+1}}$ are irreducible rational fractions ($p,q$ are natural), then  $\frac{p_{n+1}}{q_{n+1}} = \frac{\sqrt{p_n^2+q_n^2}-q_n}{p_n}=\frac{p_n}{\sqrt{p_n^2+q_n^2}+q_n}$
I deduce $p_{n+1} \leq p_n ,\ q_{n+1} \leq p_n$.
So from $y_0=\frac{p_0}{q_0}$ rational , there are at most $p_0^2$  rational consecutive terms .
Finally in the $y_n$ sequence there is an infinite number of irrational terms ... q.e.d.
 :cool:


Find the smallest value $x$ such that $3^{2548}|x^{2005}+1$


let G be the centroid of triangle ABC.prove that:

                                                   if AB+GC=AC+GB,then we have :

                                                                                ABC is isoceles.

$a\le b\le c\Longleftrightarrow m_a\ge m_b\ge m_c.$

$4m^2_a=2(b^2+c^2)-a^2,\ 4(m^2_b-m^2_c)=3(c^2-b^2)\ a.s.o.$

$AB+GC=AC+GB\Longleftrightarrow 3(c-b)=2(m_b-m_c)\ |\ \odot\ 2(m_b+m_c)\Longleftrightarrow$

$2(c-b)(m_b+m_c)=c^2-b^2\Longleftrightarrow (c-b)[2(m_b+m_c)-(b+c)]=0.$

$\frac 23 m_b+\frac 13 m_c > \frac 12 c,\ \frac 23 m_c+\frac 13 m_b >\frac 12 b\Longrightarrow 2(m_b+m_c) > b+c\Longrightarrow b=c.$


Evaluate:
\[\sin 1^o\cdot\sin 2^o\cdot\sin 3^o\cdot...\cdot\sin 245^o\]
:rotfl:

Masoud Zargar

ok so i was working on this for a whle and had to ask a friend for advie.
well the answer is 0.
this is because sin180 is 0.
this is what my friend pointed out, i had overlooked this point, even when i was mainly focusing on sin90, sin180, and sin275 to see how they would help. well there is your answer 0.
nice question.

Well, $\sin 0^o=0$, and since $180^o$ is on the opposite side of $0^o$, we say they are equal: $\sin 180^o=\sin 0^o=0$.  Since we have $\sin 180^o=0$ as one of the terms, our solution is $0$.  What I would like to know is what is the product of sine functions up to any $n<180^o$, such as $S=\sin 1^o\cdot \sin 2^o\cdot\sin 3^o\cdot...\cdot\sin 179^o$?

[quote="boxedexe"][quote="riddler"]just look on google, I know that stuff.[/quote]
Hmmm, that's odd...Oh, it's calculated in radians  Btw, I was just giving the solution.

Masoud Zargar[/quote]
oh yeah I didnt notice.

$\sin {180}=-0.801152636\\
\sin {180^\circ} = 0$

http://www.google.com/search?hl=en&lr=&q=sin+%28180+degrees%29&btnG=Search

[quote="Gyan"][quote="riddler"]:huh:  :huh:  :huh:  :huh:  :huh:  when you look in the google bar it says sin180 is something weird.[/quote]
It does give the correct result, remember $ 180^o$ is NOt the same as $180$ (radians) which google is using...
Try sin (pi) in google if you need 180 degrees.. (or  x*pi/180 - if you need x degrees)[/quote]

You said .. (or  x*pi/180 - if you need x degrees).  You're supposed to say, if you need (y) radians. Because,

$x \frac{\pi}{180} $ will give us x degrees in radians.

[quote="Gyan"][quote="boxedexe"]W  What I would like to know is what is the product of sine functions up to any $n<180^o$, such as $S=\sin 1^o\cdot \sin 2^o\cdot\sin 3^o\cdot...\cdot\sin 179^o$?

Masoud Zargar[/quote]... This is from your other thread ...:)

[hide="Hint"]Well if this was  $S=\sin 1^o\cdot \sin3^o\cdot\sin 5^o\cdot...\cdot\sin 179^o$
the answer would be[hide="big Hint "] $\frac 1 {2^{89}} $ [/hide][/hide][/quote]

lol, thanks, so if we say $n=\text{last sine function}$, our answer would be $\frac{1}{2^n}$, correct?

EDIT: no it doesn't work.  Your big hint doesn't work?  Are you sure that is it?

i compared the answer that you gave to the answer my calculator gave and they are very different. are you sure you are right?

Well, I don't know if you're referring to my post, but as you can see, I previously wrote that the hint is incorrect.

I can't believe I punched in all of those numbers..

But yes, that answer is correct. Makes sure to note that he is only doing the sine product of odd degrees...

Masoud, you're approaching this completely the wrong way.  The idea of this problem is to exploit the symmetry of trigonometric functions.

You're assuming that $\prod^{n}_{i=1} \sin i$ has a nice closed form for any $n$, but that is far too much to hope for.

Boxedexec : Yes   $x^o = x \frac {\pi}  {180} (radians) $
No the problem is not that easy , and yes I belive the answer given in the hint  is correct (Check it with a  calculator)

[hide] (Another  example try $sin 10^o sin 30^o sin 50^0 \dots sin 170^o $ and answer should  be $\frac 1 {2^8} $
Hope this gives a  big hint. .. anyway including all the terms make it  a little harder ...

By the way  this has been a favorite topic to give problems ; for example poblems similer to
  $sin 10^o sin 50^o sin 70^o $ (or) ($cos 20^o cos 40^o cos 80^o$ )  has appeared many times  just in AoPS forms!, not to mention in several  tests) (This is nothing but the the above problem) [/hide]

ok thanks for explaing that. the hint now makes more sense. i just didnt notice that he only had odd terms.

Now, would you please explain how they derive the use of $\frac{1}{2^x}$ in the product of odd sine functions? Thanks in advance.

I see no one jumping in so here is one  solution, Peek only if you  want to. Needless to say there are many different ways one can attack this. Here is a simple  one  that I like.
[hide]$\sin 1^o\cdot\sin 3^o\cdot\sin 5^o\cdots \sin179^o $
 $ = \sin 1^o\cdot\sin 3^o\cdot\sin 5^o\cdots \sin179^o \frac { \sin 2^o\cdot\sin 4^o\cdot\sin 6^o\cdots \sin178^o }{ \sin 2^o\cdot\sin 4^o\cdot\sin 6^o\cdots \sin178^o }$
$= \frac {\sin 1^o\cdot\sin 2^o\cdot\sin 3^o\cdots \sin179^o }{\sin 2^o\cdot\sin 4^o\cdot\sin 6^o\cdots \sin178^o}\ = \frac {\sin 1^o\cdot\sin 2^o\cdot\sin 3^o\cdots \sin179^o } {(2 \sin 1^o \cos 1^o)(2 \sin 2^o \cos 2^o) (2 \sin 3^o \cos 3^o) \cdots (2 \sin 89^o \cos89^o) }\\ = \frac {\sin 1^o\cdot\sin 2^o\cdot\sin 3^o\cdots \sin179^o } {2^{89}(\sin 1^o\cdot\sin 2^o\cdot\sin 3^o\cdots \sin 89^o) (\cos 1^o\cdot\cos 2^o\cdot\cos 3^o\cdots \cos89^o) }\ = \frac {\sin 1^o\cdot\sin 2^o\cdot\sin 3^o\cdots \sin179^o } {2^{89}(\sin 1^o\cdot\sin 2^o\cdot\sin 3^o\cdots \sin 89^o) (\sin 91^o\cdot\sin 92^o\cdot\sin 93^o\cdots \sin 179^o)}\ =\frac 1 {2^{89}}$

Hope this is fun.

There are a few  other methods which are also very different and nice and powerful. Needless to say  many problems having the same theme has appeared in many contests.. [/hide]

what about sin1 * sin 2 * sin3...... sin 179?

[quote="maokid7"]i would assume that there is also a patter for sin2sin4sin6... as well and prob with the same method show by Gyan.[/quote]

Okay no activity here, assuming there is still some interest, here is  a hint, (actually a solution:)

First one of my favorite formula:

$\sin (nx) = 2^{n-1} \sin (x) \sin (x+k) \sin (x+2k) \sin (x+3k) \dots \sin (x+(n-1)k)$ where $k = \pi/n$
Substiture,  $n=90$ so $k=2^o (= \pi/90)$ and  $x=1^o$  you get  one answer

$1=2^{89} \, \sin 1^o \, sin 3^o\, sin 5^o\,\dots \,sin 179^o$

Now for the second part ..

And if you put  $n=180$  and $x=A \ really \ very \ tiny \ \alpha \ which \ is \ really \ very \ small$  we get

$\frac{\sin (180 \alpha)} {sin \alpha} = 2^{179} \, \sin 1^o \, sin 2^o\, sin 3^o\,\dots \,sin 179^o$

$180 \cdot 2^{-179} = \sin 1^o \, sin 2^o\, sin 3^o\,\dots \,sin 179^o = \frac{45} {2^{177}}$

Hope that is fun..

nice. thanks i was trying to figure out a way to just do $sin2sin4sin6in8...sin178$ for a really long time.

i only have one problem w/ your proof i am not able to follow that last step. [quote]Now for the second part ..

And if you put n=180 and $x=A \ really \ very \ tiny \ \alpha \ which \ is \ really \ very \ small we get \frac{\sin (180 \alpha)} {sin \alpha} = 2^{179} \, \sin 1^o \, sin 2^o\, sin 3^o\,\dots \,sin 179^o 180 \cdot 2^{-179} = \sin 1^o \, sin 2^o\, sin 3^o\,\dots \,sin 179^o = \frac{45} {2^{177}}$[/quote]

i am hopeing that you will be able to explain this a little. o and in the first part i assuem that you set $x$ equal to 1.

Yes, $x=1^o$ for the first case (I edited it now).

You set  $x=\alpha$  and take the limit when $\alpha$ tends to zero ... for the second case.

$(\alpha + 1^o), (\alpha + 2^o)$ becomes $1^o , 2^o$ etc ..

and  limit of alpha tends to zero of $\frac{\sin (180 \alpha)} {\sin \alpha }$ is,of course, $180$

By the way, this is the source of many such problems ...

[quote="maokid7"]ok one last question i hope  :blush:
but y do u divide one side by $\sin\alpha$?[/quote]

i think it is because $\sin\alpha\rightarrow 0$ so if you leave it to multiply with other terms then they would all become $0$ and you cant find anything then .


[hide]The square numbers have to end in 4,5 or 1,8. I got lucky and tried 4,5 first. I got 28 and 35.[/hide]

[quote="vas2007"]can you help me in finding the solution of the equation in integers
                    $x^{2}+y^{2}=2009$.[/quote]

[hide]There are multiple answers to this problem.  $a$ could be either $28,35,-28, or-35$  The same goes with $b$, such that $b$ is a different integer than $a$[/hide]

If it's mathcounts, it usually asks for the sum... but this question has 12-ish answers?


In a very hotly fought battle, at least 70% of the combatants lost an eye, at least 75% an ear, at least 80% an arm, and at least 85% a leg. What can be said about the percentage who lost all four members?

Well, between 10% and 70% of the participants lost all four body parts. (30% didn't lose an eye, 25% didn't lose an ear, 20% didn't lose an arm, and 15% didn't lose a leg. 100 - (30 + 25 + 20 + 15) = 10)

actually it goes up to 100%, because all the stats were [i]at least[/i] quantities.

It also might be expected (if the losses are randomly distributed and not affected by whether or not someone already has an injury) that rougly 70%*75%*80%*85%=35.7% lost all four members.


Let $ p_n$ be the series of prime integers.

We can make a bijection between $ \mathbb{N}$ and an infinite sequence of $ \mathbb{N} \cup \{ 0 \}$ as follows: $ x \equal{} \prod_{j \equal{} 1}^{\infty} p_j ^{a_j}$

Define $ S \subset \mathbb{N} \; : \; x \in S \implies a_j < 10 \forall j$.

Then, each integer in $ S$ corresponds with one infinitely long list of digits: $ (a_1, a_2, \ldots)$, and vice versa.

Each infinitely long list of digits corresponds with a number between $ 0$ and $ 1$: $ 0. \overline{a_1 a_2 \ldots}$, and vice versa.

We can draw a bijection between the interval $ [0,1)$ and the real numbers.

Hence we can draw a bijection between a subset of the natural numbers and the set of real numbers.

[b]Edit:[/b] [i]Thanks JBL  :oops: [/i]

[quote="The Zuton Force"]We can make a bijection between $ \mathbb{N}$ and an infinite sequence of $ \mathbb{N} \cap \{ 0 \}$ as follows: $ x \equal{} \prod_{j \equal{} 1}^{\infty} p_j ^{a_j}$[/quote]

What's the image of the sequence $ (1, 1, 1, 1, 1, \ldots )$ under this bijection?  ;)  (Unless you really did mean $ {\bf N} \cap \{0\}$ instead of $ {\bf N} \cup \{0\}$.)

Oh wow, ok, thanks.

So that's a bijection between a subset of N and a subset of Q then... ok that makes more sense.


group theory
abstract algebra
geometry
geometric transformation
reflection
puzzles

let G denote the set of non-zero real numbers. for a and b in G define a*b=

ab if a>0
a/b if a <0

(G,*) is a group now using the group of this problem let J={ 3^n| n in the integers} U {-3^n| n in the integers}

Prove that J is a subgroup of G under *.

This is seriously out of place.

I'll answer anyway.
One way to think of $G$ is as the dihedral group on $\mathbb{R}^+$, the group of positive reals under multiplication. This is generated by the "translations" $x$ with $x>0$ and the reflection $-1$. Each negative element is its own inverse.

As for $J$ being a subgroup- there's only a few things to check.
It turns out that $J$ is the dihedral group on the infinite cyclic group of powers of 3.


inequalities
inequalities unsolved
Ukraine

It is not hard:
$\sum \frac{a^3}{b+2c}=\sum \frac{a^4}{ab+2ac} \geq \sum [(a^2+b^2+c^2)\frac{a^2+b^2+c^2}{3(ab+bc+ca)}]\geq \frac{a^2+b^2+c^2}{3}$
Because $a^2+b^2+c^2 \geq ab+bc+ca$

WLOG, let us assume that $a\leq b\leq c.$ From here we get  $a^{3}\leq b^{3}\leq c^{3}$ and $\frac{1}{b+2c}\leq\frac{1}{c+2a}\leq\frac{1}{a+2b}.$ Applying Chebyshef's inequality we have:

 \[ \frac{a^{3}}{b+2c}+\frac{b^{3}}{c+2a}+\frac{c^{3}}{a+2b}\geq(\frac{a^{3}+b^{3}+c^{3}}{3})(\frac{1}{b+2c}+\frac{1}{c+2a}+\frac{1}{a+2b}).  \]

  But, now using again Cauchy( or whatever you like  :D ) we have:

                 \[ \frac{1}{b+2c}+\frac{1}{c+2a}+\frac{1}{a+2b}\geq\frac{9}{3(a+b+c)}=\frac{3}{a+b+c}.  \]
  Now, we only need to prove that:

                 \[ (a+b+c)(a^{2}+b^{2}+c^{2})\leq 3(a^{3}+b^{3}+c^{3}),  \]

  which is Chebyshef. Done! :)

[quote="Cezar Lupu"]WLOG, let us assume that $a\leq b\leq c.$ From here we get  $a^{3}\leq b^{3}\leq c^{3}$ and $\frac{1}{b+2c}\leq\frac{1}{c+2a}\leq\frac{1}{a+2b}.$[/quote]

This is not true if $2b>a+c$. Take for example $a=1, b=3, c=4$. Then the last inequality is wrong and the solution is not valid.

Sorry to bring this up after so long, but I have just made similar mistake in another inequality. It is easy to make such a mistake, because
$\frac{1}{b+c}\leq\frac{1}{c+a}\leq\frac{1}{a+b}$ is true.

[quote="zibi"]$a,b,c >0$ Prove that :
\[ \frac{a^{3}}{b+2c}+\frac{b^{3}}{c+2a}+\frac{c^{3}}{a+2b}\geq\frac{a^{2}+b^{2}+c^{2}}{3} \][/quote]
The following inequality are also true.
1. Let $a$, $b$ and $c$ be positive numbers. Prove that:
\[ \frac{a^{3}}{b+2c}+\frac{b^{3}}{c+2a}+\frac{c^{3}}{a+2b}\geq\frac{a^{3}+b^{3}+c^{3}}{a+b+c} \]

2. Let $a$, $b$ and $c$ be positive numbers such that $a^5+b^5+c^5=3$.  Prove that:
\[ \frac{a^{3}}{b+2c}+\frac{b^{3}}{c+2a}+\frac{c^{3}}{a+2b}\geq1 \]

I am not sure about this one:
[quote="arqady"]
1. Let $a$, $b$ and $c$ be positive numbers. Prove that:
\[ \frac{a^{3}}{b+2c}+\frac{b^{3}}{c+2a}+\frac{c^{3}}{a+2b}\geq\frac{a^{3}+b^{3}+c^{3}}{a+b+c} \]
[/quote]
$\sum \frac{a^3}{b+2c}=\sum \frac{a^6}{a^3b+2a^3c}$
Now we use Cauchy Schwarz to get:
$\left(\sum(a^3b+2a^3c)\right)\left(\sum \frac{a^6}{a^3b+2a^3c}\right)\ge (a^3+b^2+c^3)^2$
Hence,
$\sum \frac{a^3}{b+2c}\ge \frac{(a^3+b^2+c^3)^2}{\left(\sum(a^3b+2a^3c)\right)}$
Now, by Vasc's known inequality(I may be wrong here)
$\frac{(a^2+b^2+c^2)^2}{3}\ge (a^3b+b^3c+c^3a)\text{ and }\frac{2}{3}.(a^2+b^2+c^2)^2\ge 2(ab^3+bc^3+ca^3)\implies (a^2+b^2+c^2)^2\ge \left(\sum(a^3b+2a^3c)\right)$
We get,
$\sum \frac{a^3}{b+2c}\ge \frac{(a^3+b^2+c^3)^2}{\left(\sum(a^3b+2a^3c)\right)}\ge \frac{(a^3+b^2+c^3)^2}{(a^2+b^2+c^2)^3}$
Hence it remains to prove $\frac{(a^3+b^2+c^3)^2}{(a^2+b^2+c^2)^2}\ge \frac{a^{3}+b^{3}+c^{3}}{a+b+c}$
Which is obviously true by Cauchy Schwarz:
$(a^3+b^3+c^3)(a+b+c)\ge (a^2+b^2+c^2)^2$
arqady, please check my proof.

[quote="nicusorz"]1.\[ \frac{a^{3}}{b+2c}+\frac{b^{3}}{c+2a}+\frac{c^{3}}{a+2b}\geq\frac{a^{2}+b^{2}+c^{2}}{3} \]

Using C-B-S, we have :
  $\sum \frac{a^{3}}{b+2c}=\sum \frac{a^{4}}{ab+2ac}\geq \frac{(\sum a^{2})^{2}}{3\sum ab}\geq \frac{\sum a^{2}}{3}\Leftrightarrow \sum a^{2}\geq \sum ab   $[/quote][quote="Gibbenergy"]It is not hard:
$\sum \frac{a^3}{b+2c}=\sum \frac{a^4}{ab+2ac} \geq \[(a^2+b^2+c^2)\frac{a^2+b^2+c^2}{3(ab+bc+ca)}\geq \frac{a^2+b^2+c^2}{3}$
Because $a^2+b^2+c^2 \geq ab+bc+ca$[/quote]
same solutions

[quote="Ashutoshmaths"]
arqady, please check my proof.[/quote]
Your proof is true and nice!

[quote="Ashutoshmaths"]I am not sure about this one:
[quote="arqady"]
1. Let $a$, $b$ and $c$ be positive numbers. Prove that:
\[ \frac{a^{3}}{b+2c}+\frac{b^{3}}{c+2a}+\frac{c^{3}}{a+2b}\geq\frac{a^{3}+b^{3}+c^{3}}{a+b+c} \]
[/quote]
$\sum \frac{a^3}{b+2c}=\sum \frac{a^6}{a^3b+2a^3c}$
Now we use Cauchy Schwarz to get:
$\left(\sum(a^3b+2a^3c)\right)\left(\sum \frac{a^6}{a^3b+2a^3c}\right)\ge (a^3+b^3+c^3)^2$
Hence,
$\sum \frac{a^3}{b+2c}\ge \frac{(a^3+b^3+c^3)^2}{\left(\sum(a^3b+2a^3c)\right)}$
Now, by Vasc's known inequality(I may be wrong here)
$\frac{(a^2+b^2+c^2)^2}{3}\ge (a^3b+b^3c+c^3a)\text{ and }\frac{2}{3}.(a^2+b^2+c^2)^2\ge 2(ab^3+bc^3+ca^3)\implies (a^2+b^2+c^2)^2\ge \left(\sum(a^3b+2a^3c)\right)$
We get,
$\sum \frac{a^3}{b+2c}\ge \frac{(a^3+b^3+c^3)^2}{\left(\sum(a^3b+2a^3c)\right)}\ge \frac{(a^3+b^3+c^3)^2}{(a^2+b^2+c^2)^2}$
Hence it remains to prove $\frac{(a^3+b^3+c^3)^2}{(a^2+b^2+c^2)^2}\ge \frac{a^{3}+b^{3}+c^{3}}{a+b+c}$
Which is obviously true by Cauchy Schwarz:
$(a^3+b^3+c^3)(a+b+c)\ge (a^2+b^2+c^2)^2$
arqady, please check my proof.[/quote]
Now is right
$\sum \frac{a^3}{b+2c}\ge \frac{(a^3+b^3+c^3)^2}{(a^2+b^2+c^2)^2}\ge\frac{a^{3}+b^{3}+c^{3}}{a+b+c}$

[quote="nikoma"][hide]
It's equivalent to \[\sum_{cyc} (6a^5 + 10a^4b) \geq \sum_{cyc} (4a^3b^2 + 2a^2b^3 + ab^4 + 6a^2b^2c + 3a^3bc)\] which is bunch of AM-GMs
[/hide][/quote]
@nikoma..which one are you talking about?
$\frac{(a^3+b^3+c^3)^2}{(a^2+b^2+c^2)^2}\ge\frac{a^{3}+b^{3}+c^{3}}{a+b+c} $
is just C-S.

I'm talking about the main inequality of this thread (Yes I know it has great CS solutions.)


[quote="Tiger100"]if $ xyz \equal{} (1 \minus{} x)(1 \minus{} y)(1 \minus{} z)$ where $ 0 \le x,y,z \le 1$ . Prove that:

$ x(1 \minus{} z) \plus{} y(1 \minus{} x) \plus{} z(1 \minus{} y) \ge \frac {3}{4}$[/quote]
If $ xyz \equal{} 0,$ so the inequality is obviously true.
Let $ xyz\neq0$,  $ \frac {1 \minus{} x}{x} \equal{} a$, $ \frac {1 \minus{} y}{y} \equal{} b$ and $ \frac {1 \minus{} z}{z} \equal{} c$.
Hence, $ abc \equal{} 1$ and $ x(1 \minus{} z) \plus{} y(1 \minus{} x) \plus{} z(1 \minus{} y) \ge \frac {3}{4}\Leftrightarrow\sum_{cyc}(a \plus{} ab \minus{} 2),$
which is true by AM-GM.

[quote="Tiger100"]if $ xyz \equal{} (1 \minus{} x)(1 \minus{} y)(1 \minus{} z)$ where $ 0 \le x,y,z \le 1$ . Prove that:

$ x(1 \minus{} z) \plus{} y(1 \minus{} x) \plus{} z(1 \minus{} y) \ge \frac {3}{4}$[/quote]
Is this correct? -
set :
$ x \equal{} \frac {a}{a \plus{} b}$
$ y \equal{} \frac {b}{b \plus{} c}$
$ c \equal{} \frac {c}{c \plus{} a}$

the inequality beomes -
$ \sum_{cyc} \frac {a^2}{(a \plus{} b)(a \plus{} c)} \ge \frac {(a \plus{} b \plus{} c)^2}{a^2 \plus{} b^2 \plus{} c^2 \plus{} 3(ab \plus{} bc \plus{} ca)} \ge \frac {3}{4}$
the first inequality is C-S,
the second inequality is equivalent to -
$ a^2 \plus{} b^2 \plus{} c^2 \ge ab \plus{} bc \plus{} ca$ which is true by AM-GM. :P
equality for $ a\equal{}b\equal{}c$ or $ x\equal{}y\equal{}z\equal{}\frac{1}{2}$


[quote="Makaveli"]x,y and z are positive reals with sum 1. Prove that
(x/(y^2+z))+(y/(z^2+x))+(z/(x^2+y))>=9/4[/quote]
$\sum_{cyc}\frac{x}{y^{2}+z}=\frac{x^{2}}{y^{2}x+xz}\geq\frac{(x+y+z)^{2}}{xy+xz+yz+x^{2}z+y^{2}x+z^{2}y}.$
Thus, remain to prove that
$4(x+y+z)^{3}\geq9(x+y+z)(xy+xz+yz)+9(x^{2}z+y^{2}x+z^{2}y).$
But $4(x+y+z)^{3}\geq9(x+y+z)(xy+xz+yz)+9(x^{2}z+y^{2}x+z^{2}y)\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}\left(4x^{3}+3x^{2}y-6x^{2}z-xyz\right)\geq0\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(x^{3}-xyz)+3\cdot\sum_{cyc}(x^{3}+x^{2}y-2x^{2}z)\geq0\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(x^{3}-xyz)+3\cdot\sum_{cyc}(y^{3}+x^{2}y-2y^{2}x)\geq0\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(x+7y+z)(x-y)^{2}\geq0.$


It is well known that D^2 is not retractable to S^1, i.e., a disk is not retractable to a circle. However, I found the following map f: D^2 -> S^1, which seems to be a retraction. Anything wrong here?

Suppose we have (x, y) in D^2, then f(x, y) = (R*x/sqrt(x^2+y^2), R*y/sqrt(x^2+y^2)) if (x, y) <> (0, 0); and f(x, y) = (R, 0) if (x, y) = (0, 0).

[quote="t0rajir0u"]You don't have continuity at the origin.  What happens when you approach it from different directions?[/quote]

I see, a retraction has to be a continuous map.


inequalities
trigonometry
geometry proposed
geometry

[quote="manlio"]For $ A,B,C$ angles of a triangle

$ \sum sinA\geq 2\sqrt {3}(\sum sin\frac {A}{2}sin\frac {B}{2})$[/quote]
$ <\equal{}> \frac{1}{\sin(\frac{A}{2})}\plus{}\frac{1}{\sin(\frac{B}{2})}\plus{}\frac{1}{\sin(\frac{C}{2})} \le \frac{2s}{r\sqrt{3}}$ ($ s$ - semiperimetre )

$ \sin(\frac{A}{2})\equal{}\frac{r}{\sqrt{r^2\plus{}x^2}}$ .... $ (s\equal{} x\plus{}y\plus{}z)$

$ <\equal{}> \sqrt{r^2\plus{}x^2}\plus{}\sqrt{r^2\plus{}y^2}\plus{}\sqrt{r^2\plus{}y^2} \le \frac{2}{\sqrt{3}}s$
$ sr^2 \equal{} xyz$

$ \sqrt{r^2\plus{}x^2}\plus{}\sqrt{r^2\plus{}y^2}\plus{}\sqrt{r^2\plus{}y^2} \equal{} \sqrt{x}\sqrt{\frac{yz}{s}\plus{}x} \plus{}\sqrt{y}\sqrt{\frac{zx}{s}\plus{}y}\plus{}\sqrt{z}\sqrt{\frac{xy}{s}\plus{}z} \le \sqrt{(x\plus{}y\plus{}z)(\frac{xy\plus{}yz\plus{}zx}{s}\plus{}x\plus{}y\plus{}z)} \le \frac{2}{\sqrt{3}}(x\plus{}y\plus{}z)$

Thank you very much, Sergic.

Can you please remember me the formula for $ \sum sin\frac {A}{2}sin\frac {B}{2}$ in term of $ R,r,s$ and how to deduce it.

Thank you very much.


geometry
perimeter
FTW
area of a triangle
Heron\u0027s formula
Pythagorean Theorem

What is the number of square units in the area of a triangle
whose sides measure 5, 5 and 6 units?

We draan altitude from where 5 and 5 meet to 6, forming two 3-4-5 triangles, or one 3 x 4 rectangle. Thus, the area is 3 x 4 =12.

[asy]size(1.5inch);
draw((0,0)--(6,0)--(3,4)--(0,0));
draw((3,0)--(3,4));[/asy]

the base is 6, so the half of it is 3.
the hypotenuse is 5, so altitude is 4 by pythagorean theorem.
$ 6\times4\div2 \equal{} 12$

answer : 12

EDIT : mehh beaten to it

Or you could use Heron's.

$ A \equal{} \sqrt{s(s\minus{}a)(s\minus{}b)(s\minus{}c)}$

Note: In Heron's formula, $ s$ is the semi-perimeter, and $ a$, $ b$, and $ c$ are the side lengths. There's two nice proofs on Wikipedia, one that uses the Pythagorean Theorem, and one that uses some trigonometry.

EDIT: Fantasy, please don't spam. This was just a side note, that you could also use Heron's Formula. It's good to know different ways to solve a problem. Locked.

......

ernie, if you were playing FTW and you got this problem, would you use heron's formula rather than just drawing a simple picture?


analytic geometry
geometry
geometry proposed

You are given circumscribed square ABCD, where coordinates of vertexes are B(9,9) and D(-1,3). Find centre of circle(coordinates O(x,y))?

 [geogebra]d4fff6de077adb26020e789dd50018dc0513d423[/geogebra]

Please solve!!!!!!!!!!

Don't you think the centre is the midpoint of $ \overline {BD}$?

Best regards,
sunken rock

If midpoint of BD will be centre of circle, AB=10 and BC=6
So it will not be a square

If the midpoint of BD is not the center of the incircle, you have some wrong calculation then!
You should get $ O(4,7)$, while $ A$ and $ C$ will be the solutions of the equations $ 5x\plus{}3y\equal{}41$ and $ (x\minus{}4)^2\plus{}(y\minus{}7)^2\equal{}34$.

Best regards,
sunken rock


I'm confused  :(
Is $ u_n$ a previously defined sequence, are $ u_n$ and $ x_n$ the same sequence, is $ x_n$ previously defined, or what?
Please clarify

[quote="Victory.US"][quote="Victory.US"]let $ u_n$ given by $ u_0 \equal{} a \ge 0$ and $ u_{n \plus{} 1} \equal{} \frac {1}{2 \minus{} \sqrt u_n}$

determine $ a$ such that  $ u_n$ has limit.[/quote]

i checked it. now who have idea?

no one/? :)[/quote]

Let $ f(x)\equal{}\frac 1{2\minus{}\sqrt x}$

It's easy to see that, for $ x\in[0,4)$, $ f(x)$ is strictly increasing and that :

$ \forall x\in[0,1)$, $ f(x)>x$
$ f(1)\equal{}1$
$ \forall x\in(1,\frac{3\plus{}\sqrt 5}2)$, $ f(x)<x$
$ f(\frac{3\plus{}\sqrt 5}2)\equal{}\frac{3\plus{}\sqrt 5}2$
$ \forall x\in(\frac{3\plus{}\sqrt 5}2,4)$, $ f(x)>x$

And so :
If $ a\in[0,\frac{3\plus{}\sqrt 5}2)$, $ u_n$ has a limit $ 1$
If $ a\equal{}\frac{3\plus{}\sqrt 5}2$, $ u_n$ has a limit $ \frac{3\plus{}\sqrt 5}2$
If $ a>\frac{3\plus{}\sqrt 5}2$, $ u_n$ is not defined for any $ n$ (for $ n$ great enough, $ u_n\geq 4$ and $ u_{n\plus{}1}$ or $ u_{n\plus{}2}$ is undefined).

Hence the answer : $ \boxed{a\in[0,\frac{3\plus{}\sqrt 5}2]}$


trigonometry
AMC
AIME
inequalities
trig identities
Law of Cosines
AMC 10

How hard was it in comparison to previous years? i think it was pretty easy (even though i failed)

[color=green]Added poll. -mod[/color]

Well, this year's #25 didn't have a super easy solution: last year's #25 could have been solved with law of cosines in no time at all.
Overall, it's about the same

I thought this AMC 10 was extremely easy. For me, #25 was possibly the easiest #25 ever, if you took away the answer choices. I don't see what's so tricky about it...

I didn't think it was that difficult. I doubt the cutoff will be under 120.

i thought it was extremely easy compared to previous years (last year i got a 126, this year, a 144...then again, i was stupid last year...)

cut off will probably be 120 on the dot.

the only amc's i've ever taken besides this one were the 2004 and 2003 amc10b's
this one was significantly easier
cutoff will probably ridiculously high though...
i hope the aime is hard.

[quote="funtwo"]cutoff will probably ridiculously high though...[/quote]

It's always $ 120$ [b]or[/b] top $ 1\%$.

[quote] It's always  or top . [/quote]

...nope, check amc 10 b 2003....[/quote]

Hm...I thought it was quite a bit easier than past years...but that was mainly just the last problems in general...like #1-20 were about the same or maybe a little harder than usual in general

It seemed like trickiness was quite common in this test (#12 (triangle inequality) and 22 (eliminate dice)) for example.

I think the test wasn't that hard, or that super easy. The AIME cutoff is probably still going to be 120.

[quote="DataBox"]...nope, check amc 10 b 2003....[/quote]

Umm... what's your point? Qualification was different then. Under the current rules, it's 1% or 120.

I think it was like super easy compared to last year (even tho I still failed :-/).

But maybe that's cuz I was SUPER SUPER nubbish last year. xD


Let c be a circle and ABCD be a quadrilateral inscribed in c (i.e. A,B,C and D are point on the circumference of c). Let P, Q and R be the intersection points of (AB,DC), (AD,BC) and (AC,BD) rispettivamente. Prove that OR is orthogonal to PQ, where O is the center of c.

PS
The proof is straightforward with use of some results of projective geometry. I have found  (and I ask you to find)  a classical euclidean proof.

[quote="sprmnt21"]Let c be a circle and ABCD be a quadrilateral inscribed in c (i.e. A,B,C and D are point on the circumference of c). Let P, Q and R be the intersection points of (AB,DC), (AD,BC) and (AC,BD) rispettivamente. Prove that OR is orthogonal to PQ, where O is the center of c.[/quote]

Indeed, the advanced proof of this is on http://mathlinks.ro/viewtopic.php?t=5326 . I remember a very beautiful but quite long elementary proof by Juan Carlos Salazar.

  Darij

Indeed, the advanced proof of the statment is due to the fact(*)  that R is the pole of the polar  PQ wrt circle c. Then by (one of possible) definition of pole OR is orthogonal to PQ.

(*)

for this fact see at

http://www.maths.gla.ac.uk/~wws/cabripages/quad.html

Remark.

From the facts illustrated at this page follow quite strightforward that the centers of the inscribed and circumscribed circles of a bicentric quadrilateral are collinear with the intersection of diagonals.

Could you, Darij, give exact reference of the Salazar's proof, please ?

Here

http://olimpiadi.ing.unipi.it/modules.php?op=modload&name=Forums&file=viewtopic&topic=2374&forum=5&start=20

you can find my proof.

see th.5 of the message posted  in   04-06-2004 h 11:44

unfortunately is in italian. But, if necessary, I can try to translate in inglish (or some think like inglish)  and post here.


What is the smallest value of $x$ such that $|5x-1|=|3x+2|$? Express your answer as a common fraction.

[hide]
There are 2 possibilities:
1. $5x-1=3x+2 \Longrightarrow 2x=3 \Longrightarrow x=\frac{3}{2}$
2. $5x-1=-(3x+2) \Longrightarrow 5x-1=-3x-2 \Longrightarrow 8x=-1 \Longrightarrow x=-\frac{1}{8}$
So the answer is $\boxed{-\frac{1}{8}}$ :D [/hide]

You could just graph it and find the intersections :P.

[hide]Two possibilities:

$1-5x=3x+2\implies 8x=-1\implies x=-\frac{1}{8}$
$5x-1=3x+2\implies 2x=3\implies x=\frac{3}{2}$.

-1/8 is clearly the smaller.[/hide]

[quote="samath"][quote="cat810930"] I  got -1/8 too.
I did it a little differently and I would like to show my work, but I don't know how to download LaTeX...
Help!!Anyone?? [/quote]

You don't need to download LaTex.  Just type it in.  Use the link on the side of this page to see how to use it.[/quote]

All you have to do is put your code in between two dollar signs.


Let $ p$ be a given constant number. Find $ x_n$ such that $ x_{n\plus{}1}\equal{}px_n\plus{}1,\ x_1\equal{}0\ (n\equal{}1,\ 2,\ \cdots)$.

We have $ x_n\equal{}n\minus{}1$ iff $ p\equal{}1$ and $ x_n\equal{}\frac{p^{n\minus{}1}\minus{}1}{p\minus{}1}$ otherwise...


Express the base 7 repeating decimal 0.023023023023... as a base 10 fraction in lowest terms.

Solution in spoiler:

[hide]0.023023023023..._7 is found in a similar fashion as a repeating base 10 fraction is found.  We let x=0.023023023023..._7, and 10000_7x=230.023023023023..._7, so 6666_7x=230_7, and so x=230_7/6666_7=119/2400.

I hope this is correct.

[/hide]

-interesting_move

Be careful of your multiplication, there -- the idea is right, the execution is a bit off.

Just to lay out the problems you all faced:

Interesting_Move multiplied wrong, introducing an extra 0 into the number.

H_s_potter assumed that $1000_7 - 1_7 = 999_7$.

Soulsmischief realized that 1000 - 1 = 666, but then didn't convert his answer out of base 7 -- in fact, $\frac{23_7}{666_7} = \frac{17_{10}}{342_{10}}$, the correct answer.

Could someone please show me how to do the question backwards so
Express the base 10 fraction 17/342 as a base 7 repeating decimal

Any way you do it is going to amount to base-7 long division:

[hide]For example, simply start subtracting powers of 1/7.  You notice, for example, that 1/7 is too big, but that 2/49 is small enough and that then 3/343 is small enough but then that 1/7^4 is too big and so on.  You notice after 4 or 5 terms that it the remainders have begun to repeat -- they have to, since it is a rational number.  Then you get the repeating decimal you need.

Alternatively, convert them into base-7 numbers and try long-dividing.  17 = 23 and 342 = 666, and you could note right away that, analogous to base-10, when you divide something by 666....666 in base 7, you just get that thing (perhaps padded with 0's) repeated.  Alternatively, there are 0 666's in 230, subtract, and there are 2 666's in 2300, subtract, and there are 3 666's in 3020 and so on, and you again find the repeating remainders.  I think it's easier the first way, personally, as you get to operate in base 10.  They are essentially the same method, of course. [/hide]


geometry
circumcircle
symmetry
geometry unsolved

Hi guys,

I got stuck in this question: A straight line through the circumcenter of ABC intersects AB, BC, CA at R, Q, P. Show that circles diameter AP, BQ, CR have 2 common points, one on circumcircle and one on the nine-point circle. And also show that the common chord passes through H. I have done almost except that the second common point is on the nine-point circle. Anyone can help?

Thanks,

Jas.

If $D,E,F$ are the feet of the $A,B,C$-altitudes of $ABC$ respectively, then the power of $H$ wrt the circles with diameters $AP,BQ,CR$ is $HA\cdot HD=HB\cdot HE=HC\cdot HF$. In other words, $H$ has the same power wrt our three circles. If we show that the three circles are concurrent in a point $S$ on the circumcircle (denoted by $(ABC)$ from now on), the rest follows: the three circles are concurrent in one point and have collinear centers (the Newton line of the quadrilateral $ABST$), so they are actually concurrent in two points, $P$ and $Q$. Now, $H$ lies between $S$ and $T$ and $HS\cdot HT=HA\cdot HD$, which means that $T$ lies on the image of $(ABC)$ through the inversion of pole $K$ and negative power $-HA\cdot HD$, and this image is the nine-point circle (because the mentioned inversion takes $A,B,C$ onto $D,E,F$ respectively).

The above means that all we have to show now is that the three circles meet in some point on $(ABC)$. In order to achieve this, it suffices to show that the circles with diameters $BQ,CR$ (which we denote by $(BQ),(CR)$) and $(ABC)$ are concurrent (then, by symmetry, the pairs $(AP),(BQ)$ and $(AP),(CR)$ also meet on $(ABC)$, and it follows from the fact that $(AP),(BQ),(CR)$ are coaxal that the three points on $(ABC)$ must actually coincide).

Let $B',C'$ be the antipodes of $B,C$ in $(ABC)$, and let $U=B'Q\cap C'R$. By applying the converse of Pascal’s Theorem for the hexagon $BC'UB'CA$, we conclude that $U$ lies on the circumcircle of $ABC$, or, in other words, $\angle B'UC'=\angle BAC\ (*)$.

If $O,O_B,O_C$ are the centers of $(ABC),(BP),(QR)$ respectively, and $BU_B,CU_C$ are the common chords of the pairs $(ABC),(BP)$ and $(ABC),(QR)$, we have, on the one hand, $\angle O_BOO_C=\pi-\angle(BU_B,CU_C)$, and, on the other hand, $\angle OO_BO_C=\angle B'UC'$, because $OO_B\|UB',OO_C\|UC'$. Now $(*)$ shows that $\angle(BU_B,CU_C)=\pi-\angle BAC$, and we get $U_B=U_C$, both on $(ABC)$, as desired.


In a $ \text{\LaTeX}$ document, suppose you are in text mode and you say something like "20th century" or "23rd place" only you want it to look something like "$ 20^{\text{th}}$ century" or "$ 23^{\text{rd}}$ place"?

(Only that doesn't look quite right. Probably not the right spacing and shape.)

Some word processors do this automatically. What's the appropriate syntax in $ \text{\LaTeX}$?

How about 20\textsuperscript{th} century which will look like something $ \text{20\textsuperscript{th} century}$ though it looks better in a document?

You may also prefer
\usepackage{engord}
...
\engordnumber{20} century

Thanks. I tried both in a document. Either the results are identical or the differences are so slight that I couldn't spot them at 400% magnification. So I went with \textsuperscript.


percent Support Harvard college search Princeton Yale

Anyone have a list of good scholarships or would like to recommand a book on them?

I happen to realize that a disproportionate number of folks on this site go to or plan on going to a top-notch school, the expensive kind ;-)

One thought about scholarships is that middle-class kids who apply to top colleges should make sure to apply to "need-blind" schools that promise to meet 100 percent of an applicant's demonstrated financial need. That promise may be fulfilled partly by loans, or partly by term-time work, but at least what you pay out of pocket should be bearable if both you and your parents do their part. You can get an idea how one of those promises (a no-loan support of 100 percent of need promise) works by plugging in your family's numbers into [url=http://www.princeton.edu/pr/aid/estim.shtml]Princeton's financial aid estimator[/url].

Of course cool extracurricular programs like the science fairs and some math competitions also have scholarships as prizes, but I wouldn't focus on that aspect as a reason for participating in them.

I'm applying need-blind, but also some schools have merit-based scholarship apps, Wash U and U Chicago in particular.

[quote="mirebeka"]I'm applying need-blind[/quote]

Oh actually that brings up a question. If you apply to a need-blind school (say, Harvard) then you don't have to worry about getting scholarships?

[quote="MithsApprentice"]Oh actually that brings up a question. If you apply to a need-blind school (say, Harvard) then you don't have to worry about getting scholarships?[/quote]

What SOME colleges say they will do, for applicants who are admitted, is meet 100 percent of the admitted student's calculated financial need. How much a student's financial need is figured to be depends on the family's assets and income. You can read about some of the policies involved in a [url=http://www.collegeboard.com/prod_downloads/highered/fa/Economics-Primer-2004.pdf]document prepared for financial aid officers[/url] by the College Board. You can read about Harvard's new financial aid initiative on Harvard's [url=http://www.news.harvard.edu/gazette/daily/0402/28-finaid.html]Web section about the initiative[/url].

In most cases, having a third-party scholarship means that your third-party scholarship is just one more part of your financial aid package: you have to pay the same amount out-of-pocket regardless of whether you get school aid (discounts from list price) or third-party aid. But how third-party aid is treated varies from school to school, so don't be afraid of winning the Science Talent Search or some other prestigious program that includes scholarship money as part of the prize. If your family is prosperous, you will have to pay some money out of pocket--but surprisingly rich families can still be eligible for partial financial aid.

One of the best reasons to apply to more than one college is to be able to compare financial aid offers. Make sure you get all the financial aid forms together the year you apply to college, and see what the offers look like before deciding where to go.

Hope this helps!

To answer the original question,

1. The bad news is that those scholarship books are a waste
of time.  The scholarships are mostly small sums of money
with restrictive eligibility criteria. It takes time to apply, too;
your expected gain is probably higher if you just apply that
time elsewhere.  If you are an active member
in some religious, ethnic or local community you may have a shot
(e.g. a Korean churchgoer in Texas), but you'd have probably
heard of those scholarships through the relevant grapevine
anyway, as something that older students received in the
last few years.

2. The good news is that if you're at or near the top of
the U.S. high school graduates in your year (e.g. math
olympiad summer attendee, IMO/Westinghouse champion,
published papers in a refereed journal) then there is a ton
more money available than used to be the case.
Harvard, Princeton and Yale have been reducing
or eliminating student loans.   Harvard and Princeton have
extra money for admittees in the top group, over and above
financial aid. There are special "honors"
programs at some state schools (some with full tuition + stipend),
a Gates-funded program at Duke (several Putnam/IMO types
have gone there), and I think there was a math scholarship
at Wash U St. Louis.

3.  There are some standard scholarships that begin with the
PSAT/SAT, i.e. National Merit scholarship and Presidential
Scholarship, where you will be notified of eligibility as a result
of the test scores.  The presidential one has a lot of money attached.

It's not unknown for, say, Harvard to confiscate the entirety
of a five-figure Westinghouse prize in the name of "financial aid".

If your family is low-income, don't buy the hype about saving
for college.   In all likelihood you'd just be subsidizing the
school.


geometry
ARML
probability
email
AwesomeMath
summer program
calculus

This is Bowei
Bowei was stupid and accidentally unsubscribed from woot
Bowei needs help getting back to woot
please help bowei

Yes bowei knows bowei sounds like ubemaya
ubemaya and other aops-savvy people please help bowei

thank you

i think you should ask one of the admins instead of the bay area forum which is pretty much composed of spammers(me, davids, i like pie, etc ) lurkers(ccy, bookaholic) and self proclaimed stupids(ubemaya)

it magically got fixed

but now ive got another problem
you know when you got to forum home it lists the big forums and then sub-forums and then subsubforums like national and local communities -> north america -> bay area
well i accidentally closed the "online school" forum and i cant see anything under it anymore

any ideeas on how to fix this?

thx

SPAMMERS! LURKERS! Self-proclaimed stupids?  :huh:

This forum does kinda suck... nothing really ever happens. Just spam.

[quote="i_like_pie"][quote="junggi"]spammers(me, davids, i like pie, etc ) lurkers(ccy, bookaholic) and self proclaimed stupids(ubemaya)[/quote]
WTF :?:

First off, Ubemaya isn't stupid, and I think we all know that. The middle category I don't know much about and won't say anything. But spammers? Seriously... The SFBA forum might be filled with spam, but don't call me a spammer.[/quote]
hm
you used to spam a lot
havent seen your posts much lately
and i said ubemaya is self proclaimed stupid
not stupid

yeah this forum sucks a lot

[quote="bookaholic"]Don't worry, I'll do my best to get Paly and JLS to take this forum over...

Bowei, did you solve your problem?

Random question: does anyone know where to order T-shirts? Like, where did the ARML organizers go to for team T-shirts?[/quote]
oh god paly t shirts

[quote="junggi"]havent seen your posts much lately[/quote]
Then apparently you don't look in the mathematical parts of this forum.

Anyway, do you want those t-shirts to say something like "we're from paly, and we'll own u @ cubing"?

If so, I might (with probability 0.01%) get back into cubing and own them. :P

[quote="bookaholic"]
Random question: does anyone know where to order T-shirts? Like, where did the ARML organizers go to for team T-shirts?[/quote]

I don't know; Ted takes care of that.  You can always do Cafepress, but I think unless you're doing a large order, you get substandard quality for a high price.  I'd email Ted and ask him.

[quote="i_like_pie"][quote="junggi"]havent seen your posts much lately[/quote]
Then apparently you don't look in the mathematical parts of this forum.
[/quote]
oh snap
that is pretty true
i go there like twice a week

[quote="bookaholic"]Don't worry, I'll do my best to get Paly and JLS to take this forum over...

Bowei, did you solve your problem?

Random question: does anyone know where to order T-shirts? Like, where did the ARML organizers go to for team T-shirts?[/quote]
Hmm Awesomemath ordered theirs here: http://www.customink.com/ ...

[quote="davidyko"]i love how davids is plural.[/quote]

So do I. I hate English class... using plurals when they shouldn't be used is fun.  :ninja:

[quote="sonny"][quote="archimedes1"]well maybe if I had known that I could go up to get a prize more than once... xP

ARML starts after AIME scores are received, correct?[/quote]

Oh after all right cool. I may do it then.

Oh yeah and the Marathon is dying partially because junngi is posting problems that even Altheman is getting wrong.

HMM maybe I should do BAMO this year

and dont forget SMT, its awesomely fun[/quote]
agreed
SMT was beast

DUDE THEY ARE #1S ON IMOS
ITS
EASY
AS
HECK
WTF

[quote="Sly Si"]Yay, SMT!  You should all stay for tiebreakers and awards this year.  I think I'm even going to add in a super relay if there's time.  And if I can dig up the really cool one I wrote last spring.

Also, there are brand-new contests going on.  One is a series run by Tim Sanders (if you remember him from ARML two years ago) that will go on all season, leading up to a championship meet in April at Stanford.  The other, if it gets off the ground this year, is a contest run by ARML and formatted like ARML, but run at local sites across the country.[/quote]
hmm when does tim's thing start?
(link please =D)

[quote="not_trig"][quote="bookaholic"][quote]DUDE THEY ARE #1S ON IMOS
ITS
EASY
AS
HECK
WTF[/quote]
You scare me. I feel an impending sense of doom in terms of relative math ability. :([/quote]

err I kind of agree with junggi here... well, there's always one trivial problem on IMO (one that takes <20 minutes to solve for me...)[/quote]
ie 2007 IMO #4

btw actually the porblems above are #1 and #4 from the same year('95)

[quote="junggi"][/quote][quote="not_trig"][quote="bookaholic"][quote]DUDE THEY ARE #1S ON IMOS
ITS
EASY
AS
HECK
WTF[/quote]
You scare me. I feel an impending sense of doom in terms of relative math ability. :([/quote]

err I kind of agree with junggi here... well, there's always one trivial problem on IMO (one that takes <20 minutes to solve for me...)[/quote][quote="junggi"]
ie 2007 IMO #4

btw actually the porblems above are #1 and #4 from the same year('95)[/quote]

that one was trivial

I was like "trig bash, ptolemy, done in 5 minutes"

not even IMO-caliber, that #4

[quote="not_trig"][/quote][quote="junggi"][/quote][quote="not_trig"][quote="bookaholic"][quote]DUDE THEY ARE #1S ON IMOS
ITS
EASY
AS
HECK
WTF[/quote]
You scare me. I feel an impending sense of doom in terms of relative math ability. :([/quote]

err I kind of agree with junggi here... well, there's always one trivial problem on IMO (one that takes <20 minutes to solve for me...)[/quote][quote="junggi"]
ie 2007 IMO #4

btw actually the porblems above are #1 and #4 from the same year('95)[/quote][quote="not_trig"]

that one was trivial

I was like "trig bash, ptolemy, done in 5 minutes"

not even IMO-caliber, that #4[/quote]
haha yeah

Wait, expect for what? What competition is this?

[quote="hunter34"]Mathleague. You can tell by the domain.[/quote]
Dude, if you knew just how many competitions there were titled "so-and-so Math League"...

[quote="Sly Si"]The site for Tim's stuff is [url]http://www.mathleague.org[/url].  There's no info on the California contests yet, though, probably because nothing's set in stone.  But the site gives you kind of an idea of what to expect.[/quote]
oh
i thought mathleague.org was an old test...ie not new this year

@hunter
no its not ust mathleague
its mathleague.org
there's another mathleague called mathleague.com

[quote="hunter34"]Well it is just mathleague.[/quote]
 Still there are a lot of "Math League" competitions.
EDIT: what junggi said


Find a vector, oppositely directed to $ v \equal{} (\minus{}2, 4, \minus{}1)$, haveing terminal point $ Q(2, 0, \minus{}7)$.

This question is rather poor in my opinion. The notion of "terminal point" doesn't really make sense if your definition of vector is something with length and direction.

Well, its from an old book. The answer at the back of this book is (0,4, - 8) which is quite confusing.  The fact that its is opposite direction of $ (\minus{}2,4,\minus{}1)$, shouldn't it be negated?


Prove that there exists a unique sequence $u_0,u_1,u_2,\dots$ of positive integers such that \[u_n^2 = \sum_{r = 0}^n {{n + r} \choose {r}}u_{n - r}\] for $n = 0,1,2,\dots.$

It is easy to see that $u_0=1,\, u_1 = 2$. Let first prove that $u_n = 2^n,\, n=0,1,\cdots $ satisfy the conditions. It is enought to prove:

(1) $ \sum_{r=0}^n \binom{n+r}{r} 2^{-r} = 2^n $.

Let denote the LHS of (1) with $a_n$.

$ a_n = 1+ \sum_{r=1}^n \binom{n-1+r}{r} 2^{-r} + \sum_{r=1}^n \binom{n+r-1}{r-1} 2^{-r} = $

$= a_{n-1} + \binom{2n-1}{n}2^{-n} + \frac{1}{2} a_n - \binom{2n}{n}2^{-(n+1)} \, \Rightarrow$

$a_n = 2 a_{n-1}$.

From $a_0=1$, we get $a_n = 2^n$.

Now using (1), the uniqueness of the sequence $u_n = 2^n$ can easily be proved by induction.