Numerele de la 1 la 100 sunt asezate in cele 100 de patrate ale unei retele 10x10.Aratati ca exista doua patrate adiacente cu suma mai mare decat 105.

[hide="idee"]folosim metoda reducerii la absurd[/hide]

[hide="solutia unei persoane care nu prea stie combinatorica, deci fiti ingaduitori"]1. daca patratul ce contine $100$ nu este pe marginea retelei, atunci problema se rezolva usor ( in jurul lui ar fi $1$,$2$,$3$ si $4$, pt ca altfel concluzia ar fi evidenta. urmarim pozitia lui $99$. in jurul lui vor fi cel putin doua patrate adiacente, cu minimul $5$ si $6$).
2.daca patratul ce contine $100$ este pe margine, deosebim alte doua cazuri:
a)este in coltul retelei. atunci numerele patratelor adiacente sunt mai mici decat $5$. din nou urmarim pozitia lui $99$ etc. se ajunge ca ultimul colt sa fie ocupat de $97$ si patratele adiacente sa contina numerele $8$ si $7$, de unde concluzia.
b)este pe margine, dar nu in coltul retelei. se procedeaza la fel ca a)[/hide]

[quote="icx"][hide="idee"]Cred ca este util sa cunosti numarul de elemente ale acestei multimi :

*) $M= \{1, 2, ..., 100\}$
i) $x,$ $y$ $\in$ $M$, $x \neq y$
ii) $x+y>105$ $\forall$ $x,$ $y$ $\in$ $M$

care cred ca este 48, adica $\{53, 54, 55, ... , 100 \}$ Pui numerele astea in casute astfel incat doua din ele sa nu fie adiacente si arati ca oricum ai incerca sa aranjezi numerele ramase, exista casutele pe care le cauti. Trebuie putin dezvoltata.
[/hide]

mychrom, nu am inteles prea multe din ce ai zi tu acolo  :oops:[/quote]

mda, cred ca nu prea m-am exprimat matematic. eu cu figura in fata am inteles, dar voi... daca noaptea asta o sa am inspiratie, ostez dimineata o alta solutie (sau imbunatesc una de mai sus).
ideea era sa 'urmarim' pozitiile lui 100, dupa care 99, 98 si 97.
e, oricum ideea mea era putin asemanatoare cu a ta.

[b]Soluţie.[/b] Dintre numerele de la $1$ la $100$ ne interesează doar $49$ de numere. Acestea sunt:
\[52, 53, ..., 100 \]
si au fost selectate deoarece suma oricăror două dintre ele este cel putin egală cu 105.
Mai departe, împărţim reţeaua $10 \times 10$  în patru reţele de $5\times 5$  fiecare. Atunci, conform principiului cutiei, există o reţea din cele $4$, în care vor fi plasate cel puţin $17$ din cele $49$ de numere alese.
În cazul cel mai nefavorabil, primele numere din cele $17$ pot fi plasate ca în figura $1$ sau ca în figura $2$.
[url=http://imageshack.us][img]http://img441.imageshack.us/img441/3881/untitledts8.jpg[/img][/url]
În prima situatie, oricum am plasa încă un număr, vom obţine două pătrate adiacente cu suma cel putin egală cu $105$.  Analog pentru a doua situaţie.
Acum trebuie să plasăm şi celelalte numere; în primul caz $4$ şi în al doilea caz $5$.
În oricare din cazuri, vom obţine cel putin $11$ adiacenţe de sumă cel puţin egală cu $105$. Astfel, vom avea sumele corespunzătoare patratelor adiacente cel puţin egale cu
\[105, 106, ..., 115. \]
Problema este complet demonstrată.

Comentariu. Observăm că numărul $105$ din ipoteză poate fi înlocuit cu $114$.

Foarte frumoasa solutie, o singura observatie: problema spune "bla bla bla mai mare ([b]strict[/b]!) decat 105". Deci nu ne intereseaza decat 48 de numere, adica fara acel 52.

nu cred ca sunt de acord cu acel principiu al cutiei   :blush:

problema pe care a propus o bogdan este destul de accesibila, dar ea naste o problema foarte interesanta.care este numarul maxim cu care poate fi inlocuit105?solutia lui mychrom rezolva problema-cu 105-chiar daca pare cam ,,babeasca''(intr adevar poate ca nu a explicat foarte bine dar are dreptate).solutia lui Marius Damian este cam,,complexa'' pentru a rezolva cazul cu 105.eu nu am reusit sa demonstrez decat pana la 110.Marius te rog sa explici cum demonstrezi cu 114?


If you use an obscure theorem, you should really include a proof of the obscure theorem. :D

But then it gets lengthy and loses direct focus.  I think it would be better to just cite a proof of the theorem.

Just make sure that it really is a well-known theorem, not simply some resource that has information that would trivialize the problem  :wink:

These two threads should tell you most of what you need to know:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=mathworld&t=16727[/url]
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=mathworld&t=59192[/url]

[quote="SnowStorm"]Hi,
If we use obscure theorems or internet sources to find information that helps solve the problem, is it necessary that a bibliography is included?[/quote]
As a grader, what I like to see with an obscure theorem is the theorem quoted in its entirety. That way I can see why you chose to use it and whether you applied it properly.

A bibliography would be nice if we decided to use your solution as an official solution. But I have no time to look up any references while I am grading. My only interest in grading a bibliographical reference would be to note that you are thorough, which would increase your chance of a commendation. (We try to make standards for points objective, but we leave the standards for commendations more subjective.)

You yourself should research any theorem you use to verify that it is true and properly understood, because basing your solution on a false or misunderstood theorem would ruin the solution. You don't have to show us that research.

Erin Schram
USAMTS grader

If you're worried about it sounding indirect, try using a lemma


ceiling function
pigeonhole principle
induction
graph theory
combinatorics proposed
combinatorics

Let $ n,k$ be given positive integers satisfying $ k\le 2n \minus{} 1$. On a table tennis tournament $ 2n$ players take part, they play a total of $ k$ rounds match, each round is divided into $ n$ groups, each group two players match. The two players in different rounds can match on many occasions. Find the greatest positive integer $ m \equal{} f(n,k)$ such that no matter how the tournament  processes, we always find $ m$ players each of pair of which didn't match each other.

Clearly, (if I have understood the problem correctly), $ f(n, 2n\minus{}1) \equal{} 0, f(n, 2n\minus{}2) \equal{} 2$.
For smaller $ k$, each player misses at least $ 2n \minus{} k \minus{} 1$ players, we have to ensure the presence of a $ K_m$.
Does it have a Ramsey theoretic flavour?

Claim: $ f(n,k) \equal{} \begin{cases}\lceil \frac {2n}{k \plus{} 1}\rceil, & k\text{ is odd} \\
\lceil \frac {2n}{k}\rceil, & k\text{ is even}\end{cases}$.

Lemma: To create a set of $ q$ players where each player has played each other player in the set, a minimum of $ q \minus{} 1$ rounds are required when $ q$ is even and a minimum of $ q$ rounds are required when $ q$ is odd, and this minimum is achievable.
Proof: Clearly we need to have $ \frac {q(q \minus{} 1)}{2}$ matches total. Each round can have $ \frac {q}{2}$ matches if $ q$ is even and $ \frac {q \minus{} 1}{2}$ if $ q$ is odd. The lower bound follows.
To show that the minimum is achievable, we will first work with $ q$ odd. Draw a regular $ q$-gon, then draw a line through a vertex and through the opposite side. Then, in the first round, the matches whose corresponding lines are perpendicular to this drawn line will play. Doing this one round for each vertex yields a construction for the lower bound.
Now consider $ q \equal{} 2p$. Let $ g(p)$ be the required number of rounds for $ p$ players. Use the first $ g(p)$ rounds to complete $ p$-cliques for each of two halves, then use the next $ p$ rounds to connect each of the players in the first half to the players in the second half (connect $ a_i$ to $ b_{i \plus{} j}$ in round $ j$ where indices are taken mod $ p$). If $ p$ is odd, we can eliminate one of these rounds by connecting $ a_i$ to $ b_i$ in each of the first $ g(p)$ rounds. Thus, the lemma follows.

Now, it is clear by the lemma that if $ k$ is odd, we can make $ \lceil \frac {2n}{k \plus{} 1}\rceil$ cliques of size at most $ k \plus{} 1$ and have some extra edges. The most we might possibly be able to choose here is $ \lceil \frac {2n}{k \plus{} 1}\rceil$ since we can only choose one from each clique, so $ f(n,k) \leq \lceil\frac {2n}{k \plus{} 1}\rceil$. Similarly for $ k$ even, $ f(n,k) \leq \lceil\frac {2n}{k}\rceil$.

Now consider the graph $ G$ where the vertices are the players and there is an edge between the vertices if two players were matched. We see immediately that the maximal degree of this graph is at most $ k$. Thus, it can be $ k \plus{} 1$-colored, which immediately shows that $ f(n,k) \geq \lceil\frac {2n}{k \plus{} 1}\rceil$ since at least one of the colors contains at least $ \lceil\frac {2n}{k \plus{} 1}\rceil$ vertices by the pigeonhole principle. Since this is a vertex-coloring, the vertices of this color must form an independent set, so $ f(n,k) \geq \lceil\frac {2n}{k \plus{} 1}\rceil$ for all $ n, k$. This settles the case for $ k$ odd.

For $ k$ even, we will consider each connected component $ C$. We will split into three cases:
1: C is an odd cycle
If C is an odd cycle, then the chromatic number of $ C$ is $ 3$. We also know that we cannot make an odd cycle in the case $ k \equal{} 2$, so $ k \geq 3$. Thus, $ \chi(C) \leq k$.
2: C is a complete graph
We know from the lemma that we are unable to make a $ k \plus{} 1$-clique using $ k$ rounds, so if $ C \equal{} K_m$, then $ m \leq k$. Thus, $ \chi(C) \equal{} m \leq k$.
3: C is neither an odd cycle or a complete graph
Then [url=http://en.wikipedia.org/wiki/Brooks%E2%80%99_theorem]Brooks' Theorem[/url] gives that $ \chi(C) \leq k$, since the maximal degree of $ C$ is at most $ k$.

Thus, we have shown that $ \chi(C) \leq k$ for any connected component $ C$ of $ G$. Since the chromatic number of $ G$ is simply the maximum of the chromatic numbers of the connected components, we know that $ \chi(G) \leq k$ as well. It follows that $ G$ can be $ k$-colored and thus $ f(n,k) \geq \lceil\frac {2n}{k}\rceil$ for even $ k$.

We thus arrive at our result: $ f(n,k) \equal{} \begin{cases}\lceil \frac {2n}{k \plus{} 1}\rceil, & k\text{ is odd} \\
\lceil \frac {2n}{k}\rceil, & k\text{ is even}\end{cases}$

This problem pretty much follows from the base cases and induction. The answers are $ f(n,k) =\begin{cases}\lceil\frac{2n}{k+1}\rceil, & k\text{ is odd}\\ \lceil\frac{2n}{k}\rceil, & k\text{ is even}\end{cases} $.

I'll use Lemma 1 of the above solution in my solution.

We'll induct on $n$ (it's not really an induction). Ok, our base case is basically in Lemma 1. To be more explicit for the $k$ even case, if we have $k+1$ vertices, we know that (by Lemma 1) we can't have all vertices having degree $k$. So say $v_1$ has degree $< k$. Then by choosing $v_1$, we can't choose $d(v_1)+1 \le k < k+1$ of the vertices anymore. So we can still pick some vertex, so we get 2 vertices chosen.
For the $k$ odd case, say we choose some vertex $v_1$ to be in our independence set. Then we remove $d(v_1)+1 \le k+1$ vertices from also being in the set. Similarly, each vertex chosen knocks out $k+1$ from being other possible vertices, so we can choose at least ${\lceil}\frac{2n}{k+1}{\rceil}$ vertices to be in our set. Construction is just a bunch of ${K}_{k+1}$'s and  another complete graph with the remainder.

For the $k$ even case, we need to be a little more sharp in our estimates. If the graph is disconnected, we're done by induction on the components. (Use that $\lceil{x}\rceil+\lceil{y}\rceil \ge \lceil{x+y}\rceil$) So we're going to choose vertices to appear in our independence set again. Say $v_1$ is in our set. Then we can't pick any vertices in $N(v_1)$ or $v_1$. $v_1$ eliminates $k+1$ vertices from consideration. Ok, now choose a vertex $v_2$ that is in $N(N(v_1))$ but not eliminated from consideration yet (this is possible by connectivity). Then $v_2$ eliminates AT MOST $k$ vertices from consideration (it's neighbors, and itself, but it shares a neighbor with $v_1$, who is already eliminated). Continuing this process (looking at $N(N(v_2))$ etc.), assume that $m < {\lceil}\frac{2n}{k}{\rceil}$. We get that $k+1+k \cdot (m-1) \le k+1+k \cdot {\lceil}\frac{2n}{k}{\rceil} -k \le k+1+2n-2-k < 2n$. ($k \cdot (m-1) < 2n-2-k$ because both $k$ and $2n$ are even, so our remainder can be at most $k-2$). So we can still add another vertex. So this gives us our set. Construction for the $k$ even case is the same as the odd case.

The case for $k$ odd can be tackled by applying Turan's theorem on the number of pairs which were not matched.If we give an edge to a pair who were not matched,itis to be seen that the least number of edges in the $2n$ graph is $n(2n-1-k)$.(at least)
Now  if we apply Turan's theorem on these number of edges then,we get that $K_{r+1}$ may be guaranteed where  $r-1<\frac{2n}{k+1}\le r$ and the construction of the Turan graph will give a plausible tournament if $k$ is odd.But if $k$ is even then the Turan graph cannot satisfy the tournament.Then we have the colour the vertices(this time a vertex given to a matched player) and show the chromatic number to be at most $k$.


probability calculus rotation geometry trigonometry

Three points are chosen randomly and independently on a circle. What is the probability that all three pairwise distances between the points are less than the radius of the circle?

$ \textbf{(A)}\ \frac{1}{36} \qquad
\textbf{(B)}\ \frac{1}{24} \qquad
\textbf{(C)}\ \frac{1}{18} \qquad
\textbf{(D)}\ \frac{1}{12} \qquad
\textbf{(E)}\ \frac{1}{9}$

It's probably worth linking this [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=247005]related thread[/url] for:
1. A solution to this problem (post 4)
2. A related problem and outline of its solution (posts 1, 6, 7)
3. A random calculus problem

The other solution made a slight error I think. Here's another try.

[hide]
WLOG let the first point be on the bottom of the circle. (We don't lose generality because we can just rotate the circle if we don't get what we want.)

Then the second point can fall 1 unit in either direction of the first point. So to the right the furthest distance is where a 1 unit line intersects the circle. Since the circle has a 1 unit radius, 2 radii and the unit line make an equilateral triangle. The point can fall either to the right or left of the point so that's 60*2=120 total degrees of the circle out of 360, so that's 1/3.

The third point depends on where the second point landed. That's a good hint to use geometric probability.

Here is an awesome diagram:

[img]http://img365.imageshack.us/img365/8964/thing.png[/img]

I think the diagram makes it pretty clear. Anyway, the black is the area we don't want, so P(wanted)/P(total) = 3/4. But don't forget to multiply by 1/3 again (for the chance of getting near the first dot) = 3/12 or 1/4.

Then P(all dots close) = 1/3*1/4 = 1/12

[/hide]

I solved it completely different but I'm not sure whether my approach is rigorous enough. Here it is:
First, applying Laws of Cosines into each of the three triangles having one vertex at center $ O$, the other points are 2 points on the circle:
We have:$ AB^2 \equal{} 2R^2 \minus{} 2R^2.cos(AOB) < R^2$
Similar to the other two sides of this triangle, $ BC$ and $ AC$, we get the restriction:$ cos(AOB)$,$ cos(AOC)$ and $ cos(BOC) > 1/2$.
We see that $ cos(BOC) \equal{} cos(AOB \plus{} AOC) > 1/2$, which means $ cosAOB.cosAOC \minus{} sinAOB.sinAOC > 1/2$.
Now drawing the unit circle and choosing two points A,C such that angle AOC and AOB satisfying these 3 inequalities. We get the are of that arc and dividing it by $ 2pi$, we get 1/12. Is it correct to solve it on unit circle?? Note that we can choose either 2 points out of 3, so we must count the other 2 points also.

I understand the probability when the second point is placed but how do we know the probability of the third point. I see that the third point is dependent on the first and second point.

[quote="AlanLiangTheAsian"]errr any other solutions?[/quote]

Ok...
1. First point goes wherever.
2. second point must be within 60 degrees from the first point or Else it would be too far away. You can have it 60 degrees on both sides of Your first point. This maks it 120 degrees, which is 1/3 of the circle.
3. The third point is trickier. It must stay within 60 degrees from both the previous points. But where did that second point really go? One extreme is that the second point is on the first point, then we have 120 degrees to put the last point. The other extreme is that the second point is one radius, or 60 degrees, away from the first point. Then we only have 60 degrees to place the last point. (between the two points).
So, if the second point is within a radius from the first point, then we have a range of 60-120 degrees to place the last point. Then we expect the range to be in the middle of 60 and 120, which is 90! The probability of placing that last point correct must therefore be 1/4, sine 90 degrees is 1/4 of a full circle.

So: The probability of getting all Three points within one radius mus be 1*1/3*1/4=1/12

[b]Solution:[/b] It should be clear that two points must be within $60$ degrees of each other in order for the distance between them to be less than the radius. Let the first point be wherever. Let the angle between the first point and second point be $\alpha$ (where $0 \le \alpha \le 60$, in degrees) ($\alpha$ degrees clockwise or $\alpha$ degrees anticlockwise). You can also realize that the third point has a $2(60 - \alpha) + \alpha = 120 - \alpha$ degree region where it can be placed. Not me trying to create a function out of this and then integrating :skull:. Anyways that's not the right way to go about finding the average value of the probability of it. Instead, we notice that the second point has a region of $\frac{1}{3}$rd of the circumference it can be placed in. For the third point, try noticing that $\frac{120 - \alpha}{360}$ is linear so the average value will be when the $\alpha = 30$ degrees. This equates to a probability of $\frac{1}{3} \cdot \frac{1}{4} = \boxed{D}$.


Find the smallest positive integer which when divided by 10 gives remainder 9,when divided by 9 gives remainder 8 and so on till when divided by 2 to gives remainder 1.

[quote="eminem"]2521?

lc.m of 10, 9 , 8.... 2  +1[/quote]

2521 doesn't leave a remainder of 9 when divided by 10.

[hide]I did the same thing as eminem except I subtracted 1, so the answer is 2519.[/hide]

sorry  :blush: ... forgot that .... thanks ;)


$ (a,b,c,d,e)\in{\mathbb{R}}$.

prove that:

$ a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}d^{2}\plus{}e^{2}\ge a(b\plus{}c\plus{}d\plus{}e)$

[quote="reason"]$ (a,b,c,d,e)\in{\mathbb{R}}$.

prove that:

$ a^{2} \plus{} b^{2} \plus{} c^{2} \plus{} d^{2} \plus{} e^{2}\ge a(b \plus{} c \plus{} d \plus{} e)$[/quote]

For

$ LHS \minus{} RHS$
$ \equal{} (b \minus{} \frac {a}{2})^2 \plus{} (c \minus{} \frac {a}{2})^2 \plus{} (d \minus{} \frac {a}{2})^2 \plus{} (e \minus{} \frac {a}{2})^2 \ge 0$


Determine the value of $\sum^{n-1}_{k=1}\sin\left(\cos\frac{k\pi }{n}\right)$, where $n \in \mathbb{N}$

[hide="Solution"]\begin{eqnarray*}\sin\left(\cos{k\pi\over n}\right)+\sin\left(\cos{(n-k)\pi\over n}\right) &=& \sin\left(\cos{k\pi\over n}\right)+\sin\left(-\cos{k\pi\over n}\right)\\ &=& \sin\left(\cos{k\pi\over n}\right)-\sin\left(\cos{k\pi\over n}\right)\\ &=& 0\end{eqnarray*}

Therefore the sum is equal to the middle term. If $n$ is odd, there IS no middle term, hence $S(n)=0$. If $n$ is even, the middle term is obtained for $k={n\over 2}$ and is equal to $\sin\left(\cos{\pi\over 2}\right)=\sin 0=0$. Hence the sum is always zero.[/hide]


Let $ a,b,c > 0$ then proove that

$ a \ln a \plus{} b \ln b \plus{} c \ln c \geq \ln(abc) \frac {a \plus{} b \plus{} c}{3} \plus{} a \plus{} b \plus{} c \minus{} 3\sqrt [3]{abc}$

:(

I think it is easy :wink: ,can you post your solutions?
Thank you


geometry
trapezoid
ratio
circumcircle
analytic geometry
geometric transformation
homothety

In trapezoid $ ABCD$ with $ \overline{BC}\parallel\overline{AD}$, let $ BC\equal{}1000$ and $ AD\equal{}2008$. Let $ \angle A\equal{}37^\circ$, $ \angle D\equal{}53^\circ$, and $ m$ and $ n$ be the midpoints of $ \overline{BC}$ and $ \overline{AD}$, respectively. Find the length $ MN$.

Extend AB and CD to meet at E. By some calculations, we can deduce that angle BEC is 90 degree. Since M is the midpoint of BC, EM=BM=CM=500, we can make use of similar triangle ratio to find MN.

            500/(500+MN)=1000/2008
                          MN=504

Hmmm... I think this problem seems a bit familiar...

http://www.usamts.org/Solutions/Solution2_2_18.pdf

Continuing off of what zhou said,

[hide]Angle E is 90 degrees, implying that $ BC$ and $ AD$ are diameters of the circumcircles of $ EBC$ and $ EAD$ respectively. This indicates that $ EM \equal{} 500$, and $ EN \equal{} 1004$. Therefore, $ MN \equal{} 1004 \minus{} 500 \equal{} 504$.[/hide]

Analytic geometry for the win!

[hide]
Let us put $ ABCD$ on a coordinate plane such that $ A$ is on origin and $ D$ is at $ (2008,0)$
Now, if we let $ B$ be at point $ (x, h)$, then since $ AD \parallel{} BC$, we know that $ C$ is at $ (x \plus{} 1000, h)$

Then we know $ M \equal{} (x \plus{} 500, h)$, and $ N \equal{} (1004,0)$
Now, from $ B$ and $ C$, drop perpendiculars $ BE$ and $ CF$. It is easy to see that
$ E \equal{} (x,0)$, and $ F \equal{} (x \plus{} 1000,0)$. Also, drop perpendicular $ MO$ as well

Now, we have $ AE \equal{} x$, $ EO \equal{} 500$, $ ON \equal{} 504 \minus{} x$ and $ FC \equal{} 1008 \minus{} x$
From the angle information, we know
$ \frac {BE}{AE} \equal{} \frac {DF}{FC}$

$ \frac {h}{x} \equal{} \frac {1008 \minus{} x}{h}$

$ h^2 \equal{} 1008x \minus{} x^2$

Now we know triangle $ MON$ is right, so we get that
$ MN^{2} \equal{} (504 \minus{} x)^2 \plus{} h^2$

and the answer follows immediately $ \fbox{504}$.

[/hide]

[hide="Diagrams (from wiki)- Edit, surprised I didnt see that after drawing the circumcircle"]
Extend $ \overline{AD}$ and $ \overline{BC}$ to meet at a point $ E$. Then $ \angle AED \equal{} 180 \minus{} 53 \minus{} 37 \equal{} 90^{\circ}$.
[asy]size(220);
defaultpen(0.7+fontsize(10));
real f=100, r=1004/f;
pair A=(0,0), D=(2*r, 0), N=(r,0), E=N+r*expi(74*pi/180);
pair B=(126*A+125*E)/251, C=(126*D + 125*E)/251;
pair[] M = intersectionpoints(N--E,B--C);
draw(A--B--C--D--cycle);
draw(B--E--C,dashed);
draw(M[0]--N); draw(N--E,dashed);
draw(rightanglemark(D,E,A,2));
picture p = new picture;
draw(p,Circle(N,r),dashed+linewidth(0.5));
clip(p,A--D--D+(0,20)--A+(0,20)--cycle);
add(p);
label("\(A\)",A,SW);
label("\(B\)",B,NW);
label("\(C\)",C,NE);
label("\(D\)",D,SE);
label("\(E\)",E,NE);
label("\(M\)",M[0],SW);
label("\(N\)",N,S);
label("\(1004\)",(N+D)/2,S);
label("\(500\)",(M[0]+C)/2,S);[/asy]
Since $ \overline{BC} \parallel \overline{AD}$, then $ BC$ and $ AD$ are [[homothetic]] with respect to point $ E$ by a ratio of $ \frac {BC}{AD} \equal{} \frac {125}{251}$. Since the homothety carries the midpoint of $ \overline{BC}$, $ M$, to the midpoint of $ \overline{AD}$, which is $ N$, then $ E,M,N$ are collinear. As $ \angle AED \equal{} 90^{\circ}$, note that the midpoint of $ \overline{AD}$, $ N$, is the center of the [[circumcircle]] of $ \triangle AED$. We can do the same with the circumcircle about $ \triangle BEC$ and $ M$. It follows that
\[ NE \equal{} ND \equal{} \frac {AD}{2} \equal{} 1004, \quad ME \equal{} MC \equal{} \frac {BC}{2} \equal{} 500
\]
Thus $ MN \equal{} NE \minus{} ME \equal{} \boxed{504}$. $ \ \blacksquare$

[asy]size(220);
defaultpen(0.7+fontsize(10));
real f=100, r=1004/f;
pair A=(0,0), D=(2*r, 0), N=(r,0), E=N+r*expi(74*pi/180);
pair B=(126*A+125*E)/251, C=(126*D + 125*E)/251;
pair[] M = intersectionpoints(N--E,B--C);
pair F = foot(B,A,D), G=foot(C,A,D), H=foot(M[0],A,D);
draw(A--B--C--D--cycle);
draw(M[0]--N); draw(B--F,dashed); draw(C--G,dashed); draw(M[0]--H,dashed);
label("\(A\)",A,SW);
label("\(B\)",B,NW);
label("\(C\)",C,NE);
label("\(D\)",D,NE);
label("\(F\)",F,S);
label("\(G\)",G,SW);
label("\(M\)",M[0],SW);
label("\(N\)",N,S);
label("\(H\)",H,S);
label("\(x\)",(A+F)/2,S);
label("\(h\)",(B+F)/2,W);
label("\(h\)",(C+G)/2,W);
label("\(1000\)",(B+C)/2,NE);
label("\(1008-x\)",(G+D)/2,S);[/asy]
Let $ F,G,H$ be the feet of the perpendiculars from $ B,C,M$ onto $ \overline{AD}$, respectively. Let $ x \equal{} AF$, so $ FG \equal{} 1000$ and $ DG \equal{} 2008 \minus{} x \minus{} 1000 \equal{} 1008 \minus{} x$. Also, let $ h \equal{} BF \equal{} CG \equal{} HM$.

By AA~, we have that $ \triangle AFB \sim \triangle CGD$, and so
\[ \frac {BF}{AF} \equal{} \frac {CG}{DG} \Longleftrightarrow \frac {h}{x} \equal{} \frac {1008 \minus{} x}{h} \Longrightarrow h^2 \plus{} x^2 \minus{} 1008x \equal{} 0.
\]
Since $ AH \equal{} 500 \plus{} x$ and $ AN \equal{} 1004$, it follows that $ HN \equal{} AN \minus{} AH \equal{} 504 \minus{} x$. By the Pythagorean Theorem on $ \triangle MON$,
\[ MN^{2} \equal{} (504 \minus{} x)^2 \plus{} h^2 \equal{} 504^2 \plus{} h^2 \plus{} x^2 \minus{} 1008x \equal{} 504^2
\]
and the answer is $ MN \equal{} 504$. $ \ \blacksquare$
[/hide]

Are you kidding me?

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=137689&start=80[/url]
2nd post down.

You don't need similar triangles since it's a right triangle, which immediately gives us $ \frac{1}{2}(2008\minus{}1000)$

[quote="the future"]yea lol that was my old account and I posted that problem about a year ago (May 1, 2007 for 2007 ARML practice)-referring to the link on the above post.  It basically uses the EXACT same idea from this AIME problem (just different angles and side lengths).  So for all those indiana people that dont do the problem marathon...[/quote]

darn.
it.
 :!:

[hide="not very rigorous trig/coordinate bashing"]if you drop perpendiculars from B and C to AD and call the points where they meet AD E and F respectively and call FD = n and EA = 1008-n , then you can solve an equation in tangents. since $ \angle{A} \equal{} 37$ and $ \angle{D} \equal{} 53$, then you can solve the equation:

$ \tan{37} \times (1008\minus{}n)$ = $ \tan {53} \times n$
now just do some messy trig and convert $ \tan{37}$ into $ \frac{\sin{37}}{\cos{37}}$ then use the 90-x functions for sin and cos and cross multiply and you end up with

$ \frac{1008\minus{}n}{n} \equal{} \frac{sin^2{53}}{\cos^2{53}}$ and if you do some simplification, you end up with

$ n \equal{} 1008\cos{53}$, but we want 1008-n, so we do $ 1008(1\minus{}\cos^2(53)) \equal{} 1008\sin^2{53}$, now you just co-ordinate bash it. and use distance formula. but first we must find the height so we just use our trig again.

$ \tan{37} \times 1008\sin^2{53}$
=$ \tan{37} \times 1008 \cos^2{37}$
=$ 1008 \cos{37}\sin{37}$
=$ 504\sin{74}$

now we can use distance formula

find the square root of the entire expression

$ (1008\sin^2{53} \plus{} 500 \minus{}1004)^2 \plus{} (504\sin{74})^2$

just keep simplifying it and you end up with

$ 504\sqrt{sin^2{\theta} \plus{} \cos^2{\theta}}$

which is just $ \fbox{504}$ cuz $ \sin^2{x} \plus{} \cos^2{x} \equal{} 1$.[/hide]

Does anyone else aside from me ever find it tempting to brute force geometry problems rather than look for elegant geometry solutions?
How do you get out of the habit? [hide="Crude Coordinate Bashing"]
If we extend the legs of the trapezoid and rotate the resulting right triangle such that one leg of the right triangle lies on the x axis and the other lies on the y axis, we can find the coordinates of the midpoints of the original bases of the trapezoids easily with basic right triangle trigonometry in terms of the sines and cosines of the given angles. Applying the distance formula and then using basic Pythagorean identities to simplify the resulting expression, we find the distance $ 504$. [/hide]

EDIT: cognos posted as I typed this and our solutions our similar

If you didn't know how to do this, you might have noticed that a 37-53-90 degree triangle is approximately a 3-4-5 right triangle  :) The problem can then be effortlessly put away with Stewart's, which gives the right answer.

In fact, if you used this approximation, you could be confident that your answer is within about 0.95, which on the AIME, guarantees that you have the right answer.

How exactly would we use Stewart's to approximate the answer? Also, how do we know the answer is within about 0.95?

I liked how the height is the same no matter what the two angles are, so long as they are complementary. So really the only "trig" you needed was definition of tangent and realization that

$ \frac{1}{tan{\theta}} \equal{} tan(90 \minus{} \theta)$

[hide="Answer"]
Drawing a diagram simplifies this problem.

[asy]size(250);
defaultpen(0.7+fontsize(10));
real f=100, r=1004/f;
pair A=(0,0), D=(2*r, 0), N=(r,0), E=N+r*expi(74*pi/180);
pair B=(126*A+125*E)/251, C=(126*D + 125*E)/251;
pair[] M = intersectionpoints(N--E,B--C);
draw(A--B--C--D--cycle);
draw(B--E--C,dashed);
draw(M[0]--N); draw(N--E,dashed);
draw(rightanglemark(D,E,A,2));
picture p = new picture;
clip(p,A--D--D+(0,20)--A+(0,20)--cycle);
add(p);
label("\(A\)",A,SW);
label("\(B\)",B,NW);
label("\(C\)",C,NE);
label("\(D\)",D,SE);
label("\(E\)",E,NE);
label("\(M\)",M[0],SW);
label("\(N\)",N,S);
label("\(1004\)",(N+D)/2,S);
label("\(500\)",(M[0]+C)/2,S);[/asy]

We extend $ AB$ and $ CD$ to $ E$. $ E = 90$; $ M$ is the midpoint of $ BC$, so $ BM = CM = EM$, which we can see is 500. By similar triangles, $ \dfrac{500}{500 + MN} = \dfrac{1000}{2008}$. Therefore, $ MN$ = 504.[/hide]

What is "Stewart's"?  I've never heard of it.

[hide="Solution for resources section"]
We extend segments $ AB$ and $ CD$ to meet at $ E$, which is a $ 90^\circ$ angle.
[asy]size(250);
defaultpen(0.7+fontsize(10));
real f=100, r=1004/f;
pair A=(0,0), D=(2*r, 0), N=(r,0), E=N+r*expi(74*pi/180);
pair B=(126*A+125*E)/251, C=(126*D + 125*E)/251;
pair[] M = intersectionpoints(N--E,B--C);
draw(A--B--C--D--cycle);
draw(B--E--C,dashed);
draw(M[0]--N); draw(M[0]--E,dashed);
draw(rightanglemark(D,E,A,2));
picture p = new picture;
clip(p,A--D--D+(0,20)--A+(0,20)--cycle);
add(p);
label("\(A\)",A,SW);
label("\(B\)",B,NW);
label("\(C\)",C,NE);
label("\(D\)",D,SE);
label("\(E\)",E,NE);
label("\(M\)",M[0],SW);
label("\(N\)",N,S);
label("\(1004\)",(N+D)/2,S);
label("\(500\)",(M[0]+C)/2,S);[/asy]
Since $ M$ and $ N$ are midpoints, $ M$ is the circumcenter of $ \triangle EBC$ and $ N$ of $ \triangle EAD$. Thus, $ EN \equal{} ND \equal{} 1004$ and $ EM \equal{} MC \equal{} 500$, so $ MN \equal{} EN \minus{} EM \equal{} \boxed{504}$.[/hide]

[quote="Wickedestjr"]For this problem, you need to extend the two legs, and they should meet at a 90 degree angle and then find the median of the larger triangle and subtract the median of the smaller triangle. I read a transcript of a mathjam solving this problem.[/quote]...and that's the solution I just posted.

worthawholebean your really shutting down wickedstrjr  :P

A simpler version of similar triangles, in response to Mersenne's post
$ 500 / MN \equal{} 1004 / ( MN \plus{} 500)$
MN = 500 because it is the median to the hypotenuse of tri. BMC and therefore half the hypotenuse
Plugging in we have MN + 500 = 1004 MN = [b]504[/b]

I have a little question: I see that sometimes you use a LaTeX code (or whatever it is  :lol: ) in order to obtain geometrical figures. How and where can I learn this very practical use of LaTeX?

Thanks
GM

Sorry, I didn't have much time to read what you read. I just showed my solution. And I am also curious about learning how to use LaTex. Does it cost money?

pythago11 ur a beest, 15 usamo arent u in MS?

I believe 15 was his goal, not what he got. (I know what he got, and he probably wouldn't mind me saying it, but I won't say it anyway.)

[quote="pythag011"]No.

[url]http://www.artofproblemsolving.com/Wiki/index.php/Latex[/url]

for graphics:

[url]http://www.artofproblemsolving.com/Wiki/index.php/Asymptote_%28Vector_Graphics_Language%29[/url][/quote]

Thank you!

[hide=53+37=90]Extend $AB$ and $DC$ to meet at $X$. Set $AD=a$ and $BC=b$\\
since $\triangle{XAD}\sim \triangle{XBC}$ we have  $\angle{AXD}=90^{\circ}$, then,\\
$\frac{XM}{XN}=\frac{AM}{BN}=\frac{a}{b}$\\
also we have $\frac{AX}{XB}=\frac{a}{b}$\\
so take $AX=ak,XB=bk$\\
notice $XD=a\sqrt{1-k^2}, XC=b\sqrt{1-k^2}$\\
so $XM$ from Apollonius theorem is  $\frac{a}{2}\implies XN=\frac{b}{2}$\\
$MN=XN-XM=\frac{a-b}{2}=\frac{2008-1000}{2}=\boxed{504}$[/hide]


Let z be a complex number, z=a+bi, we know also that (a+bi)^8=a-bi, find all possible values of the expression z^2001.

[hide="Addendum"] $ z^8 \equal{} \bar{z}$ implies $ |z| \equal{} 1, 0$, so $ a^2 \plus{} b^2 \equal{} 1, 0$. [/hide]

Nice guys, but i still don`t see a solution. There are just 4 possible values of the ekspresion including z=0.
I also donnot know why there must be a^2+b^2=1,0  please explain it to me, because i am not very familiar with complex numbers.

[hide]Clearly,$ (a \plus{} bi)^8 \equal{} z^8$ and $ a \minus{} bi$ is the conjugate of $ z^8$, in otherwords $ \bar{z}$. The only values for which that is satisfied is z=0,1. $ a \plus{} bi \equal{} z\rightarrow{a^2 \plus{} b^2 \equal{} z^2}$,by definition so $ a^2 \plus{} b^2 \equal{} 0,1$.[/hide]
EDITED.

[quote="xpmath"]and $ a \minus{} bi$ is the conjugate of $ z$[/quote]

$ z^8$, perhaps. :wink: Usually, $ a\minus{}bi$ is the conjugate of $ a\plus{}bi$.

[quote="xpmath"]The only values for which that is satisfied is z=0,1. $ a \plus{} bi \equal{} z\rightarrow{a^2 \plus{} b^2 \equal{} z}$,by definition so $ a^2 \plus{} b^2 \equal{} 0,1$.[/quote]

No, $ |a \plus{} bi| \equal{} a^2 \plus{} b^2$ :) so that would have to be $ z^2$.

The first 9th root of unity raised to the 8th power would be 16π/9 radians around the unit circle, with the original being 2π/9 radians. Such angles on the plane, adding up to 2π, would thus identify two complex conjugates. So that, 1 and 0 would also work. The 9th root number four also satisfies this condition, as does the second, sixth, and eighth, third, seventh, fifth, and ninth (a.k.a. 1). All ninth roots of unity and zero fit this description. 2001 = 3 mod 9. Thus the cubes of all these numbers fit the description. The cubes of the ninth roots of unity turn out to be the third roots of unity, each repeated three times. (1±i√3)/2, 1 and 0 are the only possible values.

the roc would you please explain your solution again, because i don`t get it, why are you using the roots of unity.Nobody said that a^2+b^2=1. Can sombody please send an elegant solutin of this problem, because i solved it, but my solution contain the polar descripcion of a complex number and the moivre`s formula and it`s too long.

[quote="primoz2"]Nobody said that a^2+b^2=1. [/quote]

[hide="Proof"] $ (a \plus{} bi)^8 \equal{} a \minus{} bi \implies$
$ | (a \plus{} bi)^8 | \equal{} | a \minus{} bi | \Leftrightarrow$
$ (a^2 \plus{} b^2)^4 \equal{} \sqrt {a^2 \plus{} b^2} \Leftrightarrow$
$ (a^2 \plus{} b^2)^8 \equal{} a^2 \plus{} b^2 \Leftrightarrow$
$ a^2 \plus{} b^2 \equal{} 1, 0$

(Remember that $ a^2 \plus{} b^2$ is real and nonnegative).

[b]Remark:[/b]  If you are unfamiliar with this notation, $ |a \plus{} bi| \equal{} \sqrt {a^2 \plus{} b^2}$, called the [b]norm[/b] (sometimes the square root of the norm) or the [b]absolute value[/b] of $ z \equal{} a \plus{} bi$, and it has the property that $ |zw| \equal{} |z| |w|$. [/hide]
Edit:

[hide="Second solution"] $ (a \plus{} bi)^9 \equal{} a^2 \plus{} b^2$, which is purely real; it follows that $ a \plus{} bi$ is some real multiple of a ninth root of unity.  Let $ a \plus{} bi \equal{} re^{ i \frac{2 \pi k}{9} }$.  Then

$ r^9 \equal{} r^2 \Leftrightarrow r \equal{} 0, 1$ [/hide]


Solve the equation $x\sqrt{1-x^2}-y\sqrt{1-y^2}=1$

 :)

We have $|x|, |y| \le 1$.

Put $x = \cos X$, $y = \cos Y$.  Then $\sqrt{1-x^2} = \sinh$.

It becomes $cosXsinX - cosYsinY = 1$.

or $\sin 2X - \sin 2Y = 2$.

Thus $\sin 2X = 1$, $\sin 2Y = -1$, where it is easy to find X,Y and thus x,y as $2^{-1/2}, -2^{-1/2}$.


In the figure below, points A, B, C are collinear, and the points D, E, F. The two lines ABC and DEF are parallel.

[img]http://img504.imageshack.us/img504/9803/cals.jpg[/img]

Are A1, A2 and A3 areas of regions highlighted in the figure, we can say that:

A) A2 = 2A1 = 2A3
B) A2 = A1+A3
C) A2 > A1+A3
D) A2 < A1+A3
E) A2² = A1.A3

[hide]B

Triangles ADE and BDE have the same area, as do BEF and CEF. (same bases, same altitudes)[/hide]

So does that mean that in a trapezoid, those two areas are always equal?  :what?:

[quote="ChenXiao"]whoa i never realized this before. so the 2 areas are always equal right?[/quote]Yeah, same here. It seems like they are.


I believe the AoPS Wiki Words of the Week need to be updated.  (It's past the 11th, and the current words are supposed to run through the 11th).

Indeed. And I took care of "Math Day at the Beach," that being something which is always my responsibility anyway.;


EDIT: lmao, I find the need to post this.

[hide="a screeny"]
[img]http://img.photobucket.com/albums/v259/drspyder/666posts.jpg[/img]
[/hide]

it might be labeled similarly to your smoking one (which i believe was taken down)
edit: noo!!! another firefox user!


The Introduction to Counting and Probability course has been integrated with Alcumus - the Alcumus problems supplement the course and homework, but do not replace it.

This makes the Introduction to Counting and Probability course significantly more valuable, as there are over a thousand extra practice problems in Alcumus.  I don't really think it's overkill to use both.

Other topics haven't been implemented into Alcumus yet.

A lot of the answer depends on what type of learner you are.  We are developing offerings in many different media (books, online classes, Alcumus) because people learn in very different ways.  For different people, different combinations of approaches are optimal.  I encourage you to try each (class, book, Alcumus), and find what combination works best for you.  As Alcumus develops, we expect to integrate it highly with our classes, so that it is a natural supplement to the online classes.

I took the Intro to Counting Probability class in 6th grade, when preparing for MATHCOUNTS. Its a great course. The book has a wonderful supply of problems. Alcumus wasn't made when i took the class. So, I just went to Alcumus during my 7th and currently 8th grade years. Alcumus is made up of problems similar to the problems from the intro to c&p book and contest problems. I say it is better to take the course, then Alcumus. And, it really helps as a review, but that's just my opinion. :)

[quote="pircdefense"]Are the regular courses designed to work in tandem with the corresponding Alcumus? Or is taking both overkill? Or does each have its place, and it's recommended to both take the course and work through the Alcumus section?[/quote]
Intro. to Counting and Probabily has more information than Alcumus alone. I have done Alcumus, currently own the book, and plan to take the course in the fall. It a great way to get the concepts "stuck" in your brain. :)


Ever heard of inputting random numbers on a calculator or computer program?

What question are you asking?

Of COURSE you can put anything into a calculator.
That's what makes it more useful than a table of logarithms or something..

I think this dude was trying to post a response to question that would be relevant, but accidentally hit the "New Topic" button. This post makes no sense other wise...

Yeah, that's true. But i dont remember what thread it was supposed to be in.


So like most of time, people who are preparing for academic contests can grasp most of the concepts of a problem very quickly, but there's this one snag that takes a lot of thought and time to resolve. So then you finally get it, after an hour of work, and then your friend asks you a few days later to help him with a similar problem. In like 5 minutes of discussion with him, you can isolate all of his misconceptions, and point out some things you've noticed, and he's now where you are, plus some more 'cause he saved time.

Those who write problem solving books do a great service to everyone, but to write a book outlining all the strategies you've learned while you're still competing is usually suicidal. Face-to-face discussion even more so.

It would be best if both sides could participate, and help each other so that everyone could save time. So perhaps the best strategy is to help others for a while, and see if they help you. But a lot of the time, people refuse to cooperate. But then not helping is selfish.

A lot of people on AoPs are amazing in the amount of help they give to others, and I'm sometimes ashamed I feel bad giving help to people. But I also feel terrible when they make the same progress in 5 minutes as I made in an hour.

So yeah, what do you think about helping your competitors?

I feel that by explaining something to someone else in an undestandable way, I help myself understand it more.
Moreover, competition should not just be about doing better than others, but about getting better.  Is the ultimate end of math competitions to do better than others do, or just to do well?  Overall, I do mathematics because I love it.
As far as helping on this forum...I help those whom I can help, but there's always someone better than I am willing to help me.  Eventually, those who are the best pass out of competition, right?  (based on age, etc).  So at first, you're a competitor, but then you might pass on your knowledge so that you can keep things going for the general good.  Those who are extremely good on this forum do math because they love it, and are always helpful.  I can't really imagine that they think about math in a competitive manner: they simply love it.  Besides, the more good people there are to compete with, the more fun it is.

Just my two cents.

EDIT: Moreover, know that people always have to work to understand things in math.  There's rarely such a thing as a trick you can simply see and then understand completely.  As the saying goes, "The teacher can open the door, but only the student can enter."  The more you use a tecnique and explain it to others, the more you understand it yourself.

[quote="Nerd_of_the_Ages"]Well, you can give people some guidance, but I think in the end it is best if they discover things on their own.[/quote]

Right, what I would do is point them in the right direction, and have them find the answer themselves (they're not going to learn anything by you telling them the answers).

[quote="K81o7"]I feel that by explaining something to someone else in an undestandable way, I help myself understand it more.
[/quote]

This is so true.  I've learned a ton of math while writing various books and classes, and basically coasted through classes in college by teaching my peers - they often challenged me much more than the problem sets did.

GrapeSharp, dude, no offense, but if you feel terrible about others learning in five minutes what you spent one hour after, you have a very serious problem indeed. Its good that you feel ashamed, but if you really wanna turn back, do it fast. After an age, its hard to get rid of this, and it will make you extremely unpopular. What's more, you will keep on feeling ashamed and yet be unable to do anything about it.

Forget competitions, boy, and wake up to the society around you. Make friends - you need them. AoPS may help you in that. Learn to give unconditional help to anyone and everyone. Otherwise you'll grow up to be incomplete, at least as far as my conceptions of a complete human being go.

Be fast. Time's running out. You have my blessings. :)

I'm trying to understand what you mean:
Because of your motivation and passion for problem solving, you do a lot of exploration and find things out the hard way, and you feel like if you help someone, it's like they're using you to short-cut ahead, right?

As K rightly pointed out, whether or not you know it, you are learning by helping your friends; it'll reinforce what you've learned, give you new perspectives, and help you retain them.

So here's how (I think) you can try optimize this. When you help someone, try to give them hints rather than directly giving them the innovations. It takes a lot of practice to do this well, but done right, it can help both people enormously.
[u]- How does this help them?[/u] It allows them to take the time and think it through and figure it out themselves. This way, you can help them without making them "shortcut" ahead.
[u]- How does it help you?[/u] By isolating the pointers that might help someone find the solution on their own, you are familiarizing yourself with the core principles of that concept, which is much more helpful than familiarizing yourself with a particular solution. You are also familiarizing yourself with the problem solving process and mindset. This can help you solve new problems much more quickly in a timed environment. It also gives people a good impression of you  :)

I usually help my peers in competition, but I only give them basic knowledge, expecting them to find out on their own.
Example:
Solve for $ x$ $ 4x^{3}\plus{}4x^{2}\plus{}4x$ I just tell them to factor it out.

since competition is rather scarce, I help them as much as possible in order to make tournaments more fun and challenging.

Excuse me? How is competition scarce? Competition is never scarce. If competition is scarce, you should move up to a higher level. I'm assuming by your listed location that you live in Scandinavia (or at least Europe?), and I'm sure there are [b]plenty[/b] of good math kids (and math opportunities) there.

Well, in most places, competition is scarce with most schools, or locally (local contests). Especially school, I wouldn't mind helping them, since there are so few it's kinda boring.

[quote]Yes, duty above success
Yes, if they try to help me
Only if they actually help me
No, people have to work for themselves [/quote]

I think pretty much all of my viewpoints have been stated. I would like to comment, though, that I couldn't agree less with the premise of your poll.

Your first choice, "duty above success," implies that by helping your friends catch a concept, you are directly doing yourself a disfavor. Remember that the majority of your competition is not Bob from your school, but Joe from across the country (or, if you're good enough,  Liang on the other side of the world). The fact is that you'll hardly come into a situation such as, "I didn't make the USAMO's because Joe from my school moved the index just barely out of my reach." In fact, if you work together with people from your school, then you can all strengthen each-other against the majority of the competition, which is outside of your school.

The second choice sounds bad but has a bit of truth to it. If someone else isn't willing to try help you when you need it, chances are they might be suffering from the same mindset as you. In that case, you should try to discourage that person from feeling that way (although, be subtle about it).

The third choice is just about the most selfish choice on the list.

I partially agree with the fourth choice. While you should help people, they still have to figure it out on their own. If you do the figuring for them, you deny them the experience of solving the problem by themself.

[b]@ pianoforte:[/b] Sure, competition can be low within a school or a small region. That doesn't mean competition is scarce, since the more significant competition (with whom one competes in more meaningful competitions) is globally distributed. Competition is always a reality; it's just a matter of where you set your expectations. There's nothing like a hard math contest to humble the "straight A+'s in Calc BC in freshman year" kid. Furthermore, I don't think this applies to his case.

[quote="The Zuton Force"]I partially agree with the fourth choice. While you should help people, they still have to figure it out on their own. If you do the figuring for them, you deny them the experience of solving the problem by themself.[/quote]
Agreed. Often I find myself begging for people not to give me hints. It feels so good when you finally solve a difficult problem by yourself. And anyways, math is fun, so why would you want to spoil it  :D . If the person as little or no clue, I guess if they request it you can give them some guidance.


analytic geometry
function
ratio
geometry
3D geometry
tetrahedron
perpendicular bisector

[b]Many thanks to Orl[/b] for communicating to me a lot of old olympiad and shortlist problems. Among the many data, I find a remark which may be interesting to some members of this forum:

[quote]Internal political problems meant that the IMO 1980 did not take place in Mongolia as planned. Luxembourg, which had never participated in an IMO before 1980, offered to organize a replacement IMO to be held in Mersch, a small city in the centre of the country. Four European countries [Belgium, Great Britain, Netherlands, Yugoslavia, Luxembourg - these are five, I guess Luxembourg doesn't count] accepted the invitation and sent an official team to Mersch. A similar replacement IMO was organized in Mariehamn (Finland) at the same time.[/quote]

  dg

Hi Darij,

but I think there are different opinions why there was no IMO in 1980.

Some people say due to a civil war in Mongolia it had to be cancled there. Therefore replacement IMOs were carried out as a substitute. But it it not true that there were only local replacement IMOs in Finnland and Luxembourg. In order to investigate this matter I had a look at some early volumes of Crux Mathematicorum ( http://journals.cms.math.ca/CRUX/ ). And I realized that there were further replacement IMOs. I think Crux also published the problems of all these replacement IMOs but I am not sure now. If anybody is interested I will post all these problems from the various replacement IMOs in 1980.

Other people say that some countries tried to introduce the tradition that IMO should only take place every two years. Mostly communistic countries were interested in that idea. They wanted to reduce costs as it would not be an annual event then. Moreover they could train and prepare their students more intensively. Last but not least it would increase the value of an IMO medal. But I am sure that Germany was not interested in that idea because there was a German TST in 1980.

But in my opinion I think it is a excellent idea that the IMO takes place annually. What do the others think ? I think the designated IMO2004 members of the forum would not appreciate it too much not to have an IMO this year, wouldn't you ?  :D

[quote="darij grinberg"]Thanks for the information, IMO history seems to be a very interesting topic :-)

Of course, I am glad to have the IMO taking place anually.

  Darij[/quote]

Who would not be :D

I think without IMO, olympiad stuff would not be arranged enough even to open a site like this :D

Hey Orl, it may be a good idea to post the problems like you said. Can you do this please? :) I didn't really find them all together but I'm sure you know good locations for that ;)

Hi Peter,

I will look for them and post them in due course.  :)

Since there was not any International Mathematical Olympiad in 1980, some regional European olympiads took place. By the way it was the first break of this annual competition after 20 consecutive years. Up to now I only can find the problem sets of these three replacement IMOs: Austrian-Poland, Mersch (Luxembourg) and Mariehamn (Finland).

The "Organisation Of The Teachers Of Mathematics, Physics And Chemistry Of Finland (MAOL) invited the teams of England, Hungary and Sweden to show their skills in Mariehamn (Finland).

[b]1st day, July 1, 1980, 4 hours[/b]

[b]F1[/b]: Let $\alpha, \beta$ and $\gamma$ denote the angles of the triangle $ABC$. The perpendicular bisector of $AB$ intersects $BC$ at the point $X$, the perpendicular bisector of $AC$ intersects it at $Y$. Prove that $tan(\beta) \cdot tan(\gamma) = 3$ implies $BC= XY$ (or in other words: Prove that a sufficient condition for $BC = XY$ is $tan(\beta) \cdot tan(\gamma) = 3$). Show that this condition is not necessary, and give a necessary and sufficient condition for $BC = XY$.

[b]F2[/b]: Define the numbers $a_0, a_1, \ldots, a_n$ in the following way:

\[ a_0 = \frac{1}{2}, \quad a_{k+1} = a_k + \frac{a^2_k}{n} \quad (n > 1, k = 0,1, \ldots, n-1).  \]

Prove that \[ 1 - \frac{1}{n} < a_n < 1.\]

[b]F3[/b]: Prove that the equation \[ x^n + 1 = y^{n+1}, \] where $n$ is a positive integer not smaller then 2, has no positive integer solutions in $x$ and $y$ for which $x$ and $n+1$ are relatively prime.

[b]2nd day, July 2, 1980, 4 hours[/b]

[b]F4[/b]: Determine all positive integers $n$ such that the following statement holds: A convex polygon with with $2n$ sides $A_1, A_2 \ldots A_{2n}$ is inscribed in in a circle and $n-1$ of its $n$ pairs of opposite sides are parallel which means if the pairs of opposite sides

\[(A_1 A_2, A_{n+1} A_{n+2}), (A_2 A_3, A_{n+2} A_{n+2}), \ldots , (A_{n-1} A_n, A_{2n-1} A_{2n})\] are parallel, then the sides \[A_n A_{n+1}, A_{2n} A_1\] are parallel as well.

[b]F5[/b]: In a rectangular coordinate system we call a horizontal line parallel to the $x$ -axis triangular if it intersects the curve with equation \[y = x^4 + px^3 + qx^2 + rx + s\] in the points $A,B,C$ and $D$ (from left to right) such that the segments $AB, AC$ and $AD$ are the sides of a triangle. Prove that the lines parallel to the $x$ - axis intersecting the curve in four distinct points are all triangular or none of them is triangular.

[b]F6[/b]: Find the digits left and right of the decimal point in the decimal form of the number \[ (\sqrt{2} + \sqrt{3})^{1980}. \]

Another replacement IMO was held in Mersch (Luxembourg). The four countries Belgium, Great Britain, Netherlands and Yugoslavia accepted the invitation and sent an official team to Mersch. It was possible to send observing students who were allowed to participate without being classified. France did not send an official team but rather four observing students.

[b]1st day, July 10, 1980, 4 hours[/b]

[b]L1[/b]: (GB-4): The function $f$ is defined on the set $\mathbb{Q}$ of all rational numbers and has values in $\mathbb{Q}$. It satisfies the conditions $f(1) = 2$ and $f(xy) = f(x)f(y) - f(x+y) + 1$ for all $x,y \in \mathbb{Q}$. Determine $f$.

[b]L2[/b]: (LU-3):
Remark: By definition, the $n-th$ row of Pascal's triangle contains the $n+1$ binomial coefficients $\binom{n}{k}$ with $k = 0,1, \ldots ,n$.

[b]2nd day, July 11, 1980, 4 hours[/b]

[b]L4[/b]: (BE-3): Two circles $C_1$ and $C_2$ are (externally or internally) tangent at a point $P$. The straight line $D$ is tangent at $A$ to one of the circles and cuts the other circle at the points $B$ and $B$. Prove that the straight line $PA$ is an interior or exterior bisector of the angle $\angle BPC$.

[b]L5[/b]: (NL-2): Ten gamblers staredt playing with the same amount of money. Each turn they cast (threw) five dice. At each stage the gambler who had thrown paid to each of his 9 opponents $\frac{1}{n}$ times the amount which that opponent owned at that moment. They threw and paid one after the other. At the Three points $A,B,C$ are such that $B \in ]AC[$. On the side of $AC$ we draw the three semicircles with diameters $[AB], [BC]$ and $[AC]$. The common interior tangent at $B$ to the first two semi-circles meets the third circle in $E$. Let $U$ and $V$ be the points of contact of the common exterior tangent to the first two semi-circles. Denote the are of the triangle ABC as area(ABC). Then please evaluate the ratio $R = area(EUV)/area(EAC)$ as a function of $r_1 = \frac{AB}{2}$ and $r_2 = \frac{BC}{2}$.

[b]L3[/b]: (YU-1): Let $p$ be a prime number. Prove that there is no number divisible by $p$ in the $n-th$ row of Pascal's triangle if and only if $n$ can be represented in the form $n = p^sq - 1$, where $s$ and $q$ are integers with $s \geq 0, 0 < q < q$.
10th round (i.e. when each gambler has cast the five dice once), the dice showed a total of 12, and after payment it turned out that every player had exactly the same sum as he had at the beginning. Is it possible to determine the total shown by the dice at the nine former rounds ?

[b]L6[/b]: Find all pairs of solutions $(x,y)$:

\[ x^3 + x^2y + xy^2 + y^3 = 8(x^2 + xy + y^2 + 1). \]

[b]Replacement IMO Austria-Poland[/b]

[b]1st day, July 3, 1980, 4.5 hours[/b]

[b]AP1[/b]: Given three infinite arithmetic progressions of natural numbers such that each of the numbers 1,2,3,4,5,6,7 and 8 belongs to at least one of them, prove that the number 1980 also belongs to at least one of them.

[b]AP2[/b]: Let $\{x_n\}$ be a sequence of natural numbers such that \[(a) 1 = x_1 < x_2 < x_3 < \ldots; \quad (b) x_{2n+1} \leq 2n \quad \forall n.\] Prove that, for every natural number $k$, there exist terms $x_r$ and $x_s$ such that $x_r - x_s = k.$

[b]AP3[/b]: Prove that the sum of the six angles subtended at an interior point of a tetrahedron by its six edges is greater than 540.

[b]2nd day, July 4, 1980, 4.5 hours[/b]

[b]AP4[/b]: Prove that $\sum \{\frac{1}{i_1i_2 \ldots i_k}\} = n$ is taken over all non-empty subsets $\{i_1,i_2, \ldots, i_k\}.$

[b]AP5[/b]: Let $A_1A_2A_3$ be a triangle and, for $1 \leq i \leq 3$, let $B_i$ be an interior point of edge opposite $A_i$. Prove that the perpendicular bisectors of $A_iB_i$ for $1 \leq i \leq 3$ are not concurrent.

[b]AP6[/b]: Given a sequence $\{a_n\}$ of real numbers such that $|a_{k+m} - a_k - a_m| \leq 1$ for all positive integers $k$ and $m$, prove that, for all positive integers $p$ and $q$, \[|\frac{a_p}{p} - \frac{a_q}{q}| < \frac{1}{p} + \frac{1}{q}.\]

[b]3rd day, July 5, 1980, 4 hours[/b]

[b]AP7[/b]: Find the greatest natural number $n$ such there exist natural numbers $x_1, x_2, \ldots, x_n, a_1, a_2, \ldots, a_n$ satisfying the following sum of equations for $i =1,2,\ldots,n-1$: \[ x_1x_2 \ldots x_n = 1980 \quad \text{and} \quad x_i + \frac{1980}{x_i} = a_i .\]

[b]AP8:[/b] Let $S$ be a set of 1980 points in the plane such that the distance between every pair of them is at least 1. Prove that $S$ has a subset of 220 points such that the distance between every pair of them is at least $\sqrt{3}.$

[b]AP9[/b]: Let $AB$ be a diameter of a circle; let $t_1$ and $t_2$ be the tangents at $A$ and $B$, respectively; let $C$ be any point other than $A$ on $t_1$; and let $D_1D_2. E_1E_2$ be arcs on the circle determined by two lines through $C$. Prove that the lines $AD_1$ and $AD_2$ determine a segment on $t_2$ equal in length to that of the segment on $t_2$ determined by $AE_1$ and $AE_2.$

I have uploaded the file with all the IMO 1980 problem sets I could find up to date. BTW this year in greece there are three competitions days at IMO with 9 problems as at the replacement IMO of Austria and Poland. Did you know that yet ?  :D

[quote="orl"]BTW this year in greece there are three competitions days at IMO with 9 problems as at the replacement IMO of Austria and Poland. Did you know that yet ?  :D[/quote]

Hh, was that serious?

  dg

Hmm, it was rather a joke. But it is worth to be taken into account. It makes it easier to evaluate and distinguish the participants' performances.  :)


A cylindrical bucket without a top has a circumference of 6 inches and a height of 4 inches. On the inside of the bucket, one inch from the top, is a drop of honey; on the opposite side of the bucket, one inch from the bottom, on the outside, is an ant. What is the shortest distance the ant could crawl to reach the honey?

A. $ 5$
B. $ 5\sqrt{2}$
C. $ 7$
D. $ \sqrt{13}$

[hide="Solution."]
Not taking into consideration the bottom of the bucket yet, you can represent the sides of the bucket as a $ 6$ by $ 4$ rectangle, with the horizontal distance between the ant and the honey as $ 3$ and the vertical distance as $ 2$.  By the Pythagorean Theorem, the distance between them comes out to $ \sqrt {13}$.  However, this does not take into account the possibility that the ant traverses the bottom of the bucket.  The shortest such path involves climbing down $ 1$ inch, walking across the diameter of the bucket ($ \frac {6}{\pi}$ inches), and climbing up to the honey ($ 3$ inches).  This is greater than the previous solution, so the answer is $ \boxed{\sqrt {13}}$.[/hide]


geometry
limit
trigonometry
inequalities
triangle inequality
complex analysis

this was a sort of practice rmo for me. and in the first of the 6 q's , i think there was some problem  :maybe: . it said "perpendicular from B to BC". i cud do only 3 out of the rest 5 which were 4,5,6 . Combinatorics question was too easy , similarly whichever questions i did were quite easy i think.

what did  u guys do? :icecream:

sorry to ask but what was the answer to combinatorics question

mine was 310 but my friends differ(i must have made claculation mistake)

i could solve 4,5 completely(if 310 is right) and some other half answers
i ran out of time later

and a unique thing,i read ques 6 as to find all obtuse angled triangle
i missed to read isosceles :(

Skand, I'm afraid mine and a friend's match at $ 490$ :( . That could be wrong though. Maybe I didn't remember well and later thought mine was the same when I heard from my friend.

Anyway the questions are posted online, for ppl who haven't sat for RMO:
[url=http://www.mathlinks.ro/viewtopic.php?p=1311928]Problem 1[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=238344]Problem 2[/url]
[url=http://www.mathlinks.ro/viewtopic.php?p=1311934]Problem 3[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=238346]Problem 4[/url]
[url=http://www.mathlinks.ro/viewtopic.php?p=1311919]Problem 5[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=238348]Problem 6[/url]

I did well but was so stupid that I overlooked the isosceles cond. in the last and thought on it for an hour.

i did so and gave a lot of time on it

futile task

i must be wrong  :(

man..... i couldnt do a single question fully........
i dont know wat went wrong...... geometry question was simple but i just missed on one point.....

ya i got it.......a typo............ :rotfl:  :rotfl:

i realize that i did a silly mistake  :wallbash_red: due maybe lack of practice or ignorance towards combi. but its quite frustrating to do those silly mistakes because even if u know the concept you dont get to the answer.

I got 4 correct.
But, this was my first time.
Could you please tell me how much my chances are of getting selected if i score around 65/100

well it depends probably on the region u r living in i suppose..........

but still ...........65 or 4 crrect are marks which can make u cross the regional olympiad.......... :P  :D

Well Arnav, I hope it is not 65 mark cutoff for 9th stds.

I didn't do that well....I am getting something like 55 marks.
I got 2 fully + 2 partial.

I did OK. Comparatively easier conceptually as compared to last year..

Not to discourage or disappoint you people, but having experienced it myself, i have to say that it does not matter how many questions you got. Even if you got one question and you gave a rigorous solution to it can make you through rmo or even inmo. Last year the highest in my region was 50 and it was 51 or something in orrisa region also. The cutoff was between 25 and 30. That shows that you hardly ever get cent per cent in a particular question.

[hide]That being said, i wrote 4 solutions that may be worth noticing. :P
When i mean worth noticing, it can be right or wrong, i just made it look worth noticing, if you know what i mean  :wink: [/hide]

yep hell ever is right.
dont expect a lot of marks for partial solutions :rotfl:  :rotfl:( Don't get dejected for this because last year i had only 1 problem perfectly right and of coz many partial and still made it to the INMO. I am telling this out of my experience.)
yeah but i think the cut off will still be the usual 20-25.

i posted my solutions for 2,3,6. many got 4 and 5.
i did not try doing 1 after seeing akashnil's brilliant construction.
but since the question was printed wrong maybe bonus for everybody

we will hav to revert to sm other language

but i do kno a bit of bengali :P  :rotfl: "tumi chool watever :rotfl: "

Hey what does "tumi chool" mean :P ? "tumi"="you", "chool"="hair". And together:
[hide]"you are hair" :P  :D  :rotfl: ("are" is generally omitted in bengali)[/hide]
BTW Where did you learn Bengali ???

hey! don't u guys think that this time there is a little delay in the release of official solutions!!!

right... thr is...
evry1 wants to kno whether thr answers r officially rong or rite!!

[quote="rituraj007"]we will hav to revert to sm other language

but i do kno a bit of bengali :P  :rotfl: "tumi chool watever "[/quote]

Hey i am sure u know absolutely nothing about bengali.
You just know that "tumi" word!

[hide="to rituraj"]Tumi fool. Tumi jhool! :rotfl:

And i would request akashnil or any other bengalee not to translate, please! :P [/hide]

Moreover rituraj, r u some sorta secret agent or something like that?
Or r u in the line for James Bond movies after Daniel Craig's tenure??! :rotfl:

[quote="apratimdefermat"]And i would request akashnil or any other bengalee not to translate, please! :P [/quote]
Why?? :maybe:
Coming back to the topic, I don't think they are late to publish the results yet.

Apratim, r u that same guy who was there at  KV inmotc in delhi last year from probably kolkata region.

[quote="keshav_57"]akashnil... ,
not late for results...
but late for solutions...[/quote]Oh. I misread it. Yes they should have published by now. Maybe they'll put up the best solutions from the answer-sheets? :rotfl:

actually i kno a lot abt bengali :P
not the language though
[hide]i [b]m :P [/b]bond's junior assistant :rotfl: [/hide]

the soln r too late this yr...

Is there anybody from Chattisgarh region

Anybody frm M.P. here??? Atleast i wud know the name of one getting selected in this year's RMO :P

does anybody know if rmo results are out yet in Kolkata?

Hey! guys, official solutions are out, check them and post ur expected marks!!!!
http://www.isibang.ac.in/~statmath/crmosol_08.pdf

who guys are selected for INMO??

What are the most important topics that i hav to prepare???

its urgent


i have another question about online recommendations.

it says status : notification sent to recommender.

does the status change to something like "letter receieved" when the recommender replies to the mathcamp's notification and actually sends in the recommendation letter?

im just wondering because it's getting close to deadline and i have emailed both my recommenders, they have not emailed me back,and the statuts still displays notification sent.. not "letter received" or something.

[quote="Susah"]
does the status change to something like "letter receieved" when the recommender replies to the mathcamp's notification and actually sends in the recommendation letter?[/quote]

Yes. It changes to "Received from _____"... and you also get a notification by e-mail.

[quote="mcalderbank"]Online, I filled in information for academic recommender, and the database sent the email to my teacher, but he could not open the link; the link was dead. Knowing that this is the best place to ask questions:  What is going on?[/quote]

The database was down for a couple of stretches yesterday and the day before -- it's should working now, and we'll be keeping an eye on it.


For any odd integer  $\large a_{0} \geq  5$, consider the sequence $\large a_0, a_1, a_2,. . .$ where:
$\large a_{n+1}= a^2_{n}-5$ if $\large a_{n}$ is odd
          $\large \frac{a_{n}}{2}$ if $\large a_{n}$ is even
for all $\large n \geq  0$
Prove that this sequence is not bounded.

That's false. If we take $a_{0}=5$, we have $a_{1}=20$, $a_{2}=10$,$a_{3}=5$, and so on, which is obviously bounded.


Suppose that g1(x) is a right end behavior model for f1(x) and that g2(x) is a right end behavior model for f2(x). Explain why this makes g1(x)/g2(x) a right end behavior model for f1(x)/f2(x)

No calculus and no homework problems, please.


geometry
combinatorics proposed
combinatorics

A rectangularly paper is divided in polygons areas in the following way: at every step one of the existing surfaces is cut by a straight line, obtaining two new areas. Which is the minimum number of cuts needed such that between the obtained polygons there exists $251$ polygons with $11$ sides?

[hide="Answer"]$2007$ :) [/hide]
[hide="Solution"]
Indeed, we can obtain the polygons from $2007$ cuts: we first obtain $251$ quadrilaterals from the intial one with $250$ cuts. Observe that each $n$-gone can be decomposed in a $(n+1)$-gone and a triangle (cut the interior of two consecutive sides). So each quadrilateral can be processed up to a $11$-gon with $7$ cuts. We have a total of $250+251 \cdot 7 = 2007$ cuts.
For the second part of the proof (that $2007$ is minimum), denote $n$ the minimum number of cuts needed to obtain the polygons, $v$ the total number of vertices and $k$ the number of polygons which are not $11$-gons. So $n = 250+k$, and each of the $n$ cuts adds at most $4$ vertices, hence $4n+4 \geq v \geq 3k+251 \cdot 11$, which leads to $4n+4 \geq 3 \cdot (n-250)+251 \cdot 11$ $\Leftrightarrow$ $n \geq 2007$. This completes our proof.
[/hide]


Take square ABCD

Would it have 4 diagonals or 2?

4 if:

AC,BD,CA,DB

2 if:

AC,BD

Does this change when there is another polygon?

No, its two.
They have to form a region thats area is greater than zero.
One example:  (Ignore the other stuff.)
[img]http://www.mathacademy.com/pr/minitext/number/numb20.gif[/img]
If you have further questions, click [url=http://en.wikipedia.org/wiki/Diagonal]Here[/url]

[b][i]-minicon[/i][/b]

Mods, please delete the other one.  :D

For any polygon with $ n$ sides, where $ n\ge{4}$, there are $ \frac {n(n \minus{} 3)}{2}$ diagonals.

[hide="proof"]You can pick a vertex in $ n$ ways. You can attach it to another vertex in $ n \minus{} 3$ ways (besides the two adjacient ones, and the original). But we count each twice, for each vertex. So it's $ \frac {n(n \minus{} 3)}{2}$.[/hide]

Isn't segment AC not only congruent to CA, but also the very same segment?
What about BD and DB ? Hm...

Indeed, that's why there  are only 2 diagonals in quadrilaterals.


algebra
polynomial
absolute value
algebra unsolved

A polynomial $ P(x)$ of degree $ n\geq5$ with integer coefficients has $ n$ distinct integer roots $ x_1 \equal{} 0$, $ x_2, x_3, \ldots, x_n$. Find all integer roots of $ P(P(x)).$

[quote="pluristiq"]what do you mean "solution of $ P(P(x))$"? roots?[/quote]

Yes, roots of $ P(P(x))$. I edited the problem.

all solutions of $ P(P(x))\equal{}0$ come from the equations

$ P(x)\equal{}0$

$ P(x)\equal{}x_2$
...
$ P(x)\equal{}x_n$

the solutions of the first are $ 0, x_2, ..., x_n$, which are all integers. now we'll show that the other equations have no integer solutions. We'll show it for $ P(x)\equal{}x_2$; the reasoning for the others is the same. Assume, for a contradiction, that $ P(r)\equal{}x_2$ for some integer $ r$. we then have $ P(r)\equal{}ar(r\minus{}x_2)(r\minus{}x_3)\cdots (r\minus{}x_n)\equal{}x_2$ for some integer $ a$. then $ r$ must be strictly between 0 an $ x_2$, otherwise either $ r$ or $ r\minus{}x_2$ has absolute value greater than $ x_2$ and the above equality is impossible. now, for $ r$ between $ 0$ and $ x_2$, $ \text{min}_{r\in \mathbb{Z}} |r(r\minus{}x_2)|\equal{}|x_2|\minus{}1$. since the $ x_i$ are distinct integers and $ n\geq 5$, at least one of $ |r\minus{}x_3|,...,|r\minus{}x_n|$ must be at least 2. therefore, $ |x_2|\equal{}P(r)\geq 2(|x_2|\minus{}1)\equal{}2|x_2|\minus{}2$, so $ |x_2|\leq 2$. but since $ r, r\minus{}x_2, ...,r\minus{}x_n$ are all distinct non-zero integers and $ n\geq 5$, $ |x_2|\equal{}|P(r)|\equal{}|r(r\minus{}x_2)\cdots (r\minus{}x_n)|\geq 12$, a contradiction.


function
calculus
derivative
calculus computations

In some textbooks, it states that a stationary point $ x_0$ of a univariable function $ f(x)$ is a strict minimizer if $ f'(x_0)\equal{}0, \ldots, f^{(2n\minus{}1)}(x_0)\equal{}0$ and $ f^{(2n)}(x_0)>0$.

for example,

http://books.google.com/books?id=UculnMy2GBoC&lpg=PP1&dq=Mathematical%20Analysis%20Explained&pg=PA74

It seems that this proposition has a multivariable version, that is, a stationary porint $ \mathbf{x_0}$ of a multivariable function $ f(\mathbf{x})$ is a strict minimizer if $ d f(\mathbf{x_0})\equal{}0, \ldots, d^{(2n\minus{}1)}f(\mathbf{x_0})\equal{}0$ and $ d f^{(2n)}(\mathbf{x_0})>0$, where the symbol $ `>`$ here means positive definiteness of the differentiation.

The proof seems similar and straightforward, but I want to find a book containing such statements, Can anyone tell me? Many thanks.

I'm not sure anybody would bother including such statement into his book as a theorem. It is correct, no doubt, but almost completely useless without an efficient way to check positive definiteness of the 4th and higher order forms. The best you can hope for is to find it as a side remark.


analytic geometry
FTW
geometry
3D geometry
prism
Pythagorean Theorem

[quote="Jongao"]What does it mean when a coordinate is (0,0,9) or (6,5,4)?[/quote]

I guess it's also height of it. In 3-D, like the z-axis.

It's hard to explain

When you have a regular coordinate in the 2-dimensional plane, it looks like (3, 4).  This tells us to go 3 units in the positive x-direction, and 4 units in the positive y-direction.  This coordinate is equivalent to the coordinate (3, 4, 0), which tells us to go 3 units in the positive x-direction, 4 units in the positive y-direction, and 0 units in the positive z-direction.  The coordinate (3, 4, 6) tells us to go 3 units in the positive x-direction, and 4 units in the positive y-direction, and 6 units in the positive z-direction.

For example, if I was piloting a helicopter, and I started at the origin (0,0,0) and you told me to go to the point (1, 2, 3) I would go right 1 unit, forward 2 units, then up in the air to a height of 3 units.

I thought the x axis was forward, the y axis was horizontal, and the z axis was vertical.

It doesn't really matter, that's just convention.

And I would just like to add...

To find the distance of (a,b,c) from the point of origin, you compute:

$ a^2 \plus{} b^2 \equal{} x^2$

$ x^2 \plus{} c^2 \equal{} (\text{distance of origin})^2$

For example (an FTW problem), what is the longest thing (I forgot what it was, but 1 dimensional) that you can place inside a prism that measures 3 feet by 4 feet by 12 feet?

$ a \equal{} 3, b \equal{} 4, c \equal{} 12$ (order does not matter)

$ 3^2 \plus{} 4^2 \equal{} x^2$

$ x \equal{} 5$

$ 5^2 \plus{} 12^2 \equal{} d^2$

$ d \equal{} 13$

Therefore, the answer is $ \boxed{13}$.

[quote="ernie"]

For example (an FTW problem), what is the longest thing (I forgot what it was, but 1 dimensional) that you can place inside a prism that measures 3 feet by 4 feet by 12 feet?
[/quote]

What is the longest, rigid, thinnest [insert 1000000 adjectives here] rod that one can place inside a prism that measures 3 feet by 4 feet by 12 feet?

Or something like that.

ernie, you could just generalize the pythagorean theorem to $ n$-dimensions.  If we want to find  the (euclidean) distance from $ (0,0,0\ldots)$ to $ (a_1, a_2, \ldots, a_n)$ we just compute $ \sqrt{\sum_{i\equal{}1}^n a_i^2}$.  In three dimensions, this means the length of the longest diagonal of an $ x$ by $ y$ by $ z$ prism is $ \sqrt{x^2\plus{}y^2\plus{}z^2}$.


\[ x^3\plus{}21y^2\plus{}5\equal{}0\] for integer values of $ x,y$.

[hide]Consider mod 7 - No solutions[/hide]

[hide="More specifically..."]
$ 21y^2 \equiv 0 \pmod 7$
$ x^3 \equiv 0,1,6 \pmod 7$
$ 5 \equiv 5 \pmod 7$
$ x^3 \plus{} 21y^2 \plus{} 5 \equiv 5,6,4 \pmod 7$

So they can never equal 0 mod 7, and since 0 is equivalent to 0 mod 7, there are no solutions.
[/hide]

[color=darkred]Edit: fixed :)[/color]

Yongyi781, you forgot something:
[hide]$ x^3$ can be $ \equiv 1\pmod 7$ as well, making $ x^3\plus{}21y^2\plus{}5\equiv 6\pmod 7$.  Either way, you still can't get 0.[/hide]


geometry
3D geometry
AMC
AIME
search
function

A $154 \times 324 \times 375$ rectangular solid is made by gluing together $1 \times 1 \times 1$ cubes. An internal diagonal of the rectangular solid passes through the interiors of how many of the cubes?

Can someone offer a detailed explanation? I'm stumped :?

AIME 1996 #14

http://www.artofproblemsolving.com/Forum/topic-73967-126.html
http://www.artofproblemsolving.com/Forum/topic-81929.html

Thank you JBL, unfortunately the search function was going very slow for me :?


HMMT
search
geometry
geometric transformation
FTW

Depends a lot on what you want to do with it.  Intellectual satisfaction or a practical purpose (and if practical, do you intend on doing business or traveling)?

I would like the satisfaction, yes. But I would like to, say, communicate with people at HMMT, for example.(If I make it, at IMO as well.)

Why not learn German?  I hear that a lot of work in math and the hard sciences is written in German, so you could get a lot of mileage out of it that way if you wanted.  Plenty of good literature, plenty of good music (and not just the Wagner variety-- one of my fellow students won a huge research award for his work on the German punk movement!)  It would also open the possibility of studying at a German university.  Plus, German, Germany, and Germans are just generally pretty snazzy.   :lol:

For scientific purposes, Latin is the way to go.

For entertainment purposes, Spanish is fun.

Hmm if you want satisfaction, don't suffer through Spanish. Though I guess I am the only person who thinks Spanish is ugly. For practical purposes though, Spanish is sadly the way to go. French and Italian are cool. French is also very useful for math. Or something unusual, just for the sake of being unusual :).

Just pick a language you're excited about.  It's total misery to slog through months or years of studying a language you don't like.

If you want something unusual, and don't mind taking summer classes...

Indiana University offers summer classes in languages of Central and Eastern Europe, the Caucuses, and Central Asia:
[url]http://www.indiana.edu/~iuslavic/swseel/[/url]

The University of Wisconsin-Madison offers summer classes in South Asian languages (including Sanskrit and Tibetan):
[url]http://sasli.wisc.edu/index.htm[/url]

Middlebury College offers summer immersion classes in most mainstream languages (but these are very expensive and competitive):
[url]http://www.middlebury.edu/academics/ls/[/url]

The University of Texas at Austin offers a summer class in beginning ancient Greek:
[url]http://www.utexas.edu/depts/classics/UG_Page/sumintensegk.html[/url]

If you don't mind teaching yourself, the University of Texas at Austin's Linguistics Research Center has online lessons in a number of dead Indo-European languages, like Hittite, Old Norse, and so on (free!):
[url]http://www.utexas.edu/cola/centers/lrc/eieol/armol-0-X.html[/url]

And how about Portuguese? Not only it is rich in sounds but also the writing can be quite difficult - this because the same letter can be used for different sounds and in the same word. As an example, consider the word "Souselas" (the name of a small village) - all three "s" have different sounds.

[quote="Jb05"]Though I guess I am the only person who thinks Spanish is ugly.[/quote]

You are not.

[quote="1=2"][quote="7h3.D3m0n.117"]Or to help you in the future, you can learn Chinese, which I heard will be very useful since many businesses will have some Chinese connection (unless you already know that language).[/quote]

That's what Dad is making me learn. One look at Simplified or Traditional Chinese, and I decided to go with Pin Yin for now. :)[/quote]

I hate Pin Yin! thats not even chinese...traditional is the real chinese

[quote="Field"]Why you guys think spanish is nasty? :([/quote]

I don't.   :)

IMHO, don't learn a language for practical use. I hardly know anyone that succeeds. Especially through school... you simply won't master any language enough, except for English, to use it practically. In the real world (that is, the imaginary real world of my mind), basically, you'll be able to say "hi", and then the native speaker will say something that was not one of the standard responses you learned, and then you will be totally lost.

It takes serious effort to learn a language to be able to use it practically.

go with traditional chinese!! cuz if u learn that u can read simplified, but if u only know simplified it's harder to read traditional. plus me learns traditional! :D

I'm learning hanguk (Korean) for K-(Fashion,Pop,Drama,...)  :lol:
I Love KOREA

Latin, (ancient) Greek, French, German...  If you learn both Latin and German, you can really see where English comes from.  (Plus, they're both awesome!)

[quote="elevate"]I'm obsessed with Asian languages :)
I know Cantonese and simplified Mandarin, and I'm shooting for...Korean, maybe?[/quote]
Japanese FTW

In terms of langauges, learn the ones that are fun  for you. I'm learning Japanese and all of my family says, "Learn Chinese, it's more practical, you'll never use Japanese".
But what's the point of learning Chinese if I won't have fun doing so.

My brother learned Chinese in school, even though everybody said "It was too hard".

Just look up a language on wikipedia or on google. Learn some sentences. If you like the language, then start learning it. If not, there's only 54256465464564564656345235 other languages to try. You'll find one you like eventually (even if it's something like !Xianese)

[quote="1=2"]See topic title and description.[/quote]
Korean's a good language. It's awesome. (Well, im not really good at korean)

I'd go for Latin, actually.  Latin influenced many languages, and so learning Latin could give you a start in Spanish, Italian, etc.

http://www.allthelyrics.com this will bring you a lot of fun

Finnish is my favourite language. :)

It’s a language many people think difficult. Perhaps the most notorious feature of Finnish is its plurality of cases – some dozen or so of them in the modern language. Another difficult aspect of Finnish grammar is the set of rules on consonant gradation. But then rules are what you have to expect when learning any new language. The thing with Finnish grammar is that it is very regular: one you’ve learnt the rules you can apply them with very few exceptions.

And its high grammatical regularity extends to its pronunciation. Not only is Finnish a very sonorous language, its pronunciation is also extremely easy to learn. The stress is always on the first syllable, and there is a bijection :roll: between its sounds and its spelling: Finnish words are always pronounced as they are spelt, and spelt as they are pronounced.

Finnish is a great language!  I'm learning it, too.  Some things about it are surprisingly easy (no gender, no future tense, no difficulty whatsoever with pronunciation), while other things (like all the case endings, and like the fact that there are hardly any cognates with other languages) take a long time to master.

If you don't already know this resource, here is a link to daily news in a simplified version of Finnish; I've been reading and listening to this daily for a couple years now; it's really great for building up a basic vocabulary and developing listening comprehension.

http://www.yle.fi/selkouutiset/index.php?1334

Ok, So you want to learn a language...
Biased Opinion Here, but go Bosnian! Pretty easy to learn in my opinion, and many languages are similar (Romanian, Russian, Croatian, Serbian, other Eastern European Countries)
On a more realistic note, Latin is really good if you want to learn other languages faster because of the roots.

If you don't  really want/need to speak for business purposes, Ancient Greek would be really cool...

[quote="sunny2000"]Korean is a great language. Because Asian power and influence is getting bigger, it's great to learn Korean. Languages like Chinese not only have words to pronounce but they have characters that each mean a different thing. Korean is hard i guess, but it's a great way to get used to Asian culture. Also, you can finally understand songs like Gangnam Style and songs by Bigbang or G Dragon. :P[/quote]
I've heard korean is one of the easiest asian language. Also, spanish is cool too :D. I know English, Korean, and Spanish.

Again, just find a  thread and follow it. Check out some languages, see what you like. Maybe find some stuff on Google Translate (as inaccurate as it is) to just check out if you like how it sounds. Personally, I want to learn Italian and Greek, and I intend to take French in school. Each language has a different nuance and melody, find your favorite, and then try to learn it!

[quote="Nehushtan"]Finnish words are always pronounced as they are spelt, and spelt as they are pronounced.[/quote]
Well, I pronounce for example the word "Tervetuloa" (=welcome) more like "Tervettuloa" and the same with some other words. Also, one spells the letter "g" differently if there is "n" before "g" or not. For example, g in words "gaselli" (=gazelle) and "kuningas" (=king) are pronounced differently. Also "np" is pronounced like "mp", for example the word "kunpa" (=if only). There might be other special cases but nothing particular springs to my mind right now.

I recommend japanese!

they have some really good stories. Granted, a lot of the light novels are pretty trashy but if you dig deep enough you'll definitely find something you like.
(If you want my recommendations, send me a PM and I'll reply back)

i'm considering learning it myself. right now, I have to wait for other people to translate the novels into English first. The more popular ones get translated pretty quickly, but for the others...

plus, if you're into anime then you won't have to deal with subs or poorly dubbed versions


Find the largest possible number of elements of a subset of {1, 2, 3, ... , 1989} with the property that no two elements of the subset have difference 4 or 7. obviously, a proof is needed here

Is-it [hide]996 [/hide]?

If yes I can post my solution (which is a bit verbose   )

this was around a while ago. i think i solved it (correctly). try a search.

The solution is really easy and just requires the observation that you can find a repeating unit.

taken off the kalva site:

<i>13.   can anyone see a neat way to do this? </i>

Apparently its not that easy now is it?


solve:
$ 4(x \plus{} y \plus{} z) \equal{} xyz$

EDIT: Find all possible solutions (x,y,z)

[quote="ajai"]solve:
$ 4(x \plus{} y \plus{} z) \equal{} xyz$[/quote]
[hide="solution"]
I'll assume that i'm supposed to be solving for $ x$
$ 4x \plus{} 4y \plus{} 4z \equal{} xyz$
$ 4x \minus{} xyz \equal{}\minus{} 4y \minus{} 4z$
$ x(4 \minus{} yz) \equal{}\minus{} 4y \minus{} 4z$
$ x \equal{} \frac {4y \plus{} 4z}{yz\minus{}4},yz \ne 4$
[/hide]

[url=http://www.mathlinks.ro/viewtopic.php?t=171813]For $ x,y,z \in \mathbb{Z}^{\plus{}}$.[/url]

[quote="mihail911"]i don't know why you would assume that. i mean you could just as easily be asked to solve for y or z. could you make the directions more explicit ajai?[/quote]
I assumed that because it's the first variable, and it's the same thing for any of them, so by convenience i did $ x$.

[quote="AstroPhys"][url=http://www.mathlinks.ro/viewtopic.php?t=171813]For $ x,y,z \in \mathbb{Z}^{ \plus{} }$.[/url][/quote]
[hide="solution"]
$ z \equal{} \frac {4(x \plus{} y)}{xy \minus{} 4}$
To find the answers, start plugging in integers for $ x$ and find an $ xy$ with an integer $ y$ that would make $ z$ an integer. Keep doing so until $ y$ or $ z$ is less than $ x$, in which case it means that it's repeating itself but in a different order.
when $ x$ is $ 1$, $ y$ could be $ 8$, in which case the denominator is $ 4$, and cancels out the factor in the numerator.
then the ordered triple is $ (1,8,9)$
$ y$ could also be $ 6$, in which case the denominator is also a definite factor of the numerator. Then the ordered triple is $ (1,6,14)$
$ y$ could also be $ 5$, in which case the denominator is $ 1$ and the numerator is $ 24$. Then the ordered triple is $ (1,5,24)$

When $ x$ is $ 2$, $ y$ could be $ 4$, in which case the ordered triple is $ (2,4,6)$. $ y$ could also be $ 3$, in which case the ordered triple is $ (2,3,10)$
When $ x$ is $ 3$, $ y$ could be $ 4$, but that doesn't work. Anything above won't work either because then $ xyz$ will always be greater than $ 4$ times the sum since increasing $ y$ or $ z$ any further will increase the product more than the sum.
The solutions collected so far are $ (1,8,9), (1,6,14), (1,5,24), (2,4,6), (2,3,10)$. There are $ 6$ ways to arrange each solution, so there are 30 solutions:
$ (x,y,z) \equal{} \begin{tabular}{c c c c c} (1,8,9) & (1,6,14) & (1,5,24) & (2,4,6) & (2,3,10) \\
(1,9,8) & (1,14,6) & (1,24,5) & (2,6,4) & (2,10,3) \\
(8,1,9) & (6,1,14) & (5,1,24) & (4,2,6) & (3,2,10) \\
(8,9,1) & (6,14,1) & (5,24,1) & (4,6,2) & (3,10,2) \\
(9,1,8) & (14,1,6) & (24,1,5) & (6,2,4) & (10,2,3) \\
(9,8,1) & (14,6,1) & (24,5,1) & (6,4,2) & (10,3,2) \\
\end {tabular}$
[/hide]


The electric company charges Elka 11 cents per kwh for the first 320 kwh, 10 cents per kwh for the next 100 kwh and 17 cents per kwh for anything over that. By how much does Elka's electric bill go up if her usage increases from 350 kwh to 450 kwh?

[hide="BIG HINT"]
$.11(320)+.10(350-320)$ for the first amount of kWh usage.
$.11(320)+.10(100)+.17(450-420)$ for the second amount of kWh usage.
Subtract.
[/hide]

[quote="alkjash"][hide]
Before the increase, Elka pays $.11(320)+.10(30) = 38.20$
After the increase, Elka pays $.11(320)+.10(100)+.17(30) = 50.30$
So her electric bill increases by $\boxed{\$22.10}$
You might as well have just solved it, AstroPhys.
[/hide][/quote]

Almost.  $50.30-38.2 = 12.1 \neq 22.1$


prove
a/(pb+qc) + b(pc+qd) + c/(pd +qa) + d/(pa+qb) >= 4/(p+q)for positive numbers a,b,c,dp,q satifuing p>=q
Does the inequality hold for p<q

Cezar, PLEASE [u]DON'T[/u] do that again.

Harazi just told you - http://www.mathlinks.ro/Forum/viewtopic.php?t=48926

Why were you ignoring his comment?


I'm not good in inequality so explain it :
Let $ x,y,z \in \mathbb{R}^\plus{}$ prove that: \[ \sqrt{\frac{2x}{x\plus{}y}}\plus{}\sqrt{\frac{2y}{y\plus{}z}}\plus{}\sqrt{\frac{2z}{x\plus{}z}} \le 3\]

Well, it's not true, let $ z \equal{} 1000$, $ x \equal{} y \equal{} 1$, then the third term clearly is larger than $ 3$.

Edit: Sorry, I thought that the inequality was symmetric. NVM.

[quote="tonaminet"]What if we flip the inequality?  Instead of less than equal to 3, greater than equal to 3[/quote]
Let x approach to zero and y=z, we have: LHS=1+sqrt(2)<3, so we can't flip the sign of the inequality above.

[quote="mathson"]I'm not good in inequality so explain it :
Let $ x,y,z \in \mathbb{R}^ \plus{}$ prove that:
\[ \sqrt {\frac {2x}{x \plus{} y}} \plus{} \sqrt {\frac {2y}{y \plus{} z}} \plus{} \sqrt {\frac {2z}{x \plus{} z}} \le 3
\]
[/quote]
[hide="proof"]By Cauchy-Schwartz we obtain:
$ \left(\sum_{cyc}\sqrt{\frac{2x}{x\plus{}y}}\right)^2\leq\sum_{cyc}2x(y\plus{}z)\sum_{cyc}\frac{1}{(x\plus{}y)(y\plus{}z)}.$
Thus, it remains to prove that $ 8(xy\plus{}xz\plus{}yz)(x\plus{}y\plus{}z)\leq9(x\plus{}y)(x\plus{}z)(y\plus{}z),$ which is obvious.
[/hide]

[quote="arqady"][quote="mathson"]I'm not good in inequality so explain it :
Let $ x,y,z \in \mathbb{R}^ \plus{}$ prove that:
\[ \sqrt {\frac {2x}{x \plus{} y}} \plus{} \sqrt {\frac {2y}{y \plus{} z}} \plus{} \sqrt {\frac {2z}{x \plus{} z}} \le 3
\]
[/quote]
[hide="proof"]By Cauchy-Schwartz we obtain:
$ \left(\sum_{cyc}\sqrt {\frac {2x}{x \plus{} y}}\right)^2\leq\sum_{cyc}2x(y \plus{} z)\sum_{cyc}\frac {1}{(x \plus{} y)(y \plus{} z)}.$
Thus, it remains to prove that $ 8(xy \plus{} xz \plus{} yz)(x \plus{} y \plus{} z)\leq9(x \plus{} y)(x \plus{} z)(y \plus{} z),$ which is obvious.
[/hide][/quote]
Don't you see the rofler's post, arquady?? He already shows the true counter-example for this inequality, but now you prove it(!?) :P

the rofler's post is not true. see here:
http://www.mathlinks.ro/viewtopic.php?p=120395#120395

That counterexample is incorrect, note that the inequality is asymmetric.

[quote="arqady"][quote="ghjk"]
Don't you see the rofler's post, arquady?? He already shows the true counter-example for this inequality, but now you prove it(!?) :P[/quote]
Yes, I have proved this inequality. :P[/quote]
Oopps I made a wrong calculation.Sorry arquady! :oops: . Let wait for me to give another proof as a"reward" :wink:


Let x1+x2+x3+...+xm=1.
Find the max and min of:
A= \sum x^n- \sum x^n+1 where m,n is integer which is \geq 1. :?

i think you should let xi>=0. (i=1->m)
for the max of A
i found that A max when and only when x3,x4,...,xm=0 and x1+x2=1.
so we need to find the max of x1x2(x1^n+x2^n) with x1+x2=1.
this' quite difficult for me.
when n=1-> 7 i can find the max but for n>7 i can't
so we have to solve the new problem
find the max of x1x2(x1^n+x2^n) where x1,x2 \geq 0 and x1+x2=1.


Correct :) But the full answer should include $a_{1}=0$


function
algebra
polynomial
limit
domain
complex analysis
complex analysis unsolved

How to prove one-to-one entire function can only be 1-order polynomial?

I'm sure there is a way to do this without Picard...but just for the fun of it.

Suppose the singularity at infinity is essential, then in a neighborhood $ f$ attains every value (except possibly one) infinitely many times, a contradiction.

Thus $ f$ is a polynomial.

If it has degree $ n$, then it is locally n-to-1. Thus it must be a polynomial of degree 1.

Picard is hugely overkill.

Since $ f^{\minus{}1}$ exists everywhere and is continuous, we have that $ f^{\minus{}1}$ of any compact set is compact. That shows that $ \lim_{|z|\to\infty}|f(z)|\equal{}\infty.$

That shows that $ f$ has a pole at infinity, and such an entire $ f$ is a polynomial. (Or cite Casorati-Weierstrass. Or assume WLOG that $ f(0)\equal{}0$ and study $ g(z)\equal{}[f(1/z)]^{\minus{}1}.$)

And for polynomials, cite the Fundamental Theorem of Algebra on the existence of $ n$ roots. Hence one-to-one requires $ n\equal{}1.$

[quote="Kent Merryfield"]
Since $ f^{ \minus{} 1}$ exists everywhere [/quote]
Erm... "One-to-one" normally means "injective", not "bijective"... :?

...The argument doesn't seem to use that the domain of $ f^{\minus{}1}$ is all of $ \mathbb{C}$ (since this probably isn't true). Only that the domain contains arbitrarily large compact sets and that it is well-defined which is what 1-1 gives. The arbitrarily large compact sets can be argued since the image of an entire function is dense.

It is subtler than that. The key step is that the pre-image of [b]every[/b] compact set is compact (or, at least, bounded). I do not see how that can be easily derived from the density of the image alone (though I should confess that I can hardly see anything after midnight :P). Do you? Keep in mind that, for all we know, the image doesn't need to be closed.


function
LaTeX
inequalities
algebra
system of equations

assume that positive integers a, b, c, x, y, z satisfy cy+bz=a; az+cx=b and bx+ay=c. Find the minimum value of the function f(x,y,z)=$ \dfrac{x^2}{1\plus{}x}\plus{}\dfrac{y^2}{1\plus{}y}\plus{}\dfrac{z^2}{1\plus{}z}$.
i solved x,y, and z, now what do i do

[hide]Apparently, this problem also appears on Vietnamese Mathematics and Youth Magazine last year. And you're on the right track though!
Here's my solution:
From the system of equations, we easily get(as you did):
y = \frac {a^ - b^2 + c^2}{2ac}
x = \frac {b^2 + c^2 - a^2}{2bc}
z = \frac {b^2 + a^2 - c^2}{2ab}
Now adding any 2 equations at the beginning, we easily infer that (e.g: and ,so )
From this one, we can easily show that a,b,c are sides of a triangle(can you show it?? Smile )
Plugging in y = \frac {a^ - b^2 + c^2}{2ac} into we get:
=[Unparseable or potentially dangerous latex formula.Error: 7]
Doing exactly the same for the other 2 fractions, we finally get:
\frac {(a^2 - b^2 + c^2)}^2 \2ac.(a + c - b)(a + c + b)
Using Cauchy-Schwartz's inequality we have:
[Unparseable or potentially dangerous latex formula.Error: 7](a^2 + b^2 + c^2)^2} \4(a^2c^2 + a^2b^2 + b^2c^2) + 2ac(a^2 + c^2) + 2ab(b^2 + a^2) + bc(c^2 + a^2)(1)
Now we will show that this , which is equivalent to:
a^4 + b^4 + c^4 + abc(a + b + c)\geq\ab(a^2 + b^2) + bc(b^2 + c^2) + ac(a^2 + c^2), which is obviously true followed by the 4-degree Schur's inequality:
a^2(a - b)(a - c) + b^2(b - c)(b - a) + c^2(c - a)(c - b)\geq0 with
Therefore, the minimum value of P is:$ 1/2$. The equality occurs when a=b=c=x=y=z=1/2 Wink
By the way, can you send me the problems of this contest? Chinese contest is so hard to find, especially on these regional contests.[/hide]


[quote="Ligouras"]Let $ a,b,c$ positive reals. Prove that

$ \frac {4a^2 \plus{} 9bc}{(b \plus{} c)^2} \plus{} \frac {4b^2 \plus{} 9ca}{(c \plus{} a)^2} \plus{} \frac {4c^2 \plus{} 9ab}{(a \plus{} b)^2}\geq\frac {39}{4}$[/quote]

 The following inequality is equivalent to

$ 16\sum{a^4(a \minus{} b)(a \minus{} c)} \plus{} 48\sum{a^3(b \plus{} c)(a \minus{} b)(a \minus{} c)} \plus{} 25(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2$

$ \plus{} 8\sum{b^2c^2(a \minus{} b)(a \minus{} c)} \plus{} 152abc\sum{a(a \minus{} b)(a \minus{} c)}$

$ \plus{} 40abc\sum{(b \plus{} c)(a \minus{} b)(a \minus{} c)}\geq 0$

$ \frac {a^2 \plus{} kbc}{(b \plus{} c)^2} \plus{} \frac {b^2 \plus{} kca}{(c \plus{} a)^2} \plus{} \frac {c^2 \plus{} kab}{(a \plus{} b)^2}\geq\frac {3(1 \plus{} k)}{4}$

$ k_{max} \equal{} \frac {5}{2}$

$ k \equal{} \frac {5}{2}$

$ \Longleftrightarrow$
$ 8\sum{a^4(a \minus{} b)(a \minus{} c)} \plus{} 24\sum{a^3(b \plus{} c)(a \minus{} b)(a \minus{} c)} \plus{} 11(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2$
$ \plus{} 72abc\sum{a(a \minus{} b)(a \minus{} c)} \plus{} 16abc\sum{(b \plus{} c)(a \minus{} b)(a \minus{} c)}\geq 0$


You have 23 kids in a class. Would you bet 1,000,000 dollars that any two kids will have the same birthday? Why?

No, regardless of whether 23 is the threshold of it being a good bet (it is, IIRC).  Assuming that your personal finances are not such that 1 million dollars is a small sum of money, the expected value of your profits through playing the game repeatedly are not as good as you would think, because you can't continue betting after you're broke...

Assume you have a quadrillion dollars than, and you play continuously. ;)

Well, I can tell you what I would do under those circumstances, but that would be spoiling it for someone else  :)

[hide]Yep, I would bet it.
$\frac{365\cdot 364\cdot 363\cdot ...\cdot 343}{365^{23}}$, which is the probability that everyone has a different birthday,  is slightly less than half.  This means the probability that at least two people have the same birthday is slightly over half, which means I would make money on the bet.[/hide]