induction
algebra
polynomial
function
combinatorics proposed
combinatorics

Two numbers are written on each vertex of a convex $100$-gon. Prove that it is possible to remove a number from each vertex so that the remaining numbers on any two adjacent vertices are different.
[i]F. Petrov [/i]

I suppose that numbers written in one vertex are different?

Of course, numbers in each vertes do differ

There is much easer solution:
If there are no two adjacent vertexes such that one number from one vertex differs from both numbers of the other vertex, then some numbers (a,b) are written in every vertex, and because this polygon has even number of vertexes, we can leave simultaniously $a,b,a,b,...a,b$.
Suppose that there aren't such vertexes. Then in one vertex (wlog $A_{1}$) is a number $a_{1}$ and in (wlog) $A_{100}$ numbers $c$ and $d$ are written, and $a_{1}\not =c$ and $a_{1}\not =d$. One number from some vertex can't be equal with both numbers in adjecent vertex, so there is number in $A_{2}$ different from $a_{1}$ (let's say $a_{2}$), and there is number in $A_{3}$ different from $a_{2}$ and so on... there is number from $A_{100}$ different from $a_{99}$, but both numbers from $A_{100}$ are different from $a_{1}$.This is the end, no boring case chasing :)
Bye

Very nice solution with Combinatorial Nullstellensatz. I've become familiar with it only now :oops: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=19496[/url]
I'm just curious isn't this problem obvious. Isn't it enough to take the case with the same $ (a,b)$ in each vertex? I know that couldn't be a proof, the complete one would be like [b]SpongeBob[/b]'s, but if we think logically i think it's clear that the "worst" case will be this. I'm just saying that Combinatorial Nullstellensatz is a very useful theorem, but the problem is just too simple for it.

[quote="lorincz"]Isn't it enough to take the case with the same $ (a,b)$ in each vertex? I know that couldn't be a proof, the complete one would be like [b]SpongeBob[/b]'s, but if we think logically i think it's clear that the "worst" case will be this.[/quote]

No: the list-coloring number is always at least as large as the chromatic number for every graph.  Your argument suggests that the list-coloring number is always equal to the chromatic number, when it fact it is not bounded by any function of the chromatic number (i.e. for any function $ f$, we can find a graph $ G$ such that the list-chromatic number is larger than the value of $ f$ taken at the chromatic number).  So, in fact, using the same colors on each vertex is a best-case, not a worst-case scenario (and indeed, it is how you prove that $ \chi_\ell \geq \chi$ for all graphs).

See http://en.wikipedia.org/wiki/List_coloring for more.

Thank you for these new stuff, i think i've got it. And I've become more convinced that it's the worst-case.
[quote="JBL"]So, in fact, using the same colors on each vertex is a best-case, not a worst-case scenario (and indeed, it is how you prove that $ \chi_\ell \geq \chi$ for all graphs).[/quote]
I don't understand this. I think the worst case(i.e the only case that needs to be analyzed) is when we take the list-coloring number the smallest possible, which is $ 2$ in this problem, i.e when we have the $ (a,b)$ pairs equal. But because the chromatic number is $ 2$ in this problem($ 100$ even), we have these equal, so we don't get a contradiction. Actually if there had been an odd number of vertices, then we would have got an odd cycle(the whole polygon), so we have the chromatic number$ \ge 3$(it will be $ 3$), so the only case where we get the contradiction is when the list-coloring number is $ 2$(all vertices $ (a,b)$), so here the worst case is the only case where the statement isn't true(we can see this easily, not necessary with graph stuff).
But if we just think clear, the problem asks for choosing somehow [u]different[/u] numbers from vertices, so the I say that clearly the worst case is when the $ (a,b)$ pairs at the vertices are the same. I think that it's pretty logical but hard to write. I understand that in a competition you get $ 0$ points for this. But I consider this unfair. I mean if I told this to a man who knows math, I really don't think he would argue with me, and accept it easily. But I'm really curious how many contestants did make arguments like this and NOT get $ 0$ points([b]Pedrov[/b] if you could provide me with an approximate value I would be very happy). Maybe none, because it could just be solved easily like [b]SpongeBob[/b] did or something similar. I did it the same, just I felt like it's unnecessary. I know it isn't that relevant, but I'm considering this as a general dilemma. Should we always try to find a rigorous proof, and are the logical, a-little-evasive, hard-to-write   explanations excluded? If they are, how does this affect our way of thinking(I think that in a negative way).

I don't think so:  for any graph $ G$, if I give to each vertex the lists $ \{1, 2, \ldots, \chi(G)\}$ I can choose one color from each list to get a proper coloring.  (This is what $ \chi(G)$ means, after all.)  However, there exist graphs such that $ \chi_\ell(G) > \chi(G)$; in other words, there exists a graph $ G$ such that we can assign lists of length $ \chi(G)$ to each vertex and it is not possible to choose one color from each list to get a proper coloring.  Thus, the fact that a graph can be properly list-colored with lists of length $ n$ when all lists are the same does [i]not[/i] imply that the list-chromatic number is $ n$ or less.  In fact, the reverse is true: if the list-chromatic number of a graph is $ n$, we know that it is properly list-colorable with identical lists for some value smaller than or equal to $ n$.

Is it not right?

Sorry I'd completely misunderstood what list-coloring is :oops: . I've read it more carefully now. Yes you are right in the general case. But as you said, to prove that $ \chi_\ell\ge\chi$ you use the same lists in each vertex, so it can't be a best-case scenario. I'm saying that if it's true that $ \chi_\ell=\chi$ then the case with the same lists becomes the worst-case scenario. So in this problem, these won't tell us anything :) , because the chromatic number will be equal to the list-chromatic number(this is what we have to prove).I guess you are right and can't just assume that the worst case is this in this graph, yet I'm still not very convinced(we can have "troubles" only for cycles, which is the whole polygon, but if I explain this more clearly it will always resume to the solution by SpongeBob). I'm really happy we are talking about these delicate things, although it might be the wrong topic for it. And I wanted to make a point in my previous post. Maybe this problem isn't the best example for it, but I hope it's clear of what "general dilemma" I'm talking about.

[quote="mszew"]Induction over n for a 2n-gon:

3 dif numbers)
b;(c;b);(a;b);a -> b;c;b;a
b;(a;c);(a;b);a -> b;a;b;a
b;(a;b);(c;b);a -> b;a;b;a
b;(a;b);(a;c);a -> b;a;c;a
b;(c;b);(c;b);a -> b;c;b;a
b;(a;c);(a;c);a -> b;a;c;a

[/quote]
Is there not a mistake here? What about b;(b,c);(c,a);a? Surely the induction doesn't work.
I have another question. If say we have 4 numbers on each vertex, is it possible to choose 2 from each vertex so that adjacent vertices have no numbers in common?

Hey, sorry for the revive. I'm new to Combinatorial Nullstellensatz. Can anybody clarify how it is used in Fedor's solution?
[quote=Fedor Petrov]take a plynomial $ (x_{1}-x_{2})(x_{2}-x_{3})\dots (x_{2n}-x_{1})$. Its coeficient at $ x_{1}x_{2}\dots x_{2n}$ equals $ 2$, so we may take $ x_{i}$ from the 2-elements set $ A_{i}$ such that the value does not vanish.[/quote]
Also, what is the significance of $ x_{1}x_{2}\dots x_{2n}$ and how does it help to solve the problem?

[quote=anantmudgal09][quote=Fedor Petrov]Trivial, CN says that if a coefficient in $\prod x_i$ of the polynomial $f(x_1,\dots,x_n)$, $\deg f=n$, does not vanish, then for any sets of numbers $A_i$, $|A_i|=2$, there exist representatives $c_i\in A_i$ such that $f(c_1,\dots,c_n)\ne 0$.[/quote]

Wow, even the official solution uses the Nullstellensatz? Or is it something elementary?[/quote]

Of course, official solution was elementary, something like SpongeBob's.

[quote=mszew]Induction over n for a 2n-gon:

n=1
Almost trivial for a diameter
There are only three cases only two different number, three different number and four different numbers.
I)   vertex 1 (a,b) ; Vertex 2 (a,b) -> a;b is the solution
II)  (a,b);(a,c) -> a;c
III) (a,b);(c,d) -> a;c

Given 2k+2 vertex, without loss of generality we solve for the 2k first vertex picking
a for the first vertex , and b for the k-th vertex.

There are only five cases for the remaining two vertex
two diferent numbers to six different numbers:
2 dif numbers)
b;(a;b);(a;b);a -> b;a;b;a

3 dif numbers)
b;(c;b);(a;b);a -> b;c;b;a
b;(a;c);(a;b);a -> b;a;b;a
b;(a;b);(c;b);a -> b;a;b;a
b;(a;b);(a;c);a -> b;a;c;a
b;(c;b);(c;b);a -> b;c;b;a
b;(a;c);(a;c);a -> b;a;c;a

4 dif numbers)
b;(c;d);(a;b);a -> b;c;b;a
b;(c;b);(d;b);a -> b;c;b;a
b;(c;b);(a;d);a -> b;c;d;a
b;(a;d);(c;b);a -> b;a;b;a
b;(a;d);(a;c);a -> b;a;c;a
b;(a;b);(c;d);a -> b;a;d;a
b;(c;d);(c;d);a -> b;c;d;a
b;(c;d);(b;c);a -> b;c;b;a
b;(c;d);(a;c);a -> b;d;c;a

5 dif numbers)
b;(a;c);(d;e);a -> b;a;d;a
b;(c;b);(d;e);a -> b;c;d;a
b;(c;d);(a;e);a -> b;c;e;a
b;(c;d);(e;b);a -> b;c;b;a
b;(c;d);(c;e);a -> b;c;e;a

6 dif numbers)
b;(c;d);(e;f);a -> b;c;e;a[/quote]

I have a problem with mszew solution in 3 different number part what about :
b ,, (b,c),, (a,c),,a

I think we you haven't said that    :huh:
Can anyone else explain?

[quote=Fedor Petrov]Well, when inventing this problem I mean the following solution using Combinatorial Nullstellensatz:

take a plynomial $ (x_{1}-x_{2})(x_{2}-x_{3})\dots (x_{2n}-x_{1})$. Its coeficient of $ x_{1}x_{2}\dots x_{2n}$ equals $ 2$, so we may take $ x_{i}$ from the 2-elements set $ A_{i}$ such that the value does not vanish.[/quote]

Very nice work

Combo Null  :rotfl:

Let $f(x_1,x_2,x_3,...,x_{100}) = (x_1-x_2)(x_2-x_3) \cdots (x_{100}-x_1)$. By [b]Combinatorial Nullstellensatz[/b] for any sets of real numbers $|S_1| = |S_2| = \cdots =|S_{100}| = 2$. There exists $s_1 \in S_1, s_2 \in S_2, ..., s_{100} \in S_{100}$ where $f(s_1,s_2,...,s_{100}) \not = 0$, which corresponds to every pair of adjacent numbers being different.

Denote by $S_1,...,S_{100}$ the pairs on the $100$ vertices. \\
Consider the polynomial $P(x_1,..,x_{100}) = (x_1-x_2)(x_2-x_3)....(x_{99}-x_{100})(x_{100}-x_1)$ where the coefficient of $x_1x_2...x_{100}$ is $2$ and consequently non-zero. \\
Then by the Combinatorial Nullstellensatz on sets $S_1,...,S_{100}$ with\\ $t_1 = ... = t_{100} = 1$, we have that there are $s_1 \in S_1, ..., s_{100} \in S_{100}$ such that $P(s_1,...,s_{100}) \neq 0$ and we therefore get that $s_i \neq s_{i+1}$ where we take $s_1 = s_{101}$ for any $i \in \{1,..,100\}$ which consequently satisfies the condition. $\blacksquare$ \\

Assume to the contrary that it is impossible to remove a number from each vertex so that the remaining numbers on any two adjacent vertices differ. Define the polynomial \[ f(x_1, \ldots, x_{100})=(x_1-x_2)(x_2-x_3)\cdots(x_{100}-x_1). \]
Let $S_i$ be the set of numbers on vertex $i$ for $1 \le i \le 100$. Then, $f$ vanishes on $S_1 \times \cdots \times S_{100}$. However, the coefficient of $x_1x_2\cdots x_{100}$ in $f$ is $2$, so the combinatorial Nullstellensatz yields a contradiction, as desired.


A is a complex n x n matrix for n > 1. A' is the complex conjugate of A (each element is the complex conjugate of the corresponding element of A). Show that AA' = 1 iff A = S(S')^-1 for some S

If $ A=S(S')^{-1}$ then $ A'=S'S^{-1}$ and it is easy to see that $ AA'=I$

Now if $ AA'=I$ it is obvious that $ A^{-1}=A'$ so $ A$ is invertible.
For the reverse direction I am having some trouble but am thinking along the lines of;

Express $ A$ as $ A=QDQ^{-1}$ where $ D$ is a diagonal matrix with each entry being an eigenvalue of $ A$, and $ Q$ is formed by making the columns eigenvectors.


geometry
combinatorics solved
combinatorics

Given is a convex polygon with $n\geq 5$ sides. Prove that there exist at most $\displaystyle \frac{n(2n-5)}3$ triangles of area 1 with the vertices among the vertices of the polygon.

There is no hypothesis about the area of the polygon?

Nope, this is the problem. Only one student succesfully solved it. Another one was pretty close. Both can re-take the Junior TSTs next year.

ok, let's go :

Let $A,B$ be two given vertices of the $n$-gon.
If $M$ is another vertex such that $AMB$ has area 1. Then $M$ belongs to one of the two lines parallel to the line $AB$ at distance $ \frac {2} {AB}$ from it. Using the convexity of the $n$-gon, we deduce that there are at most 4 such points $M$, two on each side of the line $AB$.

- There are exactly $n$ choices of $A,B$ such that $AB$ is a side of the $n$-gon. For such a choice, there are at most two points $M$ as above, so there are at most 2 triangles of area 1.

- There are exactly $n$ choices of $A,B$ such that there is exactly one another vertex between $A$ and $B$, in the clockwise sense. For such a choice, there are at most three triangles of area 1.

- There are exactly $ \frac {n(n-1)} {2} - 2n$ choices of $A,B$ such that there are at least 2 vertices of the n-gon in both sides of the line $AB$. For such a choic, there are at most 4 triangles of area 1.

Summing, it leads to a number $T$ of triangles of area 1 such that :
$T \leq 2n + 3n + 4( \frac {n(n-1)} {2} - 2n) = n(2n-5)$.
Moreover, since a triangle has exactly three sides  :P  each triangle of area 1 is counted three times in the above reasoning according to the choice of the height.
Thus the number of triangles of area 1 is at most $ \frac T 3$, and we are done.

Pierre.

I think you have missed a condition concerning the area ;)

No, he didn't. Basically you only need to prove that there exist at most $n(2n - 5)/3$ triangles with vertices among the vertices of the $n$-gon, the areas of which are equal.
I agree that it's a weird way to state the problem, but it is correct.

Arne could you post your solution? I am really curious about this. thanks

I didn't say I solved the problem [I don't have time right now anyway]. I just looked at the problem statement and I figured out that it is actually correct.

I might think of a solution later.


OH MY GOD I LOVE THIS TOPIC

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is someone going to lock this topic, becuase its starting to get out of control.

I need to get my post counts up, so what do I do?

Post pointless posts in this forum like this one to get it up!

But instead of posting random pictures, let's annalyze this geometry diagram

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i said this because i felt like it

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ok now i know that something is wrong with you

GO BANANAS

yea, seriously, you got the spacing messed up! come on now!

And I know what is wrong with this topic: it's still unlocked.

Problem fixed :D


calculus
integration
trigonometry
complex analysis

Prove that for $ n$ natural we have:

$ \int _{0} ^{2\pi} \frac {(1 \plus{} 2\cos \phi)^n \cos n \phi}{3 \plus{} 2cos \phi} d\phi \equal{} \frac {2\pi}{\sqrt {5}}(3 \minus{} \sqrt{5})^n$

$ \int_0^{2\pi} \frac {(1 \plus{} e^{it} \plus{} e^{ \minus{} it})^n\, e^{int}}{(3 \plus{} e^{it} \plus{} e^{ \minus{} it})\, e^{it}}\, i\, e^{it}\, dt \equal{} \oint\limits_{\partial U_1(0)} \frac {\left(1 \plus{} z \plus{} \frac {1}{z}\right)^n\, z^n}{3z \plus{} z^2 \plus{} 1}\, dz$

$ \equal{} 2\pi i\, \left.\text{Res} \frac {(1 \plus{} z \plus{} z^2)^n}{1 \plus{} 3z \plus{} z^2}\right|_{z \equal{} \frac {\sqrt {5} \minus{} 3}{2}} \equal{} 2\pi i\, \frac {\left(1 \plus{} \frac {\sqrt {5} \minus{} 3}{2} \plus{} \frac {5 \minus{} 6\sqrt {5} \plus{} 9}{4}\right)^n}{3 \plus{} 2 \frac {\sqrt {5} \minus{} 3}{2}}$

$ \equal{} 2\pi i\, \frac {(3 \minus{} \sqrt {5})^n}{\sqrt {5}}$


The difference of the roots of $ x^2 \minus{} 7x \minus{} 9 \equal{} 0$ is:

$ \textbf{(A)}\ \plus{} 7 \qquad\textbf{(B)}\ \plus{} \frac {7}{2} \qquad\textbf{(C)}\ \plus{} 9 \qquad\textbf{(D)}\ 2\sqrt {85} \qquad\textbf{(E)}\ \sqrt {85}$

[hide="Click for solution"]
The roots are $ \frac{7 \pm \sqrt{49\plus{}36}}{2}$ so their difference is either $ \frac{7\plus{}\sqrt{85}}{2}\minus{}\frac{7\minus{}\sqrt{85}}{2}\equal{}\sqrt{85}$ or $ \frac{7\minus{}\sqrt{85}}{2}\minus{}\frac{7\plus{}\sqrt{85}}{2}\equal{}\minus{}\sqrt{85}$.  Of the given choices, $ \boxed{\textbf{(E)}}$ is correct.
[/hide]

[hide="Alternate Solution"]
Let $ r$ and $ s$ be the roots of the quadratic, with $ r > s$. We have

$ (r \minus{} s)^2 \equal{} r^2 \plus{} s^2 \minus{} 2rs \equal{} (r \plus{} s)^2 \minus{} 4rs \equal{} 49 \plus{} 36 \equal{} 85\implies r \minus{} s \equal{} \boxed{\sqrt{85} \text{ (E)}}$.
[/hide]

$ ax^2\plus{}bx\plus{}c\equal{}0$
Difference of Quadratic's Roots : $ \frac{\sqrt{D}}{a}$, where $ D\equal{}b^2\minus{}4ac$
 :D


if p then q   & if -q then -p are?

 A. not logically equivalent
 B. a contradiction
 C. logically equivalent
 D. a conditonal statement


algebra
polynomial
superior algebra
superior algebra unsolved

1 - Find all $P \in \mathbb C [X]$ such that $\forall x \in \mathbb R : P(x) \in \mathbb R$
2 - Find all $P \in \mathbb R [X]$ such that $\forall x \in \mathbb Q : P(x) \in \mathbb Q$

for the first : $\mathbb{R}[x]$ and for the second $\mathbb{Q}[x]$

this can be seen with lagrange interpolation

a much nicer problem is this : which rational polynomials map each integer to an integer

For the integer case: see http://www.mathlinks.ro/Forum/viewtopic.php?t=40407

One can ask also the following question: find all polynomials $f(x)$, such that it maps integers into integers and satisfies the
following condition: if $x,y \in \mathbb{Z}$, $m \in \mathbb{N}$ and $x \equiv y ( mod \ m)$, then $f(x) \equiv f(y) (mod \ m)$.

Also another problem I had sent and has been solved, Find all polynomials $p(x),q(x)$ which if $p(x)$ is rational then $q(x)$ is rational for each $x\in\mathbb R$


1. There is at least one two-digit number such that when it is added to the two-digit number having the same digits in reverse order, the sum is a perfect square. Find the sum of all such two-digit numbers.

This one I couldn't even FIND a number that was a perfect square.
I represented the numbers as 11x+11y. No idea how to do this one.

Sorry if any of these have been posted already.  :blush:

Saw the first was was already posted. Removed it, sorry.

[hide="1."][url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=133629[/url][/hide]

[hide="2"]These sums can be expressed in the form $10a+b+10b+a=11a+11b$. For that to be a perfect square, $a+b$ must be $11$. There are 8 such 2 digit #s, which, when added to the one on the opposite end of the list, yields $121$. So, the final answer is $4\times121=\boxed{484}$[/hide]


Anne donates 4$ \%$ of her total income each year to charity. If her
2005 donations totaled $ \$1,\!308$, what was her total 2005 income, in
dollars?

If 4%=1308, and you are looking for 100%, multiply 1308 by 25 and get 32700.

You could also do $ 1308 \div 0.04 \equal{} \boxed{32700}$.

Which is basically the same thing, multiplying by $ x$ is the same as dividing by $ \frac{1}{x}$.


$ (a_n)_{n \equal{} 1}^\infty$ is defined on real numbers with $ a_n \not \equal{} 0$, $ a_na_{n \plus{} 3} \equal{} a_{n \plus{} 2}a_{n \plus{} 5}$ and $ a_1a_2 \plus{} a_3a_4 \plus{} a_5a_6 \equal{} 6$. So  $ a_1a_2 \plus{} a_3a_4 \plus{} \cdots \plus{}a_{41}a_{42} \equal{} ?$

$\textbf{(A)}\ 21  \qquad\textbf{(B)}\ 42  \qquad\textbf{(C)}\ 63  \qquad\textbf{(D)}\ 882  \qquad\textbf{(E)}\ \text{None}$

I don't think you have the final sum correct.

The last term in the desired sum should be $ a_{41}a_{42}$

[hide]By the given conditions, we have

$ a_na_{n\plus{}3}\equal{}a_{n\plus{}2}a_{n\plus{}5}\equal{}a_{n\plus{}4}a_{n\plus{}7}$

$ a_{n\plus{}1}a_{n\plus{}4}\equal{}a_{n\plus{}3}a_{n\plus{}6}$

Multiplying the following:

$ a_na_{n\plus{}3}\equal{}a_{n\plus{}4}a_{n\plus{}7}$

$ a_{n\plus{}1}a_{n\plus{}4}\equal{}a_{n\plus{}3}a_{n\plus{}6}$

we get

$ a_na_{n\plus{}1}\equal{}a_{n\plus{}6}a_{n\plus{}7}$

Hence the desired sum is $ 7(a_1a_2\plus{}a_3a_4\plus{}a_5a_6)\equal{}42$, so the answer is $ B$[/hide]


Let A be a thousand-digit number divisible by 9. Let B denote the sum of the digits of A and let C denote the sum of the digits of B. Determine the sum of the digits of C.

Well this would just be 9.  Finding the sum of the digits is just (mod9) and with three iterations on a number in the thosands, it will reduce to a one digit number.


algebra
polynomial
induction
algebra proposed

Consider the polynomial $P_{1}(x)=x^2-1$ and $P_{k+1}(x)=P(P_{k}(x))$. Find the number of dictinct real roots of the equation
     $P_{k}(x)=0$
where k is a positive integer.
{ It is easy !}

Answer: $k+1$.prove by induction
 The problem is in MYM (Math and Young Magazine) in Vietnam in Feb-2006 .


group theory
abstract algebra
superior algebra
superior algebra unsolved

Given that $ |G|\equal{}4n\plus{}2$, show that $ G$ contains a subgroup of order $ 2n\plus{}1$.

Work so far:
If $ 2n\plus{}1$ is prime, then by Cauchy's Theorem, there is an element in $ G$ that has order $ 2n\plus{}1$, i.e. there is an element $ x\in G$ such that $ x^{2n\plus{}1}\equal{}e$. So the subgroup of order $ 2n\plus{}1$ is the subgroup generated by $ x$.

How do I do this when $ 2n\plus{}1$ is composite, because then I can't use Cauchy's Theorem directly? Any hints?

Use the Cayley-embedding $ G\hookrightarrow S_G$ and the signum to obtain a homomorphism $ G\to\{\pm1\}$. Prove that there is a $ g\in G$ whose image is -1 and you're done: The desired subgroup is the kernel of the homomorphism.


superior algebra
superior algebra theorems

p is an element from z[i*sqrt{3}]. If Z[i*sqrt{3}]/(p) is a field that means that p is prime in Z[i*sqrt{3}] ?? Thanks

You do need to distinguish between prime/irreducible elements and prime ideals. That condition is equivalent to the ideal being prime; the element may be irreducible even if the ideal it generates isn't prime.

An example: $ 2$ is irreducible in this ring. $ (2)$ is not prime, since $ 4\equal{}(1\plus{}\sqrt{\minus{}3})(1\minus{}\sqrt{\minus{}3})$.


geometry inequalities HMMT Harvard college MIT ratio

Every post I make in this thread will be very long (or pretty long). That's because I'll try to introduce new concepts and give examples to it. Also, like I promised, this topic (geometry) was chosen by the poll last week, and it will contain problems that are as far as early AIME.

After completely reading this post, you may ask any questions (currently, I won't post any questions because I don't have time to find the good questions yet)

[b][u]Lesson I: The Pythagorean Theorem, Similar Triangles, and Triangle Inequality[/b][/u]

The three most basic concepts that problem solvers approach in early geometry are the Pythagorean Theorem, Similar Triangles (I just like to capitalize this word because it's pretty important), and Triangle Inequality. Although all three of them are taught in the school, the geometry class only goes in just enough for you to solve problems in SAT. This is actually bit obvious since you only get one year of geometry. That's the reason why most students became frustrated when they are encountering complex geometry problems in math contest because they never saw anything ugly like that.  :)

[color=blue]The Pythagorean Theorem

In $\triangle ABC$ with $\angle B = 90^\circ$, we have:

\[ a^2+b^2 = c^2  \]

where $a$ lies oposite of $\angle A$ and so forth. [/color]

The proof of this theorem can be easily found in google. Therefore, I won't post here (and there are many of them).

I'll start with this similar problem which appeared in HMMT (Harvard-MIT Math Tournament). Let's see this as the first example for Pythagorean Theorem.

Ex. 1: A 25 foot ladder rests against the side of a building with the base of the ladder 7 feet from the building. How far will the base of the ladder be from the building if the top of the ladder slides down 4 feet?

Let's call the height of building when ladder is 7 feet away as $x$ (we're only counting the height as the top of ladder reaches). Then:

$25^2 = 7^2+x^2\\ x = 24$

Note that we used the fact that building is perpendicular to the ground to use Pythagorean Theorem. Now:

$25^2 = (24-4)^2 + y^2\\ 25^2 = 20^2+y^2\\ y = 15$

where $y$ is the distance from the bottom of ladder to the building. We want $y$ so our answer is 15.

You have to understand that all we used was simple algebra and Pythagorean Theorem twice. Next problem was originally from AHSME but I'm going to use Mock AMC C's wording from last year.

Ex. 2: Triangle $PAB$ and square $ABCD$ are in perpendicular planes. Given that $PA = 3, PB = 4, AB = 5$, what is the length of $PD$?

You'll need to draw a picture to follow along but this is only if you need it. I personally think drawing diagram (whether it's a work of art or just sketch), it helps you on solving and understanding the problem. It even helps you assume "false" things in the geometry problem.

Anyway, in $\triangle ABP$, we note that $AB$ is the hypotenuse (side opposite of right angle) and hence:

$PA^2+PB^2 = AB^2\\ 3^2+4^2 = 5^2$

This 3-4-5 triple is called primitive triple because $GCD(3,4,5) = 1$. The above fact, however, isn't much help (I just wrote that to explain what primitive triple means). The key was to see since $AB$ is 5 and side of square, we know $AD = 5$. Also, since $AD \perp AP$, we get another equation (this time, one that we WILL use) using Pythagorean Theorem as:

$5^2+3^2 = PD^2\\ PD = \sqrt{34}$

Ex. 3: This is more complex problem in some way. I'll put problem and algebraic work but no detailed explanation like above two examples.

(UNC Charlotte 2004 Comprehensive) The hypotenuse of an isosceles right triangle has a length of $b$ units. What is the area of the triangle?

$x^2+x^2 = b^2\\ 2x^2 = b^2\\ x = \dfrac{b \sqrt 2}{2}$

Area = $\dfrac{b \sqrt 2}{2} \cdot \dfrac{b \sqrt 2}{2} \dfrac{1}{2} = \dfrac{2b^2}{8} = \dfrac{b^2}{4}$.

[color=green]Similar triangles

If $\triangle ABC$ and $\triangle A'B'C'$ are similar, we have

\[ \dfrac{AB}{A'B'} = \dfrac{BC}{B'C'} = \dfrac{AC}{A'C'} = k  \]

where $k$ is some ratio. [/color]

The ways to prove two triangles are similar appear in the school geometry text. Therefore, I won't try to prove or list those similarities here.

Ex. 4: (MA $\theta$ 1990) In $\triangle ABC$, $\angle ACB$ is a right angle and $DE||CB$ with $D$ on $AC$ and $E$ on $AB$. If $AE = 3$ and $EB = 7$, find the ratio of the area of $\triangle ADE$ to the area of the trapezoid $BCDE$.

Only thing I didn't tell is that area of $\triangle$ is $b \cdot h \cdot \dfrac{1}{2}$. Hopefully, you know what they stand for (base x height x 0.5). Other than this, you know everything to solve this problem. We want:

\[ \dfrac{[\triangle ADE]}{[BCDE]} = \dfrac{[\triangle ADE]}{[\triangle ABC] - [\triangle ADE]}  \]

where $[z]$ denotes the area of something.

The second equation is the trick. We can find $\dfrac{[\triangle ADE]}{[\triangle ABC]}$ by our similarity ratio (make sure you know why they're similar!).

$\dfrac{[\triangle ADE]}{[\triangle ABC]} = \dfrac{AD \cdot ED \cdot \dfrac{1}{2}}{AC \cdot BC \cdot \dfrac{1}{2}} = \dfrac{AD}{AC} \cdot \dfrac{ED}{BC}$

Note from our similarity definition, we know that the above quantity is $\dfrac{3}{10}$ (NOT $\dfrac{3}{7}$!) $\cdot \dfrac{3}{10}$ which is $\dfrac{9}{100}$. So, if we say that $[\triangle ADE] = 9$ then this means that $[\triangle ABC] - [\triangle ADE] = 100-9 = 91$. Hence, $\dfrac{[\triangle ADE]}{[BCDE]} = \dfrac{9}{91}$.

This is all for similarity.

[color=darkblue]Triangle Inequailty

In $\triangle ABC$ where $AC > AB, BC$, we have $AC < AB+BC$. [/color]

Last example of this lesson..

Ex. 5 (COP 2nd Round 1) What are all possible integer lengths for unknown side of a triangle if other two sides have length of 3, and 5? List all the possible values.

$3+5 > x\\ 5+x > 3\\ 3+x > 5$

These 3 inequalities are possible choices. Second one is obviously true since $5 > 3$ so solving first and third inequailty give $2 < x < 8$. Thus, the answers are 3,4,5,6,7.

I think I talked enough for this lesson. I'll let you guys work on problems but I'll post them whenever they're ready.

[b][u]Problem Set for Lesson 1[/b][/u]

This set is incomplete. I'll add more problems if I can but to keep you guys from getting bored, I'll post some problems that I found.

1. (AHSME 1984) Figure $ABCD$ is a trapezoid with $AB || CD$. If $AB = 5, BC = 3 \sqrt{2}, \measuredangle BCD = 60^\circ,$ and $\measuredangle CDA = 60^\circ$, find the length of $CD$.

2. (AHSME 1984) A right triangle $ABC$ with hypotenuse $AB$ has side $AC = 15$ ($BC > AC$). Altitude $CH$ divides $AB$ into segments $AB$ and $HB$, with $HB = 16$. Find the area of $\triangle ABC$.

3. (AHSME 1996) In rectangle $ABCD$, $\angle C$ is trisected by $\overline{CF}$ and $\overline{CE}$, where $E$ is on $\overline{AB}$, $F$ is on $\overline{AD}$, $BE = 6$, and $AF = 2$. Find the integer multiple of 10 that is closest to the area of the rectangle $ABCD$.

4. (AHSME 1995) A pentagon is formed by cutting a triangular corner from a rectangular piece of paper. The five sides of the pentagon have lengths 13,19,20,25, and 31, although this is not necessarily their order around the pentagon. Find the area of the pentagon.

5. (AHSME 1995) The sides of a triangle have lengths 11,15, and $k$, where $k$ is an integer. For how many values of $k$ is the triangle obtuse?

6. (Lehigh 2001) Each side of triangle $ABC$ has length 2. A circle with center at $A$ and radius 1 cuts $AB$ at $M$. A tangent to the circle from $B$ and lying outside the triangle meets the circle at $P$. What is the area of the region bounded by $BP, BM,$ and the minor arc $MP$? Express your answer in form $\dfrac{a \sqrt b - \pi}{c}$.

7. ([i]Mathematical Circles[/i]) The distance from Leningrad to Moscow is 660 kilometers. From Leningrad to the town of Likovo it is 310 kilometers, from Likovo to Klin it is 200 kilometers, and from Klin to Moscow is 150 kilometers. How far is it from Likovo to Moscow?

8. ([i]Challenging Problems in Geometry[/i]) On sides $\overline{AB}$ and $\overline{DC}$ of rectangle $ABCD$ ($AB || CD$ and $AD || BC$), points $F$ and $E$ are chosen so that $AFCE$ is a rhombus. If $AB = 16$ and $BC = 12$, find $EF$.

9. (AHSME 1988) A park is in the shape of a regular hexagon 2 km on a side. Starting at a corner, Alice walks along the perimeter of the park for a distance of 5 km. How many kilometers is she from her starting point?

I am very curious as to how to solve #8.  I have seen it in more than one Math Competition.  However only the final answer is given and not a detailed solution.

[hide="8"]
so, we draw a diagram including the desired rhombus, as well as its diagonals AC and EF, intersecting at G.  $AFCE$ is a rhombus, so its diagonals are perpendicular and bisect each other.  By the 3-4-5 pythagorean triple, $AC=20$, so $GC=10$.  Labeling sides, we know that $EC=AE=FA=CF=x$.  we also have $DE=16-x$.  thus, by the pythagorean theorem, $12^{2}+(16-x)^{2}=x^{2}$.  solving this gives $x=\frac{25}{2}$.  now, we can use the pythagorean theorem to get $GE^{2}=(\frac{25}{2})^{2}-10^{2}=\frac{225}{4}$.  So, $GE=\frac{15}{2}$ and $EF=2GE=\boxed{15}$[/hide]

[hide="#2"]
The altitude seperates the triangle into two similar triangles $\triangle{CBH}$ and $\triangle{ACH}$ such that $AC \sim CB$, $CH \sim HB$, $AH \sim CH$

So

$\frac{AH}{CH}= \frac{CH}{16}\Rightarrow 16AH = CH^{2}$

since the median also divides the triangle into two right angles triangles we can use the fact that
$CH^{2}+AH^{2}= 15^{2}$ and sub in our known value of $CH^{2}$ so the equation becomes:
$AH^{2}+16AH = 225$
$(AH+8)^{2}= 289$
$AH =-8 \pm 17$
$AH = 9$

so AB = 25. We can now find CB
$CB = \sqrt{25^{2}-15^{2}}= 20$

and now that we know CB we can find the area

$A = \frac{1}{2}\cdot 20 \cdot 15 = \boxed{150}$
[/hide]

[hide="#9 with Pythagorean Theorem"]Sketch the hexagon $ABCDEF$. She should have walked from point $A$ in direction $A \rightarrow B$ until she reached the midpoint of $CD$. Let that point be $K$. Find $AC=2\sqrt{3}$ by dividing $\triangle ABC$ into two 30-60-90 triangles. $\angle ACD=90^\circ$, so we have a right triangle $ACK$ with $AC=2\sqrt{3}$ and $CK=1$. So the length of the hypotenuse is $\sqrt{12+1}=\boxed{\sqrt{13}}$ (km).[/hide]

I personally find Law of Cosines faster for this problem

[quote="Glen Mann"]I am a little confused about the first one.
If $\angle{BCD}= \angle{CDA}= 60^{\circ}$ then since $AB\|CD$, then $\angle{DAB}= \angle{CBA}$ should equal $60^{\circ}$ as well, leaving us with two similar equilateral triangles formed from the diagonals of our trapezoid. Lets call the point where the diagonals intersect $I$, so we have a larger equilateral triangle $\triangle{ABI}$ and a smaller one $\triangle{CDI}$

If $BC = 3\sqrt{2}$ then so does $AD$..

Since $AB = 5$, then the sides of the $\triangle{ABI}$ must all be equal to 5 as well.

Heres where I am confused, the sidelength is only part of the diagonal, but $5 > 3\sqrt{2}$, so this makes no sense...

can someone show me where ive gone wrong?[/quote]

I don't understand what you said, but I think $\angle{DAB}= \angle{CBA}$ should equal $120^{\circ}$, not $60^\circ$.

[hide="#1"]Drop two altitudes from $A$ and $B$ and let it $H_{1}$ and $H_{2}$, respectively.

Then $H_{1}H_{2}=5$.

We can combine two triangles $ADH_{1}$ and $BCH_{2}$ - and we have an equilateral triangle.

So $DH_{1}+H_{2}C=3\sqrt{2}$.

Therefore $DC=5+3\sqrt{2}$.[/hide]

[hide="#3"]
Two 30-60-90 triangles are formed, BCE and DCF.

In $\triangle{BCE}$ since BE = 6, BC must be $6\sqrt{3}$

this means that $DF = 6\sqrt{3}-2$, because of this DC must be
$\sqrt{3}(6\sqrt{3}-2) = 18-2\sqrt{3}$

Area = $DC * CB = (18-2\sqrt{3})(6\sqrt{3}) = 108\sqrt{3}-36$

making the greatest interger multiple of 10 that is closest to the area $\boxed{150}$
[/hide]

[hide="#5"]
we have the inequalities
$15+11 > k$
$k+11 > 15$
thus:
$4 < k < 26$

But our triangle must be obtuse, which means the main angle between sides 11 and 15 must be > 90, so by pythagorans
$k > \sqrt{11^{2}+15^{2}}= \sqrt{346}$
so $\sqrt{346}< k < 26$ which means k has integer values from 19-25.

Thus there are $19-25+1 = 7$ values for k
[/hide]

im trying to solve #7, but I dont know much about quadrilaterals.. Do the diagonals of a quadrilateral form 2 sets of similar triangles? I tried searching around but couldnt find any answer.. thanks

[hide="#6"]
right angle triangle ABP is formed. since AP = 1 and AB = 2, then this is a 30-60-90 triangle. Thus BP = $\sqrt{3}$

So the area of the triangle is $\frac{1}{2}*1*\sqrt{3}= \frac{\sqrt{3}}{2}$

The arc MP is equal to 60 degrees and thus the area of the polygon AMP is equal to 1/6th the area of the entire circle, or $\frac{\pi}{6}$

so the area of the region bounded by BP, BM and the arc MP is:
$\frac{\sqrt{3}}{2}-\frac{\pi}{6}\Rightarrow \boxed{\frac{3\sqrt{3}-\pi}{6}}$
[/hide]

oh wow I had no idea that this thread was a 2 year old bump!

guess I'll stop posting, wow lol.

@ 1=2, are the number of letters in each 'day' equal to the integer length of the sides?

[quote="1=2"]Maybe we can make this into a Geometry Marathon? Because SilverFalcon hasn't posted for a long time(I think), and he abandoned this, so I'll start:

PROBLEM 1:

Let Sunday Monday Tuesday Wednesday Thursday Friday Saturday be a heptagon. Find the angle measure of Sunday Wednesday Saturday.[/quote]

 :dry:

please dont revive old topics


superior algebra
superior algebra solved

Hi,

    I know this is probably not that hard, but for some reason I am stuck and would appreciate some assistance. I'm trying to prove the following:

    I'll use + to represent the external direct product.  Let G1 and G2 be groups. Then show G1 + G2 is isomorphic to G2 + G1. I thought a mapping could be phi : (g1,g2) -> (g2,g1). But I'm not sure how to show 1 to 1 and onto. I think to show operation preserving you could do phi : ((g1,g2)*(g2,g1)) = phi : (g1g2, g2g1) = phi(g1g2) phi(g2,g1). Is that right? Thanks.

I think the bijection is obvious.
For preserving the operation, we need to show that
$\varphi((a,b))\varphi((c,d))=\varphi((a,b)(c,d))$.


trigonometry
geometry
circumcircle
geometry proposed

The side segment $BC$ of a triangle $\bigtriangleup ABC$ is fixed and its vertex $A,$ is variable on a constant line ($\ell$), parallel to $BC.$

We define two points $D,$ $E$ on $BC,$ such that $\angle BAD = \angle BCA$ and $\angle CAE = \angle CBA.$

Determine the point $A$ on ($\ell$), in order to be $BC = 3(DE)$ and find the maximum distance between ($\ell$) and $BC,$ of which the problem is valid.

Kostas Vittas.

We have $\triangle ABD\sim \triangle CBA\Rightarrow BD=\frac{AB^{2}}{BC}=\frac{c^{2}}{a}$ similarly we have $CE=\frac{AC^{2}}{BC}\frac{b^{2}}{a}$ thus
$DE=BC-CE-BD=\frac{BC}{3}\Leftrightarrow a-\frac{c^{2}}{a}-\frac{b^{2}}{a}=\frac{a}{3}\Leftrightarrow b^{2}+c^{2}=\frac{2}{3}a^{2}$
Because $A$ is variable on a constant line $(\ell)$, parallel to $BC\Rightarrow S_{ABC}=S=const$ we have
$cot A=\frac{b^{2}+c^{2}-a^{2}}{4S}=-\frac{a^{2}}{12S}=const\Rightarrow \angle A$ thus we can determine$A$.
But I don't understand when $(\ell)$ is fix why can we find the maximum distance between $(\ell)$ and $BC$  :maybe:

[quote="gemath"] ....
But I don't understand when $(\ell)$ is fix why can we find the maximum distance between $(\ell)$ and $BC$  :maybe:[/quote]

I have found that the problem is valid, when $d\leq \frac{\sqrt{3}(BC)}{6}$ $,(1)$ where $d,$ is the distance between ($\ell$) and $BC.$

So, I think that there is not a solution of the problem for distance between ($\ell$), $(BC),$ bigger than $(1).$

Kostas Vittas.

Oh ok I understood.I can prove your result from
$\cot A=-\frac{a^{2}}{12S}=-\frac{a^{2}}{6ad}=-\frac{a}{6d}\Rightarrow d=-\frac{a}{6\cot A}$
But we have $2bc \le b^{2}+c^{2}=\frac{2}{3}a^{2}\Rightarrow \cos A=\frac{b^{2}+c^{2}-a^{2}}{2bc}=\frac{-\frac{a^{2}}{3}}{2bc}\le-\frac{\frac{a^{2}}{3}}{\frac{2}{3}a^{2}}=-\frac{1}{2}\Rightarrow A\ge 120^{0}$
thus we have
$d=-\frac{a}{6\cot A}\le-\frac{a}{6\cot 120}=\frac{\sqrt{3}BC}{6}$ we are done.

Thank you dear [b][size=100]gemath[/size][/b], for your interest.

As another approach, instead of $\angle A = constant,$ we can show that $AM = \frac{\sqrt{3}(BC)}{6},$ where $M,$
is the midpoint of the segment $BC.$

So, we draw the circumcircles $(O_{1}),$ $(O_{2}),$ of the triangles $\bigtriangleup ABE,$ $\bigtriangleup ACD$ respectively and because of $\angle BAD = \angle BCA,$ $\angle CAE = \angle CBA,$ we conclude that the sidelines $AB,$ $AC$ of $\bigtriangleup ABC,$ are tangent lines of $(O_{2}),$ $(O_{1})$ respectively, at point $A.$

Hence, we have that $(BA)^{2}= (BD)\cdot (BC)$ $,(1)$ and $(CA)^{2}= (CE)\cdot(CB)$ $,(2)$

From $(1),$ $(2)$ $\Longrightarrow$ $(AB)^{2}+(AC)^{2}= (BC)\cdot (BD+CE) = (BC)\cdot (BC-DE)$

$\Longrightarrow$ $(AB)^{2}+(AC)^{2}= (BC)^{2}-(BC)\cdot (DE)$ $,(3)$

Apply the first [b][size=100]median’s theorem[/size][/b] in triangle $\bigtriangleup ABC,$ we have that $(AB)^{2}+(AC)^{2}= 2(AM)^{2}+\frac{(BC)^{2}}{2}$ $,(4)$

From $(3),$ $(4)$ $\Longrightarrow$ $2(AM)^{2}+\frac{(BC)^{2}}{2}= (BC)^{2}-(BC)\cdot (DE)$ $,(5)$

From $(5),$ for $BC = 3(DE)$ $\Longrightarrow$ $AM = \frac{\sqrt{3}(BC)}{6}$ $,(6)$

Hence, we can define the point $A,$ as the problem states, as the intersection point of the line ($\ell$) from the circle $(M),$ centered at midpoint $M$ of $BC,$ with radius given by $(6)$ and the solution is completed.

It is clear that when the distance between ($\ell$) and $(BC)$ is bigger than the radius of $(M)$ given by $(6),$ then there is not any solution of the problem.

Kostas Vittas.


trigonometry
calculus
calculus computations

1 pt) A mass is oscillating on the end of a spring. The distance, y , of the mass from its equilibrium point is given by the formula

y=5zcos(6wt)
where y  is in centimeters, t  is time in seconds, and z  and w  are positive constants.

(b) How many oscillations are completed in 1 second?

I need some help here. Thanks.

an oscillation means that the spring completes one cycle. a cycle on the spring, for example, would be from the highest point to the lowest point and back to the highest point on the spring. one cycle is $2\pi$ radians or $360$ degrees.
by the way, the answer to your problem doesn't depend on $z$.


Can anyone give me advice on how to prepare for the KCATM?

Um, once again, this isn't MC-forum appropriate. :wink:   Can you post this in the Kansas forum, cuz it has nothing to do with MathCounts?


Show that $ L \equal{} \{a^nb^n \mid n \geq 0\}$ is a CFL

You will have to define a CFL in the context of your problem here.

Posting homework problems?  What a drag.  This would have been better placed in the CS forum (but don't double-post it there, please).

Since $ L$ is not regular, it's awfully unclear to me how you're supposed to use the closure under regular operations in this question, unless you already have examples of CF languages that you can apply them to.  In fact, of course, it's trivial to show that your language $ L$ is context-free: the grammar $ S \to \epsilon \mid a S b$ works nicely, and it's not hard to write down the relevant PDA, either.


For $a,b,c>0$ and $a+b+c = 1$ prove that \[\frac{a}{bc}+\frac{bc}{a}+\frac{b}{ca}+\frac{ca}{b}+\frac{c}{ab}+\frac{ab}{c}\geq 10\]


Both the numerator and the denominator of the fraction (omitted, should be 5/4) are to be increased by two. What is the positive difference between the new fraction and the original fraction? Express your answer as a common fraction.

simplify the question into 7/6-5/4.
Equalize denominators and get 30/24-28/24=2/24=1/12.


symmetry
geometry
geometric transformation
reflection

3.1.20. (This problem is rather long~ :( )
A projectile is launched from the very center of the floor of a rectangular room which is 40 feet wide with a very high ceiling. The projectile hits the wall at a height exactly 10 feet above the floor, reflects off this wall (obeying the "angle of incidence equals angle of reflection" rule), hits the opposite wall, and reflects again, finally landing back exactly where it was launched, without hitting the ceiling. When the projectile is at its hightest point, how high above the floor is it?

[hide="answer"]
14.8 ft
[/hide]
Could someone check whether my ans is right?
Thanks,

if we assume the projectile motion without any obstical (i.e wall) then the range of the projectile is 20 + 40 + 20 = 80 by symmetry. then we use the formula of equation of trajectory

y = xtanß(1-x/R)                where ß= angle of projection

take y = 10 and x = 20   R = 80
we get ß = arctan(2/3)
now by the formula of range we can find the value of u^2/g and then put it in the formula of height we get the maximum height.

by my solution you get the answer  = 13.333.

so please check your solution and also if some mistake in my solution than inform me.


$ a,b,c,d>0$ and $ abcd\equal{}1$
Prove
if $ a\plus{}b\plus{}c\plus{}d>\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{d}\plus{}\frac{d}{a}$, then

$ a\plus{}b\plus{}c\plus{}d<\frac{b}{a}\plus{}\frac{c}{b}\plus{}\frac{d}{c}\plus{}\frac{a}{d}$

[u]Simple Solution:[/u]:

From Cauchy-Schwarz:
\[ (ab \plus{} bc \plus{} cd \plus{} da)\left(\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{d} \plus{} \frac {d}{a}\right) \geq (a \plus{} b \plus{} c \plus{} d)^2
\]

\[ (ab \plus{} bc \plus{} cd \plus{} da)(a \plus{} b \plus{} c \plus{} d)\geq (a \plus{} b \plus{} c \plus{} d)^2
\]

\[ \boxed{ab \plus{} bc \plus{} cd \plus{} da \geq a \plus{} b \plus{} c \plus{} d}
\]
Again using Cauchy Schwarz
\[ (b^2cd \plus{} c^2da \plus{} d^2ab \plus{} a^2bc)(c^2ab \plus{} d^2bc \plus{} a^2cd \plus{} b^2ad) \geq (bc \plus{} cd \plus{} da \plus{} ab)^2
\]

\[ \left(\frac {b}{a} \plus{} \frac {c}{b} \plus{} \frac {d}{c} \plus{} \frac {a}{d}\right) \left(\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{d} \plus{} \frac {d}{a}\right) \geq (ab \plus{} bc \plus{} cd \plus{} da)^2
\]

\[ \left(\frac {b}{a} \plus{} \frac {c}{b} \plus{} \frac {d}{c} \plus{} \frac {a}{d}\right)(a \plus{} b \plus{} c \plus{} d) \geq (ab \plus{} bc \plus{} cd \plus{} da)^2
\]
And after using the boxed inequality above:
\[ \left(\frac {b}{a} \plus{} \frac {c}{b} \plus{} \frac {d}{c} \plus{} \frac {a}{d}\right) (a \plus{} b \plus{} c \plus{} d) \geq (a \plus{} b \plus{} c \plus{} d)^2
\]

\[ \frac {b}{a} \plus{} \frac {c}{b} \plus{} \frac {d}{c} \plus{} \frac {a}{d} \geq a \plus{} b \plus{} c \plus{} d
\]
Proof is complete.

[quote="shaaam"]

\[ \boxed{ab \plus{} bc \plus{} cd \plus{} da \geq a \plus{} b \plus{} c \plus{} d}
\]
[/quote]
It's wrong. :wink:

Arqady, can you please explain why is that wrong? I mean, i applied Cauchy-Schwarz correctly. We have:

$ (ab \plus{} bc \plus{} cd \plus{} da)\left(\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{d} \plus{} \frac {d}{a}\right) \geq (a \plus{} b \plus{} c \plus{} d)^2$

Since, it is given that:
$ a \plus{} b \plus{} c \plus{} d \geq \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{d} \plus{} \frac {d}{a}$
we have:

$ (ab \plus{} bc \plus{} cd \plus{} da)(a \plus{} b \plus{} c \plus{} d) \geq (ab \plus{} bc \plus{} cd \plus{} da)\left(\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{d} \plus{} \frac {d}{a}\right) \geq (a \plus{} b \plus{} c \plus{} d)^2$

Therefore:
$ ab \plus{} bc \plus{} cd \plus{} da \geq a \plus{} b \plus{} c \plus{} d$

If you have a counterexample, then can you clarify what step i did wrong?

[quote="jbmorgan"][quote="shaaam"]
Again using Cauchy Schwarz
\[ (b^2cd \plus{} c^2da \plus{} d^2ab \plus{} a^2bc)(c^2ab \plus{} d^2bc \plus{} a^2cd \plus{} b^2ad) \geq (bc \plus{} cd \plus{} da \plus{} ab)^2
\]

\[ \left(\frac {b}{a} \plus{} \frac {c}{b} \plus{} \frac {d}{c} \plus{} \frac {a}{d}\right) \left(\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{d} \plus{} \frac {d}{a}\right) \geq (ab \plus{} bc \plus{} cd \plus{} da)^2
\]

\[ \left(\frac {b}{a} \plus{} \frac {c}{b} \plus{} \frac {d}{c} \plus{} \frac {a}{d}\right)(a \plus{} b \plus{} c \plus{} d) \geq (ab \plus{} bc \plus{} cd \plus{} da)^2
\]
[/quote]

How these three lines come?[/quote]

First is cauchy schwarz.
Second is after dividing both parenthesis by $ abcd$. (you can divide since $ abcd \equal{} 1$)
Third is from given condition:

$ a \plus{} b \plus{} c \plus{} d \geq \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{d} \plus{} \frac {d}{a}$

Very nice proof, shaaam!   :lol:  I am sorry.

[quote="shaaam"]

$ (ab \plus{} bc \plus{} cd \plus{} da)\left(\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{d} \plus{} \frac {d}{a}\right) \geq (a \plus{} b \plus{} c \plus{} d)^2$

Since, it is given that:
$ a \plus{} b \plus{} c \plus{} d \geq \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{d} \plus{} \frac {d}{a}$
we have:

$ (ab \plus{} bc \plus{} cd \plus{} da)(a \plus{} b \plus{} c \plus{} d) \geq (ab \plus{} bc \plus{} cd \plus{} da)\left(\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{d} \plus{} \frac {d}{a}\right) \geq (a \plus{} b \plus{} c \plus{} d)^2$

Therefore:
$ ab \plus{} bc \plus{} cd \plus{} da \geq a \plus{} b \plus{} c \plus{} d$  [/quote]

Splendid!

My solution simpler :lol:
We have

$ a\plus{}b\plus{}c\plus{}d>a/b\plus{}b/c\plus{}c/d\plus{}d/a\equal{}a^2/ba\plus{}b^2/bc\plus{}c^2/cd\plus{}d^2/ad\ge\frac{(a\plus{}b\plus{}c\plus{}d)^2}{ab\plus{}bc\plus{}cd\plus{}da}$
$ ab\plus{}bc\plus{}cd\plus{}da>a\plus{}b\plus{}c\plus{}d$(1)
We will prove that
$ 2(a\plus{}b\plus{}c\plus{}d)<a/b\plus{}b/c\plus{}\plus{}c/d\plus{}d/a\plus{}b/a\plus{}c/b\plus{}d/c\plus{}a/d$
$ a/b\plus{}b/c\plus{}\plus{}c/d\plus{}d/a\plus{}b/a\plus{}c/b\plus{}d/c\plus{}a/d\equal{}(ac\plus{}bd)(ab\plus{}bc\plus{}cd\plus{}da)$ (We use that $ abcd\equal{}1$)
$ 2(a\plus{}b\plus{}c\plus{}d)<(ac\plus{}bd)(ab\plus{}bc\plus{}cd\plus{}da)$
$ (ab\plus{}bc\plus{}cd\plus{}da)>a\plus{}b\plus{}c\plus{}d$ and $ ac\plus{}bd\ge2\sqrt{abcd}\equal{}2$
we proved, that $ 2(a\plus{}b\plus{}c\plus{}d)<a/b\plus{}b/c\plus{}\plus{}c/d\plus{}d/a\plus{}b/a\plus{}c/b\plus{}d/c\plus{}a/d$
But $ a\plus{}b\plus{}c\plus{}d>a/b\plus{}b/c\plus{}c/d\plus{}d/a$
We have that $ a\plus{}b\plus{}c\plus{}d<b/a\plus{}c/b\plus{}d/c\plus{}a/d$ :D  :D  :D

Russian_Man, do you have any other problems from Russia TST?


trigonometry
induction
strong induction
Irrational numbers

Show that $\cos 1^{\circ}$ is irrational.

Like the other ones that ask such questions: http://www.problem-solving.be/pen/viewtopic.php?t=213 .

Let $ \cos 1^{\circ}$ be rational. Then so is $ \cos 2^{\circ}$. Again we have $ \cos (n\plus{}1)^{\circ}\plus{}\cos (n\minus{}1)^{\circ}\equal{}2\cos n^{\circ}\cos 1^{\circ}$.
So using strong induction we get that $ \cos n^{\circ}$ is rational for all integers $ n\geq1$.
This is absurd as $ \cos 30^{\circ}$ is not rational.
Hence our assumption is not correct. So the proof follows.

[quote="manjil"]Let $ \cos 1^{\circ}$ be rational. Then so is $ \cos 2^{\circ}$.[/quote]Why?

Edit: because $ \cos(2x)\equal{}2\cos(x)^2\minus{}1$, never mind.


group theory
abstract algebra
advanced fields
advanced fields theorems

Let $ S \equal{} \{(a,b) \in R^2|2a \minus{}3b \equal{} 0\}$. Prove that $ S$ is a subgroup of $ <R^2,\plus{}>$

Checking what part of the definitions do you have problems with¿

I already have a proof. Please check if its correct.
By the subgroup theorem. We need to show that:

1.) The identity $ (0,0)$ is in $ S$ because it satisfies $ 2(0)\minus{}3(0)\equal{}0$
2.) Let $ (a,b)$ and $ (c,d)$ be elements of $ S$. Then, $ (a,b)\plus{}(c,d)\equal{}(a\plus{}c, c\plus{}d)$ is in $ S$ because $ 2(a\plus{}c)\minus{}3(b\plus{}d) \equal{} (2a \minus{} 3b)\plus{}(2c \minus{} 3d)$. Both terms of the sum are elements of $ S$ so closure holds.
3.) For every $ (a,b)$ in $ S$ then $ (\minus{}a,\minus{}b),$ $ 2(\minus{}a)\plus{}3(\minus{}b) \equal{} 0?$. Now, there's the trouble. How do I show that the inverse is in $ S$?

Your part 2 is not complete -- you've stated that the sum of two elements of $ S$ is the sum of two elements of $ S$ -- you haven't shown that this sum is again an element of $ S$.  (Perhaps this was just an oversight in writing it up.)

For number 3: Assume $ (a, b) \in S$, so $ 2a \minus{} 3b \equal{} 0$.  Then $ \minus{} 2a \plus{} 3b \equal{} 0$.  But $ \minus{} 2a \plus{} 3b \equal{} 2( \minus{} a) \minus{} 3( \minus{} b)$.  Therefore ....

Yup, I was in a hurry so  I didn't have the time to type it. There is this another question though.

If $ A$ and $ B$ are subgroups of a group $ G$, how do you show that the intersection of $ A$ and $ B$ is a subgroup of $ G$.

By the way, I am trying to answer the exercises in the books to prepare for my exam this coming August 3. But I don't have anybody to look if my answers are correct.

Thats good that you are studying this early =) And this forum is an awesome place to ask about anything.
Anyways:
Naturally we check for the identity (do you denote it as $ e$ as well?)
Let $ A \equal{}\bigcap A_i$. For your example of subgroups $ A,B$ let $ A\equal{}A_1, B\equal{}A_2$. So I will prove it for an arbitrary collection of subgroups.
Since $ e \in A_i$ for [i]each[/i] $ i$ (since each $ A_i$ is a subgroup) we have that the intersection has the identity.

Now we know $ A$ is non-empty. So we take $ a,b \in A$. Then $ a,b \in A_i$ for each $ i$, and since they are all subgroups already $ ab \in A_i$ for each $ i$, thus $ ab \in A$.

Try to do the inverse property on your own. Naturally take $ g \in A$. It shouldn't be longer than the second property proof.

[quote="Fermat's Little Turtle"]Thats good that you are studying this early =) And this forum is an awesome place to ask about anything.
Anyways:
Naturally we check for the identity (do you denote it as $ e$ as well?)
Let $ A \equal{} \bigcap A_i$. For your example of subgroups $ A,B$ let $ A \equal{} A_1, B \equal{} A_2$. So I will prove it for an arbitrary collection of subgroups.
Since $ e \in A_i$ for [i]each[/i] $ i$ (since each $ A_i$ is a subgroup) we have that the intersection has the identity.

Now we know $ A$ is non-empty. So we take $ a,b \in A$. Then $ a,b \in A_i$ for each $ i$, and since they are all subgroups already $ ab \in A_i$ for each $ i$, thus $ ab \in A$.

Try to do the inverse property on your own. Naturally take $ g \in A$. It shouldn't be longer than the second property proof.[/quote]

Thank you very much! :)

[quote="guile"]By the way, I am trying to answer the exercises in the books to prepare for my exam this coming August 3. But I don't have anybody to look if my answers are correct.[/quote]  In this case, it is probably sensible for you to make posts that contain both the question and your proposed solution, and ask others to confirm whether it is correct.

I agree with him. Most of the time the discussion itself with others is more insightful than solving the actual problem. All of the different perspectives make things easier, and fun xD


function
algebra
polynomial
calculus
derivative
factorial
binomial coefficients

In Wade, he defines the "pth order differential" in order to state taylor's theorem for multivariable functions as follows : $ D^pf(a;h) : \equal{} \sum_{i_1\equal{}1}^{n} ... \sum_{i_p\equal{}1}^{n}\frac{\partial^p}{\partial x_{i_1}...\partial x_{i_p}}(a)h_{i_1}...h_{i_p}$ for $ h\equal{}(h_1,h_2,...,h_n)$

I thought all was fine till I was asked to compute the 3rd degree taylor polynomial of a function of two variables; how do some of the pieces of this sum work out when there is no $ x_{i_3}$ to take a partial of?

Thanks in advance!

The $ i_k$'s are not necessarily distinct. Each $ i_k$ runs from $ 1$ to $ n$ independently of the others. So, actually, the same term repeats quite a few times. Thus, if you bring the equal partial derivatives together, you'll get
\[ D^3f(a,h)\equal{}f_{x_1x_1x_1}(a)h_1^3\plus{}f_{x_2x_2x_2}(a)h_2^3\plus{}3f_{x_1x_1x_2}(a)h_1^2h_2\plus{}3f_{x_1x_2x_2}(a)h_1h_2^2\]
for the function of 2 variables.

My favorite pathway through this notational swamp involves introducing yet more notation: the notion of a multiindex.

A [i]multiindex[/i] $ \alpha\equal{}(\alpha_1,\alpha_2,\dots,\alpha_n)$ is an $ n$-tuple of integers. In most applications, including anything relating to Taylor polynomials, we restrict this to nonnegative integers.

The [i]degree[/i] of a multiindex is $ |\alpha|\equal{}\alpha_1\plus{}\alpha_2\plus{}\cdots\plus{}\alpha_n.$

We say $ \alpha\ge\beta$ iff $ \alpha_i\ge\beta_i$ for all $ 1\le i\le n.$ Another way to say that we're restricting to nonnegative integers is to say that $ \alpha\ge\mathbf{0}.$

The [i]factorial[/i] of a mulitiindex is $ \alpha!\equal{}\alpha_1!\alpha_2!\cdots\alpha_n!.$

That means that the fraction $ \frac{|\alpha|!}{\alpha!}$ is a [i]multinomial coefficient[/i]. If $ n\equal{}2,$ (as it is in the two variable example above) a multinomial coefficient is a binomial coefficient - you do notice the $ 1,3,3,1$ in fedja's post.

If $ \mathbf{x}\equal{}(x_1,x_2,\dots,x_n)\in\mathbb{R}^n,$ then we define a basic monomial as $ \mathbf{x}^{\alpha}\equal{}x_1^{\alpha_1}x_2^{\alpha_2}\cdots x_n^{\alpha_n}.$ Note that the degree of this monomial is $ |\alpha|,$ the degree of $ \alpha.$

Assume now that we have a function that is sufficiently smooth for us to be assured that in all of the mixed partial derivatives we compute, the order does not matter.

Define $ D^{\alpha}f\equal{}\frac{\partial^{|\alpha|}f}{\partial x_1^{\alpha_1}\partial x_2^{\alpha_2}\cdots\partial x_n^{\alpha_n}}.$

In this notation, we can write the $ k$th degree Taylor polynomial of $ f$ as follows:
\[ f(\mathbf{x}\plus{}\mathbf{h})\equal{}\sum_{0\le|\alpha|\le k}\frac1{\alpha!}D^{\alpha}(\mathbf{x})\mathbf{h}^{\alpha}\plus{}o(|\mathbf{h}|^k)\]
We can separate this sum out by degree:
\[ f(\mathbf{x}\plus{}\mathbf{h})\equal{}\sum_{j\equal{}0}^{k}\frac1{j!}\sum_{|\alpha|\equal{}j}\frac{|\alpha|!}{\alpha!}D^{\alpha}(\mathbf{x})\mathbf{h}^{\alpha}\plus{}o(|\mathbf{h}|^k)\]
In particular, we can match this back up with the notation max_tm wants to use by defining
\[ D^jf(\mathbf{x},\mathbf{h})\equal{}\sum_{|\alpha|\equal{}j}\frac{|\alpha|!}{\alpha!}D^{\alpha}(\mathbf{x})\mathbf{h}^{\alpha}\]
Note the multinomial coefficients in this last sum - binomial coefficients in the special case of two variables.

If I can, I would much rather treat a multi-variable function as several single-variable functions and do three separate Taylor expansions, and then combine the results at the end.  For example, if we need a Taylor expansion of something like $ f(x,y,z)\equal{}e^{x\plus{}y\plus{}z}$, we can just treat it as $ f(x,y,z)\equal{}e^{x}e^{y}e^{z}$, do an expansion in each variable, and then multiply at the end.

Of course, this doesn't always work but when it does, it makes the problem less of a notational eyesore.


calculus
integration
logarithms
trigonometry
calculus computations

3.
$\int {\frac{x}{{\cos ^2 x}}} dx$
$=xtanx + Ln|sinx|+C$


Find all functions $ f: \mathbb{R} \rightarrow \mathbb{R}$ for which the equality
\[f(f(x)\minus{}y)\equal{}f(x)\plus{}f(f(y)\minus{}f(\minus{}x))\plus{}x\]
holds for all real $x,y$.

My gut says this solution is flawed, and my gut is usually right. :( But here we go anyways.

$ x \equal{} y \equal{} 0$ gives $ f(f(0)) \equal{} f(0) \plus{} f(0) \equal{} 2f(0)$. $ x \equal{} 0, y \equal{} f(0)$ gives $ f(0) \equal{} f(0) \plus{} f(f(f(0)) \minus{} f(0))$, or $ 0 \equal{} f(f(f(0)) \minus{} f(0)) \equal{} f(2f(0) \minus{} f(0)) \equal{} f(f(0)) \equal{} 2f(0)$. Thus $ f(0) \equal{} 0$.

Put in $ x \equal{} 0$ and conclude that $ f( \minus{} y) \equal{} f(f(y))$. Take $ f$ of both sides to get $ f(f( \minus{} y)) \equal{} f(f(f(y)))$, or $ f( \minus{} ( \minus{} y)) \equal{} f(y) \equal{} f(f(f(y)))$.

Put in $ y \equal{} 0$ and conclude that $ f(f(x)) \equal{} f(x) \plus{} f( \minus{} f( \minus{} x))$. Or, $ f(f(x)) \equal{} f(x) \plus{} f(f(f(f(x))))$. Or, $ f(f(x)) \equal{} f(x) \plus{} f(f(x))$. Thus $ f(x) \equal{} 0$ for all $ x$.

MellowMelon $ f(x) \equiv 0$ does not work...
ThiagoSPinheiro's solution is ok.

You edited in the $ \plus{} x$ after I started my post...


modular arithmetic
complex numbers
number theory unsolved
number theory

Find all pairs of natural numbers $(a,b)$ such that :
$a^{2}+2=3^{b}$

Very nice problem, where does it come from?

[hide="solution"]Anyway, (1,1) and (5,3) are the only solutions.

Let $j^{2}=-2$

3=(1+j)(1-j), where $1+j$ and $1-j$ are primes.

$(a+j)(a-j)=(1+j)^{b}(1-j)^{b}$

Let $x=(a+j,a-j)$

$x|(a+j)-(a-j)=2j|8$ and $1+j|9=8+1$ so $1+j\nmid (a+j,a-j)$ and, similarly, $1-j\nmid (a+j,a-j)$

So $(a+j)=(1+j)^{b}$ or $(a-j)=(1+j)^{b}$

In both cases the imaginary part of $(1+j)^{b}$ is odd, so:

(modulo 4 b is odd, let's say b=2c+1)
\[\sum_{i=0}^{c}{{2c+1}\choose i}(-2)^{c-i}\equiv{{2c+1}\choose c}\equiv 1 \pmod 2 \]
So $2c+1=2^{k}-1$ for some $k\in\mathbb{N}$

k=1,2 give us the only two solutions.

If $k\ge 3$ then:

$3a^{2}+5=3^{2^{k}}-1$ so $3a^{2}\equiv-5\pmod{41}$ but $\left(\frac{3}{41}\right)=-1$ while $\left(\frac{-5}{41}\right)=1$ ABSURD[/hide]

It just came to mind while I was thinking to another problem . your solution is very advanced . I can't understand how a complex number divide a natural number ? could u explain to me ?

Consider the ring $\mathbb{Z}[\sqrt{-2}]$, that is to say the set of the numbers of the form $a+b\sqrt{-2}$ with $a,b\in\mathbb{Z}$

Define $N(a+b\sqrt{-2})=a^{2}+2b^{2}$

There is unique factorization, because for every $x,y\in \mathbb{Z}[\sqrt{-2}]$ there are $q,r\in \mathbb{Z}[\sqrt{-2}]$ such that:

$x=qy+r$ with $N(r)<N(y)$

We say that y|x iff there is $q\in \mathbb{Z}[\sqrt{-2}]$ such that x=qy (of course, since $\mathbb{Z}\subset\mathbb{Z}[\sqrt{-2}]$ sometimes a "complex" number divides a natural number)

For instance, $1+\sqrt{-2}|9$, because $(1+\sqrt{-2})(3-3\sqrt{-2})=9$

Instead we say that a number $p\in \mathbb{Z}[\sqrt{-2}]$ is a prime iff there aren't $x,y\in \mathbb{Z}[\sqrt{-2}]$ s.t.:

$p=xy$ with $N(x)<N(p)$ and $N(y)<N(p)$

For example $3=(1+\sqrt{-2})(1-\sqrt{-2})$ wich are primes, do you see why?

If there is still something you don't understand, ask me.

However, it looks like you have a "less advanced" proof. Can you post it please? (or send me by pm)


geometry
perimeter
rectangle
floor function

The perimeter of the rectangle is an integer, which is  inscribed in a circle with   radius 10 . Find the least perimeter of the rectangle.

The rectangle with the least perimeter corresponds to the rectangle with the largest area, which is that of a square. It is easy to see then that the perimeter of the square is $ 4 \cdot 10 \sqrt {2} \equal{} \boxed{40 \sqrt {2}}$.

EDIT: Answer is integer, so we have
\[ \lfloor 40\sqrt {2} \rfloor \equal{} \boxed{56}.
\]
EDIT: Never mind that either. $ 40$ gives a degenerate line, so we have $ \boxed{41}$.

(Thanks azjps for catching all my mistakes.)

123456789 tripped up relatively badly.

So be mindful (everyone) and it is more accurate to stop
posting the subjective "it is easy to see then..." phrase
or some other variation of it.

Just post "such and such follows." or "this leads to..."
or the like.

You'll notice I never use coming-off-as-arrogant, subjective
phrases such as these in spelling out mathematics solutions or
proofs, and I have been tutoring mathematics for 30 years.

The posting of the mistakes of the work is welcome and
expected from users as do-overs, but the arrogance in
choice of words is avoidable.


Question1 :
a) Show that if A is nonsingular and diagonalizable then A^-1 is diagonalizable
b) Show that if A is diagonalizable then A^T is diagonalizable

Question 2
Show that if A is similar to B and B similar to C, then A is similar to C.

Just use the definitions and properties of basic matrix operations. For example, if $ A\equal{}C^{\minus{}1}DC$ with $ D$ diagonal, then $ A^{\minus{}1}\equal{}C^{\minus{}1}D^{\minus{}1}C$, where $ D^{\minus{}1}$ is diagonal as well.


How come typing in $ \text{\LaTeX}$ makes you automatically bypass the post length rule? See here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=262307&start=160 posts 172-175

Also, bolding, underlining, and italicizing does the same thing.

Bold/underline/italics are so because the BBCode for those take up as much space as you need to take up for it to be long enough.

[quote="eggylv999"][b]hi[/b][/quote]

9 letter post that makes the requirement

I'm going to think about this, and how to change the implementation.

[quote="Brut3Forc3"][quote="r15s11z55y89w21"][quote="dnkywin"]actually, you can do this to have a blank post:
[code][b][/b][b][/b][b][/b][b][/b][/code]
I think BBCcode/Latex should not be counted in the word count.[/quote]

I believe it should still be counted in the word count. That way, there won't be those messy messages where people attached "This message is too small...[blah blah blah]..." You can just have clean-looking posts without that redundant addition.[/quote]

You could always say "10chars."[/quote]

Wait, what's 10chars? I've never heard of it. Please explain.

[quote="Bugi"]char=character. Have you ever used Pascal?[/quote]

No. And I can certainly, albeit my highly diminutive brain, know that "char" is short for character. I just don't know what "10char" refers to. Apparently it's a program?

He means you could say "10chars" instead of "This message is too long..." or empty BBCode tags.

Just type "10chars"...?

Well, in that case, what's the difference?

It still makes the post look ugly.


A motorist who drives through the same intersection each day observes that a traffic light shows red as often as green, but shows yellow only one-sixth as often as it shows red and one-sixth as often as it shows green.  Based on these observations, what is the probability that a motorist approaching the traffic light encounters a yellow light?  Express your answer as a common fraction in reduced form.

[hide]x = the probability that the light will be red or green

x+x+x/6 = 1

13x/6 = 1

13x = 6

x = 6/13

Since 1/13 is 1/6 of 6/16, the probability that the light will be yellow is 1/13[/hide]

[hide] Call the probability that a red light will appear p. Since the probability that a green light will appear is the same as red, the prob. is p. Since yellow comes up one sixth of the time, the probability of yellow is p/6

Since all of the probabilities must add up to one we get the equation: p+p/6+p=1... so 13p/6 =1... so p=6/13. SInce the probability of yellow appearing is p/6, and p=6/13, the probability of a yellow light is 1/13[/hide]

[color=white]More than 1/13, as the motorist can see the yellow light before the instant he goes under it.[/color]


Another problem:

1. Let x,y,z are positive numbers, prove that:
$ \frac {{2x}}{{x^6 \plus{} y^4 }} \plus{} \frac {{2y}}{{y^6 \plus{} z^4 }} \plus{} \frac {{2z}}{{z^6 \plus{} x^4 }} \le \frac {1}{{x^4 }} \plus{} \frac {1}{{y^4 }} \plus{} \frac {1}{{z^4 }}$

2. Let x,y,z,t are positive numbers such that $ xyzt \equal{} 1$, prove that:
$ \frac {1}{{x^3 (yz \plus{} zt \plus{} ty)}} \plus{} \frac {1}{{y^3 (xz \plus{} zt \plus{} tx)}} \plus{} \frac {1}{{z^3 (xy \plus{} tx \plus{} yt)}} \plus{} \frac {1}{{t^3 (yz \plus{} zx \plus{} xy)}} \ge \frac {4}{3}$

3. $ x,y,z$ are real numbers such that $ xy \plus{} yz \plus{} zx \equal{} 1$. Find the minximum value of $ S \equal{} \sqrt {\frac {{(1 \plus{} x^2 )(1 \plus{} y^2 )}}{{1 \plus{} z^2 }}} \plus{} \sqrt {\frac {{(1 \plus{} y^2 )(1 \plus{} z^2 )}}{{1 \plus{} x^2 }}} \plus{} \sqrt {\frac {{(1 \plus{} z^2 )(1 \plus{} x^2 )}}{{1 \plus{} y^2 }}}$

[quote="L_Euler"]Another problem:

1. Let x,y,z are positive numbers, prove that:
$ \frac {{2x}}{{x^6 + y^4 }} + \frac {{2y}}{{y^6 + z^4 }} + \frac {{2z}}{{z^6 + x^4 }} \le \frac {1}{{x^4 }} + \frac {1}{{y^4 }} + \frac {1}{{z^4 }}$

2. Let x,y,z,t are positive numbers such that $ xyzt = 1$, prove that:
$ \frac {1}{{x^3 (yz + zt + ty)}} + \frac {1}{{y^3 (xz + zt + tx)}} + \frac {1}{{z^3 (xy + tx + yt)}} + \frac {1}{{t^3 (yz + zx + xy)}} \ge \frac {4}{3}$
[/quote]
I only solved problem 1 and 2 :(

1.
$ \sum\limits_{cyc} {\left( {\frac {1}{{x^4 }} + \frac {1}{{y^4 }}} \right)} = \sum\limits_{cyc} {\left( {\frac {{x^2 }}{{x^6 }} + \frac {1}{{y^4 }}} \right)} \ge \limits^{Svac} \sum\limits_{cyc} {\left( {\frac {{(x + 1)^2 }}{{x^6 + y^4 }}} \right)} \ge \sum\limits_{cyc} {\frac {{2x}}{{x^6 + y^4 }}} \to QED$

2.
$ \sum {\frac {1}{{x^3 (yz + zt + ty)}}} = \sum {\frac {1}{{x^2 \left( {\frac {1}{y} + \frac {1}{z} + \frac {1}{t}} \right)}}} \ge \frac {1}{3}.\left( {\frac {1}{x} + \frac {1}{y} + \frac {1}{z} + \frac {1}{t}} \right) \ge \frac {4}{3}.\sqrt [4]{{\frac {1}{{xyzt}}}} = \frac {4}{3}$

Anyone could solve P3 :( Thanks so much!

if $ a,b,c$ are positive the minimum value is $ 1$,
just factorize the expression inside root by writing $ 1$ as givven to get $ S \equal{} 2(x \plus{} y \plus{} z) \ge 1$


algorithm
number theory
greatest common divisor
Euclidean algorithm
real analysis
real analysis theorems

For every sequence of nonnegative integers $a_{0}, a_{1}, a_{2}, ...$, define $x_{n}= a_{0}+1/(a_{1}+1/(...+1/(a_{n-1}+1/a_{n})))$. Prove that any positive real number is the limit of a sequence of continued fractions of the same form as $\{x_{n}\}$.

Any ideas?

For a give real number $x,$ the $a_{i}$ can be constructed using the Euclidean algorithm. Start the algorithm with $x$ and $1.$

Are you referring to the Euclidean algorithm for determining the gcd of two integers? If so, then how can we use the algorithm on $1$ and $x$, a real number?

Yes, that's exactly what I mean! If $x$ is rational, the algorithm terminates; if $x$ is irrational, it does not terminate. What you want is the sequence of quotients:

$\frac{\sqrt{5}+1}2=1+\frac1{1+\frac1{1+\frac1{1+\cdots}}}$

$\pi=3+\frac1{7+\frac1{15+\frac1{1+\frac1{292+\frac1{1+\cdots}}}}}$


function
number theory proposed
number theory

Suppose $ f$ is a function on the positive integers which takes integer values with the following properties:
(a) $ f(2) \equal{} 2$
(b) (multiplicative) $ f(mn) \equal{} f(m)f(n)$ for all positive integers $ m$ and $ n$
(c) (increasing) $ f(m) > f(n)$ if $ m > n$
Find $ f(1983)$.

Generalized problem : Find all functions $ f$ on the positive integers which takes [b]real values[/b] and satisfying (b) and (c).

I think the condition (a) is necessary...
anyway a function $ f$ satisfying (a),(b),(c) is the single function $ f(n)\equal{}n$...

That's a well-known result from Erdös that if $ f: \mathbb{N^*} \rightarrow \mathbb{R}$ is multiplicative and non-decreasing then $ f$ is constant or of the form $ f(n) \equal{} n^u$ for some positive real $ u$.

Pierre.

[quote="redkimchi"][Australian 1984 (or 1983) p. 7] Suppose $ f$ is a function on the positive integers which takes integer values with the following properties:
[list](a) $ f(2) \equal{} 2$
(b) (multiplicative) $ f(mn) \equal{} f(m)f(n)$ for all positive integers $ m$ and $ n$
(c) (increasing) $ f(m) > f(n)$ if $ m > n$  [/list]
Find $ f(1983)$.

Generalized problem : Find all functions $ f$ on the positive integers which takes [b]real values[/b] and satisfying (b) and (c).[/quote]

First, we have $ f(n)\neq 0$ $ \forall n>0$ (else we would have $ f(2n)\equal{}f(n)\equal{}0$ and so contradiction with (c)).
Then, we have $ f(1)\equal{}1$ since $ f(1)\equal{}f(1\times 1)\equal{}f(1)^2$ and $ f(1)\neq 0$ (see above).
Then, we have $ f(n)>1$ $ \forall n>1$. It's true since we have then $ f(n)>f(1)\equal{}1$ $ \forall n>1$ (applying (c)).

Let then $ u,v$, such that $ u\neq v$, $ u>1$, $ v>1$.
Since $ f(u)$ and $ f(v)$ are positive, $ \ln(f(u))$ and $ \ln(f(v))$are defined. Assume then $ \frac{\ln(f(u))}{\ln(u)}\neq\frac{\ln(f(v))}{\ln(v)}$.
Then we have $ \frac{\ln(f(u))}{\ln(f(v))}\neq\frac{\ln(u)}{\ln(v)}$ WLOG say $ 0<\frac{\ln(f(u))}{\ln(f(v))}<\frac{\ln(u)}{\ln(v)}$.
Then, it exists two positive integer $ p,q$ such that $ 0<\frac{\ln(f(u))}{\ln(f(v))}<\frac{p}{q}<\frac{\ln(u)}{\ln(v)}$
$ \frac{\ln(f(u))}{\ln(f(v))}<\frac{p}{q}$  implies $ f(u^q)<f(v^p)$
$ \frac{p}{q}<\frac{\ln(u)}{\ln(v)}$ implies $ v^p<u^q$
And so we have a contradiction with  (c).

So $ \frac{\ln(f(u))}{\ln(u)}\equal{}\frac{\ln(f(v))}{\ln(v)}\equal{}c$ for some real $ c>0$

So $ \boxed{f(n)\equal{}n^c}$ $ \forall n>0$

If we add the condition (a), we have $ c\equal{}1$ and $ f(n)\equal{}n$


[b]1. Pick as many numbers as you can out of 1, 2, 3, ..., 50 so that none of the numbers picked has the half of it among the numbers picked.

2. Pete left his house at 3:30 pm to go to soccer practice. He checked his analog watch and realized that when he arrived home that day from school the minute hand of his watch was also in a "vertical" position, but it had a 15 degree less angle with the hour hand than now at 3:30 pm. What time could Pete get home from school that day?

3.

In a summer camp Pete played a ping-pong match with 25 other children, and nobody in the camp played against more children than him. We know also that if two children played ping-pong with the same number of children then those two children did not have a mutual opponent. Prove that one of Pete's opponents played ping-pong against Pete only.

[/b]

[hide="Problem 1"]

Just pick all the odd numbers  : 25 numbers

EDIT: typo

[/hide]

Sorry, No Soup For You, but their is still a component of the problem you are missing......

[hide="1"]Pick all the odd numbers (25) and pick the numbers divisible by 4.  So we get 25+12=37.[/hide]

I think I got it this time:

[hide="Number 1"]Odd numbers : $\frac {50}{2} = 25$
Numbers divisible by 4 (therefore, multiples of even numbers) : $[\frac {50}{4}] = 12$
$25+12=37$[/hide]

Lets try putting in the multiples of 4: 4, 8, 12, 16, 20, 24.....ERR!!!  4 is half of 8....... 8 is half of 16...... and 12 is half of 24......
 :wacko:

Put in the numbers that are :equiv: 4 (mod 8), which is 4-44->6 of them. 6+25=31.

[quote="NoSoupForYou"][quote="eryaman"]Wow I feel stupid... :blush: [hide]So we have the 25 odd numberes, then we can pick 4,16,24,40 for a totoal of $29$.[/hide][/quote]

Sry, but you're wrong again. Bob123 is right: get the numbers that are congruent to $4$ mod $8$. There's six of 'em: $4,12,20,28,36,44$

$25+6=31$[/quote]

No, I was wrong, I just realized that. You can have 32 and 48 in there as well, making a total of 33.

[hide]For number 1, I think the answer is 31.  THat's all the odd numbers plus all the multiples of four that are multiplied against odd numbers.  That's 4*1 to 4*12.  There's six odd number pairs in there.  25+6=31.
I think 2 is 2:00 P.M.  3:30 forms a 75 degree angle.  We want 75-15=60 degree angle.  That's the degree difference between the minute hand on the twelve and the hour hand on the 10 or 2.  I don't think school gets out at ten.
OK, Pete challenged 25 people.  We'll call Pete and the group he challenged Group X.  Now, nobody in group X could have challenged the same amount of people because if they did, they wouldn't have a common opponent which they obviously do, Pete.  Therefore, it is still possible to have someone in Group X to challenge 25 people if they didn't challenge anyone in Group X except for Pete.  But you can't have 3 people in Group X with that quality, because then, there's a common foe between the two people other than Pete, namely, Pete.  For any other number of challenges, there can only be one person with that many in Group X.  Therefore, we go from 25 challenges all the way to someone with only 1 challenge.  That person only challenged Pete.[/hide]

I got 33 as well.

Here are the list:

[hide="Key"]
1
3
5
7
9
11
13
15
17
19
21
23
25
27
29
31
33
35
37
39
41
43
45
47
49
4
12
16
20
28
36
44
48

25 odd is right. Unless I'm missing anything, I'm pretty sure that's right.[/hide]

I understand it now:

A number can be in the set:

If it a multiple of 1,
But not if it is a multiple of 2,
Unless it is a multiple of 4,
But not if it is a multiple of 8,
Unless it is a multiple of 16,
But not if it is a multiple of 32.

Therefore, there are 8 even numbers in the set: 4, 12, 16, 20, 28, 36, 44, 48.

Therefore, the answer is $33$.

For those who are interested in how to solve the problem.....[hide] First u pick the 25 odd numbers, because each odd number can be expressed in the form on 2x+1, it has no halve.  But were not done yet.  We can still work with the even numbers.  Sinch some even numbers like 2 and 6 have halves that are odd numbers, we can only play with the even numbers that are multiples of 4.  There are 12 multiples of 4, which means we can take 4*( 1, 3, 5, 7, 9, 11) all the ones that produce odd numbers when divided by four, which is a total of 6. We are lest over with the even ones 4* ( 2, 4, 6, 8, 10, 12), the numbers that produce even numbers when divided by four, we keep the multiples of 4 which are 4, 8 and 12 and can oly keep 2 of them (i. e 4 * 4 which is 16 and 4 * 12 which is 48. Thus, we have 25+6+2=33. Hope that wasn't too confusing! :P  [/hide]

Actually, I found another way of thinking about it.  25 odd numbers+6 odd multipliers of 4+2 odd multipliers of 4^2.
Yeah, you're right, the answer is 33.

nice solution soup, i approached the problem by classifying numbers into odds and evens, i didnt however use the solution you did classifying by powers of 2. :P

Number 2:[hide]The angle the hands make at 3:30 is 75 degrees; from 3 to 6 is 90 degrees, but at 3:30, the hour hand is half way to 4, so 5/6*90 is 75 degrees.

This angle, 75 degrees, minus 15 degrees, is 60 degrees. Combining this with the fact that the minute hand is at 12 or 6 (the time is at the hour or + half), your get possibilities as 10:00 am and 2:00 pm. If the minute hand is at 6, then the angle formed cannot be 60 degrees.[/hide]

Correct! :P  But I'd would like to think he returned at 10pm otherwise he would have satyed up quite late.....

[hide="1"]I also get 33. All of the odd numbers as well as 4, 12, 16, 20, 28, 36, 44, 48. [/hide]

[quote="bob123"][quote="G-UNIT"][hide]hmmmm.... thats quite possible, but i obtained 25 odd numbers, and 8 multiples of 4: 4, 12, 20, 28, 36, 44 and 16 and 48.. which altogether is 8.(I can explain how i found it if im right) Thus, 25+8=33......mabye im wrong.....[/hide][/quote]

I think you're right... Having 16 or 32 is the same thing. You can have either but not both. This also applies for any of the other multiples of 4 that are double each other, like 20 and 40. You can have either, but not both. So what you do is you find all the pairs. (4,8) (12,24) (16,32) (20,40) (24,48) (28) (36) (44). For the ones with only one number in it, it is because the double of that number is not between 1 and 50. So now you have 25 odd numbers and also any 8 out of those 13 numbers. 25+8=33.[/quote]

I got 33 too, that has to be right...

[hide="problem1"]1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50

I did it a different way than just listing them, but I'm just showing you why it's 37.  You are all going at it the wrong way--odds don't work in this problem.[/hide]

mathgeek ur solution in wrong... u included numbers 1, 2 ,4 ,8 5, 10.   According to the problem no number can be half of any other but, 1 is half of 2, 4 is half of 8, 5 is half of 10, etc.


What is the value of the product $ \sqrt{8} \cdot \sqrt{18}$?

Simplify the radicals to 2 SQRT2 and 3 SQRT2. Then multiply them together to get 6*SQRT2*SQRT2=6*2=12

or you can just multiply:
\begin{align*}
\sqrt{8}\cdot \sqrt{18} &= \sqrt{8\cdot 18} \\
&= \sqrt{144} \\
&= \boxed{12} \\
\end{align*}

multiply the radicands inside the radical and them take the square root of that


Find all pairs $ (n,k)$ of non-negative integers such that $ \binom{n}{k}\equal{}nk$.

[hide="?"]
$ \binom{n}{1}$ always works, so $ (n,1)$ is a possibility

And it's possible to go from the expression to $ \binom{n}{k} \minus{} k^2 \equal{} \binom{n \minus{} 1}{k}$ but I don't know how to continue.

I smell induction.
[/hide]

I know of one other pair (one not found by CircleSquared), but I can't find any more.

I think the equation can be turned into a $ k$-degree polynomial for any given $ k$, and then solved for $ n$. Can it be shown which $ k$ lead to integral $ n\geq k$?


[quote="xzlbq"]$ a,b,c > 0$,

find:

$ a^2/b \plus{} b^2/c \plus{} c^2/a \geq (3^t\frac {a^t \plus{} b^t \plus{} c^t}{3})^{1/t}.$
BQ[/quote]

$ t\equal{}\frac{13}{2}$  the inequality  holds.

$ t\equal{}7$ the inequality doesn't hold.


geometry
geometric transformation
reflection
trigonometry
search
projective geometry
geometry proposed

Let $ \triangle ABC$ be a triangle and $ \zeta$ its incircle,that tangent to $ BC$ at point $ D$.Denote $ G\equal{}AD\cap\zeta$,$ K\equal{}\zeta\cap BG$ and $ L\equal{}\zeta\cap CG$.Let $ l$ be the tangent line to $ \zeta$ at point $ G$.If $ M\equal{}l\cap BC$ then prove that points $ K,L,M$ are colliner.

[quote="vittasko"]It is also true ( I think ) for any point $ G,$ between $ A,$ and $ D.$

I will search....

kostas Vittas.[/quote]
Maybe,i'll try to prove it too.BTW do you solve the first one? :wink:

[quote="April"]I guess that you are trying to find another solution for problem O53, which was proposed by Cosmin Pohoata, in Mathematical Reflections 3 - 2007  :wink:. Your problem is not new, and I have a very nice solution for it. But I will post it later, because I want to see other proofs by some other Mathlinkers.[/quote]
You are right.I know Cosmin Pohoata's problem.But i solved this and his problem too.I just want to know other solutions,without using trig.Maybe you will post it. :wink:
This problem is very easy.
[hide="Solution"]
Let $ E$,$ F$ be the point of tangency $ \zeta$ with $ AC$ and $ AB$ respectively.From La Hire's theorem $ M,E,F$ are collinear.Since $ AD,BE,CF$ are concurrent.Hence four point $ M,B,D,C$ are harmonic division.So we must just prove that $ GD,CK,BL$ are concurrent.But it is obviously with using trigonometry.
[/hide]
I suppose that you've already known that this problem is the half of his problem.I saw pohoata's problem,when it was posted by nobody1.
[hide="Here is my solution:"]
[size=100]
[quote="nobody1"][color=red] Let $ ABC$ be a triangle and let $ \zeta$ be its incircle. Denote by $ D$, $ E$, $ F$ the intersections of $ \zeta$ with $ BC$, $ CA$, $ AB$, respectively. Let $ T = AD\cap\zeta$, $ M = BT\cap\zeta$, $ N = CT\cap\zeta$. Let $ \zeta_{1}$ be a circle tangent to $ \zeta$ at $ T$, and $ \zeta_{2}$ a circle tangent to $ \zeta$ at $ D$, so that $ \zeta_{1}$ and $ \zeta_{2}$ intersect on chord $ (XY)$. Prove that $ X$, $ Y$, $ M$, $ N$ lie on the same circle.[/quote][/color]
Not hard problem,but really nice.
[color=blue]Solution:
Let $ l_{1}$ and $ l_{2}= BC$ be the tangent lines to $ \zeta$ at points $ T$ and $ D$ respectively.Denote $ T' = l_{1}\cap l_{2}$.At first i'll prove that $ E,F,T'$ are collinear,it follows from:
La Hire theorem.$ A$ lies on polar of $ T'$ w.r.t $ \zeta\Rightarrow T'$ lies on polar of $ A$ w.r.t $ \zeta\Rightarrow E,F,T'$ are collinear.
Now i'll prove that points $ X,Y,T'$ are collinear.It obviously since $ l_{1},l_{2},XY$ are radical axes of $ \{\zeta_{1},\zeta\},\{\zeta_{2},\zeta\},\{\zeta_{1},\zeta_{2}\}$ respectively.
Now using previous problem,we obtain that $ T',M,N$ are collinear.
$ \Rightarrow T'X\cdot T'Y = T'M\cdot T'N = TD^{2}$.
Thus we are done.
[/color][/size]
[/hide][/img]

I rewrite your problem as follow:

[b]PROBLEM.[/b] Let $ ABC$ be a triangle and let $ w$ be its incircle. Denote by $ D$, $ E$, $ F$ the intersections of $ w$ with $ BC$, $ CA$, $ AB$, respectively. Let $ T\equal{} AD\cap w$, $ M\equal{} BT\cap w$, $ N\equal{} CT\cap w$. Then we have $ MN$, $ BC$ and $ EF$ are concurrent.

[b]SOLUTION.[/b] Denote by $ S$ the intersection of $ EF$ with $ BC$.

We have the quadrilateral $ DETF$ is harmonic, therefore $ ST$ is tangent to $ w$ at $ T$. On the other hand, the division $ (BDCS)$ is harmonic, therefore the pencil $ T(BDCS)$ is harmonic. By intersecting it with the circle $ w$, it follows that the quadrilateral $ MDNT$ is harmonic. Therefore $ MN$ passes through $ S$. Hence $ MN$, $ BC$ and $ EF$ are concurrent.

Best regards.

[quote="BaBaK Ghalebi"]also we can prove that $ AD,CK,BL$ are concurrent...[/quote]
I used this in my solution.Look above. :wink:

[quote="BaBaK Ghalebi"]oh sorry I didnt notice that...
but how did you prove that?I know a solution using harmonic division which I think you used in the rest of the solution...
anyway if this is not your solution (harmonic division) then I would post it...[/quote]
I proved this with using trigonometry,it isn't hard.
Solution with using harmonic division was posted by April.Again look above.
I was stupid,that don't use harmonic division in the end of my proof. :blush:

[b][size=100][color=DarkBlue]GENERAL  LEMMA. – Let $ SA,$ $ SB$ be, two tangent lines of a given circle $ (O),$ from an arbitrary point $ S,$ outwardly to it. We define an arbitrary pair of points $ M,$ $ N$ on the line segment $ SB,$ as the harmonic conjugates of $ S,$ $ B$ $ ($ $ M,$ between $ S,$ $ B$ $ ).$ The lines connecting an arbitrary point $ P,$ on the line segment $ AB$ with $ M,$ $ N,$ intersect the circle $ (O)$ at points $ K,$ $ L,$ respectively. Prove that the points $ S,$ $ K,$ $ L,$ are collinear.[/color][/size][/b]

[b][size=100]PROOF.[/size][/b] – We denote the points $ K\equiv (O)\cap PM,$ $ Q\equiv AB\cap SK$ and also the points $ L\equiv (O)\cap PN,$ $ L'\equiv SK\cap PN.$

It is enough to prove that $ L\equiv L'.$

$ \bullet$ Because of the points $ S,$ $ M,$ $ B,$ $ N,$ are in harmonic division, we conclude the same about the points $ S,$ $ K,$ $ Q,$ $ L'.$ $ ,(1)$

As the line segment $ AB,$ is the polar of $ S$ with respect to $ (O),$ we conclude that the points $ S,$ $ K,$ $ Q,$ $ L,$ $ ,(2)$  are in harmonic division.

From $ (1),$ $ (2)$ $ \Longrightarrow$ $ L\equiv L'$ and the proof is completed.

Kostas Vittas.

PS. It is easy to show your problem as the particular case of the above [b][size=100]Lemma.[/size][/b] Also that it is true for any point $ G,$ between $ A,$ $ D.$

Also, from $ (1)$ we conclude that the point $ R\equiv KN\cap LM,$ lies on the line segment $ AB$ and so, in your configuration, the line segments $ AD,$ $ CK,$ $ BL,$ are concurrent as [b][size=100]BaBak Ghalebi[/size][/b] mentioned and proved by [b][size=100]April[/size][/b]

I didn't check it but here is a discussion for this problem (Iran 2002 TST )
http://www.mathlinks.ro/viewtopic.php?search_id=1141184650&t=2619


How many digits does it take to write the numbers 1 to 1999 ?

A  6880	  B.6882	     C 6884	 D.      6885           E.    6889

[hide] There are $ 9$ $ 1$digit numbers, and hence $ 9$ digits. There are $ 90$ $ 2$digit numbers, and hence another $ 180$ digits. There are $ 900$ $ 3$digit numbers, and hence another $ 2700$ digits. The numbers from $ 1000$ to $ 1999$ produce another $ 4000$ digits. So, $ 4000\plus{}2700\plus{}180\plus{}9\equal{}6889$. Thus, the answer is E.[/hide]

[hide=" I think this is right"]
For the numbers 1 - 9, you need [b]9[/b] digits
For the numbers 10 - 99, you need $ 90 \times 2$, or [b]180[/b] digits
For the numbers 100 - 999, you need $ 900 \times 3$, or [b]2700[/b] digits
For the numbers 1000 - 1999, you need $ 1000 \times 4$, or [b]4000[/b] digits
$ 9 \plus{}180 \plus{} 2700 \plus{} 4000 \equal{} 6889$ digits, so 6889, or [b]E[/b][/hide]


[hide="9th grade:"][b]1st[/b] day:
[url=http://www.mathlinks.ro/viewtopic.php?p=1155578#1155578]1st problem[/url]

[url=http://www.mathlinks.ro/viewtopic.php?p=1155582#1155582]2nd problem[/url]

[url=http://www.mathlinks.ro/viewtopic.php?t=209616]3rd problem[/url]

[url=http://www.mathlinks.ro/viewtopic.php?p=1155595#1155595]4th problem[/url]

[b]2nd[/b] day:
[url=http://www.mathlinks.ro/viewtopic.php?p=1155601#1155601]5th problem[/url]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=205955]6th problem[/url]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=209596]7th problem[/url]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=1155614#1155614]8th problem[/url][/hide]

[hide="10th grade:"][b]1st[/b] day
[url=http://www.mathlinks.ro/viewtopic.php?p=1155578#1155578]1st problem[/url]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=209598]2nd problem[/url]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=209609]3rd problem[/url]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=209597]4th problem[/url]

[b]2nd [/b]day:
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=1155621#1155621]5th problem[/url]

[url=http://www.mathlinks.ro/viewtopic.php?t=209610]6th problem[/url]

[url=http://www.mathlinks.ro/viewtopic.php?t=209458]7th problem[/url]

[url=http://www.mathlinks.ro/viewtopic.php?t=209456]8th problem[/url][/hide]

[hide="11th grade:"][b]1st[/b] day:
[url=http://www.mathlinks.ro/viewtopic.php?p=1155582#1155582]1st problem[/url]

[url=http://www.mathlinks.ro/viewtopic.php?t=209628]2nd problem[/url]

[url=http://www.mathlinks.ro/viewtopic.php?t=209238]3rd problem[/url]

[url=http://www.mathlinks.ro/viewtopic.php?t=209619]4th problem[/url]

[b]2nd [/b]day:
[url=http://www.mathlinks.ro/viewtopic.php?t=209623]5th problem[/url]

[url=http://www.mathlinks.ro/viewtopic.php?t=209617]6th problem[/url]

[url=http://www.mathlinks.ro/viewtopic.php?t=209241]7th problem[/url]

[url=http://www.mathlinks.ro/viewtopic.php?t=209239]8th problem[/url][/hide]

Thanks for the collection, Erken  :lol:

Maybe you can help by posting some more solutions  :D


Let $p(1), p(2), \ldots, p(k)$ be all primes smaller than $m$, prove that

\[\sum^{k}_{i=1} \frac{1}{p(i)} + \frac{1}{p(i)^2} > ln(ln(m)).\]

Someone have a solution to this problem?