The edges of a regular $2^n$-gon are colored red and blue in some fashion. A step consists in recoloring each edge whose neighbors are both the same color in red, and recoloring each edge whose neighbors are of opposite colors in blue. Prove that after $2^{n-1}$ steps all of the edges will be red, and show that this needn't hold after fewer steps.

This has been discussed before (as have most of the Iranian problems posted on the board these last couple of days). However, here's a stronger one:

Find those $n$ for which no matter how the edges of the regular $n$-gon are colored in red and blue, we always end up with an all-red polygon by performing a finite number of moves as the ones described in the post above.

the answer is $n=2^k$. This is almost the cellular automaton with [url=http://mathworld.wolfram.com/Rule90.html]rule 90[/url] (black cells=blue edges, white cells=red edges), only that the boundary is "cyclic". (Imagine wrapping the Sierpinski triangle around a cylinder.)
As this is [url=http://mathworld.wolfram.com/AdditiveCellularAutomaton.html]additive[/url], it essentially suffices to consider what happens if starting with just one cell being black. The necessary condition is obviously $n=2^k$ for a certain $k$ (otherwise there are always some black cells remaining in the 'middle', corresponding in a certain way to the 1's other than the leftmost one in the binary representation of $n$), and it is easy to see that this is also sufficient.

Can you,spanferkel or someone else,give easier solution or explain the theory of these rules such as rule 90,which was included in spanferkel's solution,please?


geometry
Pythagorean Theorem
IMO Shortlist

Ten gangsters are standing on a flat surface, and the distances between them are all distinct. At twelve o’clock, when the church bells start chiming, each of them fatally shoots the one among the other nine gangsters who is the nearest. At least how many gangsters will be killed?

Answer: 7 ( I will post my solution when I have free time)

it seems like 4 is possible if you have 6 gangsters in a large nearly-regular hexagon, with 4 in the middle in a small nearly-regular quadrilateral?

I think $ 3$ is least. Consider gangsters at:

$ (\minus{}100,0),(0,0),(101,0),(213,0),(0,110),(0,\minus{}111),(\minus{}120,110),(\minus{}120,\minus{}111),(121,110),(121,\minus{}111)$

(the hundred's digit gives the rough position, tens digit changes the relative positions a little bit so that desired 3 people get killed and the units digit distorts the positions slightly so that any 2 distances aren't the same)

2 can't happen can't be that hard ...

[quote="aadil"]i dont see the geometry in this problem[/quote]

In the case to prove 8 doesn't work there are some angle chasing and geo inequality things.

Hmm why is everyone in this thread except not_trig answering the question "what is the maximum number of survivors" rather than "at least how many people will die"? Anyways, here's my solution - might be wrong, though:

As before, claim that $ 3$ is the minimum, with an example helpfully provided by Akashnil. It remains to prove that we cannot have only two deaths; one or zero is trivial.

Suppose for the sake of contradiction that we could have exactly two deaths. Then let the fated ones be $ A,B$, and the other eight $ C,D,E,F,G,H,I,J$. $ A,B$ clearly must be the closest pair together, shooting each other, and each of the other eight must shoot one of $ A,B$. We claim that exactly four must shoot each; this is an easy contradiction argument - suppose five shot one ($ B$), and consider the second-closest gangsta to $ B$ ($ C$); draw a circle with center $ B$ and radius $ BC$. Then it's easy to show the projections of the other four must be spaced at least $ \pi/3$ apart on the circle, and $ A$ lies within the circle with a person being closer to $ A$ if his projection onto the circle was on a certain $ 2\pi/3$ arc, so at least one of them must shoot $ B$ rather than $ A$.

For the 4 and 4 case, let $ S_A\equal{}\{C,D,E,F\}$ shoot $ A$, $ S_B\equal{}\{G,H,I,J\}$ shoot $ B$, with $ CA<DA<EA<FA$, $ GB<HB<IB<JB$. Consider the circles with centers $ A,B$ and radii $ CA,GB$ respectively, as before, and note that again the projections of the gangstas onto their respective circles are spaced at least $ \pi/3$ apart, as before. Additionally, note both radii are greater than $ AB$ and hence circle $ A$ contains $ B$ and circle $ B$ contains $ A$. We claim that there exists either a gangsta in $ S_A$ who shoots someone from $ S_B$ or vice versa.

[b]Lemma:[/b] If there is a gangster $ A'$ in $ S_A$ which has projection onto circle $ A$ which lies on circle $ B$'s side of the radical axis of the two circles, and vice versa for gangster $ B'$, then $ B'$ will shoot $ A'$ or vice versa.
[b]Proof:[/b] (outline) It clearly suffices to consider when the projections are the same point and hence on the radical axis by the hinge theorem or law of cosines; WLOG $ A'$ is farther from $ A$ than $ B$ from $ B'$; there exist two $ B''$ on $ BB'$ such that $ AA'\equal{}BA''$; one lies past $ B$ by assumption and so can be discounted; it suffices to show that the other lies farther from $ B$ than $ B'$ by the hinge theorem again (or loc again). But this easily follows from $ AA'>BB'$ as well, so the lemma is proven.

[b]Lemma:[/b]  If two circles $ A$, $ B$ intersect and contain each other's centers, then each cuts off at least a $ 2\pi/3$ degree arc from the other.
[b]Proof:[/b] (outline) This implies the smaller circle's radius is at least half the larger one's and the distance is between the centers is less than the smaller circle's radius; draw the radical axis and find a lower bound on its length using Pythagorean theorem; use this to prove that the arc it cuts off from either circle is at least $ 2\pi/3$.

From these, the result follows.

[quote=vsathiam]Who wrote this problem -____-[/quote]
The problem was proposed by Iran.

At least three gangsters are killed.

First assume for contradiction only two gangsters, say $a$ and $b$, die. Then, those two gangsters must be the minimal distance across the $\binom{10}{2} = 45$ pairs. Assume they are a horizontal line segment $\overline{ab}$ with $a$ to the left of $b$.

Suppose two gangsters $x$ and $x'$ both shoot $a$, then we must have $\angle xax' > 60^{\circ}$ to ensure that side $xx'$ is strictly longest in that triangle. Similarly for $b$.

Now partition the eight remaining gangsters based on whether they shoot $a$ or $b$. Suppose that $a$ is shot by $b$, $x_1$, \dots, $x_k$ in counterclockwise order, and $b$ is shot by $a$, $y_1$, \dots, $y_{8-k}$ in counterclockwise order. The ``spokes'' we get are spaced at least $60^{\circ}$ apart, and so we find one of two cases: either rays $ax_1$ and $by_1$ meet, or rays $ax_k$ and $by_{8-k}$ meet. Assume WLOG we are in the first case and let $x=x_1$, $y=y_1$.

Thus $\angle xab + \angle yba < 180^{\circ}$. On the other hand: [list] 	[*]Since $ax$ is longer than $ab$, 	we have $\angle axb < \angle xba$. 	[*]Similarly, $\angle ayb < \angle yab$. 	[*]Since $xy$ is longer than $xa$, 	we have $\angle xya < \angle xay$ 	[*]Similarly, $\angle yxb < \angle xby$. [/list] Summing gives $\angle a + \angle b > \angle x + \angle y$, which is impossible.

[b]Remark[/b]:  	Here is a sort of ``equality case'' with 	two deaths if one relaxes the distinct condition 	and allows gangsters to shoot any of the nearest neighbors, 	which motivates the earlier proof. 	All edges below have unit length.

[asy] 		size(3cm); 		pair O = (0,0); 		pair A = dir(60); 		pair B = dir(-60); 		pair C = A+B; 		draw(O--A--C--(A+C)--(2*C)--(2*C+A)--(3*C)--cycle, grey); 		draw(O--B--C--(B+C)--(2*C)--(2*C+B)--(3*C), grey); 		draw(A--(A+2*C), grey); 		draw(B--(B+2*C), grey); 		dot(O); 		dot(A); 		dot(A+C); 		dot(A+2*C); 		dot(B); 		dot(B+C); 		dot(B+2*C); 		dot(3*C); 		dot(C, red); 		dot(2*C, red); 	[/asy]

There are many constructions with three deaths, for example in the earlier post.

This is my 1000th post, since my keyboard is partially [hide=busted right now]The "b" key is not working, I'm copy-pasting to type it and Shift+T is also not working, I have to caps lock then un-caps lock it[/hide]. I'll post the complete solution later.
[hide=Hidden because of length]
The case with $2018$ gangsters instead of $10$ came in Sharygin 2018 (Correspendence Round) as P10 (the word was quite different, of course).
While solving the prolem I found the generalization that, for $n$ gangsters, the answer is
\[\left.\begin{cases}2k &\text{ if } n = 9k\\ 2k + 1 &\text{ if } n = 9k+1, 9k + 2\\ 2k+2&\text{ if } n = 9k+3,9k+4, 9k+5, 9k+6, 9k+7, 9k+8\end{cases}\right\}\text{where } k\in\mathbb{N}\]

Essentially, if we draw a circle centred at the shooter and the person shot on the circle, then no gangster lies inside that circle, and then just note that $\frac{360^{\circ}}{6} = 60^{\circ}$, so, a person can be shot by only 5 gangester.
So divide, $n$ into $\left\lfloor \frac{n}{9} \right\rfloor - 1$ groups of nine. For each of these groups, place people like this:
[asy]
unitsize(2cm);
pair P_1 = (0,0), P_3 = 1.02 dir(62), P_4 = 1.04 dir(124), P_5 = 1.06 dir(186), P_6 = 1.08 dir(248);
pair P_2 = dir(0), P_7 = 1.03 dir(59) + 1, P_8 = 1.04*dir(-66) + 1;
pair P_9 = 1.04*dir(-4) + 1, P_2017 = 1.06*dir(58) + P_9, P_2018 = 1.08*dir(-4) + P_9;

draw(unitcircle, linetype("0 2"));
draw(circle(P_2, abs(P_2-P_1)), linetype("0 2"));
draw(circle(P_3, abs(P_3-P_1)), linetype("0 2"));
draw(circle(P_4, abs(P_4-P_1)), linetype("0 2"));
draw(circle(P_5, abs(P_5-P_1)), linetype("0 2"));
draw(circle(P_6, abs(P_6-P_1)), linetype("0 2"));
draw(circle(P_7, abs(P_7-P_2)), linetype("0 2"));
draw(circle(P_8, abs(P_8-P_2)), linetype("0 2"));
draw(circle(P_9, abs(P_9-P_2)), linetype("0 2"));
dot(P_3^^P_4^^P_5^^P_6^^P_7^^P_8^^P_9);
dot(P_1^^P_2, 0.7red);

label("P$_k$", P_1, dir(0), 0.7red);
label("P$_{k + 1}$", P_2, -P_2, 0.7red);
label("P$_{k + 2}$", P_3, P_3);
label("P$_{k + 3}$", P_4, P_4);
label("P$_{k + 4}$", P_5, P_5);
label("P$_{k + 5}$", P_6, P_6);
label("P$_{k + 6}$", P_7, dir(60));
label("P$_{k + 8}$", P_8, dir(-60));
label("P$_{k + 7}$", P_9, dir(0));
[/asy]
And for the final group, make it like this, if $n \mod{9} \leq 2$:
[asy]
unitsize(2cm);
pair P_1 = (0,0), P_3 = 1.02 dir(62), P_4 = 1.04 dir(124), P_5 = 1.06 dir(186), P_6 = 1.08 dir(248);
pair P_2 = dir(0), P_7 = 1.03 dir(59) + 1, P_8 = 1.04*dir(-66) + 1;
pair P_9 = 1.04*dir(-4) + 1, P_2017 = 1.06*dir(58) + P_9, P_2018 = 1.08*dir(-4) + P_9;

draw(unitcircle, linetype("0 2"));
draw(circle(P_2, abs(P_2-P_1)), linetype("0 2"));
draw(circle(P_3, abs(P_3-P_1)), linetype("0 2"));
draw(circle(P_4, abs(P_4-P_1)), linetype("0 2"));
draw(circle(P_5, abs(P_5-P_1)), linetype("0 2"));
draw(circle(P_6, abs(P_6-P_1)), linetype("0 2"));
draw(circle(P_7, abs(P_7-P_2)), linetype("0 2"));
draw(circle(P_8, abs(P_8-P_2)), linetype("0 2"));
draw(circle(P_9, abs(P_9-P_2)), linetype("0 2"));
draw(circle(P_2017, abs(P_2017-P_9)), linetype("0 2"));
draw(circle(P_2018, abs(P_2018-P_9)), linetype("0 2"));
dot(P_3^^P_4^^P_5^^P_6^^P_7^^P_8^^P_2017^^P_2018);
dot(P_1^^P_2^^P_9, 0.7red);

label("P$_1$", P_1, dir(0), 0.7red);
label("P$_2$", P_2, P_2, 0.7red);
label("P$_3$", P_3, P_3);
label("P$_4$", P_4, P_4);
label("P$_5$", P_5, P_5);
label("P$_6$", P_6, P_6);
label("P$_7$", P_7, dir(60));
label("P$_9$", P_8, dir(-60));
label("P$_8$", P_9, dir(0), 0.7red);
label("P$_{2017}$", P_2017, dir(60));
label("P$_{2018}$", P_2018, dir(0));
[/asy]
Do something similar for the third case.
(The previous two diagrams were recycled from the Sharygin solution, my keyboard is too irritating for me to draw the third diagram)

If you understood what I meant above, good. If you didn't, I don't blame you, the above "solution" is not very well-explained.
[/hide]

the problem has an older source with 50 gangsters [url=https://artofproblemsolving.com/community/c6h1422363p7997829]here[/url]

Kalva's (official ?) solution is mentioned [url=https://artofproblemsolving.com/community/c6h1482706p8662463]here[/url]

Mom I Solved A G7. Now how do I write up this obnoxious problem?

The answer is 3. A construction is provided above.

Obviously the answer cannot be 0 or 1. To prove that the answer cannot be 2, suppose that gangsters 1,2 are killed. Then, everyone must fire at one of them; this means that gangsters 1,2 have the closest distance among all pairs.

Consider $l$, the perpendicular bisector of gangsters 1,2; either there is one side of $l$ with six gangsters or each side has five gangsters.

In the former case, assume WLOG that gangster 1 is shot by 5 other gangsters (excluding gangster 2). Then, consider gangster 1 as the origin of a polar coordinate system. The range of possible arguments of those who shot gangster 1 is $(60^\circ,300^\circ)$, meaning that two of them are less than 60 degrees apart, which forces another death.

Hence, both gangsters 1,2 are shot by 4 gangsters excluding themselves. Consider the perpendiculars $l_1,l_2$ to the segment connecting gangsters 1,2 at the locations of gangsters 1,2. Suppose they divide the plane into finite width region $R$ and infinite width regions $r_1,r_2$. If any four gangsters are in $r_1$ or $r_2$, then two of them are less than 60 degrees apart with respect to one of gangsters $1,2$ Hence, there must be at least two gangsters in $R$.

Now, let the segment connecting gangsters 1,2 divide $R$ into $R_1,R_2$. Then, it's easy to see by pythag that neither can contain two gangsters. Assume that gangsters 3,4 are in $R_1,R_2$ and shot gangsters 1,2, respectively. Use letters instead of numbers from this point onward, with gangster $1$ as A and so on.

Assume $\angle CAB\ge \angle DBA$. Then, since no two gangsters are less than 60 degrees apart, the line $AC$ and its parallel through $B$ bounds another $A$-aligned gangster $E$ on the same side of segment $AB$ as $D$. We also arrive at $ED<EA$ in this case, which cannot happen.

[hide=Post #14 by mathleticguyyy]
[url=https://aops.com/community/user/409180]mathleticguyyy[/url] [url=https://aops.com/community/p25597782](view original)[/url]
Mom I Solved A G7. Now how do I write up this obnoxious problem?
[/hide]

Mood.

At least three gangsters will be killed. Exactly three can be achieved by doing the following: let $A_1$ through $A_{10}$ be the locations of the gangsters. Construct almost equilateral triangle $A_1A_2A_3$. Extend a ray of the angle bisector of $A_1$ going to the outside of the triangle, and two rays from $A_1$ that are just over sixty degrees apart from the angle bisector ray.

Pick points $A_4$ on the ray close to $A_2$, $A_5$ on the angle bisector, and $A_6$ on the ray close to $A_3$ so that they are nearly equidistant from $A_1$ and just over the length of each side in $A_1A_2A_3$. The sixty degree condition ensures that each of the gangsters at $A_4$, $A_5$, and $A_6$ shoot the one at $A_1$.

Draw the same three rays from $A_2$, but with just over thirty degrees of separation between the rays. Pick $A_7$ and $A_8$ on the rays closest to $A_1$ and $A_3$, respectively, such that they are almost equidistant from $A_2$ and their distances is just over each side length of $A_1A_2A_3$. Note that $\angle A_7A_2A_1$ is just under $120^\circ$ and $\angle A_4A_1A_2$ is just under $90^\circ$, so $A_7$ and $A_8$ will both shoot $A_2$.

Repeat on $A_3$. Note that $\angle A_8A_2A_3$ and $\angle A_9A_3A_2$ are both just under $120^\circ$ so $A_8A_9$ is not too short of a distance. Furthermore, $A_1A_2A_3$ will shoot each other, so only $A_1$, $A_2$, and $A_3$ are shot.

Now, suppose only two gangsters are shot. Clearly, the gangsters at the ends of the shortest distance will shoot each other. WLOG let them be $A_1$ and $A_2$. Let $A_i$ and $A_j$ be gangsters that will shoot $A_1$ then clearly $A_iA_j$ must be the longest length, so $\angle A_iA_1A_j>60^\circ$.

Thus, if there are at least five gangsters that shoot $A_1$, along with $A_2$ there will be an angle at most sixty degrees. Hence, there are four gangsters each that shoot $A_1$. Let them be $A_3$, $A_4$, $A_5$, $A_6$ shooting $A_1$ and $A_7$, $A_8$, $A_9$, $A_{10}$ shooting $A_2$, in that order, going clockwise.

Since each angle is at least sixty, \[\angle A_3A_1A_2+\angle A_6A_1A_2 < 180^\circ\text{ and }\angle A_7A_2A_1+\angle A_{10}A_2A_1 < 180^\circ\]
Thus, either $\angle A_3A_1A_2+\angle A_1A_2A_{10}$ or $\angle A_6A_1A_2+\angle A_7A_2A_1$ is less than $180^\circ$, resulting in one of them being less than $A_1A_2$, contradiction to $A_1A_2$ being shortest length.


geometry calculus algebra polynomial quadratics trigonometry function

In the past we have only had to wait for ICAE to take the test. According to http://www.icae.org/files/upload/home/194.pdf they have already taken it this Saturday, so I don't know when we can start discussing it. Some schools in the past have messed up and had students writing it later in the week, so my best guess would be Saturday before we can discuss.

hey, sounds cool. but can you tell me one thing (question A, 6) p.s. just in case some people didnt write it and perchance stumble upon this i dont think question A 6 is gonna make or break anyone.
Is the answer 9 or 12? I think its 9, and my friends tell me its 12.

[quote="Jenkins"]Is the answer 9 or 12? I think its 9, and my friends tell me its 12.[/quote]

It's 9. I don't know where your friends got 12 from... XD

I completely screwed up 3c and 4c in part B. I estimate my mark is around 71 or 72.

BTW, how many solutions were there for A8? I got 3: (14, 13, 13), (6, 5, 4), and (4, 1, 2).

That should be it. Everyone at my school (except me) missed the first one :p.

I estimate I got a 76 or 77  :lol: .

Here. Laugh at me for I am [b]Lord of the Idiots[/b]

I made 5 careless mistakes on Part A and didn't get 3c. However I did get all of 4 except 3c.

BTW for 4b how did you guys prove? I used Let b = 2^n since f(2^n) = 2^(n+1) - 1. I hope I didn't mess that up too.

I agree for your 8A answer by the way. Expected score: 40-45

[quote="Bacteria"]Well, b has to be odd so use b=(2^n)-1 instead; thats what I did.[/quote]

Sorry typo, I did it the same way. I meant I said b+1 = (2^n). I agree.

This year's COMC was hard. I found most answers unsure during the test.
Anyway, here are my personal answers:
Part A
1. 9
2. 4
3. 8
4. 20
5. 1024
6. 9
7. 210 and 330
8. (14,13,13), (6,5,4), (4,1,2)
Part B
1. a)54 b)13 c)30
2. a)y=-(2/5)x+43/5
    b)(9,5) c)(7,7) d)H (orthocenter)
3. a)2,2 and -1,-1 b)x^3-3x^2 c)proved Y^4+2=Y^3 can't have any real solution. So q has no real fixed points.
4. a)9,12,18,36
    b)I used b=2^(100+n)-1 in which n can be any non-negative integer
    c)k=3. as we had in a) f(9)=f(9+3). proved that k=1 and k=2 are not possible because then c(c+k)>the largest posible sum of s.

Math was pretty hard. I expect a 65+ to get at least into repechage.

[quote="Silent Soliloquy"]Hm, what was your approach for 3c?[/quote]

since q and p are commute switch q(p(x)) with p(q(x)). then use proof by contradiction so [i]assume[/i] that q has real fixed points, q(c)=c

2(p(c))^4 + 2 = (p(x))^4 + (p(x))^3

for simplification let p(x) = m

2m^4 + 2 = m^4 + m^3
m^4 - m^3 + 2 = 0

now prove this is impossible by diffrenciating and finding extreme points all positive so it cannot be that q has any fixed points.

i tried to do it without calculus but cannot ~_~ I think question 2partB shouldve been harder, it seemed like a question1partB :P

Im from ICAE, so I dont know any of this canadian stuff. What is repechage?

And what do you guys think the CMO cutoff will be? It was harder than last year, so probably 67-68?

I got everything but A8 (missed the 14,13,13 case  :rotfl: ) and B3C (see, it looked really hard so I didnt even attempt it, but apparently it was EXTREMELY trivial :wallbash_red: )

Im happy though... I got all of B4  :lol:

This is my first time with the COMC in 8th grade. I will probably lose 1 point for A8 and 3 pts for 3C, and then lets just say I will lose 7 more points for bad proofs. That leaves me with around 69-72.

@gf4848
For A7, it got only 330. 210 is extraneous, if im not mistaken.

[quote="Thunder365"]Im from ICAE, so I dont know any of this canadian stuff. What is repechage?

And what do you guys think the CMO cutoff will be? It was harder than last year, so probably 68+?

I got everything but A8 (missed the 14,13,13 case  :rotfl: ) and B3C (see, it looked really hard so I didnt even attempt it, but apparently it was EXTREMELY trivial :wallbash_red: )

Im happy though... I got all of B4  :lol:

This is my first time with the COMC in 8th grade. I will probably lose 1 point for A8 and 3 pts for 3C, and then lets just say I will lose 7 more points for bad proofs. That leaves me with around 69-72.

@gf4848
For A7, it got only 330. 210 is extraneous, if im not mistaken.[/quote]

I wasn't sure about that for A7 as well so I omitted 210. But if we change 2logx into logx^2 can't we have 210 as a solution?

sorry, i dont remember the exact question. But I think the reason 210 is extraneous has something to do with cos210 being negative, or something. Do you have the exact question?

Could someone please post the problems, or at least questions B4, B3 and A8 for those of us who didn't write the contest?

Thanks.

Ideally, through the CMOR the next top group of students should qualify. So that means that the people with a 68 on the COMC should qualify through the CMOR, as they were the next top scorers. Is this usually the case?

[quote="Complex_Ninja"][quote="jix"]Only 8? Then I guess there will be seven COMC-A6/7 level problems and one CMO+ level problem, if my difficulty gauging is accurate.[/quote]
Is the last one really that tough? CMO+[/quote]

I'm not very good at comparing problem difficulties, but I think that the last problem is usually CMO level at least. Plus/minus the difficulty of CMO. (:

guess they decided to go on holidays early this year...

Yep, I was in contact with them and the results should be ready today but there is nobody there to post it  :( . It should be posted first thing in January...

[quote="Rofler"]I need to read questions... like A2 and A8... lol A2 fail. Also, it would have helped if for the last question I actually checked all of the factors of 36...  because that would have helped for B4c...

So subtracting marks... I don't know if I technically even qualified for the CMOR, -10 for part A and I partial failed at B4a and definitely failed at B4c because of not checking all of the factors...

Lol,...lol.

Cheers,

Hunter[/quote]

Noob

[quote="Rofler"]Danny, you're a retard. The email for sunchips account, which was created yesterday, is set to your own, and you should have at least tried to put a capital letter in somewhere in the name...

Your ill-attempts at inflicting pain on me are in line with the growing sociopathic tendencies towards me coupled with an animosity which I have yet to explain. Under the facade of a normal human being lies a crater, filled with the harbored feelings of anger, guilt, and shame directed at those who have done wrong unto you, and who you have done wrong unto. All attempts at forming an actual relationship with you have ended with bitter feelings, cynicism, and an ever growing apathy towards you. This is furthered by using the screen name of my close friend against me to illicit feelings of disgrace and shame, almost as if you are trying to alleviate your own demons by inflicting them onto others. Your smile is a lie, your steely eyes tell all my, at best, acquaintance...

or something like that. That's what you get for reading English 20/30 lit. for about 5 hours before going on AOPS. Look, you're alright Danny, but seriously, ease off a bit. Just because I was in a state of inebriation from my required readings doesn't mean that you can't take something out of the above paragraph.

Anyways, are you going to Winter Camp? If you are, I hope to see you there!

Cheers,

Hunter[/quote]

Sorry dawg, can't help it. Jarno came by my house yesterday and we were bored. And nah man, I'm not going to the winter camp cuz I can't afford a flight to Toronto xD. You at Calgary? We should chill out sometimes man. How's life?

Results are IN!

http://www.cemc.uwaterloo.ca/contests/past_contests/COMCResults.pdf

Its interesting that there are more people with a score of 78-80 than last year even though this year's COMC was harder

Saroosh will make the IMO team with now a 70% chance.
And I thought my score was hilarious! Just high enough to make the results page. Best epic fail ever!

Cheers,

Hunter

Since you will obviously make IMO then we have a 70% chance of 2 people from one school making IMO  :o. I wonder if that's happened before...

That was a pretty funny epic fail. Its even better than my 145 in the Fermat...

Darn 77... i got 38/39... I didn't even get the highest in my STATE (North Carolina)... dnkywin...

Hello. What do you have to do to be put in the "official contestant" category? I met the requirements (I think). I am in Grade 10 at an official high school in Toronto. I got a 68 and qualified for CMOR. Does the fact that I did not fit the official contestant category affect whether or not I am able to pass the CMOR?

I filled out the CMO registration form today, and lets just say I failed... Filling in my first name in the last name spot, forgetting to sign, and putting all the data in the wrong place  :oops:. My math department head said something like: "They'll be able to figure it out, they expect that kind of thing from people like you"  :P.

@SnowEverywhere
To be an official contestant you have to be either a Canadian citizen or permanent resident, age eligible (<20 I think), and have not been enrolled in university. These are all the requirements I believe (though I could be wrong). Several unofficial people write CMO every year so I do not think that that will affect your qualification.


Let $a,b,c$ be three pairwisely different complex numbers. Prove that:
\[\left| \frac{a}{b-c}\right| +\left| \frac{b}{c-a}\right| +\left| \frac{c}{a-b}\right|\ge \sqrt 3\][b]Remark.[/b] If $a,b,c\in R$ then $\sum \left| \frac{a}{b-c}\right|\ge 2\ \ (*)\ \ $.
 ([i]Darij Grinberg - http://www.mathlinks.ro/Forum/topic-20869.html [/i])

[b]A interesting problem (at least for me)[/b]. The proposed problem has the following

geometrical interpretation: $\triangle ABC\Longrightarrow \left( \forall \right)P,\ \sum \frac{PA}{a}\ge \sqrt 3$.

Is wellknown (from C.S.'s inequality) that $\sum \frac{PA}{a}\ge \frac {\left( \sum \sqrt {PA} \right) ^2}{\sum a}$.

How is $\left( \sum \sqrt {PA}\right)^2$ with respect to $\sqrt 3\cdot (a+b+c)\ ?$

How shows the set of the points for which $\left( \sum \sqrt {PA}\right)^2=\sqrt 3\cdot (a+b+c)\ ?$.
(Possibly with computer !)

Oh my God, Levi! I thought I created this problem and I was intending to propose this problem for NMO 2006 for district round. :(  :(  :(


inequalities
limit
calculus
integration
algebra unsolved
algebra

Let a sequence of positive reals  $\{u_n\}^{\infty}_{n=1}$ be given. For every positive integer $n$, let $k_n$ be the least positive integer satisfying:

\[\sum^{k_n}_{i=1} \frac{1}{i} \geq \sum^n_{i=1} u_i.\]

Show that the sequence $\left\{\frac{k_{n+1}}{k_n}\right\}$ has finite limit if and only if $\{u_n\}$ does.

Let's note that $|\sum_{i=k_{n-1}}^{k_n}\frac{1}{i}-u_n| < \frac{1}{k_{n-1}}$, so $\lim_{n-> \infty}\sum_{i=k_{n-1}}^{k_n}\frac{1}{i}=\lim_{n->\infty}u_n$ provided that at least one of them has a limit (cause $k_{n-1}->\infty$ when $n-> \infty$).
From integral approximation we know that:
$ln(\frac{b+1}{a+1}) \leq \sum_{i=a}^{b}\frac{1}{i} \leq ln(\frac{b}{a})$, thus
$ln(\frac{k_n+1}{k_{n-1}+1}) \leq \sum_{i=k_{n-1}}^{k_n}\frac{1}{i} \leq ln(\frac{k_n}{k_{n-1}})$, so $\lim_{n->\infty}\sum_{i=k_{n-1}}^{k_n}\frac{1}{i}=\lim_{n->\infty}ln(\frac{k_n}{k_{n-1}})$, provided that at least one of them has a limit.
If $\lim_{n->\infty}(\frac{k_n}{k_{n-1}})->g$ then $\lim_{n-> \infty}u_n=\lim_{n->\infty}\sum_{i=k_{n-1}}^{k_n}\frac{1}{i}=\lim_{n-> \infty}ln(\frac{k_n}{k_{n-1}})=lng$.
If $\lim_{n-> \infty}u_n=g$ then $\lim_{n->\infty}ln(\frac{k_n}{k_{n-1}})=\lim_{n->\infty}\sum_{i=k_{n-1}}^{k_n}\frac{1}{i}=\lim_{n->\infty}u_n=g$, so $\lim_{n->\infty}(\frac{k_n}{k_{n-1}})=e^g$.


Given that $ a,b,c,d$ are  natural numbers such that $ a<b\le c<d$ .Also given that $ ad\equal{}bc$ and $ \sqrt{d} \minus{} \sqrt{a} \le  1$.
Prove that $ 4b\plus{}1$ is a perfect square.

Well, since no one has posted the solution despite 75 views, I am going to post my solution but will definitely like to see your solutions.

With the given conditions, $ a < b\le c < d$, we can assume $ b \equal{} a \plus{} k,c \equal{} a \plus{} l,d \equal{} a \plus{} m$ where $ k\le l < m$.
THus with the given condition that $ ad \equal{} bc$, we have $ a(a \plus{} m) \equal{} (a \plus{} k)(a \plus{} l) \equal{} \equal{} \equal{} \equal{} \equal{} > am \equal{} (l \plus{} k)a \plus{} lk \equal{} \equal{} \equal{} \equal{} \equal{} > m \equal{} (l \plus{} k) \plus{} \frac {lk}{a}$.
Since $ m$ is an integer, $ \frac {lk}{a}$ should also be an intger{postive}.  THis implies that $ a$ divides $ lk$ which implies that $ a \le lk \equal{} \equal{} \equal{} \equal{} > \frac {lk}{a} \ge 1 \minus{} \minus{} \minus{} \minus{} \minus{} > (1)$  $ m \equal{} (l \plus{} k) \plus{} \frac {lk}{a} \ge l \plus{} k \plus{} 1$.
 Also given that $ \sqrt {d} \minus{} \sqrt {a} \le 1$. this implies that  $ \sqrt {a \plus{} m} \le 1 \plus{} \sqrt {a} \equal{} \equal{} \equal{} \equal{} > a \plus{} m \le 1 \plus{} a \plus{} 2\sqrt {a} \equal{} \equal{} \equal{} \equal{} > m \le 1 \plus{} 2\sqrt {a}$.
THis implies that $ l \plus{} k \plus{} 1 \le 1 \plus{} 2\sqrt {a} \equal{} \equal{} \equal{} \equal{} \equal{} > l \plus{} k \le 2\sqrt {a}$. However by AM GM inequality  $ l \plus{} k \ge 2\sqrt {lk}$.
THis implies that $ 2\sqrt {lk}\le 2\sqrt {a} \equal{} \equal{} \equal{} \equal{} \equal{} > lk\le a \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} > \frac {lk}{a} \le 1 \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} > (2)$.

So we get from $ (1)$ and $ (2)$ that $ lk \equal{} a$. But this also implies that $ AM \equal{} GM$ in the inequality $ l \plus{} k\ge 2\sqrt {lk}$. This implies $ l \equal{} k$ which further implies that $ a \equal{} l^2$ or $ a \equal{} k^2$.

$ a \equal{} (b \minus{} a)^2 or (c \minus{} a)^2 \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} > a \equal{} b^2 \plus{} a^2 \minus{} 2ab \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} > a^2 \minus{} (2b \plus{} 1)a \plus{} b^2 \equal{} 0$.THus  $ 4b \plus{} 1$ is a perfect   square.  FRom now, we can finsh the problem by proving the required and can even develop it further.

First of all, don't be so impatient -- this forum is not here for your personal amusement, and just because no one has responded to you in two days doesn't mean you should go PMing random members of the forum.

Second of all, your LaTeX could use some serious work.  Use \Longrightarrow to get $ \Longrightarrow$ and \to or \longrightarrow to get $ \to$ or $ \longrightarrow$.  Several of your equations are too large -- leave a line of space before and after them, use display mode [code]\[
...
\][/code] instead of regular math mode, and consider breaking them up (e.g., by using a command such as align or align*, or simply by not putting so much in each equation, especially text that gets rendered difficult-to-read by math mode) so that they are more legible.

Your proof (which seems correct) can be shortened somewhat at the end: once we have that $ l \equal{} k \equal{} \sqrt{a}$, we have immediately that $ b \equal{} l^2 \plus{} l$ and so $ 4b \plus{} 1\equal{} (2l \plus{} 1)^2$.


algebra
IMO Shortlist
functional equation

Find all functions $ f: \mathbb{R}^{ \plus{} }\to\mathbb{R}^{ \plus{} }$ satisfying $ f\left(x \plus{} f\left(y\right)\right) \equal{} f\left(x \plus{} y\right) \plus{} f\left(y\right)$ for all pairs of positive reals $ x$ and $ y$. Here, $ \mathbb{R}^{ \plus{} }$ denotes the set of all positive reals.

[i]Proposed by Paisan Nakmahachalasint, Thailand[/i]

[quote="the.sceth"]
$ f(x \plus{} f(y)) > f(x \plus{} y),f(y)$ so there cannot exist a solution $ f(t)$ that is constant for sufficiently large $ t$. [Statement 1]

If $ f(m) \equal{} f(n)$ then
$ f(s \plus{} f(m)) \equal{} f(s \plus{} f(n)) \forall s \in \mathbb{R}^{ \plus{} }$
$ f(s \plus{} m) \plus{} f(m) \equal{} f(s \plus{} n) \plus{} f(n)$
$ f(s \plus{} m) \equal{} f (s \plus{} n)$.
If $ m \neq n$ then $ f(t)$ is constant for sufficiently large $ t$.
By Statement 1, $ f$ is injective. [Statement 2]
[/quote]

Lets say $ m\equal{}1$, $ n\equal{}2$, then $ f(s\plus{}n)\equal{}f(s\plus{}1)$ where $ n$ is an integer...that does not make $ f(t)$ constant.

[quote="Altheman"][quote="the.sceth"]
$ f(x \plus{} f(y)) > f(x \plus{} y),f(y)$ so there cannot exist a solution $ f(t)$ that is constant for sufficiently large $ t$. [Statement 1]

If $ f(m) \equal{} f(n)$ then
$ f(s \plus{} f(m)) \equal{} f(s \plus{} f(n)) \forall s \in \mathbb{R}^{ \plus{} }$
$ f(s \plus{} m) \plus{} f(m) \equal{} f(s \plus{} n) \plus{} f(n)$
$ f(s \plus{} m) \equal{} f (s \plus{} n)$.
If $ m \neq n$ then $ f(t)$ is constant for sufficiently large $ t$.
By Statement 1, $ f$ is injective. [Statement 2]
[/quote]

Lets say $ m \equal{} 1$, $ n \equal{} 2$, then $ f(s \plus{} n) \equal{} f(s \plus{} 1)$ where $ n$ is an integer...that does not make $ f(t)$ constant.[/quote]

I see the inadequacy in intuiting something not really trivial. This should cover it up - it can fit in the penultimate line of the quoted text.

WLOG let $ m\geq n$, $ m \minus{} n: \equal{} S$, $ f(S): \equal{} T$, a constant.
The direct implication is that $ f(s) \equal{} f(s \plus{} hS)$ where $ h$ is any positive integer.
For any positive integer $ k$ and any positive real $ a$,
$ f(a \plus{} S \plus{} f (kS)) \equal{} f(a \plus{} (k \plus{} 1)S) \plus{} f(kS)$
$ f(a \plus{} T) \equal{} f(a) \plus{} T$...  $ [*]$
$ f(S \plus{} T) \equal{} 2T$
$ f(T) \equal{} 2T$
$ f(T \plus{} f(kS)) \equal{} f(T \plus{} kS) \plus{} f(kS)$
$ f(2T) \equal{} 3T$
Using $ [*]$ as an inductive step, $ f(AT) \equal{} (A \plus{} 1)T$ for the positive integer $ A$.

If  $ A$ and $ B$ are positive integers then
$ f(AT \plus{} f(BT)) \equal{} f((A \plus{} B)T) \plus{} f(BT)$
$ (A \plus{} B \plus{} 2)T \equal{} (A \plus{} 2B \plus{} 2)T \forall B \in \mathbb{N}$
$ T \equal{} 0$ not in range of $ f$
$ S$ not in domain of $ f$
$ S \equal{} 0$
$ f$ is injective.

I think that the substitution g(x)=f(x)-x solves the problem !
 :P

1. $ f(x) > x$ for every positive $ x$.
Indeed, if there exists $ x_0$ such that $ x_0 > f(x_0)$, then we know $ f(x_0 \plus{} f(x_0) \minus{} f(x_0)) \equal{} f(x_0) \equal{} f(2x_0 \minus{} f(x_0)) \plus{} f(x_0)$ - contradiction.

2. For every $ x > y$ we have $ f(x \plus{} f(y) \minus{} y) \equal{} f(x) \plus{} f(y)$.
Let $ g(x) \equal{} f(x) \minus{} x$ as [b]msecco[/b] said. $ ( g: \mathbb{R}^{ \plus{} }\to\mathbb{R}^{ \plus{} } )$
Then for every $ x > y > 0$ we have $ g(x \plus{} g(y)) \equal{} g(x) \plus{} y$ and if we fix $ x \equal{} x_0 > y$ and we assume $ g(y_1) \equal{} g(y_2)$ for two positive numbers less than $ x_0$, we will obtain that $ g(x_0 \plus{} g(y_1)) \equal{} g(x_0 \plus{} g(y_2))\Leftrightarrow g(x_0) \plus{} y_1 \equal{} g(x_0) \plus{} y_2$ - impossible unless $ y_1 \equal{} y_2$. Therefore, $ g(x)$ is injective and from this follows that $ f(x)$ is injective function, too.

3. For every $ x > y > 0$ we have $ g(x \plus{} g(y)) \equal{} g(x) \plus{} y \equal{} g(x \plus{} g(\alpha)) \plus{} (y \minus{} \alpha) \equal{} g(x \plus{} g(\alpha) \plus{} g(y \minus{} \alpha))$, $ \Rightarrow g(y) \equal{} g(y \minus{} \alpha) \plus{} g(\alpha)$.
Since $ g: \mathbb{R}^{ \plus{} }\to\mathbb{R}^{ \plus{} }$, then $ g(x) \equal{} cx$ and after substituting, we find $ c \equal{} 1$.

(1) You got that function $ f(x) \minus{} x$ is injective, which doesn't at all mean that $ f$ is also injective.
(2) You got that $ g$ satisfies the Cauchy equation and it's defined for the set of positive real numbers and gets only positive values. This doesn't mean that $ g(x)\equal{}cx$. ;)

Your complaint in (1) is legitimate, but it was never used in the solution. You're wrong about (2). A function from $ \mathbb{R^\plus{}}$ to $ \mathbb{R^\plus{}}$ that satisfies Cauchy's equation must be linear. You can prove this by getting $ f(x) \equal{} cx$ on rationals and showing $ f$ is strictly increasing: if $ x > y$, then $ f(x) > f(y)$ because $ f(x) \equal{} f(y) \plus{} f(x\minus{}y)$ and $ f(x\minus{}y) > 0$.

Note that any noncontinuous solution to Cauchy's equations (for functions on $ \mathbb{R}$) is dense in the plane, so if you restrict everything to one quadrant you must have continuity.

Yes you're right. Sorry for my dull post :(

My Solution:

0) $ f(x \plus{} f(y)) \equal{} f(x \plus{} y) \plus{} f(y)\ \forall \ x,y\in \mathbb R^ \plus{}$.

1) $ f(x) > x\ \forall \ x\in \mathbb R^ \plus{}$.

2) Putting $ x: \equal{} x \minus{} y$ in (0) $ f(x \plus{} [f(y) \minus{} y]) \equal{} f(x) \plus{} f(y)\ \forall \ x > y\in \mathbb R^ \plus{}$.

3) In (0) Let $ x: \equal{} f(z) \minus{} z$ where $ 0 < z < y$. We have $ f(f(y) \plus{} [f(z) \minus{} z]) \equal{} f(y \plus{} [f(z) \minus{} z]) \plus{} f(y)$.
Using (2) : $ [f(f(y)) \plus{} f(z)] \equal{} [f(y) \plus{} f(z)] \plus{} f(y)\implies f(f(y)) \equal{} 2f(y)\ \forall \ y\in \mathbb R^ \plus{}$.

4) Putting $ y: \equal{} f(y)$ in (2) and using (3) we have $ f(x \plus{} f(y)) \equal{} f(x) \plus{} 2f(y)\ \forall \ x > f(y)$, $ x,y\in \mathbb R^ \plus{}$.

5) Equating (4) and (0) we have $ f(x \plus{} y) \equal{} f(x) \plus{} f(y)\ \forall \ x > f(y)$, $ x,y\in \mathbb R^ \plus{}$.

6) If $ f(a) \equal{} f(b)$ for some $ a\neq b\in \mathbb R^ \plus{}$ then putting $ y: \equal{} a$ in (0) and then putting $ y: \equal{} b$ in (0) we get $ f(x \plus{} a) \equal{} f(x \plus{} b)\ \forall \ x\in \mathbb R^ \plus{}$. That is, $ f$ becomes eventually periodic. But (5) implies that for each $ y$, all large enough $ x$s satisfy $ f(x) > f(y)$. Contradiction. So $ f$ is injective.

7) Applying (5) in RHS of (0) we get $ f(x \plus{} f(y)) \equal{} f(x \plus{} 2y)\ \forall \ x > f(y) \minus{} y$ but (6) gives $ f(y) \equal{} 2y\ \forall \ y\in \mathbb R^ \plus{}$.

This problem was proposed by Paisan Nakmahachalasint, leader of Thailand IMO team in 2008 and 2009.

@Raja Oktovin. I doubt you will see this, but this was one of the neatest solutions in my opinion. i really loved it.

can we extend this result if the function is defined from all reals to all reals? (R to R instead of R+ to R+)

[quote="Raja Oktovin"]The following solution is due to Andreas Dwi Maryanto Gunawan, from Indonesia.

For any positive real numbers $ z$, we have that \[ f(x\plus{}f(y))\plus{}z\equal{}f(x\plus{}y)\plus{}f(y)\plus{}z\]  \[ f(f(x\plus{}f(y))\plus{}z)\equal{}f(f(x\plus{}y)\plus{}f(y)\plus{}z)\] \[ f(x\plus{}f(y)\plus{}z)\plus{}f(x\plus{}f(y))\equal{}f(x\plus{}y\plus{}f(y)\plus{}z)\plus{}f(x\plus{}y)\] \[ f(x\plus{}y\plus{}z)\plus{}f(y)\plus{}f(x\plus{}y)\plus{}f(y)\equal{}f(x\plus{}2y\plus{}z)\plus{}f(y)\plus{}f(x\plus{}y)\] \[ f(x\plus{}y\plus{}z)\plus{}f(y)\equal{}f(x\plus{}2y\plus{}z)\] sehingga \[ f(a)\plus{}f(b)\equal{}f(a\plus{}b)\] and by Cauchy in positive reals, then $ f(x)\equal{}\alpha x$ for all $ x \in (0, \infty)$. Now it's easy to see that $ \alpha\equal{}2$, then $ f(x)\equal{}2x$ for all positive real numbers $ x$.

Is this OK? Please check this solution...[/quote]

i think it is wrong !!!

Because we have :

$f(x)+f(y)=f(x+y)\: \: \forall x,y\in \mathbb{R}^{+}\: ,x> y$

Let $P(x,y)$ denote the given assertion.

We have $P(x,x+f(y)) \implies f(x+f(x+f(y)))=f(2x+f(y))+f(x+f(y))=2f(y)+f(x+y)+f(2x+y)$

On the other hand, $f(x+f(x+f(y)))=f(x+f(x+y)+f(y)) \stackrel{P(x+f(y),x+y)} = f(2x+y+f(y))+f(x+y) \stackrel{P(2x+y,y)} = f(2x+2y)+f(y)+f(x+y)$

Comparing the two expressions, we get $f(y) + f(2x+y) = f(2x+2y)$ for any $x,y \in \mathbb{R^+}$ which implies $f(x) + f(y) = f(x+y)$ for any $x \neq y$.

Now suppose for some $y$, $f(y) < y$, then $P(y-f(y),y) \implies f(2y-f(y))=0$, contradiction, thus $f(y) > y$ for all $y \in \mathbb{R^+}$.

Suppose for some $a > b$, $f(a)=f(b)$, then $P(x,a)$ and $P(x,b)$ yields that $f(x) = f(x+c)$ for all $x > b$ where $c = a-b > 0$. But this is clearly a contradiction as then $f(x) = f(x+kc) > x + kc$ for any $k \in \mathbb{N}$, so $f$ is injective.

Finally, from the equation and additivity & injectivity we get: $f(x+f(y))=f(x+y)+f(y)=f(x+2y) \implies f(y) = 2y \ \forall y \in \mathbb{R^+}$.

Thus the only solution that works is $\boxed{f(x) = 2x \ \forall x \in \mathbb{R^+}}$ which clearly satisfies the given assertion. $\square$

[quote=lgkarras]$P(x, y + f(z))$ finishes the problem.

We know that if $f(x)$ is a solution of $f(x+y) = f(x) + f(y)$ then $f(x)=ax, a=f(1)$, if the following condition is satisfied:
There is $d>0$ s.t. $\forall x \in (0, d)$ we have $f(x) \geq 0$. (or the opposite ($f(x) \leq 0$))

There was no point in the whole discussion since this finishes the problem, because the condition is satisfied by the definition of $f$.

The idea behind substituting $y$ with $y + f(z)$ is to use the given relation many times ending up in sth useful. In fact, one can see that you can use the condition 3 or 4 times in a row easily.[/quote]

I got $f(x+y+2z)=f(x+y+z)+f(z)$. Then $f(x+y)=f(x)+f(y)$ for all $x>y$ or by symmetry if you want $f(x+y)=f(x)+f(y)$ for $x\ne y$. Now it's not trivial to finish, but still pretty easy.

[quote=jasperE3]
I got $f(x+y+2z)=f(x+y+z)+f(z)$. Then $f(x+y)=f(x)+f(y)$ for all $x>y$ or by symmetry if you want $f(x+y)=f(x)+f(y)$ for $x\ne y$. Now it's not trivial to finish, but still pretty easy.[/quote]

You have that $f(x)=ax$ and substituting this in the given relation yields the answer. When I said trivial, I meant that you don't have to do anything special rather than common algebraic operations.

We claim that the only function that works is $f(x)=2x$. Let $P(x, y)$ be the assertion, we begin with the following claim

[b][color=#f00]Claim:[/color][/b] $f(y)>y$, $\forall y>0$
[b][color=#00f]Proof:[/color][/b] $$P(y-f(y), y) \rightarrow f(2y-f(y))=0$$ So $y<f(y)$ since the co-domain is positive reals. $\blacksquare$
Now let $g(x)+x=f(x)$ and we have $g(x+y+g(y))=g(x+y)+y)$ and this is also on positive reals. Let $Q(x, y)$ be its assertion and now $Q(a, b)$ and $Q(b, a)$ tell us that $g$ is injective. Now,
$$Q(x, y+z) \rightarrow g(x+y+z+g(y+z))=g(x+y+z)+y+z$$
$$g(x+y+z+g(y+z))=g(x+y+z+g(y))+z$$
$$g(x+y+z+g(y+z))=g(x+y+z+g(y)+g(z))$$ This implies $g$ is linear by the injectivity of $g$ and therefore we see that $g(x)=x$ and $f(x)=2x$. $\blacksquare$

The answer is $f(x)=2x$ only, which clearly works. Let $P(x,y)$ denote the given assertion. We begin with the following claim:

[b][color=#00f]Claim:[/color][/b] $f(x)>x$ for all $x$.
[color=#00f]Proof:[/color] If $f(a)=a$ for some $a$, then $P(1,a)$ yields $f(a)=0$, which is illegal. If $f(a)<a$ for some $a$, we have $f(x+a)-f(x+f(a))=-f(y)$. Then $f(1+f(a)),f(1+a),f(1+2a-f(a)),f(1+3a-2f(a)),\ldots$ is a strictly decreasing arithmetic sequence, but it must be bounded below by $0$, which is impossible. This yields the desired result. $\blacksquare$

From the claim, we can readily define $g : \mathbb{R}^+ \to \mathbb{R}^+$ such that $g(x)=f(x)-x$. Substituting, the given assertion is equivalent to
$$g(x+y+g(y))=g(x+y)+y,$$
which is equivalent to $g(x+g(y))=g(x)+y~\forall x>y$.
By varying $y$ with $x$ fixed, $g$ is injective. Now, for any $y,z$, pick $x>y+z$, so we have
$$g(x+g(y+z))=g(x)+y+z=g(x+g(y))+z=g(x+g(y)+g(z)).$$
By injectivity this means $g(y)+g(z)=g(y+z)$. As $g : \mathbb{R}^+ \to \mathbb{R}^+$, this means $g(x)=cx$ for some $c>0$. Substituting yields $c=1$, hence $g(x)=x$ and $f(x)=2x$. $\blacksquare$

I'm an actual clown.

Let $g: \mathbb{R}^{ + }\to\mathbb{R}$ such that $f(x)=g(x)+x.$ We have $g(x+y+g(y))+x+y+g(y)=g(x+y)+x+y+g(y)+y$ which implies that \[g(x+y+g(y))=g(x+y)+y\]
If $g(x)=g(y)$ then $g(x+y)+y=g(x+y)+x$ which implies $g$ injective. Note that $\displaystyle\lim_{y\to 0}(g(x+y+g(y))-g(x+y))=0.$ If $\displaystyle\lim_{y\to 0}g(y)=c$ for some $c>0$ then $g$ is periodic which contradicts injectivity. Thus, $c=0$ and $g$ is continuous.

Since $g$ is continuous, we take $0=\displaystyle\lim_{x\to 0}(g(x+y+g(y))-g(x+y)-y)=g(y+g(y))-g(y)-y=0.$ Thus, for all $z=g(y)+y$ we have $g(z)=z.$ Thus, for all $z$ in the range of $f$ we have $f(z)=2z.$ If $z$ is in the range of $f$ then clearly $2z$ is. Since $f$ must also be continuous, any number in $[z,2z]$ is in the range of $f.$ Inductively, all numbers in $[z,\infty)$ is in the range of $f.$ However, we can pick $z$ as small as we want because $\displaystyle\lim_{y\to 0}g(y)=0.$ Thus, $f$ is surjective which implies $f(x)=2x$ for all $x.$ This is a valid function after checking.

We claim $f(x) = 2x$. Note that this works. Let $P(x,y)$ be the assertion
\[f\left(x + f\left(y\right)\right) = f\left(x + y\right) + f\left(y\right)\]
If $f(y)<y$ for some $y$, $P(y-f(y),y)$ yields a contradiction. It is also clear that $f(y)\ne y$ for all $y$, so $f(y)>y$ for all $y$.

[b]Claim:[/b] f is injective.
[b]Proof:[/b] if $f(a)=f(b)$ for $a>b$,
\[P(x,a)-P(x,b):f(x+a) = f(x+b)\]
\[f(x+(a-b)) = f(x) \forall x>b \]
\[\implies f(x) = f(x+n(a-b)) > x+n(a-b)\]
for all positive integers $n$ and positive reals $x>b$; a contradiction for $n$ large.

Fix $y$. Take x>y.
\[P(x-y,y):f(x+f(y)-y) = f(x)+f(y)\forall x>y\]
\[P(x,x+f(y)-y):f(x+f(x)+f(y)) = f(2x+f(y)-y)+f(x+f(y)-y) = f(2x)+f(x)+2f(y) \]
\[P(x,x):f(x+f(x)) = 2f(x) + f(2x)\]
\[\implies f(x+f(x)+f(y)) = f(x+f(x))+2f(y) = f(x+f(x)+2(f(y)-y))\]
\[\therefore f(y) = 2f(y)-2y \therefore f(y) = 2y.\]

Let $g:\mathbb{R}^+\rightarrow \mathbb{R}$ be with $f(x)=g(x)+x$. The given equation rewrites \[g(x+g(y))=g(x)+y\textrm{ for }x>y\]
then, notice by putting $y\rightarrow y+g(z)$ with $y>z$ and $x>y+g(z)$ \[g(x+z)+y=g(x+z+g(y))\stackrel{P(x,y+g(z))}{=}g(x)+g(z)+y\Rightarrow g(x+z)=g(x)+g(z) \textrm{ for }x>z+g(z).\]
With the last additivity, \[g(x)+g(y+z)=g(x+y+z)=g(x+y)+g(z)=g(x)+g(y)+g(z) \textrm{ for } x>\max(y+g(y),z+g(z),y+z+g(y+z))\] and thus $g$, and hence $f$, is additive. We get that $f$ is linear and substituting, we get $f(x)=2x$ as the only solution.

[b][color=#f00]Remark:[/color][/b] The motivation for $g$ isn't only from the solution $f(x)=2x$, but it's easily seen (plug $x=y-f(y)$) that $f(x)>x$, hence the codomain doesn't change when introducing $g$. This isn't directly used.

Somehow got a weird infimum solution

This is equivalent to \[ f(x + (f(y) - y)) = f(x) + f(y) \] holding for $x > y$.
Note that $f(x) \ne x$ due to there being no roots. Furthermore, $f(x) < x$ can not hold, or else $f(x - k(f(y) - y))$ would be negative for sufficiently large positive integer $k$ by the expression.

[b][color=pink]Claim:[/color][/b] Fix $y$ and let $c$ be the infimum of $f(x)$ on the interval $[y, y + f(y) - y)$. Then, $c + kf(y)$ is the infimum on the interval $[y + k(f(y) - y), y + (k+1)(f(y) - y))$.
[i]Proof.[/i] Follows since shifting by a constant preserves the infimum. $\blacksquare$
Note that for each $y$, these intervals are covering and the infimums lie on a line with slope $\frac{f(y)}{f(y) - y}$.
Then, $\frac{f(y)}{f(y) - y}$ must be constant over all $y$ or else that leads to contradiction. As such, this implies that $\frac{y}{f(y)}$ is constant, and thus $f(x) = cx$ for some $c$. Checking shows that only $f(x) = 2x$ works.

$P(0,y)$ gives $f(f(y)) = 2y$, then $P(x,f(y))$ gives $f(x + 2f(y)) = f(x + f(y)) + 2f(y) =  f(x + y) + 3f(y)$. This then implies that $f(x - f(y) + f(y)) = f(x - f(y) + y) ) + f(y), f(x - f(y) + 2f(y)) = f(x - f(y) + y) + 3f(y)$, so $f(x + f(y))  - f(x) = 2f(y)$, or $f(x + y) + f(y) - f(x) = 2f(y)$, giving $f(x + y) = f(x) + f(y)$. Now we have $f(x + f(y)) = f(x + y) + f(y) = f(x + 2y)$. Now we just need to show $f$ injective to finish. Consider two different values $a,b$ of $y$ giving the same $f(y)$. Then we have $f(x + a) =  f(x+b)$, thus we can find numbers $c,d$ with arbitrarily large difference such that $f(c) = f(d)$. Then, we can set $y = c$ and $x + f(y) = d$, which forces $f(x + y) = 0$, which is impossible. Since $f$ injective, we can easily get $f(y) = 2y$.

Wasn't thinking of posting this one since I took some hints... But someone check this please... Also the above solution is totally flawed as of now(at the time I'm posting) since we're talking $x,y > 0$.

[i]Solution:[/i] The answer is $f(x) = 2x$ for all $x \in \mathbb{R}^+$. This clearly works and we now show that this is the only possible solution. Let $P(x,y)$ denote the assertion to the functional equation.

[b][color=red]Claim:[/b][/color] $f(x) \ge x$ and $f$ is an injective function.

[i]Proof:[/i] Assume for the sake of contradiction there is some positive real $k$ such that $f(k) < k$. Then $P(k-f(k), x)$ gives $f(2x - f(x)) = 0$ which is impossible for the given co-domain. Therefore $f(x) \ge x$ for all $x$.

Once again, assume on the contrary that there exists some positive reals $a>b>0$ such that $f(a) = f(b)$. This time $P(x,a)$ and $P(x,b)$ yield us that $f$ is eventually periodic with period $T \coloneqq a-b$. Consider some huge enough $\alpha \in \mathbb{R}^+$ for periodicity. Choose a positive integer $m$ which is large enough such that $f(\alpha)< \alpha + mT$. Combining everything
        \[\alpha \le f(\alpha) < \alpha + mT \le f(\alpha + mT) = f(\alpha)\]
which is absurd. Hence the claim holds true. $\square$
-----------------
The latter part of the solution is mainly grinding substitutions until you get a nice one. Our primary goal here is to show that $f(2x + y) = f(2x) + f(y)$ for all $x > y$ which is almost a Cauchy like condition.
[list]
        [*] $P(x-y, y)$ and $P(x,x-y+f(y))$ for $x > y$ in sequence would yield us that
        \begin{align}
            f(x + f(x) + f(y)) = f(2x) + 2f(y) + f(x) \tag{1.1} \label{1.1}
        \end{align}
        [*] $P(x+f(x), y)$ for $x > y$ we will get
        \begin{align}
            f(x + f(x) + f(y)) = f(x + f(x) + y) + f(y) \tag{1.2} \label{1.2}
        \end{align}
        [*] Combine $(1.1)$ and $(1.2)$ to get
        \begin{align}
            f(x + f(x) + y) = f(2x) + f(y) + f(x) \,\, \text{for $x>y$.}\tag{1.3}\label{1.3}
        \end{align}
        [*] Consider $P(y + x, x)$ for $x > y$ to get
        \begin{align*}
            f(x + y + f(x)) = f(2x + y) + f(x)
        \end{align*}
        and finally combine this with $(1.3)$ to get
        \begin{align}
            f(2x + y) = f(2x) + f(y) \tag{1.4} \label{1.4}
        \end{align}
        for $x>y$ which is the partial Cauchy condition advertised above.
[/list]
Replace $x$ by $2y$ in $(1.4)$. This would fetch us $f(y) + f(4y) = f(5y)$. Now consider $P(3x,x)$ to get
    \[f(3x + f(x)) = f(4x) + f(x) =  f(5x) \iff f(x) = 2x\]
due to injectivity. This completes the solution. $\blacksquare$

We claim the only solution is $f(x)=2x$, which obviously works.
Conversely, let $f$ be a function that satisfies the given equation.

[b]Claim [/b]: For all $y>0$ we have $f(y)>y$.
[i]Proof :[/i] Suppose for the sake of contradiction that $f(y)\le y$ for some $y>0$.
Clearly we have $f(y)<y$, because any fixed point $y$ is such that $f(y)=0$ by simply looking at the statement, which is not possible.
We can thus take $x=y-f(y)>0$ and get $f(x+y)=0$, which is again a contradiction. $\square$

This motivates us to define a function $g:\mathbb R_+^{*}\rightarrow \mathbb R_+^{*}$ such that $g(x)=f(x)-x$ for all $x>0$. The equation that we have for $g$ is
\[\forall x,y\in \mathbb R_+,\quad g(x+y+g(y))=g(x+y)+y\quad (*),\]
and we need to prove that $g$ is the indentity function.

[b]Claim [/b]: $g$ is injective.
[i]Proof :[/i] Suppose $g(x)=g(y)$ for some reals $x,y>0$. We then have
\[g(x+y)+y=g(x+y+g(y))=g(x+y+g(x))=g(x+y)+x,\]
i.e. $y=x$. $\square$

[b]Claim [/b]: For every positive integer $n$ and for every real number $x>0$, we have $g(nx)=n\cdot g(x)$.
[i]Proof :[/i] Note that we can interprate $(*)$ as follows : For all real numbers $y>0$ and $A>y$, we have
\[g(A+g(y))=g(A)+y\quad (1)\]
In particular, note that iterating gives
\[g(A+2g(y))=g(A+g(y)+g(y))=g(A+g(y))+g(y)\]
We thus get by induction that $g(A+kg(y))=g(A)+ky$ for every positive integer $k$ and real numbers $A>y>0$. In particular, we have that for all $k\in \mathbb N_0$ and $y>0$,
\[g(B+kg(y))=g(B)+ky=g(B+g(ky)),\]
where $B>ky$ is some constant. By injectivity this implies that $g(ky)=k\cdot g(y)$, as desired. $\square$

[b]Claim [/b]: $g$ is surjective.
[i]Proof :[/i] Let $m$ be the infimum of $g$. Note that the $RHS$ of $(1)$ takes all real values greater than $m$, whereas the $LHS$ takes values of $g$ where the input is greater than $m$. It follows that $m$ must be zero by injectivity.
In particular, $g$ takes arbitrarily small values.
Given any $t>0$, we can thus take some $A>0$ such that $g(A)<t$, and choose a positive integer $k$ large enough so that
\[y=\frac{t-g(A)}{k}<A.\]
We then have $g(A+kg(y))=g(A)+ky=t$, which proves that $g$ is indeed surjective. $\square$

We're now ready to finish off the problem.
Take some $y>0$, and let $y_0>0$ be such that $g(y_0)=0$. Then, we choose a positive integer $k$ large enough, so that $A=k\cdot y_0>y$. We then have
\begin{eqnarray*}
    g(ky+g(g(y_0)))=g(A+g(y))=g(A)+y&=&g(ky_0)+y\\
    &=& kg(y_0)+y\\
    &=& (k+1)y\\
    &=& g((k+1)y_0)
\end{eqnarray*}
By injectivity, we must have
\[ky+g(g(y_0))=(k+1)y_0\iff k(y-y_0)=T,\]
where we let $T=y_0-g(g(y_0))$. Note that this holds for all $k\in \mathbb N$ large enough, thus it must be the case that $y=y_0$, which proves that $g$ is the identity.
$\blacksquare$

The solution is $f\equiv 2x$.

$P(x,y+f(z)) \implies f(x+f(y+f(z))) = f(x+y+f(z)) + f(y+f(z)) = f(x+y+z) + f(z) + f( y+ z) + f(z)$.

Also,
\begin{align*}
    f(x+f(y+f(z))) &= f(x+ f(y+z) + f(z)) \\
    &= f(x+f(y+z) +z) + f(z) \\
    &= f(x + z + f(y+z)) + f(z) \\
    &= f(x+z+y+x) + f(y+z) + f(z)
.\end{align*}

Comparing, we get that $f(x+y+2z) = f(x+y+z) + f(z)$.

Now for any $b>a>0$, set $z = a$ and $x = y = \dfrac{b-a}{2}$.

So $f(b+a) = f(b) + f(a)$. Also, $f(2x) = f(x+ \varepsilon) + f(x - \varepsilon) = f(x) + f(\varepsilon) + f(x-\varepsilon) = f(x) + f(x) = 2f(x)$.

Thus $f$ is Cauchy and linear $\implies f\equiv 2x$. :yoda:

solved with ThapaKazi
pog
[hide]
Denote $P(x, y)$ to be the plug-in of $x$ and $y$ in $f$.
[b][color=#f00]Claim[/color][/b]. $f(y) > y$ for all $y \in \mathbb{R}^+$.
[b][color=#0f0]Proof[/color][/b]. Suppose for the sake of contradiction that $y > f(y)$ for some $y \in \mathbb{R}^+$. Thus $P(y - f(y), y)$ implies:
$$f(y) = f(2y - f(y)) + f(y) \implies f(2y -f(y)) = 0$$
which is not possible, a contradiction. Thus the claim is proven.
$\blacksquare$
Now we define a function $g: \mathbb{R}^+ \to \mathbb{R}^+$ such that $g(x) = f(x) - x$. From the original equation, this gives
$$g(x + y + g(y)) = y + g(x + y)$$
Let $z = x + y$:
$$g(z + g(y)) = g(z) + y$$
Now assume the existence of $y_1 \neq y_2 \in \mathbb{R}^+$ such that $g(y_1) = g(y_2)$. Plugging in $y_1$ and $y_2$ instead of $y$:
$$g(z) + y_2 = g(z + g(y_2)) = g(z + g(y_1)) = g(z) + y_1 \implies y_1 = y_2$$
This means $g$ is injective. Now substituting $z = y + z$ in
$$g(z + g(y)) = g(z) + y$$
we obtain the series of equalities
$$g(z + g(y + z)) = g(z) + y + z = g(z + g(y)) + z = g(z + g(y) + g(z))$$
However by injectivity, this means $g(y + z) = g(y) + g(z)$, which is Cauchy, and since we are working over $\mathbb{R}^+$, $g(x) = cx \forall x$. Plugging this into $g(z + g(y)) = g(z) + y$ gives $c = 1$, thus $g(x) = x \implies f(x) = 2x \forall x$.
Easily check to see that this works.
[/hide]


That isn't true

take $x=1$

$1(e^{1}) = 1^{1+\log 1}\implies e = 1$ which isn't true

[quote="bos1234"]show that $xe^{x}$=$x^{x+logx}$[/quote]
Maybe you mean $xe^{x}$=$e^{x+logx}$, which of course is trivial (just take logaritms in both parts).

[hide]
$xe^{x}= e^{x+\log x}$
$\ln x+x = x+\log x$
$\ln x \not = \log x$

EDIT
If $\log$ meant $\ln = \log_{e}$
Then $\ln x = \log x$.
[/hide]

[quote="AstroPhys"][hide]
$xe^{x}= e^{x+\log x}$
$\ln x+x = x+\log x$
$\ln x \not = \log x$
[/hide][/quote]
Some people (more so in college than in high school) write $\log$ to mean $\ln$, rather the common base 10 log.

Who uses $\log_{10}$ anyways? $\log$ for $\ln$ is simply better notation.


can someone tell me what activities do they offer in the USA/Canada math camp.
do they have a tennis court?
thanks

Colby College (where MC is held this year) most probably does have a tennis court.

See: http://www.mathcamp.org/day.php

Does anyone think people on the waiting list can still get in?
People like me of course...

If you're on the waiting list, there still is a chance that you will get to go.  Last year the first person I met at mathcamp was notified only two weeks before camp started that she could go, so be patient!

In terms of activities at mathcamp, there are usually a lot.  Last year at UPS there were soccer fields, a pool, a weight room/gym, and I'm pretty sure there were tennis courts.  Might be different at colby though.  I ran a lot, but I guess you can do that anywhere.  (rock climbing was the best field trip ever!  just had to put that in)  And of course there are a lot of other activities besides athletics.

-Sara Sheehan

David Savitt will be hosting a Math Jam in early June and can answer some of your questions.


Find the largest positive integer x such that $ x^{2}\minus{}19x\plus{}94$ is equal to a perfect Square.

[hide]I got 13[/hide]

Simply note that $ (x\minus{}10)^{2}\equal{} x^{2}\minus{}20x\plus{}100 < x^{2}\minus{}19x\plus{}94 < x^{2}\minus{}18x\plus{}81 \equal{} (x\minus{}9)^{2}$, if $ x\ge 14$, and check that $ x^{2}\minus{}19x\plus{}94$ is a perfect square (namely, $ 2^{4}$), when $ x \equal{} 13$.

Yes we can solve
$ P(x)\equal{}x^2\plus{}ax\plus{}b$
Find $ max\{x\in N\}$ such that $ P(x)$ is a perfect square.


well most problems like this would use  PIE, but it looks pretty ugly...
any clever ideas?

you could also use a recursion i think.

[quote="aidan"]How many 7 digit-integers are there in which no digit appears 3 or more times consecutively (7777321 is not allowed, but 7732771 is allowed)?[/quote]

[hide]Let $ a_n$ be the number of $ n$-digit numbers that end in one digit different from the second to last digit and also avoid the restriction.
Let $ b_n$ be the number of $ n$-digit numbers that end in two digits that are the same and also avoid the restriction.

Then we have $ a_{n\plus{}1}\equal{}9a_n\plus{}9b_n$ and $ b_{n\plus{}1}\equal{}a_n$.

Start with $ a_1\equal{}9$ and $ b_1\equal{}0$ and get $ a_7\plus{}b_7\equal{}\boxed{8588349}$?[/hide]

Yup, i also get 8588349 by direct counting, considering k blocks of double digits and (7 - 2k) blocks of single digits and evaluating over k = 0, 1, 2, 3.


Bob and Ricky bought identical boxes of stationary. Bob used his to write 1-page letters, Ricky used his to write 3-page letters. Bob used all his envelopes and had 50 sheets of paper lefter over. Ricky used all the paper and had 50 envelopes left over. How many sheets of paper were in each box.

Solution:
[hide] let e=envelopes and p=paper. By Bob's condition, e=p-50. By Ricky's, e=p/3 +50.

e=p-50
e=p/3 +50.

Subtract equations to get 2p/3 = 100, so p=150. Then e=150-50=100. There are 150 pieces of paper and 100 envelopes.[/hide]


I hear they have changed the rules for the countdown round.  Can anyone tell me how it is now?

Thanks!

At Nationals, the top 12 made it into the Countdown Round.

1st round:

Two people from places 5 to 12 go against each other. The winner goes on to round 2. Repeat this three times.

2nd round:

Two people go against each other. The winner goes on to round 3. Repeat this three times.

3rd round:

Two people go against each other. The winner goes on to the final round. Repeat this once.

Final round:

The two people who made it go against each other. The winner is the champion.


LaTeX
Support
trigonometry
ratio
geometry proposed
geometry

Assuming the theorem of Ceva and its converse deduce the theorem of Menelaus

The equivalence is not often mentioned.The proof is a bit tricky.

I have solved your problem but today LaTeX is not okey and to write my solution using MSWord takes too much time, so I promise that I'll write my solution tomorrow(ofcource if LaTeX work).

Here is my solution(as I promissed).
Consider the triangle [tex]ABC[/tex].Let the line [tex]\ell[/tex] intersects the sides [tex]AB,BC,CA [/tex]at the points [tex]G,E[/tex] and [tex]F[/tex] respectively.
There are two major cases.
[u]Case a)[/u][tex]E[/tex] and [tex]G[/tex] are internal points of the segments[tex] BC[/tex] and [tex]AB[/tex](obviously [tex]F[/tex] is an external point of the segment [tex]AC[/tex]).
[u]Case b) [/u][tex]E,F,G[/tex] are external points of the segments [tex]BC,CA,AB.[/tex]
Lets start with [u]Case a)[/u]:
We have to prove that [tex]\frac{AF}{FC}\cdot\frac{CE}{EB}\cdot\frac{BG}{GA}=1 (1)[/tex]
 Applying Ceva's theorem we have [tex]\frac{FA}{AC}\cdot\frac{CE}{EB}\cdot\frac{BH}{HF}=1 (2)[/tex],where [tex]H={BF}\cap{CG}.[/tex]
So using [tex](2)[/tex] we can say that for proving [tex](1)[/tex] it sufficient to prove that
[tex]\frac{BG}{GA\cdot{FC}}=\frac{BH}{HF\cdot{AC}} (3)[/tex]
Next idea is the using of Van-Obel's theorem which states that
[tex]\frac{BG}{GA}=\frac{BH}{HF}+\frac{BE}{EC}.[/tex]
Puting last one in [tex](3)[/tex] and doing some simplifications we get
 [tex]\frac{FA}{AC}\cdot\frac{CE}{EB}\cdot\frac{BH}{HF}=1[/tex]
which is exactly [tex](2)[/tex].
 [u]Case b) [/u]
In this case WLOG we can assume that points [tex]G,E [/tex]are on the rays [tex]BA[/tex] and [tex]BC[/tex],and the point  [tex]F[/tex] is on the ray [tex]AC[/tex].
So here we can use the result of the Case a)(as a result of Ceva's theorem) .
It is to write Menelause's theorem for triangle [tex]BGE[/tex] and line[tex] AC[/tex](as we already proved it using Ceva's theorem).Here it is:
[tex]\frac{GA}{AB}\cdot\frac{BC}{CE}\cdot\frac{EF}{GF}=1[/tex].So it remains to show that [tex]\frac{BE}{GB}\frac{CF}{AF}=\frac{BC}{AB}\frac{EF}{GF}[/tex].This becomes obvious after using sinerule(sorry for the wrong word) for triangles [tex]ABC,BGE,CEF,AGF. [/tex]

Yes,I have.But there is no any need of Ceva's theorem here,because it is simple.
Consider triangle [tex]ABC[/tex] with points [tex]M,N,K[/tex] on the sides [tex]BC,CA,AB[/tex],respectively.Suppose that [tex]AM,BN,CK[/tex] intersect at the point [tex]G[/tex].Van-Obel's thorem states that
[tex]\frac{BG}{GN}= \frac{BK}{KA}+\frac{BM}{MC}[/tex].
Here is the prove:
[tex]\frac{BG}{GN}=\frac{[ABG]+[CBG]}{[AGC]}=\frac{[ABG]}{[AGC]}+\frac{[CBG]}{[AGC]}=
\frac{BM}{MC}+\frac{BK}{KA}.[/tex]

[quote="juancarlos"]Do you have a proof of Van-Obel's theorem with Ceva?[/quote]

Actually I have one, but curiously enough, it is much more involved than the simple proof without Ceva:

Consider a triangle ABC and three points A', B', C' on its sidelines BC, CA, AB such that the lines AA', BB', CC' concur at one point P. Then Van Aubel's (or Van Obel's) theorem states

$\displaystyle \frac{BP}{PB^{\prime }}=\frac{BC^{\prime }}{C^{\prime }A}+\frac{BA^{\prime }}{A^{\prime }C}$.

The proof using Ceva (and Menelaos) goes as follows: Use Menelaos for the triangle ABB' and the collinear points C, P, C' to get

$\displaystyle \frac{BP}{PB^{\prime }}\cdot \frac{B^{\prime }C}{CA}\cdot \frac{AC^{\prime }}{C^{\prime }B}=-1$.

Thus,

$\displaystyle \frac{BP}{PB}=-\frac{CA}{B^{\prime }C}\cdot \frac{C^{\prime }B}{AC^{\prime }}=-\frac{CA}{B^{\prime }C}\cdot \frac{BC^{\prime }}{C^{\prime }A}=-\frac{CB^{\prime }+B^{\prime }A}{B^{\prime }C}\cdot \frac{BC^{\prime }}{C^{\prime }A}$
$\displaystyle =\frac{B^{\prime }C+AB^{\prime }}{B^{\prime }C}\cdot \frac{BC^{\prime }}{C^{\prime }A}=\left( 1+\frac{AB^{\prime }}{B^{\prime }C}\right) \cdot \frac{BC^{\prime }}{C^{\prime }A}$
$\displaystyle =\frac{BC^{\prime }}{C^{\prime }A}+\frac{AB^{\prime }}{B^{\prime }C}\cdot \frac{BC^{\prime }}{C^{\prime }A}=\frac{BC^{\prime }}{C^{\prime }A}+\left( \frac{AB^{\prime }}{B^{\prime }C}\cdot \frac{CA^{\prime }}{A^{\prime }B}\cdot \frac{BC^{\prime }}{C^{\prime }A}\right) :\frac{CA^{\prime }}{A^{\prime }B}$
$\displaystyle =\frac{BC^{\prime }}{C^{\prime }A}+1:\frac{CA^{\prime }}{A^{\prime }B}$ (from the Ceva theorem applied to triangle ABC)
$\displaystyle =\frac{BC^{\prime }}{C^{\prime }A}+\frac{A^{\prime }B}{CA^{\prime }}=\frac{BC^{\prime }}{C^{\prime }A}+\frac{BA^{\prime }}{A^{\prime }C}$,

proving Van Aubel's theorem.

But since this proof also uses the Menelaos theorem, it is not of value for the proof of Menelaos using Ceva. A real proof of Menelaos using Ceva can be found on [url=http://www.cut-the-knot.org/Curriculum/Geometry/MenelausFromCeva.shtml]Alexander Bogomolny's site[/url]. But actually note that this proof uses inside-outside observations and the Pasch axiom ("It could be argued that the points D, E, F either all lie on the extensions of the sides of ABC, or only one does, while the other two lie on its sides."), whether the [url=http://www.cut-the-knot.org/Generalization/Menelaus.shtml#CfromM]proof of Ceva using Menelaos[/url] doesn't.

  Darij

That is a interesting theorem:"In a Euclidean triangle the angles' sum is  $\pi$."

The demonstration is classic... The problem now is: The theorems of Cristea,Van Obel, Ceva, Menelaos are consequences of the picture of the theorem above.

See the Cut-the Knot's site: http://www.cut-the-knot.org/triangle/VanObel.shtml

Darij,Actually I didn't know about the[u] Pasch [/u][u]axiom[/u].I thought that it was quite  simple statement(as most of the axioms :) ) and I thought that I could support it in thise way:
For example in the case when $A^{'},B^{'},C^{'} $are internal points of the segments $BC,CA,AB.$ , (using Darij's definitions) $\angle{B^{'}A^{'}C^{'}}=\pi-\angle{C^{'}A^{'}B}-\angle{B^{'}A^{'}C}<\pi,$
Is there any mistake here?

No, of course not. Your proof of the Pasch axiom is correct, but it used angles. And you must use this axiom to derive Menelaos from Ceva, so you must use angles to derive Menelaos from Ceva, whereas deriving Ceva from Menelaos does not use anything but simple algebra. That's an important matter if you are trying to do such things in more general spaces than the Euclidean one. And hence the Menelaos theorem can be said to be more fundamental than the Ceva theorem, since the Menelaos theorem implies the Ceva theorem using just algebra, while the Ceva theorem implies the Menelaos theorem only when the Pasch axiom is guarranteed to hold. (Well, this "only" is not really an "only" in the strict mathematical sense; it actually means that the proofs of the Menelaos theorem using the Ceva theorem I have seen use the Pasch axiom, but who knows, maybe someone will find a proof without using the Pasch axiom.)

  Darij

Oh!Thanks Darij for reply.Now I understand my [i]mistake[/i] at all.
 Actually I didn't understand the problem completely(using Van-Aubels theorem,
sine rule and angles). :blush:

[quote="Armo"]Now I understand my [i]mistake[/i] at all.[/quote]

Actually it's not a mistake, it's at most a misunderstanding. Every problem of the kind "Prove ... using ..." is, strictly speaking, not a mathematical problem, unless it is really exactly said what one may use and what not. In this actual problem, to prove Menelaos using Ceva, it was probably intended to find a proof like the one at [url=http://www.cut-the-knot.org/Curriculum/Geometry/MenelausFromCeva.shtml]Alexander Bogomolny's website[/url], i. e. a proof mostly using ratios of directed segments, but one time involving the Pasch axiom. But strictly speaking, it's a heuristic problem, not a mathematic one.

  Darij


Uh I randomly started writing a mock AMC. I'll be gone on July 1, so I need to grade all the answers before then. So I guess it will be on like Thursday or something. I can post it any time that day, I'm thinking 7:00PM EDT (it's 8:10PM EDT right now for your reference) but it's up to those who want to take it. Then you'll have to finish by 8:15PM and type up and send me the answers by 8:20PM.

Oh, and I'll try to make it the same difficulty as real AMC's, but those who took the mock ARML know that I'm not good at making up easy problems :P :P

So yeah sign up here.

I'll do it.  I should be able to make it at the time.

Well I won't be able to finish writing all the problems by early tomorrow, but I will let you take it as soon as I finish (which should hopefully be tomorrow afternoon).

I'll take it.

Will it be an AMC 10 or 12, or a mix of both?

I'm guessing an AMC 12.

I'll take it.

Most likely I will be at home.

[quote="Adunakhor"][quote="chess64"][quote="Adunakhor"]I'll try to take it, though it does turn out to be on Thursday, I might not since there's an AoPS class at that time.[/quote]

What would be a better time then?[/quote]

If I could take it earlier that day, or Wednesday, that'd be great.[/quote]

Ditto here; same class, so could I too take it Wed.?

I'll take it.  I'm also in the AoPS class with adunakhor and solafidefarms, so a time change would be nice.  Pretty much any other time is fine.

OK, those who can't take it today (in about 40 minutes), you have to take it on your own by tomorrow at 7:00, because that's when I will post the results and answer key and stuff.

Of course, you can always just take it on your own for fun and grade it yourself.

After you post the contest, if you post it somewhere else, can you post the link here?  I sent this link to all of the SC people.  Thanks!

I believe it has been 8 seconds already :)

Ugh, I won't be able to do it, I was busy today and tommorow's my last day of school meaning my graduation will be in the evening lasting around 4 hours (for less than 65 kids, a nearby school has 500 kids and only took 2 hours)

Can I still participate? If so, i guess i still have an hour & change left till the deadline.

EDIT: Never mind. I'll take it after the deadline...

[quote="mathgeniuse^ln(x)"]I believe I have pmd chess64 my answers.  When will we find out how good (or bad) we did.[/quote]

He pmed me back in like 20-30 minutes, IDK and I don't remember. But that's only when he's online.

[quote="xdl615x"]All the threads say MOCK AMC A, so does that mean we'll have a second MOCK AMC?[/quote]

No.

That means we might have a second mock AMC this year. The next person to write a mock AMC this year will write "Mock AMC 12 B 2007".

I hope someone does.  They make really good problems for us to solve and prepare for the AMC's.


function
Diophantine equation
number theory proposed
number theory

Indeed easy problem. Now i got the solution.

[u][b]Solution:[/b][/u]
If $f(n)=2+2\sqrt{28n^2+1}$. Let's say $f(n)=x$ and it is obvious that x is even(*). Then we have $2\sqrt{28n^2+1}=x-2$, taking the power of two we have: $4(28n^2+1)=x^2-4x+4$, $x=2k$ since (*). so we have
$4(28n^2+1)=4k^2-8k+4... 28n^2=k^2-2k$ from this equations we can easily say that k is even so $k=2m$, thus we have : $28n^2=4m^2-4m$, $7n^2=m^2-m$ ( Which is called as Pell equation) , $7n^2=m(m-1)$(*') we know that two consecutive integers are relatively, since $d|mx + (m-1)y$ for x=1 and y=-1 we have the fact $d|1$ which implies that d=1. so the equation (*') has the solutions:

1) $m=p^2$ and $m-1=7t^2$, from here we have $p^2 -7t^2=1$, which has many solutions and we don't need them because our aim is to show that $x=N^2$, we said $x=2k=4m=4p^2=(2p)^2$.

2)$m=7t^2$ and $m-1=p^2$, in this case we have $7t^2-p^2=1$,  which leads to contradiction, since $p^2$ never gives remainder 6 on division by 7. So no solutions in this case.

[Thanks for your attantion: for the first time i couldn't understand the problem, now everything is ok for me, and i thought that someone can get this useful because of that i decided to put a solution my self.]

Davron

[quote="Davron"]Let  n be a positive integer. If $2+2\sqrt{28n^2+1}$ is an integer, then it is a square.[/quote]
As a matter of interest ,such problems can be automatically solved by introducing hiperbolic functions ([b]sh[/b](x)[b],ch[/b](x)) when considering Pell.
$(2+2\sqrt{28n^2+1})\in\mathbb{N}\Longrightarrow(\sqrt{28n^2+1}=m)\in\mathbb{N}$.
$Pell: m^2-28n^2=1$
The least solution : [b]n=24,m=127[/b]; while for all the others the following holds:
$n=\frac{sh(ka)}{\sqrt{28}}$ (*),$m=ch(ka)$,for every $k\in\mathbb{N_{+}}$.
$a\in\mathbb{R_{+}}$ is determined as $ch (a)=127$

After substitution of (*) ,by algebra of hiperbolic functions we have:
$2+2\sqrt{28n^2+1}=2+2ch(ka)=4\frac{1+ch(ka)}{2}=4ch^2(\frac{ka}{2})=(2ch (k\frac{a}{2}))^2$
Furtherly,it is trivial to show  that :
if $ch^2 x \in\mathbb{N}\Rightarrow ch^2 (nx)\in\mathbb{N}$.
Since : $ch\frac{a}{2}=\sqrt{\frac{1}{2}(1+ch(a))}=\sqrt{\frac{1}{2}(1+127)}=8 \in\mathbb{N}$ ,the claim is prooved.


Prove that there is and only one value of a rational number x to satisfy the condition: x, x+5 and x-5 are rational perfect numbers!

Rational perfect number? what is it??? :huh:

He means "rational perfect square" (square of a rational number).

Can you give me a solution, Mr.Zeta X? Although I appplied the formula to solve the system of Pythagorean equation, it seems to be out of order! I found out that there is an article written about this special number on JSTOR but I can't read it thoroughly since I don't have an account!

Can anybody give me a hint about how to apply the general formula of Pythagore's equation to prove that there is just only one sets of (x,y) satisfying the system equation?I think it is a very famous and classical problem in Number theory

They used to be called congruum's back in the day. Not sure about there being only one value though

[quote="seamusoboyle"]They used to be called congruences back in the day. Not sure about there being only one value though[/quote]
OH! If your saying is true, can you show me another value besides the value I gave(41/12)! Anyway, thanks for your concerning! :)

I managed to find this not very informative link, it only deals with integers. Ive a book upstairs with the general case solved, but im in a wee rush so ill have to try dig it out later tonight

http://mathworld.wolfram.com/CongruumProblem.html

ok this is not finished cos i have no time now but i think it is a good idea:

if $ x$ is a perfect rational square we can assume $ x \equal{} \frac {a^2}{b^2}$ where $ a,b$ are integers. wlog $ a,b > 0$ and $ gcd(a,b) \equal{} 1$
then $ x\pm5 \equal{} \frac {a^2\pm5b^2}{b^2}$ and since $ x\pm5$ are perfect rational squares which implies $ a^2\pm5b^2$ are perfect integer squares.

therefore the problem is equal to finding such positive integers $ a,b$ that $ a^2 \minus{} 5b^2$, $ a^2$, $ a^2 \plus{} 5b^2$ are all perfect squares...

Sorry but I have to tell truthly that your solution is the most important thing I have ever seen in this problem! I'm very apprecited for your help if you can help me seriously!By the way, I found the general formula for this system equation and maybe, apply it with the condition 5 is a prime will help me kill this problem after a long time! :)

I'm not sure if this is what we are after, but this paper seems to prove it false.

[url]http://www.math.uconn.edu/~kconrad/ross2007/3squarearithprog.pdf[/url]

Corollary 3.6


$ a,b,c\geq 0$
$ a^2\plus{}b^2\plus{}c^2\equal{}3$
$ (2\minus{}ab)(2\minus{}bc)(2\minus{}ca)\geq 1$
thx for your nice solution  :)

[quote="Rose_joker"]$ a,b,c\geq 0$
$ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$
$ (2 \minus{} ab)(2 \minus{} bc)(2 \minus{} ca)\geq 1$
thx for your nice solution  :)[/quote]

Setting:
$ p \equal{} a \plus{} b \plus{} c;q \equal{} ab \plus{} bc \plus{} ac;r \equal{} abc$
$ \equal{} > p^2 \minus{} 2q \equal{} 3$
Inequality
$ < \equal{} > 13 \minus{} 2p^2 \minus{} r^2 \plus{} 2pr \ge 0$
$ \equal{} > f'(r) \equal{} 2p \minus{} 2r \ge 0$
We have:
$ r \ge \frac {p(p^2 \minus{} 6)}{9}$
$ \equal{} > f(r) \ge f(\frac {p(p^2 \minus{} 6)}{9}) \equal{} \frac {(p \minus{} 3)(12p^3 \plus{} 48p^2 \minus{} 162p \minus{} 486}{81} \ge 0$
It's true because $ p \le 3$
Nice  :P


Why can't I save the web pages as I did some times ago?

its probably because of the new server
but if u r facing this problem on all sites you can change your browser to FIREFOX.. ITS THE BEST


UNM-PNM second round is next weekend.  Any of you taking it?  How about your scores in the first round?  I got a 48, about what I expected, but still pretty bad.  I thought it was harder than last year's first round, even though my score was almost double (Last years 1st round was my first math contest).

I got a 52 on the UNM-PNM first round, a little better than last year, by only 2 points!

I wonder if they're ever going to come out with the 2nd round solutions and results.  It's been more than a month now.

haha

[size=200][color=blue]GO INDIANA!!!!!!!!!!!!!!!!!!![/color][/size]

For some weird reason, i never received my results!!!!
But oh well , i am still in fifth grade.....

Hmm... Now the first round is just based on the number of questions answered correctly. But the second round is atleast still going to grade work.

What's the scoring system? I just go by how many problems i get right. I got 7.

What everybody else is talking about is the old test. This year, the first round is just scored based upon number of questions correct. The second round, in a couple of weeks, is scored out of 80 (Ithink). They will grade your work, as they said in the letter sent out to RD. 2 qualifiers. I got an 8 1st RD. Good LUck all

Yay, second round scores are out! Banquet's on April 11th.

I wonder if they'll ever post 07-08 results on the website.

Wow, I did a lot better than I thought.....

I got 8 on the first round, but I only got 35 on the second round.  :(

Hi ppl!

2009-2010 UNM PNM starting up.

Nov. 7     1st Round @ Schools
Feb. 6     Lecture & 2nd Round @ Woodward Hall 101 UNM
Apr. 24     Winners Banquet @ Student Union Building UNM

http://www.math.unm.edu/math_contest/contest.html

yay good luck all. :)

Hmmm, we should list everyone's goals. Yay.

I'm gonna open a new thread

1st round 10/12. Needed 4/12 for my grade to qualify.

Tons and tons of people from my school qualified for second round.  :)

Hmm, a ton (or atleast a decent number) of people from NM are on AoPS. But only essentially three on this forum...

8/12 bleh, needed a 2/12.

how do you get 4 1/2 on the first round???(that was the 11th grade cutoff)

10/12 too darn that was fail. needed 2/12 to qualify....

hm, this year, prealgebra people made it into UNMPNM

oh, but this is for our school, aa is different....

Hey guys, this is Evan, I guess we'll be discussing the solutions at seminar on saturday.

Hmm, I can't remember from last year. I wonder if the results will be out in 1 day? 1 week? 2 weeks? 1 month? Day before award ceremony?

darn seriously when are the results coming out. last year, they came out one year from yesterday online... maybe there is the oft chance that dylan, andy, robert, me, etc. all didn't make it

I got 5th place. Up from tied 14th last year. Cannot attend banquet, though.


inequalities
number theory
prime numbers
inequalities solved

If $a,b,c,d$ are prime numbers all distinct then prove the following inequality:
$abc+bcd+cda+abd+173 \leq 2abcd$

cheers! :D :D :cool:

First divide both sides by abcd.

WLOG, a > b > c > d.

Then d \geq 2, c \geq 3, b \geq 5, a \geq 7.

Then LHS = \sum 1/a + 173 / abcd \leq 1/2 + 1/3 + 1/5 + 1/7 + 173/210 = 420/210 = 2.


When the decimal point of a certain positive decimal number is moved 4 places to the right, the new number is four times the reciprocal of the original number. What is the original number?

[hide]yes you just set up an equation and solve for x[/hide]


Prove that for any nonempty finite set $ S$, there exists a function
$ f : S\times S\rightarrow S$ satisfying the following conditions:
(a) for all $ a,b\in S$, $ f(a, b) \equal{} f(b, a)$
(b) for all $ a, b \in S$, $ f(a, f(a, b)) \equal{} b$
(c) for all $ a,b,c,d\in S$, $ f(f(a, b), f(c, d)) \equal{} f(f(a, c), f(b, d))$.

[hide]Let $ |S| \equal{} n$. $ f(a,b)$ equals the residue of $ \minus{}(a\plus{}b)$ mod $ n$ seems to work.[/hide]


algorithm
inequalities
number theory unsolved
number theory

If $a,b,c$ are positive integers such that:
$\displaystyle 0<a^2+b^2-abc\leq c$, show that $\displaystyle a^2+b^2-abc$ is a perfect square.

i think this problem is very interesting
obviously (a^2+b^2)/ab+1 <= c < (a^2 + b^2) / ab.
now if a^2+b^2/ab+1 is an integer, then a^2 + b^2 - abc = c. so in this case it just reduces to a past IMO problem ( i think 1989 q6. just look at kalva sol )
now otherwise we have (a^2 + b^2) /ab+1 < c < (a^2 + b^2) / ab. i dont think this is possible with c lying strictly between those two. i cant prove it tho. could somebody plz help?

Write it as $a^{2}-bc\cdot a+b^{2}-d=0$. We are to prove $d$ is a perfect square. Note that $0<d\leq c$.
If $b^{2}-d=0$ then we are done. Suppose $b^{2}-d\neq 0$ and $a\geq b$.

[b]Case 1.[/b] $b^{2}-d<0$. It follows that $a^{2}-bc\cdot a>0$, so $a>bc$, but then $a^{2}-bc\cdot a+b^{2}-d=a(a-bc)+b^{2}-d> bc+b^{2}-d>0$, since $c\geq d$. Contradiction.

[b]Case 2.[/b] $b^{2}-d>0$. It follows that equation $a^{2}-bc\cdot a+b^{2}-d=0$ wrt $a$ has another positive integer root $\frac{b^{2}-d}{a}<a$.

We get a descent algorithm, therefore, finally, we will come to $b^{2}-d=0$.

let $a^2+b^2-abc=t$ (1) with $0<t<c$. Now put $c$ fixed and pick up the pair $a$,$b$ with minimun sum satisfying (1). iff $a$, $b$ satisfy (1), then clearly, $ac-b$, $b$ give a good pair, and we can suppose $a>b$. Now we prove $0\leq ac-b<a$, unless $b=0$.  we have $a(c-1)<b$ is equivalent to $abc-ab-b^2<0$, wich is also equivalent to $a^2-t-ab<0$ or $a(a-b)<c$ then $(a-b)^2<c$. and $t=(a-b)^2-(c-2)ab<c-abc+2ab<0$ for $c>2$, and $a,b>0$. (the cases $c=1$, $c=2$, will be checked manually later). Now $0\leq ac-b$ is easy since $a>b$, So we find the minimal solution is always of the form $(a,0)$, and we find $t=a^2$ as we wanted.

for $c=2$ $t=(a-b)^2$. for $c=1$ we find out that $t=1$, a square!.

Fixed Myth! :P

Maybe you will fix your reply, Pascual? ;)

The inequality can be strengthened into the form: If a,b,c positive integers s.t. 0<a^2+b^2-abc<c+2 prove that middle expression is always a perfect sqaure.
This question is similar to IMO 1988 Q6. A harder question is to find which integers k such that if a^2+b^2-abc=c+k, then c+k is actually a perfect square. First few values of k=3,4,5,7,15,43,....

Bomb

[quote="Beat"]Myth, can you post your solution more clearly and with details, becouse i cannot understand some things. Please..[/quote]
It seems my solution is quite complete. I don't know what I can add to it.

to Beat: read, please!  :wink:
[quote="Myth"]
We get a descent algorithm, therefore, finally, we will come to $b^{2}-d=0$.[/quote]

What do you mean with a descent algorythim and how did you finally get to b^2-d=0

@above: Descent algorithm is Fermat's method of infinite descent

$a^2+b^2-abc=k\le c$
suppose there is set $S=\{(a,b)\in\mathbb{N}^2|a^2+b^2-abc=k\}$
$a^2-abc+b^2-k=0$
if $k=b^2$ then $a^2-abc+b^2$ is a perfect square so assume that $k\not = b^2$
there is $(A,B)\in S$ such that $A+B$ is the minimal among the other element and WLOG $A\ge B$
consider a quadratic $x^2-xBc+B^2-k=0$
note that $A$ is one of the root
and the other root $x_2=Bc-A=\frac{B^2-k}{A}$
$x_2=Bc-A$ give us information that $x_2\in \mathbb{Z}$
then if $x_2$ is negative
$0=x_2^2-x_2Bc+B^2-k\ge x_2^2+c+B^2-c>0$ a contradiction
so $(x_2,B)\in S$
because $A+B$ is the minimal we have $x_2\ge A$
$\frac{B^2-k}{A}\ge A\implies B^2-k>A^2\ge B^2$ contradiction so $k=b^2$ means that $a^2+b^2-abc$ is a perfect square


linear algebra
matrix
superior algebra
superior algebra solved

I am a bit weak with matrices, please help! :D
Let X,Y be in M_n(R) so that X^2Y^2=Y^2-X^2. Prove that: (X+I_n)Y^2X=X(X+I_n)Y^2.

:)

X^2Y^2-Y^2+X^2-In=-In => (X^2-In)(In+Y^2)=-In => (In-X^2)(In+Y^2) = In => (In-X)(In+X)(In+Y^2)=In. This should be enough if you also take into consideration that AB=In means BA=AB=In.


factorial
induction
inequalities proposed
inequalities

hello, proving by induction that
$ \frac {3}{1! \plus{} 2! \plus{} 3!} \plus{} \frac {4}{2! \plus{} 3! \plus{} 4!} \plus{} \cdots \plus{} \frac {n \plus{} 2}{n! \plus{} (n \plus{} 1)! \plus{} (n \plus{} 2)!} \equal{} \frac{1}{2}\left(1 \minus{} \frac {2}{(n \plus{} 2)!}\right)$, we get the desired result.
Sonnhard.


calculus
integration
function
algebra
polynomial
real analysis
real analysis unsolved

Let $f(x)$ be continuous function on $[0;1]$ such that $\int_{0}^{1}f(x)x^{n}dx=0$ for infinetly many positive numbers n. Prove that $f(1)=0$.

Any continuous function can be approximated uniformly on compact sets by polynomials. Fix $\epsilon >0$. Then there is a polynomial $P$ such that

$|f(x)-P(x)|<\epsilon$ for $x\in [0,1]$.

$|f(1)|=|f(1)-n\int_{0}^{1}x^{n}P(x)dx-n\int_{0}^{1}x^{n}\left(f(x)-P(x)\right)dx|$

$\leq |f(1)-n\int_{0}^{1}x^{n}P(x)dx|+\epsilon \frac{n}{n+1}$.

However it is not hard to see that

$n\int_{0}^{1}x^{n}P(x)dx\rightarrow P(1)$.

Letting $n$ converge to $\infty$ we get that

$|f(1)|\leq |f(1)-P(1)|+\epsilon\leq 2\epsilon$,
and since $\epsilon$ is arbitrary taken the result follows.

Thanks.
The original problem was for positive integers n (sorry for my typo :blush: )

if the original problem was for positive n, then its not only true that $f(1)=0$ but also true that $f(x)=0$ $\forall x\in [0,1]$, the idea is similar, uses approximation by polynomials.


[quote="bryantacademy"]If a cricket chirps 6 times in 8 seconds, how many times will it chirp in two minutes?[/quote]

[hide]$\frac{120}{x}=\frac{8}{6}$, $90 \text{ chirps}$[/hide]

[hide]In two minutes there are 120 seconds
120/8=15
15*6=90[/hide]

[quote="easyas3.14159..."][hide]Two minutes is 120 seconds and eight goes into it 14 times, so 14*6=84 chirps. Another way to solve is to set up a proportion (6/8=x/120).[/hide][/quote]
$\frac{120}{8}=15$ :P  :D

[hide="solution"]We want how many times it chirps in 40 seconds, because that is a multiple of 8 and a factor of 120. It chirps 6*5=30 times in 1/3 of 2 minutes. So the answer is 90[/hide]

[hide]
set up a proportion

120 (seconds in two minutes)/x=8/6

solve to get x=90[/hide]

[quote="daermon"][hide="answer"]90[/hide][/quote]

:spam:

Please, show your work instead of posting lonely results. Thanks.


algebra
polynomial
function
algebra solved

I've some problem.
P1: Prove that polynomial P(x,y) = 4 + x2y4 + x4y2 -3x2y2
 cannot be written as a sum of squares

P2: Let a,b,c be integer numbers, a is even and b is odd.Prove that for all nature numbers n>=1, always exits positive integer x satisfy:
 2n/(ax2 +bx+c)
P3:
Let P(x) be the real polynomial. Prove that If P(x)-P'(x)-P''(x)+ P'''(x)>=0 for all real number x then P(x)>=0 for all real numbers
Try to solv  :D

This one's for p3

Let's consider the following LEMMA:
If f is a polynomial and f(x)+f'(x)>=0 for all real x (or f(x)-f'(x)>=0) then f(x)>=0 for all real x:
f(x)+f'(x) has the same degree and the same leading coefficient as f, so f tends to infinity as x tends to infinity or to -infinity (because f+f' does the same). So f must have a minimum at x0. If f(x0)<0, because f'(x0)=0, we get f(x0)+f'(x0)<0, contradiction, so f(x0)>=0, but f(x0)is the minimum of f.

We put f=p-2*p'+p''. You can check that f+f'>=0 for any x, so f>=0. Put g=p-p', so g-g'>=0, therefore g>=0, so p>=0 for all real x.

Regarding p2

Let's assume c is odd(the other case is similar. If we look at the value of the function in 2 distinct odd values (the x we are looking for must be odd because c is odd) which are smaller than 2^n, x1 and x2, we will find that their residues modulo 2^n are different:

if 2^n/(x1-x2)*(a*(x1+x2)+b) then 2^n/x1-x2 because the other factor is odd, but x1-x2<2^n, so we got a contradiction. For an odd x, the residue of ax^2+bx+c modulo 2^n is even, and since all are different for different odd values of x smaller than 2^n, then at least one must be 0, so we found our x.

For problem 1

Just an idea (i don't know if it is work),

First: Does P(x,y) is constant sign ? (by some algebraic manipulations)

Second: If we suppose

P(x,y) = 4+x^2y^4 + x^4y^2 -3x^2y^2 is a sum of two squares

P(x,y) = q(x,y)^2 + r(x,y)^2

by consideration of partial degree on x or y can we get a contradiction ?

P1

Suppose P(x,y)= \sum f_j(x,y) 2, here f_j are polynoms with real coefficients.

\sum f_j(x,0) 2=P(x,0)=4  ==>  f_j(x,0)=c_j=const  ==>  f_j(x,y)=c_j+yg_j(x,y). By the same reasons f_j(x,y)=d_j+xk_j(x,y). It is cleary, c_j=d_j  ==> f_j(x,y)=c_j+xyh_j(x,y) and
deg h_j  \leq deg f_j - 2  \leq 1/2 deg P - 2=1.

So, x2y2(x2+y2-3)+4=x2y2\sum h_j 2 + 2xy \sum c_jh_j + \sum c_j 2  ==>
x2y2(x2+y2-3) - x2y2\sum h_j 2 = 2xy \sum c_jh_j + \sum c_j 2 - 4.

But deg <left side>  \geq 4 or deg <left side> =0 and deg <right side>  \leq 3  ==> <left side>=0, i.e. x2y2(x2+y2-3) - x2y2\sum h_j 2 = 0  ==>
x 2 +y 2 - 3 = \sum h_j 2 \geq 0 - contradiction since 0 2 + 0 2 - 3 <0

There is a similar problem to P1, posted by Alek
post subject: polynomial + squares
Prove that P(x,y)= x^2(x^2-y^2)(x^2-1) + y^2(y^2-x^2)(y^2-1) +
(1-x2^)(1-y^2) cannot be written as sum of squares.

Any solution ?


I see that in some Mathematics articles (like Forum Geom.), their images are very nice. And I hear that they use Geometer Sketchpad, so could someone help me to make a LaTex file with very nice pictures like that?

Pictures from Geometer's Sketchpad are not easy to include in LaTeX but you can do it as Valentin has posted at http://www.artofproblemsolving.com/Forum/viewtopic.php?p=339812#339812. You may find the rest of the thread useful as well.

I want to make an image like this one. Could you help me?

A very similar image should be fairly easy to create in asymptote if you first draw the outer circle and the two circles that are tangent to it, and then construct the triangle and its incircle from there.


limit
real analysis
real analysis unsolved

We have : $\lim x_n = 0$
  On the other hand , we have : $\dfrac{1}{x_{n+1}}-\dfrac{1}{x_n}=\dfrac{1}{1-x_n} \to 1$
  Applying  Cesaro's theorem , we have : $\lim nx_n =1$

For the second problem , I think the answer is $-1 \leq u_1 \leq 2$

What's Cesaro's theorem? Never heard of. Can u give a link?


group theory
abstract algebra
linear algebra
matrix
linear algebra unsolved

Prove that if $\textrm{rank} \ A = \textrm{rank} \ A^2 = t$ then $\textrm{rank} \ A^k = t$, for all $k \in \mathbb N^{\ast}$.

I have a college level solution to this problem, but I'm seeking a 11th grade level solution. (More specifically, I'm looking to avoid using linear transformations / rank-nullity theorem. It is very easy to do it with these tools.)

There is a nice characterization of such matrices: they are those matrices that belong to subgroups of $M_n(K)$. From here the conclusion follows easily.

Can you (or somebody else) explain what was meant above?

I really don't understand  :blush:

I don't know what harazi meant, but here's what you can do:

Look at the matrix $A$ as being the matrix of an endomorphism, also denoted by $A$, of $K^n$ ($K$ is the field we're working in, and $n\times n$ is the size of the matrix). The rank of an $n\times n$ matrix with entries in $K$ is just the dimension of the image of the corresponding endomorphism. In other words, $r(A)=r(A^2)$ simply says that $A(K^n)$ and $A^2(K^n)$ have the same dimension. Since this dimension is finite and $A(K^n)\supseteq A^2(K^n)$, this is equivalent to $A(K^n)=A^2(K^n)$, and it's obvious that this implies that $A^p(K^n)=A(K^n),\ \forall p$, i.e. exactly what we want.

I used the formula:  $\textrm{rank}(AB)+\textrm{dim}N(A) \cap R(B) = \textrm{rank}(B)$.

Letting $B=A^{k-1}$, we get that $\textrm{rank}(A^{k})+\textrm{dim}N(A) \cap R(A^{k-1}) = \textrm{rank}(A^{k-1})$.

The case $k=2$ tells us that $\textrm{dim}N(A) \cap R(A) = 0$.

But $R \left( A^{k}\right) \subseteq R \left( A^{k-1}\right)$ since $A^{k}x = T$ implies $A^{k-1}(Ax) = T$. Thus, $\textrm{dim}N(A) \cap R(A^{k-1}) = 0$.

The conclusion now becomes obvious. I guess this is the same as grobber's solution (the similarities are striking), but I don't know what some of the terms he used mean.

And of course, the discussion about what Harazi meant is still open.

What I was saying is that one can prove that ( at least for real matrices, but the proof can be adapted for any field) the condition $rank(A)=rank(A^2)$ is equivalent to the fact that $A$ belongs to a certain multiplicative subgroup of $M_n(K)$. But since all matrices of a multiplicative subgroup of $M_n(K)$ have same rank, the conclusion follows. I just wanted to present this nice caracterization, there was nothing subtle or impossible to find in my affirmation.

[quote="harazi"]I just wanted to present this nice caracterization, there was nothing subtle or impossible to find in my affirmation.[/quote]

For one that knows very little about groups, fields etc. it is indeed very subtle.


geometry
circumcircle
incenter
trigonometry
geometry unsolved

Hey, could you all help me in this  problem :oops: ?
[b]ABC is a triangle in which none of the angles is obtuse. The perpendicular AD from A to BC is produced to meet the circumcircle of the triangle at E. If, the incentre of the triangle is equidistant from A and E, prove that cosB+cosC = 1[/b]
Thxxxxxxx so much ^__^

[b]behdad.math.math[/b], don't spam. You have the right to remain silent.

$ H \in AD$ is orthocenter, $ R$ circumradius, $ r$ inradius. $ AD$ cuts $ BC$ at $ K.$
Given $ AH \plus{} HK \minus{} r \equal{} AK \minus{} r \equal{} KD \plus{} r \equal{} HK \plus{} r$ $ \Longrightarrow$ $ 2r \equal{} AH \equal{} 2R \cos A.$ Put it in $ R( \cos A \plus{}\cos B \plus{} \cos C) \equal{} R \plus{} r$ $ \Longrightarrow$ $ \cos B \plus{} \cos C \equal{} 1.$

[quote="dgreenb801"]How do you prove that $ \cos A \plus{} \cos B \plus{} \cos C \equal{} 1 \plus{} \frac {r}{R}$? I've seen Dr. Sonnhard Graubner post it many times.[/quote]
First introducing:
$ a \plus{} b \plus{} c \equal{} 2s$
$ ab \plus{} bc \plus{} ca \equal{} s^2 \plus{} r^2 \plus{} 4Rr$
$ abc \equal{} 4Rrs$
$ \cos A \equal{} \frac {b^2 \plus{} c^2 \minus{} a^2}{2bc} \equal{} \frac {b^2a \plus{} c^2a \minus{} a^3}{2abc}$ and so on.
So $ \cos A \plus{} \cos B \plus{} \cos C \equal{} \frac {\left ( \sum_{cyc} ab(a \plus{} b) \minus{} a^3 \right )}{2abc}$
$ a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca \equal{} (a \plus{} b \plus{} c)^2 \minus{} 3(ab \plus{} bc \plus{} ca) \equal{} s^2 \minus{} 3r^2 \minus{} 12 Rr$
$ a^3 \plus{} b^3 \plus{} c^3 \equal{} (a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca) \plus{} 3abc \equal{}$ $ 2s(s^2 \minus{} 3r^2 \minus{} 12Rr) \plus{} 12Rrs \equal{} 2s^3 \minus{} 6sr^2 \minus{} 12Rrs$
$ \sum_{cyc} ab(a \plus{} b) \equal{} (a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca) \minus{} 3abc \equal{}$ $ 2s(s^2 \plus{} r^2 \plus{} 4Rr) \minus{} 12Rrs \equal{} 2s^3 \plus{} 2sr^2 \minus{} 4Rrs$

Hence: $ \cos A \plus{} \cos B \plus{} \cos C \equal{} \frac {2s^3 \plus{} 2sr^2 \minus{} 4Rrs \minus{} 2s^3 \plus{} 6sr^2 \plus{} 12Rrs}{8Rrs}$ $ \equal{} \frac {8sr^2 \plus{} 8Rrs}{8Rrs} \equal{} 1 \plus{} \frac {r}{R}$

There might be a prettier way to do it, but this is very straight foward, and can be used to prove a lot of identities that look like that one.

$ \triangle ABC$ with circumcenter $ O$ and centroid $ G.$ $ A', B', C'$ are midpoints of $ BC, CA, AB.$  Let $ N$ be circumcenter of $ \triangle A'B'C'.$$ \triangle A'B'C' \sim \triangle ABC$ are similar with center $ G$ and coefficient $ \minus{} \frac {_1}{^2}$ $ \Longrightarrow$ $ \overline{GN} \equal{} \minus{} \frac {_1}{^2} \overline{GO}$ and $ \overline{NO} \equal{} \frac {_3}{^2} \overline{GO}$

Put equal masses at $ A, B, C.$ By Steiner theorem, moment of inertia $ \mathcal J_O \equal{} OA^2 \plus{} OB^2 \plus{} OC^2 \equal{} 3R^2$ of $ A, B, C$ WRT $ O$ is equal to moment of inertia $ \mathcal J_G \equal{} GA^2 \plus{} GB^2 \plus{} GC^2 \equal{} \frac {_1}{^3}(BC^2 \plus{} CA^2 \plus{} AB^2)$ of $ A, B, C$ WRT $ G$ plus moment of inertia $ 3GO^2$ of the total mass at $ G$ WRT $ O.$ $ \mathcal J_O \equal{} \mathcal J_G \plus{} 3 GO^2$ $ \Longrightarrow$ $ 4NO^2 \equal{} 9GO^2 \equal{} 9R^2 \minus{} (BC^2 \plus{} CA^2 \plus{} AB^2).$

Incircle $ (I, r)$ of $ \triangle ABC$ touches $ BC, CA, AB$ at $ D, E, F.$ Inversion in $ (I)$ takes $ A, B, C$ into the midpoints $ D', E', F'$ of $ EF, FD, DE$ and circumcircle $ (O, R)$ of $ \triangle ABC$ into circumcircle $ (O', R')$ of  $ \triangle D'E'F',$ where $ R' \equal{} \frac {_r}{^2}.$ Circles $ (O') \sim (O)$ are similar with center $ I$ and coefficient

$ \frac {r}{2R} \equal{} \frac {R'}{R} \equal{} \frac {R'^2 \minus{} O'I^2}{r^2} \equal{} \frac {r^2 \minus{} 4O'I^2}{4r^2} \equal{} \frac {EF^2 \plus{} FD^2 \plus{} DE^2 \minus{} 8r^2}{4r^2} \equal{}$

$ \equal{} \sin^2 D \plus{} \sin^2 E \plus{} \sin^2 F \minus{} 2 \equal{} \minus{} \frac {_1}{^2}(\cos 2D \plus{} \cos 2E \plus{} \cos 2F \plus{} 1)$

Using $ 2\angle D \equal{} 180^\circ \minus{} \angle A \Longrightarrow \cos 2D \equal{} \minus{} \cos A$ and cyclic exchange,

$ \frac {r}{R} \equal{} \cos A \plus{} \cos B \plus{} \cos C \minus{} 1$

[quote=yetti][b]behdad.math.math[/b]
$ H \in AD$ is orthocenter, $ R$ circumradius, $ r$ inradius. $ AD$ cuts $ BC$ at $ K.$
Given $ AH \plus{} HK \minus{} r \equal{} AK \minus{} r \equal{} KD \plus{} r \equal{} HK \plus{} r$ $ \Longrightarrow$ $ 2r \equal{} AH \equal{} 2R \cos A.$ Put it in $ R( \cos A \plus{}\cos B \plus{} \cos C) \equal{} R \plus{} r$ $ \Longrightarrow$ $ \cos B \plus{} \cos C \equal{} 1.$[/quote]

Apologies for bumping this thread, but I need some help in understanding [b]yetti's[/b] solution. I'm assuming first of all that he relabelled $E$ in the diagram as $K$. After this he does some calculations, but I can't understand any of it. However, in my attempt of the problem, I did note that it suffices to prove $\cos A = \frac{r}{R}$, but I would appreciate it if someone could help me understand yetti's work or contribute further.

Anyone please? The specific parst that I don't understand is how $AK -r = KD+r =HK+r$ and then when it says put in $R( \cos A + \cos B + \cos C)$. Please someone help me.


Fie $ABCD$, $A'B'C'D'$ doua paralelograme in spatiu. Sa se arate ca
   mijloacele segmentelor $[AA'], [BB'], [CC'], [DD']$ sunt mijloacele tot ale unui paralelogram.

Draguta, desi cam usoara. Dar la clasa a VII-a nu prea poti sa aduci vorba de "paralelograme in spatiu". Cred ca ar merge mai bine la clasa a VIII-a.