This must have come up at some point or other to most of you.

The soldiers of two or more nations shoot each other down, blow each other up, and somehow, this solves the world's problems? Suppose theoretically that instead of fighting real wars, the generals and national leaders use a computer program to fight virtual wars. Each country would be alloted the resources they possess in reality, and one side would reach a virtual victory, with the implication that that would be the outcome if the war were fought in real life. Clearly, this plan still has a lot of holes. For example, computer hacking, inaccurate simulations, complicated simulations, different battle strategies (harmless virtual nuclear weapons and disposable virtual soldiers), and most serious of all, reluctance to accept the outcomes. But I'm hoping that, since AoPS is filled with problem solvers, maybe the system could be implemented someday so that people would suffer a little less. Maybe? Any suggestions?

No. Let's say someone wins and the opponent says he cheated. He sends out a militia, and attacks the winner.

See? It really doesn't work.

Modern wars fight by and for economy. It seems bloodless, but kills much more efficiently than traditional wars, the victims don't even realize the fact that they are killed.

this makes no sense
war doesnt solve anything
it just sends a message: i can skool you
war doesnt solve anything
all it does is waste money, destroy reputations, and blow other ppl into smithereens

War isn't something you just win or lose.  Even if you win, you lose money and maybe even people.
So there's something you can't predict : if I( country A) attack country B, will country C be indifferent about it or not?(knowing that I'm doomed if country C gets involved).

The events of the war itself (the movement of troops, the blockades of ports, the razing of cities &c.) means little; the reconstruction effort is the War Mongers' true goal. The subjugated nation is humiliated and the entity is dissolved. A puppet government is created to serve the economic interests of the occupier, and elements which challenge the victor's authority or the consistency of its ideology are silenced. It all comes down to money and influence. The needless waste of human life is not real to those who perpetrate war. An example I find very amusing is my roommate, we both play lots of video games. A favorite genre are role-playing games, you play a character and you roam about being good or evil as you choose. He told me he never had the heart to play an evil character and slay the innocent and whatnot. But when we'd play strategy games, (you assume the role of nations at war) we'd have no problem spreading the Plague in each other's capital cities. Sure it's just a simulation, but inflicting misery is easier to swallow when you're removed from it.


We can use difference of two perfect squares:
Lets say $ 29$ is side $ a$

$ c^2 \minus{} b^2 \equal{} 29^2$

$ (c \plus{} b)(c \minus{} b) \equal{} 841$

Since we know they are consecutive numbers

$ (c \plus{} b)(1) \equal{} 841$

$ c \plus{} b \equal{} 841$

So the answer is $ \boxed{841}$

Its also interesting to note that:

$ c^2 \equal{} a^2 \plus{} b^2$
$ c^2 \minus{} b^2 \equal{} a^2$
$ (c \plus{} b)(c \minus{} b) \equal{} a^2$
now since $ c$ and $ b$ are consecutive:
$ (c \plus{} c \minus{} 1)(c \minus{} (c \minus{} 1) \equal{} a^2$
$ 2c \minus{} 1 \equal{} a^2$

so whenever $ a$ is odd, a right triangle can be formed, namely $ a, \frac {a^2 \plus{} 1}{2} \minus{} 1, \frac {a^2 \plus{} 1}{2}$
In this case, $ 29^2 \equal{} 2c \minus{} 1$
$ c \equal{} 421$
$ b \equal{} 420$
$ c\plus{}b\equal{}841$
EDIT: Supernerd, from your second step onwards you mixed up $ a$ and $ b$.
it is $ (c \plus{} b)(c \minus{} b) \equal{} 841$
$ (c \plus{} b)(1) \equal{} 841$
$ (c \plus{} b) \equal{} 841$
The answer is still right though

[quote="tennis123"]Is there a way to do[b] [i]Twenty-nine is the shortest leg of a right triangle whose other leg and hypotenuse are consecutive whole numbers. What is the sum of the lengths of the other two sides? [/i] [/b]other than guess and check?[/quote]

Would this be guess and check? If you know your Pythagorean Triples (generally useful in MATHCOUNTS) you might immediately get (20,21,29) so the sum is 41.

Of course, there is an algebraic solution. If you let $ a$ be the shorter leg, the other leg is $ a\plus{}1$ and we need $ a\plus{}(a\plus{}1)\equal{}2a\plus{}1:$

$ a^2\plus{}(a\plus{}1)^2\equal{}29^2 \Rightarrow a^2\plus{}(a^2\plus{}2a\plus{}1)\equal{}841 \Rightarrow 2a^2\plus{}2a\equal{}840 \Rightarrow a^2\plus{}a\equal{}420 \Rightarrow a^2\plus{}a\minus{}420\equal{}0 \Rightarrow (a\plus{}21)(a\minus{}20)\equal{}0$ Of course, $ a>0,$ so $ a\equal{}20$ and $ 2a\plus{}1\equal{}\boxed{41}$

[quote="tennis123"]Is there a way to do: Twenty-nine is the shortest leg of a right triangle [b]whose other leg and hypotenuse are consecutive whole numbers[/b]. What is the sum of the lengths of the other two sides? other than guess and check?[/quote]

Toad, you made $ a$ and $ b$ consecutive, but instead $ b$ and $ c$ are consecutive. The solution for that, as me and supernerd have shown, is a=29, b=420, c=421

EDIT: bleh beaten to it....


What schools typically have lots of people that pass amc 10?

troy, michigan, i have a friend that goes there.. but like my school is one of the biggest high schools in the country (over 4200 kids), but no one has ever passed aime

Just a question:
What score is considered passing?

[quote="randomdragoon"]How many kids passed the AMC at your school?[/quote]
33 last year, I'm from same school as Jason. And we do have 2 USAMO qualifiers (stupid freshman, they don't know what they're talking about  :P , and we only have 3999 [yes exactly 3999] students not 4200)

what school are you from jason and stupidkid?

[quote="jasonwingofcarmelin"]ok im not from the same school as you are probably (carmel high, indiana), also someone teold me no ones ever passed, and we are the biggest in indiana.[/quote]

Mind reading your PMs?


But, these circles form planar graph so that it can be properly colored with $4$ colours.
In another hand, choosing four circles pairwise tangents (Soddy's kiss) we see that $4$ colours are necessar.

Pierre.

sorry pbornsztein, Maybe I missed some words
"Find out the smallest number k such that for every system of finite pairwise disjoint [b]equal radius [/b] circles (tangent is ok) on the plane , one can color each circle one of k color such that no two tangent are the same color."


A six digit perfect square has the same non zero digit in the hundreds, tens, and units places. What is the tens digit?

Thanks

[hide]

I think that the answer is zero though I may be wrong.

The square root of 000000 is zero.

[/hide]

Nope, I thought it was  zero also but it is supposed to be a non Zero digit in the hundreds, tens, and units places

I give up. I can't find it... Are you sure it is a rational number?

It is a rational number but....i think it is either a 4 or 3 or 5... I don't know which one it is though

Well, it can't be $2$, $3$, $7$, or $8$, since perfect squares never end in those numbers. $317^2$ is the smallest six-digit perfect square, and $999^2$ is the largest, so you have quite a big range to work with, but you can simplify it some. I've done one like this with cubes before; for example, $192^3=7077888$, but I'm not sure of the answer for the square. You can do some simple testing to eliminate certain possibilities. For example, the units digit of $n$ (where $n^2$ is the six-digit perfect square) isn't $1$ because the tens digit of $n^2$ can only be an even number, which you can find by just testing the first values and seeing that it repeats. For $2$, however, the tens digit is also even, so you have cases where the tens digit is equal to the units digit. I'm not sure if that'll be your answer, but try finding cases like that.

[hide]The possible units digits of a perfect square are 0, 1, 4, 5, 6, 9.  0 won't work because the problem says it isn't 0.  It can't be 5 because the only way to create a perfect square and end in 5 is to square a number ending in 5.  This will never end in 55.  1 won't work because $111=3\cdot37$, so squares can't end in it.  Just try the others and when you use 4, it works.  $462^2=213444$ works  The 10's digit is $\boxed{4}$.  I need help on the reason it can't be 1, 6, and 9.[/hide]

[quote="b-flat"]I need help on the reason it can't be 1, 6, and 9.[/quote]

You already showed why it can't be $1$.

[hide="As for the others..."]The square of any number ending in a $6$ must have a units digit of $6$ but an odd tens digit, so that's obviously impossible. Similarly, for $9$ we have a units digit of $1$, $3$, $7$, $9$, and an even tens digit. Therefore, the only possibility that you're left with is $2$. Since the tens digit repeats every $50$ (i.e., $312^2$, $362^2$, $412^2$, etc.), you just need to start with the first possibility over $317^2$ and test everything. :P[/hide]

Wait is this a target or sprint type of problem can u use the calculator?

A target I assume. You're not supposed to square tons of three digit numbers really quick :what?:

[quote="indianacutiepink"]If the 6 digit number has to end with the SAME 3 numbers, then the 3 numbers must be a perfect square. For example, 111,222,333,etc. should all be perfect squares in order for the whole six digit # to be a perfect square. I took the square root of all of these and none of them were a perfect square. What I think, is that this is impossible![/quote]

well, b-flat found a solution  ;)

[quote="indianacutiepink"]If the 6 digit number has to end with the SAME 3 numbers, then the 3 numbers must be a perfect square. For example, 111,222,333,etc. should all be perfect squares in order for the whole six digit # to be a perfect square. I took the square root of all of these and none of them were a perfect square. What I think, is that this is impossible![/quote]

No, it's possible, as both b-flat and I found the solution (although I didn't post it, since I'm not a middle schooler). The last three digits don't have to be a perfect square for it to work.

biff: Yes, it probably was a target. However, you don't have to square "tons" of three-digit numbers really quick. You have an average of three minutes per problem, and if the other problem was easy, you could have as much as five minutes on this problem. Then, if you make the deductions that I did followed by those that b-flat did, then the third number you test will be the answer. ;)


A license plate in a certain state consists of 4 digits, not necessarily distinct, and 2 letters, also not necessarily distinct.  These six characters may appear in any order, except that the two letters must appear next to each other.  How many distinct license plates are possible?

$ \textbf{(A) } 10^4\cdot 26^2 \qquad \textbf{(B) } 10^3\cdot 26^3 \qquad \textbf{(C) } 5\cdot 10^4\cdot 26^2 \qquad \textbf{(D) } 10^2\cdot 26^4\\
\textbf{(E) } 5\cdot 10^3\cdot 26^3$

[hide="Solution"]Well, individually there are $10^{4}$ possible digit combinations and $26^{2}$ possible letter combinations.  You could have them arranged as @@####, #@@###, ##@@##, ###@@#, and ####@@.  So you have $5\cdot 10^{4}\cdot 26^{2}\implies\boxed{C}$.[/hide]

[hide=Solution]
Each digit has $10$ options, and each letter has $26$. So far, we have $10^4 \cdot 26^2$. Next, to reorder, note that the digits can be in any order, but think of the letters as a "block". Thus, we have $5$ "characters" to rearrange, giving us $5 \cdot 10^4 \cdot 26^2$ or $\boxed{\textbf{(C)}}$.
[/hide]

[quote name="programmeruser" url="/community/p25663310"]
Were old AMCs really this easy? This seems too simple for a P18
[/quote]

Yes, they were so easy that an AMC 12 before 2007 is not too far from the difficulty of a modern AMC 10

[quote=programmeruser]Were old AMCs really this easy? This seems too simple for a P18[/quote]

rbh I was thinking the same thing...
[s]y am I learning these complicated things if this is a question on the amc 10 again?[/s]


quadratics
polynomial
algebra
system of equations
number theory proposed
number theory

Solve the system of equations over $ \mathbb{Z}$

\[ \begin {cases}ac\minus{}bd\equal{}1 \\ad\plus{}bc\minus{}bd\equal{}0 \end {cases}\]

The two equations imply $ (a-b)(c-d)=ac.\qquad(*)$

[b]Lemma:[/b] If $ a,b,c,d\in\mathbb Z$ and $ ab=cd$, then $ \exists w,x,y,z\in\mathbb Z$ such that $ (z,x)=1$ and $ (a,b,c,d)=(wx,yz,wz,xy)$.

[b]Proof:[/b] Let $ (a,c)=w,a=wx,c=wz$. Then $ (z,x)=1$ and $ xb=zd$. Thus $ z\mid b$ and $ x\mid d$. Let $ b=yz$. Then $ d=yx$ and the conclusion follows. $ \blacksquare$

Hence by the lemma, from $ (*)$ we conclude that
\begin{align*} a-b&=wx\\
c-d&=yz\\
a&=wz\\
c&=xy\end{align*}
where $ w,x,y,z$ are integers such that $ (z,x)=1$. Hencey $ (a,b,c,d)=(wz.w(z-x),xy,y(x-z))$. But $ ac-bd=1$ implies $ wxyz+wy(z-x)^2=1$. This implies $ wy=\pm 1$ and $ z^2-zx+x^2=\pm 1$. Considering the last equation as a quadratic in $ z$ we find that the discriminant equals $ x^2-4(x^2\mp 1)=\pm 4-3x^2$, which is a perfect square iff $ x\in\{0,\pm 1\}$. Then $ (x,z)\in\{(0,\pm 1),(\pm 1,0),(\pm 1,\pm 1)\}$. These imply $ (w,y)\in\{(\pm 1,\pm 1)\}$. Considering all possible combinations we conclude that $ (a,b,c,d)\in\{(\pm 1,\pm 1,0,\mp 1)\}$ etc.

[b]HINT:[/b]
[b]1).[/b] $ ac \minus{} bd \equal{} 1\Rightarrow (a;b) \equal{} (a;d) \equal{} (b;c) \equal{} (c;d) \equal{} 1.$
[b]2).[/b] $ ad \plus{} bc \minus{} bd \equal{} 0\Rightarrow ad \equal{} b(d \minus{} c)\Rightarrow b|ad$ or $ (a;b) \equal{} 1\Rightarrow b|d.$   (1)
[b]3).[/b] $ ad \plus{} bc \minus{} bd \equal{} 0\Rightarrow bc \equal{} d(b \minus{} a)\Rightarrow d|bc$ or $ (c;d) \equal{} 1\Rightarrow d|b.$   (2)
[b]4).[/b] (1) and (2)$ \Rightarrow d \equal{} \pm{b}.$
[b]5).[/b] $ d \equal{} \pm{b}$ and $ ad \plus{} bc \minus{} bd \equal{} 0\Rightarrow ab \plus{} bc\mp{b^2} \equal{} 0\Rightarrow b.(a \plus{} c\mp{b}) \equal{} 0\Rightarrow (b \equal{} 0) \lor (a \plus{} c \equal{} \pm{b});\dots$

$ (a \plus{} b \omega)(c \plus{} d \omega) \equal{} 1 \implies (a^2 \minus{} ab \plus{} b^2)(c^2 \minus{} cd \plus{} d^2) \equal{} 1$.

t0rajir0u... How...? Divine inspiration?

Was this your first approach?

To be fair, there's a related problem that is based off of multiplication in the Gaussian integers and [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=143603]I've used multiplication in the Eisenstein integers before[/url].


Well, you could brute force expand the right.

OR

Notice on the LHS that $\sin a\cos a=\frac 12\sin 2a$ and $\sin b\cos b=\frac 12\sin 2b.$  So then the LHS is $\frac 12\left(\sin 2a+\sin 2b\right).$  It's just the sum to product formula from there.

Just expand the right hand side with the addition formula.

What's the "sum to product" formula?

Sum to product formula for sines:

$\sin(a+b)+\sin(a-b) = 2\sin a \cos b$

Thus,

$\sin (a+b) \cos (a-b) = \frac1{2}(\sin (2a) + \sin (2b)) = \sin a \cos a + \sin b \cos b$


What is the sum of all of the multiples of 7 between 100 and 200?

hello, we must calculate $ \sum_{i\equal{}0}^{28}7i$ this sum is equals $ 2842$.
Sonnhard.

Notice that the smallest multiple of 7 greater than 100 is $ 15 \times 7\equal{} 105$. The largest one less than 200 is 196. That means we're looking for the sum 105 + 112 + 119 + ... + 189 + 196. Now, notice that if we pair up the terms, we'll get the same sum from each pair. Since there are 14 pairs in all, the total sum is $ (105 \plus{} 196) \times 14 \equal{} 4214$.

hello, i don't think so we must sum up from $ i\equal{}15$ until $ i\equal{}28$ so we get
$ \sum_{i\equal{}15}^{28}7i\equal{}2107$.
Sonnhard.

[quote="isabella2296"]Notice that the smallest multiple of 7 greater than 100 is $ 15 \times 7 \equal{} 105$. The largest one less than 200 is 196. That means we're looking for the sum 105 + 112 + 119 + ... + 189 + 196. Now, notice that if we pair up the terms, we'll get the same sum from each pair. Since there are 14 pairs in all, the total sum is $ (105 \plus{} 196) \times 14 \equal{} 4214$.[/quote]

there's only 7 pairs, resulting in the answer of 2107


LaTeX
email
algorithm
combinatorics unsolved
combinatorics

Hmm I solved the other 3 in an hour, and couldn't solve this problem in 2.5 hours! Am I missing something? Is there a clean way of solving this?

Let $A_i$ be sets for $1 \leq i \leq n$:
[list]1. $\forall i, |A_i| = 3$
2. $\forall i, A_i \subset \{1,2,...,36\}$
3. $\forall i, j, A_i  \cap A_j \neq \emptyset$
4. $\forall i, j, A_i \neq A_j$
5. $ \cap_{\forall i} A_i =\emptyset $[/list]
[list] a) Show that $n \leq 100$.
b) For $n = 100$, find the number of such constructions. [/list]

PS: Edited a lot of stuff, typo notwithstanding

[quote="kueh"]Hmm I solved the other 3 in an hour, and couldn't solve this problem in 2.5 hours! Am I missing something? Is there a clean way of solving this?

Let $A_i$ be sets for $1 \leq i \leq n$ such that for all $i, j$:
[list]1. $A_i \in {1, 2, 3, ..., 36}$
2. $A_i  \cap A_j \neq$ empty
3. $ \cup_{all i} A_i =$ empty[/list]
[list] a) Show that $n \leq 100$.
b) For $n = 100$, find the number of such constructions. [/list]

PS What is empty/null set in latex?[/quote]

there is need for clarification:
do you mean that $A_i \subset \{1,2,...,36\} \forall i?$
what is 3) actually?(what you have written means that all the sets are empty)

and $\emptyset$ is \emptyset in LaTeX.

[quote="seshadri"]there is need for clarification:
do you mean that $A_i \subset \{1,2,...,36\} \forall i?$
what is 3) actually?(what you have written means that all the sets are empty)

and $\emptyset$ is \emptyset in LaTeX.[/quote]

All's been done in the question, thanks.

maybe the construction will make the problem clear:

let $a,b,c$ be distinct elements of $\{1,2,..,36\}$.the family of sets $\mathcal{A}=\cup\{a,b,*\}\cup\{b,c,*\}\cup\{a,c,*\}$ has 34+33+33=100 elements and satisfies the conditions.these are the maximal families.

another construction is $\mathcal{A}=\cup\{a,b,*\}\cup\{a,c,*\}\cup\{a,d,*\}\cup\{b,c,d\}$. Apparantly, these are the only constructions, but I don't know how to prove it. One of my ideas is that if I can prove in the four most abundant elements, every set has 2 in it, I will be done. But I can only prove every set has 1 in the four most abundant elements.

could you please fix the latex ? thanks!

[quote="kueh"]Hmm I solved the other 3 in an hour, and couldn't solve this problem in 2.5 hours! Am I missing something? Is there a clean way of solving this?[/quote]

I looked at the questions - I needed 15 minutes for the first three questions [although question 2 is known to me] and I am still struggling with question 4. Is it really that hard? I can't really imagine, I have the feeling that there is a simple way to do it but that I am just overlooking it!

It looks likes this question is for "sorting out" the really brilliant people.

Did anyone in the UK IMO squad actually solve the question?

Apparantly Martin (S'03 B'04) had a long wordy solution for (a), but he didn't get (b) right.

[quote="Arne"]But that is just weird because once you have proved a then it is just basic counting... no?[/quote]

How? If the solution is not revealing in any way...

3-subsets of $\{1, 2, ... , 36\}$ are now called points. The plane $(a)$ is the set of all points which contain $a$. The line $(a,b)$ is the set of all points which contain both $a$ and $b$. Two points which are not coplanar are called dislocate. The problem now becomes
"A set $K$ contains $n$ points which are pairwise coplanar but not all coplanar. Show that  $n\leq100$."
Let $(x,y,z) \in K$. Clearly all points in $K$ lie in $(x)$, $(y)$ or $(z)$.

Case (I) All points of $K$ lie on $(x, y)$, $(y, z)$ or $(x, z)$.
Each line contains 34 points. But $(x, y, z)$ is counted three times. Here $n\leq34*3-2=100$.

Case (II) Some point of $K$ does not lie on $(x, y)$, $(y, z)$ or $(x, z)$.
Let this point wlog be $(x, a, b)$. Now all the other points of $K$ must lie in $(y,a)$, $(y,b)$, $(z,a)$, $(z,b)$ or $(x)$.

Subcase (II) (a). Suppose we have $p \in K, \in (y,a)$, not in $(x)$, $(y,b)$ or $(z,a)$. Then $p=(y,a,c)$. All the remaining points of $K$ must lie in $(y,a)$, $(y,b)$, $(z,a)$, $(x,y)$, $(x,a)$ or $(x,c)$. Here in order to have a large value of n we must have $\geq 2$ points in at least 3 of these lines, which must then be pairwise coplanar. The cases that follow regard the selection of these lines. The four choices are $(y,a)$, $(x,y)$, $(x,a)$ (Here $(x,y,a)$ gets counted three times so we have $n\leq100$ as above), $(y,a)$, $(y,b)$ and $(y,x)$ (in which case $(y,a,b)$, $(y,a,x)$ and $(y,b,x)$ each get counted twice, but we also have $(a,b,x)$, so $n\leq100$), similar cases with $(x,y)$, $(x,a)$ and $(x,c)$ and also $(a,y)$, $(a,x)$ and $(a,z)$

Subcase (II) (b). Suppose we have $p = (y,a,b) \in K$. Very similar for the above case except for a few extra lines involving b.

Subcase (II) (c). Suppose we have $p = (y,z,a) \in K$. All the remaining points lie in $(y,a)$, $(y,b)$, $(z,a)$, $(z,b)$, $(x,y)$, $(x,a)$ or $(x,z)$. Here we can also have lines with z, but it would be a waste of my time (and an insult to your intelligence) to go into detail. Maybe I'll come back and edit it in when I've got more time.

Otherwise, all the points lie in $(x)$.
Simple eh? I told you I could do it.  :P

ohohoh! the growth idea strikes back with a vengeance! i knew the growth idea was effective, but i didn't think of it in this way :lol: -hats off- That's sweet. I was stuck with an extremal idea.

We already have $(x, y, z)$ and $(x, a, b)$.
A nicer way to do the ending is to note the following (as I said in my email):
$(y, a)$ opposes $(x, c)$
$(x, a)$ opposes $(y, b)$
$(x, y)$ opposes $(z, a)$

Now. If we choose both growths in 1 group, we can only end up with 2, 1 of each. So we can only choose 1 growth from each group. Now to show that $n \leq 100$, we note that each growth comes up with 2 extra overlaps, and 32 distinct except when it does not overlap with sets already chosen. Now these are: $(y, a)$, $(y, b)$, $(z, a)$ and $(x, c)$. But using the same argument:

$(z,a)$ opposes $(y, b)$, $(x, c)$
$(x,c)$ opposes $(y,b)$, $(y,a)$, $(z, a)$

So we cannot choose 3 non-opposing growths. So it must be 100.

Now this shows that we must also choose $(x, y)$ or $(x, a)$. Due to lack of time, I cannot finish it now, but someone else should be able to finish it from here.

Okay, I will do this clearer now: We conjecture that the solutions fall in two categories: (A) - $\mathcal{A}=\cup\{a,b,*\}\cup\{b,c,*\}\cup\{a,c,*\}$, (B) - $\mathcal{A}=\cup\{a,b,*\}\cup\{a,c,*\}\cup\{a,d,*\}\cup\{b,c,d\}$. Using the same start as Saul's proof, we assume there is $\{1, 2, 3\}, \{1, 4, 5\}$ (otherwise it is just A)

Now, we have a set which does not have $1$. WLOG, let it not have $5$ either.

Thus, every other set must fall into one of the following categories: $\{1, 2, x\}, \{1, 3, x\}, \{1, 4, x\}, \{3, 5, x\}, \{2, 5, x\}, \{2, 4, x\}, \{3, 4, x\}$.

Lemma: We must have at least $3$ categories with more than $2$ sets in them. Assume not. Then maximum, there are only $34 + 34 + 5 + 2 \leq 100$. So obviously, we must have 3 categories.

Now,
$\{1, 2, x\}$ opposes $\{3, 4, x\}$
$\{1, 3, x\}$ opposes $\{2, 4, x\}$
$\{1, 4, x\}$ opposes $\{2, 5, x\}, \{3, 5, x\}$

Note that if we take two opposites, we can only take 2 sets from each group, ie, 1 each. Then this would not be the categories.

Case 1: We choose $\{2, 5, x\}$. ($\{3, 5, x\}$ is a similar case.)
Then we note that $\{2, 5, x\}$ opposes $\{1, 3, x\}$, $\{1, 4, x\}$, $\{3, 4, x\}$.
We also note that $\{3, 5, x\}$ opposes $\{1, 2, x\}$, $\{1, 4, x\}$, $\{2, 4, x\}$.

Thus, we cannot choose both $\{2, 5, x\}$ and $\{3, 5, x\}$ because everything else opposes one of them.
Thus the three must be $\{2, 5, x\}$, $\{1, 2, x\}$ and $\{2, 4, x\}$ which corresponds to B.

Case 2: We neither choose $\{2, 5, x\}$ or $\{3, 5, x\}$. Thus, we need to choose $\{1, 4, x\}$. There are four ways to choose a pair from the rest, but these fall into categories A/B.

A - $\{1,2,x\}$, $\{2,4,x\}$ or $\{1,3,x\}$, $\{3,4,x\}$
B - $\{1,2,x\}$, $\{1,3,x\}$ or $\{2,4,x\}$, $\{3,4,x\}$

Thus it remains to count the number of ways for A and B.

A - ${36}\choose{3}$ (choose a,b,c which are equivalent, which defines it)
B - $33{{36}\choose{3}}$ (choose b,c,d which are equivalent, then a. this defines it)

i only managed to do part a... by breaking down into a few cases.  horrendous and wordy... part b looked daunting.

apart from qn 4 i dont think the qns posed much of a problem

In my opinion this problem is more familiar with algorithms ( algorithmic solution! )

[quote=Pascual2005]for each $i$ in $(1,2,3...,36)$, let $d_i$ be the number  of sets $A_j$ such that $i$ is in $A_j$. we have:

$-1<d_i<n$ and $\sum_{i=1}^{36}d_i=3n$. also, we have $\sum_{i=1}^{36}\binom{d_i}{2}>=\binom{n}{2}$

Now, $\sum_{i=1}^{36}\binom{d_i}{2}=1/2\sum_{i=1}^{36}d_i^2  - 1/2\sum_{i=1}^{36}d_i=1/2\sum_{i=1}^{36}d_i^2  -3n/2>=\binom{n}{2}$.

So we get: $\sum_{i=1}^{36}d_i^2>=n^2+2n$. Now we have to find a maximun for the l.h.s taking into acount that there are no equal $A_i$'s.

the first think to notice is that iff $d_i+d_j=n+k$ then there are at least $k$ sets having them. so for the other $d_m$, we have $d_m<n-k+2$. this will help with our bounds.

Another thing i had forgot. we have, that no 3 sets are equal so we have $n<\binom{36}{3}$ wich doesnt helps too much.

Now fix n, since x^2 is convex, we would like to maximise a sum of squares, and we doit making some bigger and some smaller, because their sum is fixed. I dont get it clear from here, so i will try to continue later.[/quote]

olsun be abi


Prove that every convex set in an euclidean space $ X$ having $ 0$ as boundary point, has a tangent $ n\minus{}1$ dimensional hyperplane passing through $ 0$ that is also a $ n\minus{}1$ dimensional subspace of $ X$.

In the usual understanding of "tangent", this is false. Every hyperplane through the origin is a supporting hyperplane of $ \{0\}$, but none is tangent.

So... what are your definitions, exactly?


number theory
greatest common divisor
least common multiple

For this kind of problems, multiply GCD of the denominators on both sides.
So for #1, u would multiply $12x$ on both sides.
For #2, multiply both sides by $a^2-7a$.

I'll do question $1$ out:

$\frac{3x-3}{4x}=\frac{6x-9}{6x}+\frac{1}{3x}$

Multiply by $12x$ as this is the lowest common multiple of each of the denominators.  So:

$3(3x-9)=2(6x-9)+4$

Multiplying out and rearranging gives:

$3x=5$

[b]$x=\frac{5}{3}$[/b]

[quote="cooly291"]I need help, I can do most problems like this, but not these two:

[img]http://img.photobucket.com/albums/v486/cooly291/math-1.jpg[/img][/quote]

[hide]Multiply two out:
$6a=(a-49)+(a-7)$
$6a=2a-56$
$4a=-56$
$a=\boxed{-14}$[/hide]

For these kinds of problems, you just multiply both sides by the LCM of all the denominators.

[hide="#1"]The denominators are $4x$, $6x$, and $3x$, and their LCM is $12x$, so we multiply by that to get
$3(3x-3)=2(6x-9)+4$.
Simplifying, we get
$3x=5$,
so $x=\boxed{\frac{5}{3}}$.[/hide]

[hide="#2"]The denominators are $a-7$, $a^2-7a=a(a-7)$, and $a$, and their LCM is $a(a-7)$, so we multiply by that to get
$6a=a-49+(a-7)$.
Simplifying, we get
$4a=-56$,
so $a=\boxed{-14}$.[/hide]

[quote]$6a=2a-56$
$4a=56$[/quote]
You messed up between these two steps.  ;)

[quote="mathgeniuse^ln(x)"]And also, if you are not a 100% sure that your answer is correct, you can always check your answer by plugging the solution into the original equation, right?[/quote]

Yes, and often this is necessary because sometimes extraneous solutions can appear in your solutions, such as in
$\frac{x+2}{x-3}=\frac{5}{x-3}$,
where multiplying it out and solving it gives you $x=3$ as the solution, but this is extraneous because plugging it back into the equation, you find that division by $0$ occurs.

for this kind of problems multiply all variables by the LCD and that rationalisez the denominator, meaning it gives you somthing you can work with. In this case multiplyh by 12x, use fundamental distributive property and algebra and you get for number 1 -3/5.


geometry
geometric transformation
rotation
linear algebra
linear algebra unsolved

prove that if AB=BA then
A and B have at least one common eigenvector.

Let $ A$ and $ B$ be rotations with $ \frac{\pi}{100}$ and $ \frac{\pi}{200}$ respectively. Then $ AB\equal{}BA$ however they have no eigenvector at all.

The result is valid over an algebraically closed field. In particular the rotations of vesszabo have exactly the same eigenvectors over the complex numbers.


Neither do I...

Anyway, Newton Sums let you figure out properties of numbers given the equation they fit. I will use a cubic as an example. Given $x^3-ax^2+bx-c=0$ and roots $r_1$, $r_2$, and $r_3$, we know that $r_1+r_2+r_3=a$, $r_1r_2+r_2r_3+r_1r_3=b$, and $r_1r_2r_3=c$. Therefore, we have, for example, $r_1^2+r_2^2+r_3^2=a^2-2b$. In the simplest example of a quadratic, $x^2-ax+b=0$ and roots $r_1$ and $r_2$, we know that a is their sum and b is their product. If we wanted to find the sum of their cubes, we would have $a^3-3ab$ (think about what ab is). That is the basic idea behind Newton Sums.

hmmm.  the given equation $12x^3-985x-1728=0$ is true for $a, b, c$.  So solving for $x^3=\frac{985x+1728}{12}$.  then we plug in $a, b, c$ to get 3 equations.

$a^3=\frac{985a+1728}{12}$
$b^3=\frac{985b+1728}{12}$
$c^3=\frac{985c+1728}{12}$

we also know $a+b+c=0$, so adding the 3 equations we get $a^3+b^3+c^3=3\cdot\frac{1728}{12}$.


geometry
geometric transformation
reflection
geometry solved

let $ABCD$ be a convex quadrilateral such that :
$\widehat{DAB}<\frac{\pi }{2}$
$\widehat{ADB}+\widehat{ACB}=\frac{\pi }{2}$
$\widehat{DBC}+2\widehat{DBA}=\pi $
Prove that : $\left( DB+BC\right) ^{2}=AD^{2}+AC^{2}$

This problem is used on the french correspondance program and has to be solved by 29st March.

Please don't discuss this problem before....

(Can you delete your solution please ?)

Edit: *Thanks ...*

I think this problem can now be discussed...

[color=blue][b]Problem.[/b] Let ABCD be a convex quadrilateral such that

< DAB < 90°;
< ADB + < ACB = 90°;
< DBC + 2 < DBA = 180°.

Prove that $\left(DB+BC\right)^2=AD^2+AC^2$.[/color]

[i]Solution.[/i] Let X be the reflection of the point D in the line AB. Then, XB = DB, AX = AD, < AXB = < ADB and < XBA = < DBA, so that

  < DBC + < DBA + < XBA = < DBC + < DBA + < DBA = < DBC + 2 < DBA = 180°;

thus, the point B lies on the segment CX, so that CX = XB + BC = DB + BC. Also, by the sum of angles in triangle XAC, we have

  < XAC = 180° - (< AXC + < ACX) = 180° - (< AXB + < ACB) = 180° - (< ADB + < ACB) = 180° - 90° = 90°;

thus, the triangle XAC is right-angled at A, and by the Pythagoras theorem, we get $CX^2=AX^2+AC^2$. Using CX = DB + BC and AX = AD, this becomes $\left(DB+BC\right)^2=AD^2+AC^2$, and the problem is solved.

Sbos, was your solution similar to this?

[b]EDIT:[/b] I am just seeing this problem solved on http://www.mathlinks.ro/Forum/viewtopic.php?t=33784 (oh yes, the solution is more than the same :D ). Italy 2000 is the source. Interesting how the French correspondance program recycles problems from Italy MO...

  darij


abstract algebra
superior algebra
superior algebra unsolved

Determine which abelian groups $ A$ fit into a short exact sequence $ 0 \rightarrow \mathbb{Z}_{p^m} \rightarrow A \rightarrow \mathbb{Z}_{p^n} \rightarrow 0$ with $ p$ prime.

It is easy to see that $ |A|\equal{}p^{n\plus{}m}$ and that $ A\equal{}\mathbb{Z}_{p^m} \oplus \mathbb{Z}_{p^n}$ fits into the sequence but what are the rest of them?

Thank you!!!!!!!!!!!!

Problem already posted, see http://www.mathlinks.ro/viewtopic.php?t=255341.


geometry
circumcircle
inradius
incenter
power of a point
radical axis
geometry proposed

In other words:
Let I be incenter of ABC(center of Q) and XYZ intouch triangle of ABC
(perform by tangency points with incrcle).Now:
Q is orthog Q(a), Q(b), so I is on radical axis of Q(a), Q(b)
Q is orthog Q(b), Q(c), so I is on radical axis of Q(b), Q(c)
Q is orthog Q(a), Q(c), so I is on radical axis of Q(a), Q(c)
Therefore I is the radical center of Q(a), Q(b), Q(c).
Also  A'B'C' and XYZ are similar triangles.
Juan Carlos

[quote="grobber"]Let Q be the incircle of ABC, and let Q(a), Q(b), Q(c) be circles orthogonal to Q, which pass through B,C ; C,A ; A,B respectively. Let A' be the second intersection pt of Q(b) and Q(c). In the same manner we get B' and C'. Show that the circumradius of A'B'C' is half the inradius of ABC.[/quote]

Nice exercise, Grobber, but I believe it becomes too easy implied the knowledge of your favorite geometric transformation...

It is well-known that one can construct the inverse of a point P in a circle k as follows: All circles passing through the point P and orthogonal to the circle k have, apart from the point P, another common point, and this point is the inverse of the point P in the circle k.

Applied to the point A as the point P and the incircle of triangle ABC as the circle k, we get that the two circles Q(b) and Q(c), both passing through the point A and orthogonal to the incircle of triangle ABC, must pass through the inverse of the point A in to the incircle of triangle ABC. In other words, the point A', being defined as the second point of intersection of the circles Q(b) and Q(c), is just the inverse of the point A in the incircle of triangle ABC.

But now, if you call X, Y, Z the points where the incircle of triangle ABC touches the sides BC, CA, AB, then it is known that the midpoint of the segment YZ is the inverse of the point A in the incircle of triangle ABC. Hence, our point A' is the midpoint of the segment YZ. Similarly, the points B' and C' are the midpoints of the segments ZX and XY. Hence, the circumcircle of triangle A'B'C' is the nine-point circle of triangle XYZ, and therefore, the circumradius of triangle A'B'C' is the radius of the nine-point circle of triangle XYZ, i. e. it equals 1/2 of the radius of the circumcircle of triangle XYZ. Finally, since the circumcircle of triangle XYZ is the incircle of triangle ABC, we may state that the circumradius of triangle A'B'C' equals 1/2 the radius of the incircle of triangle ABC.

Qed..

  Darij

Dear Mathlinkers,
the circle passing through B and C, and orthogonal to the incircle has it a name?
Where is the center of this circle?
Sincerely
Jean-Louis


1)  Daca $\ x,y,z$ sunt numere intregi pentru care $\ (x,y)=1$ si avand relatia $\ x^4-y^4=z^2$, atunci $\ |x|=1$.
2)  Daca $\ x,y,z$ sunt numere intregi si $\ x^4+y^4=2z^2$, cu
$\ (x,y)=1$, atunci $\ |x|=1$ si $\ |y|=1$.
3)  Fie $\ a,b,c$ numere intregi astfel incat $\ a^4+b^4=c^2$.Sa se arate ca $\ a=0$ sau $\ b=0$.

Problema 3)
Ppa $\exists (x,y,z)\in \mathbb{N^*}^3$ a.î $x^4+y^4=z^2$
In continuare consideråm $z$ minimal
Dacå $z$ par $\Rightarrow$ putem presupune $x$ par.
Dacå $z$ impar, $z^2 \equiv 1 (\mod 8)$. Pp $x$,$y$ impare $\Rightarrow x^4+y^4 \not\equiv z^2 (\mod 8)$, contradictie. $\Rightarrow$ putem pp si în acest caz $x$ par.
Putem pp de asemenea $(x^2,y^2,z)=1$.
Scriem $(x^2)^2+(y^2)^2=z^2$ si conform sol. tripletelor pitagoreice (si pp cå $x$ par) $\Rightarrow x^2=2mn, y^2=m^2-n^2,z^2=m^2+n^2$, cu $(m,n)=1; m\not\equiv n (\mod 2)$.
Cazul I.$m$ par, $n$ impar $\Rightarrow y^2\equiv 3(\mod 4)$, absurd.
Cazul II. $m$ impar, $n$ par. $n=2q \Rightarrow mq =(x/2)^2$.
$(m,n)=1 \Rightarrow m=w^2, q=t^2, (w,t)=1$.
$y^2=(w^2)^2-(2t^2)^2 \Leftrightarrow (2t^2)^2+y^2=(w^2)^2$.
Aplicåm din nou sol. tripletelor pitagoreice:
$\Rightarrow t^2=ab,y^2=a^2-b^2,w^2=a^2+b^2$ cu $(a,b)=1$ iar $t^2=ab \Rightarrow a=u^2,b=v^2 \Rightarrow u^4+v^4=w^2.$ Dar z minimal $\Rightarrow w\ge z \Rightarrow w^2 \ge z^2 \Rightarrow m \ge m^2+n^2$, absurd.
Deci $a=0$ sau $b=0$, ori $c=0 \Rightarrow a=0$ si $b=0$.
QED


function
LaTeX
geometry
3D geometry
absolute value
algebra solved
algebra

find all functions from the set of integers Z to itself satisfying
f(a^3 +b^3 +c^3) = f(a)^3+f(b)^3+f(c)^3
sorry abt not using latex,.. guys.
I have a hunch that for every x>3 ,x^3 can be expressed as a sum of five cubes with absolute value less that x.i have  got such a representation of odd integers but no luck with even integers

This one has been proposed many times here.
You may find a solution on
http://www.animath.fr
see section
'l'entrainement de l'olympiade franaise de mathmatiques'.

Pierre.

And what if one does not speak French ?  ;)

Just learn it   :P
Anyway, it is not difficult to understand the solution from french because the math part is clear.

Pierre.

You're absolutely right, Pierre! By the way, joyeux Noel! ( I'm sure I wrote that wrong, but you know what I mean).

No, non, you are right. C'est exactement a.
Happy Christmas too.

Pierre.


[quote="Peter"]Show that there is a unique sequence of integers $\{a_{n}\}_{n \ge 1}$ with \[a_{1}=1, \; a_{2}=2, \; a_{4}=12, \; a_{n+1}a_{n-1}=a_{n}^{2}\pm1 \;\; (n \ge 2).\][/quote]

It is obvious that $a_{3}=5$ and that $a_{n}\geq 3$ for all $n\geq 3$.

Now let us prove that there is at most one sequence that satisfys the conditions of the problem.

If there are two such a sequences, then for some $n\geq 4$ we should have two ways to define $a_{n+1}$. That means that both $\frac{a_{n}^{2}+1}{a_{n-1}}$ and $\frac{a_{n}^{2}-1}{a_{n-1}}$ are integers, then their sum should be an integer too. But $\frac{2}{a_{n-1}}<1$ can not be an integer.

Now we should only find a sequence of integers that satisfys the condition of the problem. We define $b_{n}$ as $b_{1}=1,b_{2}=2$ and $b_{n+1}=2b_{n}+b_{n-1}$.(Obviously this is a sequence of integers.)

Let us prove that $b_{2n+1}b_{2n-1}=b_{2n}^{2}+1$ and $b_{2n+2}b_{2n}=b_{2n+1}^{2}-1$.(This means that $b_{n}b_{n-2}=b_{n}\pm 1$)

We have that $b_{n+1}=2b_{n}+b_{n-1}=2b_{n}+2b_{n-2}+b_{n-3}$ then $b_{n+1}-b_{n-3}=2(b_{n}+b_{n-2})$ $\Rightarrow$ $b_{n-1}(b_{n+1}-b_{n-3})=2b_{n-1}(b_{n}+b_{n-2})=(b_{n}-b_{n-1})(b_{n}+b_{n-1})=b_{n}^{2}-b_{n-1}^{2}$. So we got that $b_{n+1}b_{n-1}-b_{n}^{2}=b_{n-1}b_{n-3}-b_{n-1}^{2}$, thus $b_{n+1}b_{n-1}-b_{n}^{2}$ is a constant for any $n$.

Finaly we have that $b_{2n+1}b_{2n-1}-b_{2n}^{2}=b_{3}b_{1}-b_{2}^{2}=1$ and $b_{2n+2}b_{2n}-b_{2n+1}^{2}=b_{4}b_{2}-b_{3}^{2}=-1$, as we clamed above.


How many real number solutions are there for the equation \[ (x^2\plus{}4x\plus{}5)^{x^2\plus{}1}\equal{}1\]

Is there a formula for this?

It's actually much simpler than you think. We can even extend it: What are the numerical solutions to the given equation?
[hide="Big Hint"]What is the unique number $ a$ such that $ a^b \equal{} 1$ for all real $ b$?[/hide]

[hide="Solution?"]
There are two cases in which $ a^b \equal{} 1$. When $ b \equal{} 0, a\not \equal{} 0$ and when $ a \equal{} 1, \minus{} 1$.

Case 1: $ b \equal{} 0$
$ x^2 \plus{} 1 \equal{} 0 \\
x^2 \equal{} \minus{}1$

No real solutions here.

Case 2: $ a \equal{} 1$
$ x^2 \plus{} 4x \plus{} 5 \equal{} 1 \\
(x \plus{} 2)^2 \equal{} 0 \\
x \equal{} \minus{} 2$

This solution works fine.

Case 3: $ a \equal{} \minus{} 1, 2|b$

$ x^2 \plus{} 4x \plus{} 5 \equal{} \minus{} 1 \\
x^2 \plus{} 4x \plus{} 6 \equal{} 0$

No real solutions here.

So the answer is $ \boxed{1}$.
[/hide]

Edit: Thanks, Patterns. (I've been doing math all day and my brain's tired.)

[hide="sunehra"]You might want to recheck Case 1.
Other than that, it's good.[/hide]


let $ a,b,c,d,m$ and $ n$ are $ \in N*$
such that : $ a^2\plus{}b^2\plus{}c^2\plus{}d^2\equal{}1989$ and  $ a\plus{}b\plus{}c\plus{}d\equal{}m^2$ and  $ n^2\equal{} max(a,b,c,d)$
find  $ m$ and $ n$

$ m^2\equal{}a\plus{}b\plus{}c\plus{}d\le 4\sqrt{\frac{a^2\plus{}b^2\plus{}c^2\plus{}d^2}{4}}<90$.
$ \sqrt{1989}>n^2\equal{}max(a,b,c,d)>\sqrt{1989/4}>22$.
Therefore $ n\equal{}5$ or $ n\equal{}6$.
1. In first case we get $ x^2\plus{}y^2\plus{}z^2\equal{}(1989\minus{}625)/4\equal{}341,max(x,y,z)\le 12$.
But $ 341\equal{}3\mod 4$, therefore $ max(x,y,z)\le 11$ and all of them odd. It mean, that at least two of them 11, then third $ z^2\equal{}341\minus{}2*121\equal{}99$ is not square.
2. $ n\equal{}6$. Let $ d\equal{}n^2$, then $ a^2\plus{}b^2\plus{}c^2\equal{}693.$ It give $ \sqrt{693}<a\plus{}b\plus{}c\le 3\sqrt{231}<46.$
Because $ a\plus{}b\plus{}c$ odd $ m^2\equal{}81$ or $ m\equal{}9$.


geometry
inradius
ratio
Euler
inequalities
geometry solved

(1) ABC is an acute-angled triangle. Prove that if a circle intersects the sides of ABC 6 times in total, then its radius must be more than the inradius of ABC.

(2) ABC is a triangle, and the angle bisectors of A, B and C meet the opposite sides at P, Q and R respectively. Prove <PQR=90 iff <ABC=120.

I have the feeling that (2) was solved here before, but I don't know where you could start searching...

[color=red][Moderator edit: Yes, it was solved here. I found the thread: http://www.mathlinks.ro/Forum/viewtopic.php?t=1336 ][/color]

Regarding (1):

Assume the circle intersects BC at A1, A2, CA at B1, B2 and AB at C1 and C2. Now draw tangents to the circle in the midpoints of the smaller arcs A1A2, B1B2, C1C2. These tangents define another triangle A'B'C' which contains ABC in its interior and whose sides are parallel to those of ABC. This means that ABC and A'B'C' are similar and B'C'/BC>1; the incircle of A'B'C' is the circle which cuts ABC in 6 points, and since the similarity ratio A'B'C'/ABC>1 we get that the inradius of A'B'C'>the inradius of ABC=> the radius of the circle>the inradius of ABC, Q.E.D.

I think it's Ok.

It's probably not complete. I autoimatically assumed the sides to be segments, but the circle can also cut the prolongations of the sides in 6 points, and my construction doesn't work then. We have to consider each case separately and when the circle cuts prolongations of sides instead of segments, I think we should compare it's radius with one of the radii of the exinscribed circles (or something like that :)).

I don't really feel like checking all of that right now..

Well, here is another approach for (1):

I don't need the fact that the circle intersects the sides in 6 points; I only will use that the circle meets all sides of the triangle, among them (at least) one side in 2 points. I don't need the assumption that ABC is acute-angled either. So I will prove the following rather general result (which I call "Generalized Euler inequality" since it generalizes the well-known Euler inequality $R\geq 2r$):

[b]Generalized Euler inequality:[/b] If ABC is a triangle, and an arbitrary circle k intersects all three sidelines of this triangle, then the radius t of k is not less than the inradius r of triangle ABC, with equality if and only if k is the incircle of triangle ABC.

[i]Proof.[/i] Let O be the center of k, and let x, y, z be the distances from O to the sidelines BC, CA, AB. In a right-angled triangle, any catet is less or equal to the hypotenuse (with equality only if the triangle is degenerate); hence, every of the distances x, y, z is less or equal t. Hence,

  $ax+by+cz\leq at+bt+ct=\left(a+b+c\right)t$.

Now, if S is the area of triangle ABC, we have ax + by + cz = 2S, since ax, by, cz are the double areas of triangles BOC, COA, AOB and these double areas add up to the double area of triangle ABC. Hence,

  $2S\leq\left(a+b+c\right)t$,   and therefore   $t\geq\frac{2S}{a+b+c}=r$.

It is easily seen that equality occurs if and only if our circle k touches all three sides BC, CA, AB, i. e. if it coincides with the incircle of triangle ABC.

If O lies in the exterior of triangle ABC, then we have not 2S = ax + by + cz anymore, but $2S = \pm ax \pm by \pm cz$ with some minus and some plus signs. In any case, this yields $2S \leq ax+by+cz$ and we still have $ax+by+cz\leq\left(a+b+c\right)t$, so that $2S\leq\left(a+b+c\right)t$ and the proof remains valid. Qed..

  Darij


logarithms
inequalities
function
inequalities unsolved

For $a,b,c>0$ .$k \in R$
Prove or Disprove:
\[ \left(\frac a{b+c}\right)^k+\left(\frac b{c+a}\right)^k+\left(\frac c{a+b}\right)^k \ge 3\cdot (\frac 1 2)^k  \]
I can only solve:$k\le 0$ and $k\ge \frac 2 3$ :)

This is an old question, still unsolved if I am not mistaken. See

http://www.mathlinks.ro/Forum/viewtopic.php?t=12763
http://www.mathlinks.ro/Forum/viewtopic.php?t=5520

By the way, concerning the Shapiro article, I have no idea what is meant either, and I don't know the proofs for $n\geq 7$...  :blush:

  Darij

Not really unsolved, but hungkhtn didn't post his solution: http://www.mathlinks.ro/Forum/viewtopic.php?highlight=fraction&t=30733
he outlined his solution though.

[quote="darij grinberg"]
By the way, concerning the Shapiro article, I have no idea what is meant either, and I don't know the proofs for $n\geq 7$...  :blush:

  Darij[/quote]
Thanks all the same,And if I know the solution I will tell you right away.So do you :)
[quote="darij grinberg"]This is an old question, still unsolved if I am not mistaken. See

http://www.mathlinks.ro/Forum/viewtopic.php?t=12763
http://www.mathlinks.ro/Forum/viewtopic.php?t=5520
[/quote]
Thanks,but I think I have solved it now. :) But ugly solution. :(
From $a=b=1,c \to 0$,We get $k \ge k \geq \log_2 \frac32$.
Then we should only prove $k = \log_2 \frac32$
[b]Theorem[/b],$a,b,c>0,k = \log_2 \frac32$
 \[ \left(\frac a{b+c}\right)^k+\left(\frac b{c+a}\right)^k+\left(\frac c{a+b}\right)^k \ge 3\cdot (\frac 1 2)^k  \]

[b]Proof[/b]We may assume $a+b+c=1,$
Let $f(t)=\frac{t^k}{(1-t)^k}$,then we should only to prove:
\[ f(a)+f(b)+f(c) \ge 3\cdot (\frac 1 2)^k  \]
$a+b+c=1$(also assume $a \ge b \ge c$)
And because,\[ f'(t)=\frac{kx^{k-1}}{(1-x)^k+1},f\"(t)=\frac{kx^{k-2}(k-1+2kx)}{(1-x)^{k+2}}  \]
Then we get $f(t)$ is concave on $(0,\frac{1-k}{2k}]$,and convex on  $[ \frac{1-k}{2k},1 )$
Then:$a \in [ \frac{1-k}{2k},1 )$(because $a \ge \frac 1 3 \ge \frac{1-k}{2k}$)

Case 1: $a,b \in[ \frac{1-k}{2k},1 )$.
Use jensen's inequality,we have \[ f(a)+f(b) \ge 2f(\frac{a+b}2)  \]
So we should only prove:
\[ 2f(\frac{a+b}2)+f(c) \ge 3\cdot (\frac 1 2)^k  \]
\[ \iff 2f(\frac{1-c}2)+f(c) \ge 3\cdot (\frac 1 2)^k  \]
\[ \iff 2\frac{(1-c)^k}{(1+c)^k}+\frac{c^k}{(1-c)^k} \ge 3\cdot (\frac 1 2)^k,(0<c \le \frac 1 3]  \]
\[ \iff 2\frac{1}{(1+2x)^k}+x^k \ge 3\cdot (\frac 1 2)^k,(x=\frac{c}{1-c},0<x \le \frac 1 2]  \]
let $g(x)=2\frac{1}{(1+2x)^k}+x^k$.
Thus,$g'(x)=x^{k-1}-\frac{4k}{(1+2x)^{k+1}}$
(Let $h(x)=x^{k-1}(1+2x)^{k+1}$,we have $h'(x)<0$,check it :) )
Then it is easy to that $g(x)$ get its minimum on the endpoints.
Thus $g(x) \ge \min\{g(0),g(\frac 1 3)\}=3\cdot (\frac 1 2)^k$.
Thus we prove the inequality in Case 1.
Case 2:$b,c \in (0,\frac{1-k}{2k}]$
Case 2.1:$b+c \ge \frac{1-k}{2k}$.
Then by karamate's inequality(In fact It can be proved directly) we have:
\[ f(b)+f(c) \ge f(b+c- \frac{1-k}{2k})+f(\frac{1-k}{2k})  \]
Then we should prove:\[ f(b+c- \frac{1-k}{2k})+f(\frac{1-k}{2k})+f(a) \ge 3\cdot (\frac 1 2)^k  \]
But $\frac{1-k}{2k},a \in \frac{1-k}{2k})$ this turns to Case 1.
Case 2.2:$b+c \le \frac{1-k}{2k}$.
Then by karamate's inequality(In fact It can be proved directly) we have:
\[ f(b)+f(c) \ge f(b+c)+f(0)  \]
So we should to prove:
\[ f(0)+f(b+c)+f(a) \ge 3\cdot (\frac 1 2)^k  \]
This is easy by AM-GM.
Thus we complete the proof.

Hope my solution is right, :) I am sorry If there have any mistake.
In fact the idea is the half-convex and half-concave theorem I post [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=64933]here[/url].Anyway,I have mention the inequality there.
But seems few people interested in it.So I didn't mention it in my proof. :)

Hi,Have anybody seen my proof?I hope someone can help me to check it,Thanks :)

My solution base on:

First, for $k \ge 1/2$ we have :
 If $a \ge b \ge c$ then $f(a,b,c) \ge f(t,t,c)$ with $t=(a+b)/2$.

Then, use the lemma:

\[ \sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}} \ge 2 \]

We have done.

Sorry for my stupidness,can you detail your solution?

It's not hard but better you should try to prove the same but more difficult inequality :

Find the minimum of :

\[ (\frac{a}{b+c+d})^k+(\frac{b}{c+d+a})^k+(\frac{c}{d+a+b})^k+(\frac{d}{a+b+c})^k \]

This one I creat in the soon minute after read again the first problem.

But I still can't understang your proof....

your new one will be easy if use the method of my solution...

[quote="zhaobin"]
let $g(x)=2\frac{1}{(1+2x)^k}+x^k$.
Thus,$g'(x)=x^{k-1}-\frac{4k}{(1+2x)^{k+1}}$[/quote]
I think this is not true.
[quote="zhaobin"]
(Let $h(x)=x^{k-1}(1+2x)^{k+1}$,we have $h'(x)<0$,check it :) )[/quote]
$h'(x)<0$ when $k<\frac{1}{3}.$ But $k>\frac{1}{3}$
[quote="zhaobin"]
Then it is easy to that $g(x)$ get its minimum on the endpoints.
Thus $g(x) \ge \min\{g(0),g(\frac 1 3)\}=3\cdot (\frac 1 2)^k$.[/quote]
If $h'(x)<0$ then $g'(x)\geq g'(\frac{1}{2})=0.$
Hence, $g(x)>g(0)=2=\frac{3}{2^k}.$
[quote="zhaobin"]
 ...the half-convex and half-concave theorem I post [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=64933]here[/url].Anyway,I have mention the inequality there.
But seems few people interested in it.[/quote]
In my opinion, this is very useful and nice theorem! :lol:

[quote="zhaobin"]Thank you very much.
[quote="arqady"][quote="zhaobin"]
let $g(x)=2\frac{1}{(1+2x)^k}+x^k$.
Thus,$g'(x)=x^{k-1}-\frac{4k}{(1+2x)^{k+1}}$[/quote]
I think this is not true.
[/quote]
Sorry,It should be $g'(x)=kx^{k-1}-\frac{4k}{(1+2x)^{k+1}}$
[quote="arqady"]
[quote="zhaobin"]
(Let $h(x)=x^{k-1}(1+2x)^{k+1}$,we have $h'(x)<0$,check it :) )[/quote]
$h'(x)<0$ when $k<\frac{1}{3}.$ But $k>\frac{1}{3}$
[quote="zhaobin"]
Then it is easy to that $g(x)$ get its minimum on the endpoints.
Thus $g(x) \ge \min\{g(0),g(\frac 1 3)\}=3\cdot (\frac 1 2)^k$.[/quote]
If $h'(x)<0$ then $g'(x)\geq g'(\frac{1}{2})=0.$
Hence, $g(x)>g(0)=2=\frac{3}{2^k}.$
[/quote]
Sorry,I find it is wrong here.Because $h'(x) \le 0,x\le \frac{1-k}{4k}$ and $h'(x) \ge 0, x \ge \frac{1-k}{4k}$
So there may be a little problem in proving:
\[ g(x)=2\frac{1}{(1+2x)^k}+x^k \ge 3\cdot (\frac 1 2)^k,0<x \le \frac 12  \]
I can't figure it now,can you prove it now :( ,can you prove it,arqady?
[quote="arqady"]
[quote="zhaobin"]
 ...the half-convex and half-concave theorem I post [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=64933]here[/url].Anyway,I have mention the inequality there.
But seems few people interested in it.[/quote]
In my opinion, this is very useful and nice theorem! :lol:[/quote]

Thanks :)[/quote]

By  the theorem it is left to prove the inequality for $a=b$ i.e
The following inequality $h(a)=F(a,a,1-2a)=2\left(\frac{a}{1-a}\right)^k+\left(\frac{1-2a}{2a}\right)^k$(use $a+b+c=1,a\ge b\ge c\Rightarrow a\in [1/3;1/2]$)
Next $h'(a)=2k\left(\frac{1}{a^{1-k}(1-a)^{1+k}}-\frac{1}{(1-2a)^{1-k}(2a)^{1+k}}\right)$
So extremas will be at $a=1/2$(singular point) or at $a_0,h'(a_0)=0$ where $a_0$ is the unique root of $g(a)=(1-2a)^{1-k}(2a)^{1+k}-a^{1-k}(1-a)^{1+k}=0$
Proof:$g(a)=t(1-2a)-t(a)$ where $t(a)=a^{1-k}(1-a)^{1+k}=a(1-a)\left(\frac{a}{1-a}\right)^k$ is increasing for $a\in [0;1/2]$
so $g(a)=0$ iff $1-2a=a$ or $a=1/3$

Blusea is right. 2 problem are quite same in proving. But in my problem with 4 variable is more difficult.

[quote="zhaobin"][quote="hardsoul"]
So extremas will be at $a=1/2$(singular point) or at $a_0,h'(a_0)=0$ where $a_0$ is the unique root of $g(a)=(1-2a)^{1-k}(2a)^{1+k}-a^{1-k}(1-a)^{1+k}=0$
Proof:$g(a)=t(1-2a)-t(a)$ where $t(a)=a^{1-k}(1-a)^{1+k}=a(1-a)\left(\frac{a}{1-a}\right)^k$ is increasing for $a\in [0;1/2]$
so $g(a)=0$ iff $1-2a=a$ or $a=1/3$[/quote]
Thanks,but you proved that $t(a)$ is increasing,not $g(a)$...
($t(1-2a)$ is not descreasing when $a \in [0,\frac 1 2]$)

[quote="Bluesea"]I think this kind of inequality is really well-known.Study carefully vasc's proof for the famous:$(ab)^k+(bc)^k+(ca)^k\leq\ max(3;\frac{3^{2k}}{2^{2k}})$-then you can do it yourself  :lol:  :lol:[/quote]
Thanks,can you tell me where is Vasc's proof?[/quote]

If $f$ is increasing function on $[a,b]$ and $f(x)=f(y)$ doesnt't it follow that $x=y$???In the example I just put $f=t,x=a,y=1-2a$ :?

For example the intersection of $y=x$ and $y=x^2-x+1$ give us a root $x=1$  of $x-(x^2-x+1)=0$

[quote="hardsoul"]

If $f$ is increasing function on $[a,b]$ and $f(x)=f(y)$ doesnt't it follow that $x=y$???In the example I just put $f=t,x=a,y=1-2a$ :?

For example the intersection of $y=x$ and $y=x^2-x+1$ give us a root $x=1$  of $x-(x^2-x+1)=0$[/quote]
But $1-2a \in [0,1]$ the function is not increasing in $[0,1]$,so I don't think you can asserct so quickly.(Maybe I am wrong)

[quote="hardsoul"]
Next $h'(a)=2k\left(\frac{1}{a^{1-k}(1-a)^{1+k}}-\frac{1}{(1-2a)^{1-k}(2a)^{1+k}}\right)$
So extremas will be at $a=1/2$(singular point) or at $a_0,h'(a_0)=0$ where $a_0$ is the unique root of $g(a)=(1-2a)^{1-k}(2a)^{1+k}-a^{1-k}(1-a)^{1+k}=0$
Proof:$g(a)=t(1-2a)-t(a)$ where $t(a)=a^{1-k}(1-a)^{1+k}=a(1-a)\left(\frac{a}{1-a}\right)^k$ is increasing for $a\in [0;1/2]$
so $g(a)=0$ iff $1-2a=a$ or $a=1/3$[/quote]
Ok,the gragh say your assert is wrong...but the inequality it true


I swear, I saw this problem less than a month ago, and we agreed it couldn't be done without calculus.
Or I might just be imagining things.

Do you know where the source is.  better yet what the answer is.

The problem was posed by another student, so I don't know where it came from.

By typing into a calculator, it is approximately 2.517.

This is the first topic:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=135807


can someone plz help me on how to get old national mathcounts test questions.  on the mathcounts website there are only school, chapter, and state questions, and they are all frmo 2006. so can someone plz help me on how to get old nationsl mathcounts test questoins or old  (like 2004) state mathcounts questoins

The Search button at the top of this forum.

but i didnt know it can give u entire tests

To get national tests, you either have to buy it or get it from a coach that went to nationals that year.

Or you can just ask someone to email you them.

[quote="jclarkemathL314159"]Our state was lucky

Mr. Findell, who was Mr. Mosca's successor at Diamond Middle School (MA), the school that always won, has every mathcounts problem ever written. Mr. Mosca actually won at nats one year (I think 99).

Anyway, Diamond got second at states this year and Meadowbrook won, so Mr. Findell gave all the problems to Mr. Bhat

Our school had like no problems, and it showed, seeing that we came in 6th at state. But our coach next year, Mr. Frost, who was the NH nat coach the last 7 years or something, has 17 years on PDF, all the years of the current format. (30 question sprints, 8 question targets with calcs, etc.)[/quote]

My coach also has every nats problem.  I only wish I had the target round for 96-98, to practice with.  I guess it helped me to get 14th Place at Nationals.

BTW, it says that in 1999 Po Ru Loh, won the Mathcounts Competition.

[quote="Ignite168"]Umm, I think he means coached a team to victory, as you wouldn't really become a teacher that's already suceeded when you just graduated from college would you?[/quote]

Yeah, I guess you are right.

I really hope that Michigan has someone who does something nationally-amazing.

That's odd. The 4shareds aren't up anymore. :(

[quote="SplashD"][quote="mathgeniuse^ln(x)"]4shareds?

What does that mean?  Where could I have found out?[/quote]

It's a secret that bpms just blew.[/quote]
They aren't secret. They're gone completely too. And mathgenius was here back then

[quote="mathgeniuse^ln(x)"]Huh?  I was?

But, waht are 4shared?  Are they shared mathematics?[/quote]
They used to be databases of old tests but they have been lost in the sands of time, only a shadow of their golden age. :D.


vector
linear algebra
matrix
LaTeX
algebra
linear equation
complex numbers

some true false questions.I make some notations but i don't know if are correct

1)is possible the vectors $ v_1 , v_2, v_3$ to be linear dependent and the vectors $ v_1 - v_2, v_1 - v_3, v_2 - v_3$ be linear independent?

we have $ v_1 - v_2$ so the one vector is dependent on the other.How can be indpendent?

2)if $ A \in R{}^n{}^x{}^1$ and $ B \in R{}^1{}^x{}^n$ then AB is non invertible

so by multiplying these vectors
we have an array nxn but how can we tell if is invertible?

3)if $ A$ is an orthogonal matrix then $ det(A) = + - 1$

4)if for matrices $ A,B \in R{}^n{}^x{}^m$ we have that $ Ax = Bx$ then $ A = B$

5)if $ A$ and $ B$ are orthogonal matrices then $ AB$ is also orthogonal

i think false

For #1, the question you ask afterwards shows that you don't understand what independence is. Vectors are not independent or dependent; rather [b]sets[/b] of vectors are independent or dependent.

As for the set $ \{v_1\minus{}v_2,v_1\minus{}v_3,v_2\minus{}v_3\},$ is it always dependent, no matter what properties the set $ \{v_1,v_2,v_3\}$ has. Why? Because $ (v_1\minus{}v_2)\minus{}(v_1\minus{}v_3)\plus{}(v_2\minus{}v_3)\equal{}0.$

On #2, have you heard of the [b]rank[/b] of a matrix?

On #3, make use of this: $ \det(AB)\equal{}\det(A)\det(B).$

#4 is false, so you should be looking for a counterexample. But #5 is true. How can you recognize an orthogonal matrix?

3) it's true since $ |\lambda_k| \equal{} 1$ for an orthogonal matrix ,  $ k \equal{} \overline{1,n}$ where $ \lambda \in \mathbb{C}$ are the eigenvalues of A and because
$ det(A) \equal{} \prod_{k \equal{} 1}^n \lambda_k$ the conclusion follows .

4) if $ x$ is a scalar i.e. $ x\in \mathbb{C^*}$ I believe that this implies $ A \equal{} B$

5) it's true .(Try prove it yourselv , having in mind the that $ (AB)^T \equal{} B^TA^T$ )

thanks!
i know that in orthgonal matrices we hav $ Q^ - ^1 = Q^T$

and i unerstand clearly only 1

for 2  don't know about rank(or i don't understand about what you are talking,because i don't speak very good english)

for 3 we have only one array A , how will i use det(AB)
@Svejk i understand your way but not kent way

for 4 , i understand Svejk

for 5 , so i must prove that $ (AB)^T = (AB)^ - ^1$
but from where to begin?
from $ ((AB)^T)^ - ^1)$ ?

5)

We know that $ AA^T\equal{}I_n$ and $ BB^T\equal{}I_n$ and we want to prove that $ (AB)(AB)^T\equal{}I_n$ $ \Leftrightarrow$ $ ABB^TA^T\equal{}I_n$ , but $ BB^T\equal{}I_n$ so we are left with $ AA^T\equal{}I_n$ which is also true .

Rank is the concept under discussion [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=185466]here[/url].

Let me get at this without using that word.

We have that $ B$ is an $ 1\times n$ row vector. ($ \text{\LaTeX}$ tip: use \times, not x.) Find an $ x$ that is perpendicular to $ B.$ That is, find a column vector $ x$ such that $ Bx=0.$ (We're solving one homogeneous linear equation for $ n$ unkonwns; this always has infinitely many solutions. Just arbitrarily pick one, with the only restriction being that we want $ x\ne 0.$)

Then $ (AB)x=A(Bx)=A\,0=0.$ Any square matrix that sends some nonzero vector to zero is not invertible. (What this shows directly is that the columns are dependent.)

For #3, orthogonal means that $ A^TA=I.$ Now do you see a product of two matrices?

You said "@Svejk i understand your way but not kent way."

He splashes around words like eigenvalue, invokes the complex numbers when they're not needed, tacitly assumes some big theorems that wouldn't have been done by the point in the course at which these problems would arise, and you think you understand that, but you don't see the direct proof? I think that makes you a mindless symbol-pusher who doesn't want to think.

For #4, Svejk is wrong, because $ x$ is not scalar. In order for $ Ax$ to be defined for an $ n\times m$ matrix $ A,$ $ x$ should be an $ m\times 1$ column vector - that is, $ x\in \mathbb{R}^m.$ And the result in #4 is false.

For #5, please do not write $ AB^T$ or $ AB^{-1}$ if that's not what you mean; use the parentheses you need. If you did that on a test of mine, I [i]would[/i] deduct points for it.

Note that $ (AB)^T=B^TA^T,$ and that $ (AB)^{-1}=B^{-1}A^{-1}.$ You should be familiar with both of those facts.


How many different ways can you place 2 different coins on the square fields of an 8x8 chess board so that their fields do not share a side or a vertex?

[quote="nat mc"][hide]If one is on the corners:
the other can't be in 4 places 64-4=60.  4 corners, 2 orders
60*4*2=480

If one is on the edge:
the other can't be in 6 places 64-6=58. 24 edge spots, 2 orders
58*24*2=2784

If one is in the middle:
the other can't be in 9 places 64-9=55. 36 middle spots, 2 orders
55*36*2=3960

480+2784+3960=7224, but divide by 2 because everything was counted twice 7224/2=3612[/hide][/quote]

  Everything looks good to me, however I think its unessecary to include the 2 orders, when your placing the second coin in terms of the first.

When i think of a chess problem it normaly involves chess pieces

[hide]3 cases:

Case 1: First coin is on a corner square: There are then 64-4=60 ways to place the second coin. There are 4 corner squares, for a total of 4x60=240 for Case 1.

Case 2: First coin is on a side square that is not a corner square: There are then 64-6=58 ways to place the second coin. There are 6x4=24 side squares, for a total of 24x58=1392 for Case 2.

Case 3: First coin is on one of the remaining interior squares: There are then 64-9=55 ways to place the second coin. There are 6x6=36 interiior squares, for a total of 36x55=1980 for Case 3.

Adding, 240+1392+1980=3612.[/hide]

You could
[hide] count what is not needed... if one coin is in a corner then there are 3 possibilities which can't take place, and since there are four corners: 4*3=12
if the 1st coin is on an edge that is not a corner there are 5 combinations that can't take place for each square: 6*4*5=120
if the 1st coin in a middle square by the same logic: 8*6*6=288
64*63-(12+120+288)=3612[/hide]

[quote="k8reindeer"]like this:

what is the probability that if 4 rooks are placed on an 8 by 8 chess board none of them is in position to take any other one?

rooks can move in a straight line as far as there are no other peices in the way and they take other peices by being able to occupy the space that the other peice is in currently. there are no peices besides the rooks on the board to worry about.

that is what I would think of as a chess problem![/quote]

just for fun, I did this problem.
[hide="the answer"]i think it's 1/16. if you have a rook in a corner (it doesn't matter where, it's just easier to visualize) then it "takes up" 15 squares, the next one takes up 13, the next one takes up 11, and the last one takes up 9. 64-15-13-11-9 is 16, so 1/16.[/hide]

I think it is
[hide]
350/1891
The first rook can go in any of the 64 spaces, but the second can not occupy the 8 vertical and horizontal one that this rook threatens (in all 15-since it is one on of them). Thus, the second one can occupy any of the 64-15=49 squares. The third one can occupy 49-13=36, and the fourth 36-11=25. Multiply these and place over the possible combinations without any restrictions...
(64*49*36*25)/(64*63*62*61)= 350/1891[/hide]

[quote="k8reindeer"]yes, but we want the probability of the second one not being in the way of the first one, and the third one not being in the way of the other 2 and the fourht one not being in the way of the other 3, and you are (I think) only considering where the last guy is open to go.[/quote]

ok, fine, I'll do it right.
[hide]there's a 100% chance that the first rook is on an "ok" square.  it makes 15 squares "not ok" to be on. there are 49 squares left. so, 49/63. it takes away 13 squares. so there's 36/62. the last one has 11 less squares to go on, so 25/61. the 64/64 for the first one automatically cancels out, so 49/63*36/62*25/61 is 350/1891.[/hide]


2 part game.

(a) If the person above you posted a word, search that word on [url=http://images.google.com]Google Image Search[/url] and post one of the images you find.

(b) If the person above you posted an image, describe the image in a single word, trying to use a different word that the image came out of.

Also:
(c) Do not post words that will obviously lead to inappropriate image results on google.
(d) If someone misspells a word, you still must use the misspelled version unless it actually returns 0 results, which is highly rare.

First word

psychedelic

thats obviously gonna lead to something inappropriate
-jorian

How so? I just searched it and there was nothing inappropriate with safesearch OFF until the 2nd page.


Siteul oficial este: http://www.mate.isjcta.ro/. Deocamdata succes tuturor participantilor, impresiile vor fi dupa.

Vreau sa multumesc organizatorilor, deoarece conditiile de cazare au fost extraordinar de bune fata de anii precedenti. Am fost cazati la hotelul Doina din Neptun. Pe urma, selectia problemelor a fost destul de bine facuta. La clasa a 11-a, de exemplu, problemele au avut un grad ridicat de dificultate, necesitand doar cunoasterea teoriei de baza. Spre deosebire de unii ani in care se dadeau doua probleme de algebra si doua probleme de analiza, anul acesta s-au dat o problema de algebra si trei probleme de analiza, ceea ce a aratat inca o data ca la o olimpiada trebuie stiuta materia in intregime. De asemenea, cele 4 probleme au avut enunturi destul de frumoase, desi doua dintre ele mi s-au parut destul de dificile.

P.S. Ii salut pe toti participantii de la ONM. M-am intalnit cu foarte multi prieteni, insa mi s-a parut ca timpul a trecut prea repede.

[color=orange][b]Subiectele de la clasa a XI-a[/b][/color]

[color=orange][b]Subiectul 1[/b][/color]

[color=darkblue]Fie $ (t_n)_n$ un sir convergent de numere reale, $ t_n \in (0,1)$ pentru orice $ n \in \mathbb{N}$ si $ \displaystyle\lim_{n\to\infty}t_n \in (0,1)$. Definim sirurile $ (x_n)_n$ si $ (y_n)_n$ prin relatiile:

$ i)$ $ x_{n \plus{} 1} \equal{} t_n x_n \plus{} (1 \minus{} t_n)y_n$

$ ii)$ $ y_{n \plus{} 1} \equal{} (1 \minus{} t_n) x_n \plus{} t_n y_n$

pentru orice $ n \in \mathbb{N}$ si $ x_0,y_0$ numere reale fixate

$ a)$ Sa se arate ca sirurile $ (x_n)_n$ si $ (y_n)_n$ sunt convergente si au aceeasi limita

$ b)$ Aratati ca daca $ \displaystyle\lim_{n\to\infty}t_n \in \{ 0,1 \}$ concluzia nu mai este adevarata[/color]

[color=orange][b]Subiectul 2[/b][/color]

[color=darkblue]Fie $ f: \mathbb{R} \rightarrow \mathbb{R}$ o functie continua cu proprietatea ca pentru orice $ x \in \mathbb{R}$, limita $ \displaystyle\lim_{h\to 0} \mid \frac {f(x \plus{} h) \minus{} f(x)}{h}\mid$ exista si este finita.

Aratati ca in orice punct din $ \mathbb{R}$ $ f$ este derivabila sau admite derivate laterale finite, de acelasi modul si semn contrar.[/color]

[color=orange][b]Subiectul 3[/b][/color]

[color=darkblue]Fie $ A,B \in M_n(\mathbb{C})$, cu $ AB \equal{} BA$ si $ \det B \not \equal{} 0$.

$ a)$ Aratati ca daca $ \mid \det (A \plus{} zB) \mid \equal{} 1$ pentru orice $ z \in \mathbb{C}$ cu $ \mid z \mid \equal{} 1$, atunci $ A^n \equal{} O_n$.

$ b)$ Ramane adevarata concluzia daca eliminam conditia $ AB \equal{} BA$?[/color]

[color=orange][b]Subiectul 4[/b][/color]

[color=darkblue]Fie functiile $ f,g,h: \mathbb{R} \rightarrow \mathbb{R}$, unde $ f$ este derivabila, $ g$ si $ h$ sunt monotone, iar $ f' \equal{} f \plus{} g \plus{} h$.

Demonstrati ca multimea punctelor de discontinuitate ale functiei $ g$ coincide cu multimea punctelor de discontinuitate ale functiei $ h$.[/color]

In continuare, 2 poze de la ONM, in prima este marea, iar in a doua este Tropaeum Traiani.


function
complex analysis
complex analysis unsolved

Let f(z) is an entire function and |f(z)|=1 on the unit circle |z|=1. Then show that for some n,  f(z)=c*z^n    where  |c|=1

Sketch: first note that our function $ f$ does not have zeros on the unit circle. Let $ a_1,a_2...a_k$(counting multiplicities) are zeros of our function which lie in the unit circle. Consider the product $ B_k (z) \equal{} \prod \frac {z \minus{} a_i}{1 \minus{} \overline{a_i}z}$ and the function $ g(z) \equal{} \frac {f}{B_k (z)}$. It's easy to check that $ |B_k (z)| \equal{} 1$ when $ |z| \equal{} 1$, thus $ |g| \equal{} 1$ when $ |z| \equal{} 1$ and $ g(z)$ is analytic function in the unit disk which does not vanish there. By maximum principle (applying to $ \frac {1}{f}$) one can get that $ g(z) \equal{} c$ where $ |c| \equal{} 1$.The rest should be clear :wink:

See also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=277820]this topic[/url] and note especially the post that fedja made near the end of it. (My post in that topic looks quite a bit like what leshik just said here.)


[quote="2006"]b)$ \frac {a^{2}}{b \plus{} c} \plus{} \frac {b^{2}}{c \plus{} a} \plus{} \frac {c^{2}}{a \plus{} b} \plus{} a \plus{} b \plus{} c\geq\frac {3}{2}\sqrt {3(a^{2} \plus{} b^{2} \plus{} c^{2})}$
         Good luck :D[/quote]
$ \frac {a^{2}}{b \plus{} c} \plus{} \frac {b^{2}}{c \plus{} a} \plus{} \frac {c^{2}}{a \plus{} b} \plus{} a \plus{} b \plus{} c\geq\frac {3}{2}\sqrt {3(a^{2} \plus{} b^{2} \plus{} c^{2})}$
<=>$ \frac {a^{2}}{b \plus{} c} \plus{} \frac {b^{2}}{c \plus{} a} \plus{} \frac {c^{2}}{a \plus{} b} \minus{} \frac{a\plus{}b\plus{}c}{2} \geq\frac {3}{2}\sqrt {3(a^{2} \plus{} b^{2} \plus{} c^{2})}\minus{}\frac{3}{2}.(a\plus{}b\plus{}c)$
<=>$ \sum_{cyc} (a\minus{}b)^2[\frac{1}{a}\plus{}\frac{1}{b}\minus{}\frac{3}{2\sqrt{3(a^2\plus{}b^2\plus{}c^2)}\plus{}a\plus{}b\plus{}c} ] \geq 0$
So $ S_a\equal{}\frac{1}{b}\plus{}\frac{1}{c}\minus{}\frac{3}{2\sqrt{3(a^2\plus{}b^2\plus{}c^2)}\plus{}a\plus{}b\plus{}c}$
$ S_b\equal{}\frac{1}{a}\plus{}\frac{1}{c}\minus{}\frac{3}{2\sqrt{3(a^2\plus{}b^2\plus{}c^2)}\plus{}a\plus{}b\plus{}c}$
$ S_c\equal{}\frac{1}{a}\plus{}\frac{1}{b}\minus{}\frac{3}{2\sqrt{3(a^2\plus{}b^2\plus{}c^2)}\plus{}a\plus{}b\plus{}c}$
Easy to prove that $ S_a ;S_b;S_c>0$
So $ S \geq0$


Solve the equation in R: $ (x\minus{}3)\sqrt{10\minus{}x^2}\equal{}x^2\minus{}x\minus{}12$

hello, after squaring both sides and putting all terms to the left hande-side of the equation i have got $ \minus{}2x^4\plus{}8x^3\plus{}24x^2\minus{}84x\minus{}54\equal{}0$.
Sonnhard.

[quote="ekledes"]to solve this equation

must be ( x + 3 )

try

x= - 3[/quote]
but (x-3) not (x+3)

Oh! I have posted this problem at Algebra forum.

[color=darkblue]Yes I think we must change $ x - 3$ by $ x + 3$.

This problem can be done by many way. Solution of [u][b]ekledes
[/b][/u] is very nice and this is a nice solution too.

Let $ t = \sqrt {10 - x^2}, t\ge 0 \Longrightarrow x^2 = 10 - t^2$.

So we get equation $ t^2 + (x + 3)t + x + 2 = 0$

$ \Longleftrightarrow  [ {\begin {array} {c} t = - 1 \\
t = - x - 2 \end {array}}$

$ \Longleftrightarrow t = - x - 2$, because $ t \ge 0$

$ \Longrightarrow \sqrt {10 - x^2} = - x - 2 \Longrightarrow \begin {cases} x \le - 2 \\
x^2 + 2x - 3 = 0 \end {cases} \Longleftrightarrow x = - 3$

Hence equation have only root is $ x=-3$[/color]

If change  x-3 by  x+3
This problem is easy
I try solve this equation with x+3
I complete my solution but It very long
From :  Vu Duc Tuong  10A CQT

Yes, i think the problem is wrong. If you can solve it, you have to do it in complex field by using Cardano's fomular

[quote="thanhnam2902"][color=darkblue]Yes I think we must change $ x - 3$ by $ x + 3$.

This problem can be done by many way. Solution of [u][b]ekledes
[/b][/u] is very nice and this is a nice solution too.

Let $ t = \sqrt {10 - x^2}, t\ge 0 \Longrightarrow x^2 = 10 - t^2$.

So we get equation $ t^2 + (x + 3)t + x + 2 = 0$

$ \Longleftrightarrow  [ {\begin {array} {c} t = - 1 \\
t = - x - 2 \end {array}}$

$ \Longleftrightarrow t = - x - 2$, because $ t \ge 0$

$ \Longrightarrow \sqrt {10 - x^2} = - x - 2 \Longrightarrow \begin {cases} x \le - 2 \\
x^2 + 2x - 3 = 0 \end {cases} \Longleftrightarrow x = - 3$

Hence equation have only root is $ x = - 3$[/color][/quote]

but -3 doesnot satisfy the original equation :maybe:

i use the gsp slove this question:
x=-0.57


range = greatest value - smallest value = $ 160 \minus{} 112 \equal{} \boxed{48}$.


quadratics
algebra
polynomial
Galois Theory
quadratic formula
Rational Root Theorem
algebra theorems

Alright, i know that there is a quadratic formula.

I knew that there was a way to solve a cubic, but I wasn't sure if it there was actually a formula. Is there? Can anybody prove that it works, if there is one?

Is there any way to solve a quartic? Is there a formula? Can anybody show that it works?

Is there any way to solve a polynomial with a degree higher than four? Is there a formula for any degree above four?

hello, for solving the cubic see here
http://mathworld.wolfram.com/CubicFormula.html
Sonnhard.

Alright, thank you.

Would about the other questions?

hello, and for solving the quartic see here
http://mathworld.wolfram.com/QuarticEquation.html
Sonnhard.

I think it was proven that there is no formula for over quartics...either that or for higher than 5th degree.

Yeah, I just read about it. It was proven by Galois.

Can anybody find the actual proof though?

I can't, but here's the formula for quartics:

http://planetmath.org/encyclopedia/QuarticFormula.html

The theorem you want is the [url=http://en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem#Proof]Abel-Ruffini theorem[/url].  You should be aware that without the prerequisite background in Galois theory the content of the proof won't help you at all.

Nevertheless, there are ways of solving polynomials of degree $ 5$ or higher, even after we have factored into irreducible components; they just don't involve a finite number of radical extractions and field operations.

So how would you solve an equation like this one;

5x^5 - 3x^4 - 20x^3 + 10x^2 + 5x - 2 = 0  ?

[quote="Wickedestjr"]5x^5 - 3x^4 - 20x^3 + 10x^2 + 5x - 2 = 0  ?[/quote]
We should be guessing numbers of the form $ \frac {a}{b}$ where $ a|2$ and $ b|5$ in the general case and check if $ P(\frac {a}{b}) \equal{} 0$, where $ P(x) \equal{} 5x^5 \minus{} 3x^4 \minus{} 20x^3 \plus{} 10x^2 \plus{} 5x \minus{} 2$. Then $ bx \minus{} a|P(x)$. In this case $ (a,b) \equal{} ( \minus{} 2,1)$ works, which means $ 5x^5 \minus{} 3x^4 \minus{} 20x^3 \plus{} 10x^2 \plus{} 5x \minus{} 2 \equal{} (x \minus{} 2) ( 5 x^4 \plus{} 7 x^3 \minus{} 6 x^2 \minus{} 2 x \plus{} 1)$. Of course these won't always work, in that case expanding $ (ax^2 \plus{} bx \plus{} c)(dx^3 \plus{} ex^2 \plus{} fx \plus{} g)$ and comparing coefficients might help. These too might not work, then sometimes I've seen trigonometric substitution helps to get closed form, if all fails, numerical approximation (newton's method) is the best bet I know of...

Knowing that (x-2) is a factor of the equation, how do you factor it?

http://en.wikipedia.org/wiki/Polynomial_long_division
In particular: http://en.wikipedia.org/wiki/Ruffini%27s_rule

Okay, thanks for showing me that. I had been wondering how to divide one polynomial by another for a long time.

A nice book on Galois Theory is (which covers the above stuff) is David Cox's [i][url=http://books.google.com/books?id=3u4RF8SrRooC&printsec=frontcover&dq=trigonometric+solution+cubic&client=firefox-a]Galois Theory[/url][/i]. It's aimed at undergraduates who've taken the equivalent of one class in abstract algebra. Plus, it includes historical perspective.

I'm guessing that that book is too advanced for me.

And here's a very good thing [b]if you work with Z[/b]:

Just divide your equation with x and you get that x is factor of 2.

Alright one more problem;

x^3 + x^2 + x + 1 = 40

Lol, I know that the solution is pretty obvious to you guys, but I want to see all the different possible solutions to equations like this.

Using the rational root theorem, we see that the possible rational roots of the cubic $ x^3\plus{}x^2\plus{}x\minus{}39\equal{}0$ are $ \pm{1}$, $ \pm{3}$, $ \pm{13}$, and $ \pm{39}$.  Checking these values in the equation, we find that $ 3$ is a solution.  Synthetically dividing $ x\minus{}3$ into the polynomial, we get the quadratic $ x^2\plus{}4x\plus{}13\equal{}0$, and applying the quadratic formula we find that $ x\equal{}\minus{}2\pm3i$.  Thus, the solutions are $ 3$ and $ \minus{}2\pm3i$.


does anyone have the program of double link list.please help.


Define the sequence $a_{1},a_{2},a_{3},...$ by $a_{1}=1,a_{2}=b,a_{n+1}=2a_{n+1}-a_{n}+2$where $b$ is a positive integer .Show that $a_{n} a_{n+1}=a_{m}$ for some $m$ is a positive integer

snake_pissken, try finding $a_n$ by $n$, then maybe done :D .

I proved this problem before and I need to have a second solution . Here is my solution:
by using induction , we have $a_{n}=(b-4)(n-1)+n^2$
 then we choose $m=n^2+(b-3)n+4-b$  and It is easy to prove $a_{n} a_{n+1}=a_{m}$
very simple !


I wasn't at nats (so close!!!!!), but how did mah state delaware do? Anyone know (like, delaware as a team, and the individuals)

if you cant do individs, can you at least do daniel wang?

You'll haver to ask someone from Delaware since they only list the top 57 kids and top 24 or so teams and no one from Deleware made either list.

[quote="karatemagic7"]Can someone from the Delaware team and/or the coach herself Please pm me with a way to contact the Delaware coach who was at Nats? Thanks a lot!!![/quote]

Um...I have the teacher's name and the phone number and extension, but no email =(

And I'd ask my friend at de, but he forgot =(

You guys can have posted in the Delaware forum, under North America in the National and Local Communities Forums.

Maybe that's where you can find all the answers to your questions... :wink:

Um...is it me, or does it go from Connecticut to Florida.


function
induction
algebra unsolved
algebra

Let $ A$ denote the set of nonnegative integers. Find all functions $ f: A\rightarrow A$ such that

$ 2f(m^2\plus{}n^2)\equal{}f(m)^2\plus{}f(n)^2$, for all $ m,n\in A$, and

$ f(m^2)\geq f(n^2)$, for all $ m,n\in A$ such that $ m\geq n$.

[quote="radio"]Let $ A$ denote the set of nonnegative integers. Find all functions $ f: A\rightarrow A$ such that

$ 2f(m^2 \plus{} n^2) \equal{} f(m)^2 \plus{} f(n)^2$, for all $ m,n\in A$, and

$ f(m^2)\geq f(n^2)$, for all $ m,n\in A$ such that $ m\geq n$.[/quote]

[hide="My solution"]

1) [u][b]the only two constant solutions are [/b][/u]$ f(m)\equal{}0$ $ \forall m\in\mathbb N_0$ and $ f(m)\equal{}1$ $ \forall m\in\mathbb N_0$
Quite obvious

2) [u][b]$ f(x)$ is non decreasing[/b][/u]
$ a\geq b$ $ \implies$ $ f(a^2)\geq f(b^2)$ $ \implies$ $ f(a)^2\plus{}f(0)^2\equal{}2f(a^2\plus{}0^2)\geq 2f(b^2\plus{}0^2)\equal{}f(b)^2\plus{}f(0)^2$ $ \implies$ $ f(a)\geq f(b)$
Q.E.D

3) [u][b]Non constant solutions are strictly increasing[/b][/u]
Let $ a$ such that $ f(a\plus{}1)\equal{}f(a)$.
Then $ f(x^2\plus{}a^2)\equal{}f(x^2\plus{}(a\plus{}1)^2)$ and so $ f(n)$ is constant, since non decreasing, over $ [a^2\plus{}x^2,(a\plus{}1)^2\plus{}x^2]$ $ \forall x$.
So, for instance $ f(a^2\plus{}4)\equal{}f(a^2\plus{}5)$ and we can consider $ a\geq 3$ (else replace $ a$ with $ a^2\plus{}4$)

Then, if, for some $ k\leq a$, $ f(n)$ is constant over $ [a^2,a^2\plus{}k^2]$, and since we know that $ f(n)$ is constant over $ [a^2\plus{}k^2,(a\plus{}1)^2\plus{}k^2]$, we get that $ f(n)$ is constant over $ [a^2,(a\plus{}1)^2\plus{}k^2]$. And, since $ a^2\plus{}(k\plus{}1)^2\leq (a\plus{}1)^2\plus{}k^2$, we get that $ f(n)$ is constant over $ [a^2,a^2\plus{}(k\plus{}1)^2]$
So, a simple induction shows that $ f(n)$ is constant over $ [a^2,a^2\plus{}(a\plus{}1)^2]$.
Since $ a\geq 3$ $ \implies$ $ (a\plus{}2)^2\leq a^2\plus{}(a\plus{}1)^2$, we get $ f((a\plus{}2)^2)\equal{}f(a^2)$
And so $ f(a\plus{}2)\equal{}f(a)$ (using $ f(a\plus{}2)^2\plus{}f(0)^2\equal{}2f((a\plus{}2)^2\plus{}0^2)$ $ \equal{}2f(a^2\plus{}0^2)\equal{}f(a)^2\plus{}f(0)^2$)

So, $ f(a)\equal{}f(a\plus{}1)$ $ \implies$ $ f(b)\equal{}f(b\plus{}1)$ for some $ b\geq 3$ and then $ f(b)\equal{}f(b\plus{}2)$ and so $ f(n)$ is constant over $ [b,\plus{}\infty)$
Then, let $ m$ such that $ m^2\geq b$ : $ f(m^2\plus{}n^2)\equal{}f(m^2\plus{}0^2)$ and so $ f(n)\equal{}f(0)$ $ \forall n$

So, if $ f(n)$ is not a constant function, $ f(a\plus{}1)>f(a)$ $ \forall a$
Q.E.D

4) [u][b]The unique non constant solution is $ f(n)\equal{}2n$[/b][/u]
$ 2f(0^2\plus{}0^2)\equal{}2f(0)^2$ $ \implies$ $ f(0)\equal{}0$ or $ f(0)\equal{}1$
If $ f(0)\equal{}1$, $ 2f(1^2\plus{}0^2)\equal{}f(0)^2\plus{}f(1)^2$ $ \implies$ $ f(1)\equal{}1\equal{}f(0)$ so $ f(n)$ constant (see point 3 above)
If $ f(0)\equal{}0$, $ 2f(1^2\plus{}0^2)\equal{}f(0)^2\plus{}f(1)^2$ $ \implies$ $ f(1)\equal{}1\equal{}f(0)$ and $ f(n)$ is constant, or $ f(1)\equal{}2$

Let then $ f(0)\equal{}0$ and $ f(1)\equal{}2$ :
$ 2f(m^2\plus{}1^2)\equal{}f(m)^2\plus{}4$ and so $ f(m)$ is even for all $ m$.

Using $ f(2m^2)\equal{}f(m)^2$, it's immediate to show with induction that $ f(u_n)\equal{}2u_n$ for $ u_n\equal{}2^{2^n\minus{}1}$

And, since $ f(n)$ is strictly increasing, even, and $ f(u)\equal{}2u$ for some $ u$ as great as we want, we obviously have $ f(n)\equal{}2n$ $ \forall n$

5) [u][b]synthesis [/b][/u]:
Solutions are :
$ f(n)\equal{}0$
$ f(n)\equal{}1$
$ f(n)\equal{}2n$

[/hide]

for $ m \equal{} n \equal{} 0$  so $ f(0) \equal{} f^2(0)$ ==>  $ f(0) \equal{} 0$ or $ f(0) \equal{} 1$

        1)if $ f(0) \equal{} 1$  ==> $ 2f(1) \equal{} f^2(1) \plus{} f^2(0) \equal{} f^2(1) \plus{} 1$ so $ f(1) \equal{} 1$

 let Un suite such that $ U0 \equal{} 1$  and $ U_{n \plus{} 1} \equal{} 2(U_n)^2$

so $ 2f(U_{n \plus{} 1}) \equal{} 2f^2(U_n)$  BY RECURRENCE YOU CAN PROVE $ f(u_n) \equal{} 1$ for all n

now you must prove that f is increasing

$ f^2(m \plus{} 1) \plus{} f^2(0) \equal{} f((m \plus{} 1)^2) \geq f(m^2) \equal{} f^2(m) \plus{} f^2(0)$ so $ f(m \plus{} 1) \geq f(m)$

for all $ x \in [U_n;U_{n \plus{} 1}] f(U_n)\geq f(x) \geq f(U_{n \plus{} 1}$ so $ f(x) \equal{} 1$==>$ f \equal{} 1$

      2)if $ f(0) \equal{} 0$ so 2f(1)=f²(1) so $ f(1) \equal{} 0or 2$

   2-a) if $ f(1) \equal{} 0$  ===> f(Un)=0 and f is increasing so f=0

   2-b) if$ f(1) \equal{} 2$   is easy to prove $ f(U_{n}) \equal{} 2U_n$ by recurrence

prove that f is strict increassing f²(m+1)=2f((m+1)²) $ \geq$ 2f(m²+1) because f is increassing so so $ f(m) \equal{} > m$

$ f^2(m \plus{} 1) \geq f(m^2) \plus{} f^2(1)$ ==>$ f(m \plus{} 1) > f(m)$

for to include f(n)=2nyou must to prove $ f(m \plus{} 1) \geq f(m) \plus{} 2$     <==>

$ f(m \plus{} 1) > f(m) \plus{} 2$ (because f(m) is integer)

$ f^2(m \plus{} 1) \equal{} 2f((m \plus{} 1))^2 \plus{} 0^2) \equal{} 2(f(m^2 \plus{} 2m \plus{} 1))$  you have f is strictly increasing

so $ 2(f(m^2 \plus{} 2m \plus{} 1))\geq 2(f(m^2 \plus{} 1) \plus{} 2m) \geq f^2(m) \plus{} f^2(1) \plus{} 2f(m) \equal{} (f(m) \plus{} 1)^2 \plus{} 3 > (f(m) \plus{} 1)^2$

so $ f(m \plus{} 1) > f(m) \plus{} 1 so f(m \plus{} 1) \geq f(m) \plus{} 2$ ce que nous permet de conclure que $ f(n) \equal{} 2n$


geometry
circumcircle
trigonometry
incenter
congruent triangles
perpendicular bisector
geometry solved

O is the circumcenter of $\triangle MB'C'$ iff $R\cos A=R\cos (180^\circ -2A)$. Solving this gives $A=60^\circ$

how would you prove that o is the center of the circle circumscribed to the triangle MB'C'????

This is very simple, just prove OM=OC'=OB' when $A=60^\circ$.
btw, the cosine stuff in the previous post actually is unnecessary. It can be replaced by proving congruent triangles instead.

The triangle $\triangle BCO$ is isosceles, because $OB = OC = R$ are radii of the circumcircle $(O)$ of the triangle $\triangle ABC$. Hence, the angles $\measuredangle BCB' \equiv \measuredangle BCO = \measuredangle CBO \equiv \measuredangle CBC'$. But then the right angle triangles $\triangle BCB' \cong \triangle CBC'$ are congruent by ASA, because the side $BC$ is common and the angles $\measuredangle BCC'= 90^o - CBC' = 90^o - \measuredangle BCB' = \measuredangle CBB'$ are also equal. This means that the lines $B'C' \parallel BC$ are parallel. The sides $B'M = C'M$ of the triangle $\triangle MB'C'$ are equal, because they are the corresponding medians of the congruent triangles $\triangle BCB' \cong \triangle CBC'$. The angles $\measuredangle BCB' = \measuredangle B'C'B$ and $\measuredangle CBC' = \measuredangle C'B'C$ are the alternate angles of the transversals $BC'$ and $CB'$ of 2 parallel lines $BC \parallel B'C'$. The angles $\measuredangle MBC' = \measuredangle MC'B$ and $\measuredangle MCB' = \measuredangle MB'C$ are also equal, because the point $M$ is the common circumcenter of the right angle triangles $\triangle BCB', \triangle CBC'$, the segments $BM = B'M = C'M = CM$ are all radii of their common circumcircle and the triangles $\triangle BMC', \triangle CMB'$ are both isosceles. It follows that the lines $BB'$ and $CC'$ are bisectors of the internal angles of the triangle $\triangle MB'C'$ at the vertices $B', C'$, thus intersecting in its incenter $O$. Of course, the perpendicular bisector of the segment $BC$ is also perpendicular to the parallel line $B'C'$ and consequently, it is the bisector of the remaining internal angle of the isosceles triangle $\triangle MB'C'$ at the vertex $M$, also passing through its incenter $O$. As a result, the point $O$ is always the incenter of the isosceles triangle $\triangle MB'C'$, regardles of the measure of the angle  $\measuredangle BAC$. If it is also its circumcenter, then this triangle has to be equilateral, because only an equilateral triangle has the incenter and circumcenter identical. But then the angle $\measuredangle BOC$ formed by its angle bisectors $B'C$ and $C'B$ is equal to $120^o$. Since this angle is the central angle of the major arc $BC$, the angle $\measuredangle BAC$ spanning the major arc$BC$ is equal to

$\measuredangle BAC = \frac{\measuredangle BOC}{2} = \frac{120^o}{2} = 60^o$ [color=white].[/color] or [color=white].[/color] $\frac \pi 3$.

[quote="pinky"]in an acute triangle ABC we denote O the center of the circumscribed circle, R the radius of the circumscribed circle and by M the midpoint of the side BC.  Let us denote B' the projection of B on CO and by C' the projection of C on BO.  prove that O is the center of the circle circumscribed to the triangle MB'C' if and only if the measure of the angle BAC is pi/3[/quote]

Another solution:

Since O is the circumcenter of the triangle ABC, by the central angle theorem, we have < BOC = 2 < BAC, so that < BOC = 2A. On the other hand, again because O is the circumcenter of triangle ABC, we have BO = CO, so that the triangle BOC is isosceles, and for its base angle < OBC we find

$\measuredangle OBC=\frac{180^{\circ}-\measuredangle BOC}{2}=\frac{180^{\circ}-2A}{2}=90^{\circ}-A$.

Thus,

< BOC - < OBC = 2A - (90° - A) = 3A - 90°.

Now, since the point O is the circumcenter of triangle ABC, it lies on the perpendicular bisector of the side BC; in other words, $OM\perp BC$ (since the point M is the midpoint of this side BC). Hence, < BMO = 90°. On the other hand, < BB'O = 90° (since the point B' is the orthogonal projection of the point B on the line CO). Consequently, the points M and B' both lie on the circle with diameter BO. Now, in one circle, two chords are equal if and only if their chordal angles are equal. Hence, in the circle with diameter BO, the chords OB' and OM are equal if and only if their chordal angles < OBB' and < OBM are equal. We write this as an equivalence of two assertions:

{We have OB' = OM}
$\Longleftrightarrow$ {We have < OBB' = < OBM}.

But in the right-angled triangle BB'O, we have < OBB' = 90° - < BOB', and since < BOB' = 180° - < BOC, we obtain < OBB' = 90° - (180° - < BOC) = - 90° + < BOC. On the other hand, < OBM = < OBC. Thus, we can continue our chain of equivalences as follows:

{We have OB' = OM}
$\Longleftrightarrow$ {We have < OBB' = < OBM}
$\Longleftrightarrow$ {We have - 90° + < BOC = < OBC}
$\Longleftrightarrow$ {We have < BOC - < OBC = 90°}
$\Longleftrightarrow$ {We have 3A - 90° = 90°} (since we showed < BOC - < OBC = 3A - 90° in the above)
$\Longleftrightarrow$ {We have 3A = 180°}
$\Longleftrightarrow$ {We have A = 60°}.

So we have seen that OB' = OM holds if and only if A = 60°. Similarly, we find that OC' = OM holds if and only if A = 60°. Thus, if A = 60°, we have OB' = OM and OC' = OM, so that OM = OB' = OC', and it follows that the point O is the circumcenter of triangle MB'C'; conversely, if the point O is the circumcenter of triangle MB'C', we have OB' = OM and OC' = OM, so that A = 60°. Thus, we have proven that the point O is the circumcenter of triangle MB'C' if and only if A = 60°. The problem is solved.

  Darij


1. How many solutions are there to the following equation?
$ (x^2 \plus{} 5x \plus{} 5)^{x^2 \plus{} 5x} \equal{} 1$

2. The number of integer pairs that satisfy the equation $ 3x^2 \minus{} y^2 \equal{} 20 \minus{} 2xy$ is?

So basically I would like to know ways to solve these types of problems, as my solutions are pretty lengthy and definitely not optimal. Thanks!

[quote="Thunder365"]1. How many solutions are there to the following equation?
$ (x^2 \plus{} 5x \plus{} 5)^{x^2 \plus{} 5x} \equal{} 1$
[/quote]

Not sure if this is considered faster, but you now that $ (x^2 \plus{} 5x \plus{} 5) \equal{} 1$, $ {x^2 \plus{} 5x} \equal{} 0$, or $ (x^2 \plus{} 5x \plus{} 5) \equal{} \minus{} 1$ and $ {x^2 \plus{} 5x}$ is even.

Then you just consider every case, and you get: $ x \equal{} 1$, $ x \equal{} 4$, $ x \equal{} 0$, $ x \equal{} \minus{} 5$, $ x\equal{}\minus{}3$, and $ x \equal{} \minus{}2$. I think....  :maybe:

Edit: Fixed.

[quote="$ LaTeX$"][quote="Thunder365"]1. How many solutions are there to the following equation?
$ (x^2 \plus{} 5x \plus{} 5)^{x^2 \plus{} 5x} \equal{} 1$
[/quote]

Not sure if this is considered faster, but you now that $ (x^2 \plus{} 5x \plus{} 5) \equal{} 1$, $ {x^2 \plus{} 5x} \equal{} 0$, or $ (x^2 \plus{} 5x \plus{} 5) \equal{} \minus{} 1$ and $ {x^2 \plus{} 5x}$ is even.

Then you just consider every case, and you get: $ x \equal{} 1$, $ x \equal{} 4$, $ x \equal{} 0$, $ x \equal{} \minus{} 5$, and $ x \equal{} 2$. I think....  :maybe:[/quote]

Hmmm, yep thats the way I did it, which now seems to be pretty fast actually.

But you did something wrong I think. $ (x^2 \plus{} 5x \plus{} 5) \equal{} 1$, $ x \equal{} \minus{} 1, \minus{} 4$. $ {x^2 \plus{} 5x} \equal{} 0$, $ x \equal{} 0, \minus{} 5$. $ (x^2 \plus{} 5x \plus{} 5) \equal{} \minus{} 1$ and $ {x^2 \plus{} 5x}$ is even. $ x \equal{} \minus{} 2, \minus{} 3$.

Thanks! Does anyone have a fast way of doing #2?

[quote="Thunder365"]2. The number of integer pairs that satisfy the equation $ 3x^2 \minus{} y^2 \equal{} 20 \minus{} 2xy$ is?[/quote]

[hide="hint"]Add $ 2xy$ to both sides and factor the LHS.[/hide]

[hide=" Partial Solution"] After doing the hint, we notice that both factors are of the same parity. So they both must have a factor of $ 2$ in them. Now you just have to decide where the $ 5$ goes. (Don't forget they can be negative.) [/hide]

[hide="2"]We have that $ 3x^{2}\plus{}2xy\minus{}y^{2}\equal{}20$. This simplifies to $ (3x\minus{}y)(x\plus{}y)\equal{}20$. The factors of 20 are $ 1,2,4,5,10,20$. Trying everything out with a lot of system of equations you should be able to find all integer solutions :D[/hide]

Notice that when you are solving each system of equations with $ 3x \minus{} y$ and $ x \plus{} y$, if you add both equations, you get $ 4x$. Therefore, just look for the pairs of factors of 20 that add up to a multiple of $ 4$. That should narrow it down for you :) .

Now instead of solving each possible system, just write out all the pairs of factors of 20, and add to see which are divisible by 4:

$ 1 \plus{} 20 \equal{} 21$ nope
$ 2 \plus{} 10 \equal{} 12$ yup
$ 4 \plus{} 5 \equal{} 9$ nope

There! you're done! We only have 2 cases (how we match the equations with the factors):

Case 1:

$ 4x \equal{} 12 \rightarrow x \equal{} 3$

$ 3 \plus{} y \equal{} 10$

$ y \equal{} 7$

Case 2:

Same value for x.

$ 3 \plus{} y \equal{} 2$

$ y \equal{} \minus{} 1$

So the only pairs are $ (3,7)$ and $ (3, \minus{} 1)$.

Note: Solving diophantines this way is very common.

[quote="Thunder365"]Thanks ckck and ihatepie that was the solution I used, but is that the fastest way?[/quote]
Well once you realize those two factors have the same parity, there are only 4 cases to try, so it should only take 1 minute max.

My way narrows it down to one case so it should take less.  :wink:

But yours takes more setup.  :P

And you have two, not one.

Well not much setup, since this is a very common shortcut. So once you learn it, it really doesn't take that long.


topology
advanced fields
advanced fields unsolved

Prove that a closed curve in the plane divides the plane in two regions.  :D

It was considered a difficult theorem, before the systematization of algebraic topology (in this case,  homology groups).   There is a proof by pure point-set topology in [i]The Topology of Plane Sets of Points[/i] by Newman and (coauthor whose name I forgot).  I think the proof in Munkres [i]Topology[/i] also avoids the use of homology, but uses the fundamental group or coverings.

I think there is a proof in the book Graphs on Surfaces , by Mohar, using planar graphs.
I ordered the book, so hope to see it in these days


Find all $ f: N \longrightarrow N \minus{}${$ 1$} such that for all $ n \in N$ :

$ f(n) \plus{} f(n\plus{}1) \equal{} f(n\plus{}2)f(n\plus{}3) \minus{} 168$

[quote="behdad.math.math"]Find all $ f: N \longrightarrow N \minus{}${$ 1$} such that for all $ n \in N$ :

$ f(n) \plus{} f(n \plus{} 1) \equal{} f(n \plus{} 2)f(n \plus{} 3) \minus{} 168$[/quote]

These problems have been posted many many many times (maybe more). ;)

$ f(n)\plus{}f(n\plus{}1)\equal{}f(n\plus{}2)f(n\plus{}3)\minus{}168$
$ f(n\plus{}1)\plus{}f(n\plus{}2)\equal{}f(n\plus{}3)f(n\plus{}4)\minus{}168$

Subtracting implies $ f(n\plus{}2)\minus{}f(n)\equal{}f(n\plus{}3)(f(n\plus{}4)\minus{}f(n\plus{}2))$ and so $ f(n\plus{}2)\minus{}f(n)\equal{}f(n\plus{}3)f(n\plus{}5)...$ $ f(n\plus{}2p\plus{}1)(f(n\plus{}2p\plus{}2)\minus{}f(n\plus{}2p))$

And since all $ f(n\plus{}3),f(n\plus{}5), ...f(n\plus{}2p\plus{}1)>1$, we need $ f(n\plus{}2p\plus{}2)\minus{}f(n\plus{}2p)\equal{}0$ for some $ p$ and so $ f(n\plus{}2)\equal{}f(n)$ $ \forall n$ and the sequence $ \{f(n)\}$ is $ f(1),f(2),f(1),f(2),f(1),f(2),...$

So we need $ f(1)\plus{}f(2)\equal{}f(1)f(2)\minus{}168$ $ \iff$ $ (f(1)\minus{}1)(f(2)\minus{}1)\equal{}169\equal{}13^2$ hence three solutions :

$ 14,14,14,14,14,...$
$ 170,2,170,2,170,2, ...$
$ 2,170,2,170,2,170, ...$


The incircle of $ \triangle ABC$ touches $ BC , CA , AB$ at $ D , E , F$ respectively . $ AD$ meet the incircle at $ Q$ .

The line passing through $ A$ and parallel to $ BC$ , meets $ DF , DE$ at $ P , R$ respectively .

Prove that $ \angle PQR \equal{} \angle EQF$ .

I always appreciate your help .  :blush:

[hide]
$ \angle RAD\equal{}180\minus{}\angle ADC\equal{}180\minus{}\angle QFD\equal{}\angle QED\equal{}180\minus{}\angle QER$ so $ AQER$ is cyclic. Then $ \angle AQR\equal{}\angle AER\equal{}\angle DEC\equal{}\angle EQD$.

Similiarily, $ \angle AQP\equal{}\angle FQD$.

$ \angle PQR\equal{}\angle AQR\plus{}\angle AQP\equal{}\angle EQD\plus{}\angle FQD\equal{}\angle EQF$ as desired.
[/hide]

[quote="dgreenb801"]Why does ADC=QFD?[/quote]

Because they both inscribe the arc QED. (Remember DC is a tangent.)


Consider two complex numbers $x$ and $y$. If $x^2+y^2=7$ and $x^3+y^3=10$, what is the greatest real value $x+y$ can have?

I am kinda stuck on this problem. I found a really nasty approach so any hints on a shorcut would be appreciated! :lol:

$x^2+y^2=7 \Rightarrow (x+y)^2-2xy=7 \Rightarrow  3(x+y)^3-6xy(x+y)=21(x+y)$
$x^3+y^3=10 \Rightarrow (x+y)^3-3xy(x+y)=10 \Rightarrow  2(x+y)^3-6xy(x+y)=20$
By subtraction of these equations you get:
$(x+y)^3=21(x+y)-20 \Rightarrow  a^3-21a+20=0$ with $a=x+y$.
The last equation has the solutions $-5,1,4$, so we have for the searched real value: $x+y=4$

This problem was from 1983(or 1982) AIME. When you post the question, try to put where you got the question from. :)


No.You can find an  error in second line.

$k=2\Rightarrow f(a,b,c)=1$

$k=3\Rightarrow f(a,b,c)=a+b+c$

This introduces some neat identities that people should know:

$\sum_{cyc} a^2b-\sum_{cyc} a^2c = (a-b)(a-c)(b-c)$

and in general

$\sum_{cyc} a^kb-\sum_{cyc} a^kc = (a-b)(a-c)(b-c)\left( \sum_{sym} k-2 \right)$

where by $k-2$ I mean all symmetric sums of degree $k-2$.

The general solution is thus

$k=0,1\Rightarrow f(a,b,c)=0$
$k=2\Rightarrow f(a,b,c)=1$

...

$k=n\Rightarrow f(a,b,c)=\sum_{sym} a^{n-2}+\sum_{sym} a^{n-3}b+\ldots$

What is $\sum_{cyc}$ and $\sum_{sym}$ ?

Maybe, this appears to be a dumb question, but I never learned this stuff, and this problem is hard for k = 2 and k = 3...

It's just shorthand for symmetric sums and cyclic sums, respectively.

For example,

$\sum_{cyc} a^2b = a^2b+b^2c+c^2a$
$\sum_{sym} a^2b = a^2b+a^2c+b^2a+b^2c+c^2a+c^2b$

and so on.

[quote="darij grinberg"][quote="kunny"]Simplify the following equation for $k=0,1,2,3$.

\[\frac{a^k}{(a-b)(a-c)}+\frac{b^k}{(b-c)(b-a)}+\frac{c^k}{(c-a)(c-b)}\][/quote]

Why only for k = 0, 1, 2, 3 ? Actually, if you multiply your expression with - (b - c) (c - a) (a - b), you get

$a^k\left(b-c\right)+b^k\left(c-a\right)+c^k\left(a-b\right)$.

For this one, see http://mathlinks.ro/Forum/viewtopic.php?t=16360 (post #8).

  Darij[/quote]

Because this problem is entrance exam for Nagoya City University 1976.

kunny


Gnosis, I think you are wrong as the inequality is at least true for $2 \ge x \ge \frac{1}{2}$.

[quote="gnosis"][quote="zhaobin"]find the value of $x$,such that for every nonnegitive number $a,b,c$ the following inequality is always ture:
\[ 3(a^xb+b^xc+c^xa) \le (a^x+b^x+c^x)(a+b+c)  \][/quote]

If $x>1$ then by taking $a\geq b\geq c$ in $(1,+\infty)$ then $a^x \geq b^x \geq c^x$ and by Cebasev we get
$3(a^xb+b^xc+c^xa) \geq (a^x+b^x+c^x)(a+b+c)$ with equality only for $a=b=c$, but by choosing initialy $a>b>c$
we get that the last inequality must be strict, thus $x\leq1$.
By doing the same thing for $x<1$ with $a,b,c\in (0,1)$ we get that $x\geq1$.

Hence, $x=1$ is the only solution I could find.[/quote]

You applied Chebyshev with sequences that aren't similarly ordered...

:oops_sign:
I'm sowwy.Inequalities sure aren't easy  :|


What is the value of $ (26^2 \minus{} 24^2 \minus{} 10)^2 \minus{} 10^2$?


induction
pigeonhole principle
Divisibility Theory

There is a large pile of cards. On each card one of the numbers $1$, $2$, $\cdots$, $n$ is written. It is known that the sum of all numbers of all the cards is equal to $k \cdot n!$ for some integer $k$. Prove that it is possible to arrange cards into $k$ stacks so that the sum of numbers written on the cards in each stack is equal to $n!$.

For $ n \equal{} 1$ this is obvious, so assume by induction it is true for $ n\minus{}1$.

We make piles of sum $ mn$, $ m < n$. [list][*]As long as there are at least $ n$ cards $ a_{1},...,a_{n}$, either the set $ \{a_{1},a_{1}\plus{}a_{2},a_{1}\plus{}a_{2}\plus{}a_{3},...\}$ contains a multiple of $ n$, or - by pigeonhole - two have the same remainder $ \mod n$, so their differences form a subset with sum divisible by $ n$. So we have such a set[*]When there are less than $ n$ left, the remaining cards form such a set[/list]
Applying the induction hypothesis with the values for $ m$ as card numbers we have $ k$ piles of $ n(n\minus{}1)! \equal{} n!$ as desired.

Hello. I understand the overall  solution, but I do not understand what do you mean exactly by by aplying the values of m as card numbers.