algebra
polynomial
function
modular arithmetic
combinatorics open
combinatorics

What is the value of $ \binom {n} {1} \plus{} \binom {n} {4} \plus{} \dots$???

in terms of n

All the questions of this type are solved using the following observation:
if $ a \equal{} cis \frac {2 \pi}{k}, k \in N$, and $ f(i) \equal{} \frac {\sum_{i \equal{} 0}^{k \minus{} 1} a^i}{k}$, then: $ k|i \implies f(i) \equal{} 1$ and  $ k \not|i \implies f(i) \equal{} 0$
because if $ k|i$, then $ a^i \equal{} (a^k)^{i/k} \equal{} 1^{i/k} \equal{} 1$ for all i, and if $ k \not|i$ then $ f(i) \equal{} \frac {a^k \minus{} 1}{k(a \minus{} 1)} \equal{} \frac {1 \minus{} 1}{k(a \minus{} 1)} \equal{} 0$
So if you have polynomial (or a generating function): $ P(x) \equal{} \sum a_i x^i$, and you want the sum $ S \equal{} \sum_{k|i} a_i$, it just equals (from similar arguments) $ \frac {\sum_{i \equal{} 0}^2 P(a^i)}{3}$.

In this case, take $ P(x) \equal{} x^2(1 \plus{} x)^n \equal{} \sum_{i \equal{} 0}^{n} x^{i \plus{} 2} \binom{n}{i} ( \equal{} \sum a_i x^i)$ and you'll get:
$ \sum_{i \equal{} 0} \binom{n}{1 \plus{} 3i} \equal{} \sum_{3|i} a_i \equal{} \frac {\sum_{i \equal{} 0}^2 P(a^i)}{3} \equal{} \frac {(1 \plus{} a)^n a^2 \plus{} (1 \plus{} a^2)^n a \plus{} 2^n}{3}$, where $ a \equal{} cis \frac {2 \pi}{3} \equal{} \frac { \minus{} 1 \plus{} \sqrt {3}i}{2}$

EDIT: and if you use the identity $ 1\plus{}a\plus{}a^2\equal{}0$, you can simply that into $ \frac{2^n\plus{}(\minus{}1)^n a^{n\plus{}1}(a^{n\plus{}1}\plus{}1)}{3}$.
If $ n \equiv 2 \pmod 3$, then $ \frac{2^n\plus{}(\minus{}1)^n a^{n\plus{}1}(a^{n\plus{}1}\plus{}1)}{3}\equal{}\frac{2^n\plus{}2(\minus{}1)^n}{3}$
else, $ \frac{2^n\plus{}(\minus{}1)^n a^{n\plus{}1}(a^{n\plus{}1}\plus{}1)}{3}\equal{}\frac{2^n\minus{}(\minus{}1)^n}{3}$


1.
Math team practices at my school are [b]very[/b] boring (we sit in our category tables and stare at the past contests) and few people ever came regularly. In an attempt to make it more fun and pleasant for my teammates, I decided to plan an "opener" for them.

I was thinking of two things: starting with a math strategy game (selected from Mathematical Circles), get everybody involved and then explain how come the game went like this or that.

Or starting with a fun logic problem... where people can give weird yet reasonable answer/solutions and then I lead them to the actual solution.

Do you have better ideas of suggestions for me? I would love to hear it!

2.
Do you have suggestions on how make make math practices more fun? instead of bored people sitting in their own chair attempting the problem if at all?

Please help me! Thanks!

I am beginning to try out SD Math Circle's "food strategy" where each person is responsible for one day.. I just hope most of them will remember/ bring it.

More suggestions please?

chocolate (it really helps you while you are doing science problems)

bring food and make sure everyone is being involved in the team.

Have people compete amongst each other on practice tests. Teach interesting random stuff sometimes. Discuss hard problems instead of just working on things individually.

And food can't hurt. ;)

[quote="mathnerd314"]What do you do at math team practice?

Now, now, before you say "math, duh": Are you mainly practicing for competitions, or are you mainly doing what?[/quote]

I don't get it...
we are mainly practicing for competitions.?

[quote="goldendomer"]Bring food and give them out to the top scorers. :)[/quote]

That will only work if most of the people are cabable of scoring about the same on whatever you give out.  If you have a few student(s) who are a lot better than the rest, they will always win the prize, which would not help with getting other people to come, if they think that they can never get the prize, etc.

Here's what happens at AAMC(Albany Area Math Circle):

We come, we have a little fun before the problems.
We sit and do the problems for 40 minutes or so.
The pizza arrives,  so we eat and write on a graph what we did and didn't answer.
We review the problems we did answer with each other.
We check the answers.
And we leave!

P.S. If you don't have End-Of-The-Year picnics, try it. It's fun!

P.P.S. Even though there can be levity in the air, please keep gravitas.


suppose that $T$ is a tree with $k+1$ vertices.and $G$ is a simple graph and $\delta \geq k$.show that $G$ has a subgraph like $Q$ such that $Q \simeq T$.
($\delta$ is the minimum degree of $G$.)

Let $Q$ be a maximal (with respect to inclusion) subgraph of $G$ isomorphic to a subtree of $T$, and assume $|Q|\le k$. There must then be a vertex $v_G$ of $Q$ corresponding to a vertex $v_T$ of $T$ through an isomorphism $f$ from $Q$ to $T$ s.t. $v_T$ has degree larger than the degree of $v_G$ in $Q$.

If $v_T'$ is a vertex in $T$ connected to $v_T$ and which does not belong to $f(Q)$, then choose a vertex $v_G'$ of $G$ connected to $v_G$ but not belonging to $Q$ (there must be such a vertex, since $v_G$ has degree $\ge k$, and there are only $\le k-1$ other vertices in $Q$), add it to $Q$ together with the edge $\{v_G,v_G'\}$, and extend the ismorphism $f$ to the new $Q$ by setting $f(v_G')=v_T'$. This shows that $Q$ was not maximal after all, so we have a contradiction.

nice solution i really enjoyed it,grobber thanks. :D


inequalities
geometry
circumcircle
inradius
inequalities proposed

let a,b,c be the lengths of the sides of triangle ABC.and R is the  circumradius ang r is the inradius of triangle ABC
Prove this Inq
$ \frac{(a\plus{}b\plus{}c)^2}{2(ab\plus{}bc\plus{}ca)\minus{}(a^2\plus{}b^2\plus{}c^2)} \plus{}\frac{2r}{R} \geq 4$
p/s:I write this for my teacher,my family and my GF.

But I have another solution for my problem.
Let x=p-a,y=p-b,z=p-c;(p=$ \frac{a\plus{}b\plus{}c}{2}$).....and you will have something which is obvious.

[quote="arqady"][quote="tuandokim"]let a,b,c be the lengths of the sides of triangle ABC.and R is the  circumradius ang r is the inradius of triangle ABC
Prove this Inq
$ \frac {(a \plus{} b \plus{} c)^2}{2(ab \plus{} bc \plus{} ca) \minus{} (a^2 \plus{} b^2 \plus{} c^2)} \plus{} \frac {2r}{R} \geq 4$
[/quote]
Let $ a \plus{} b \plus{} c \equal{} 3u,$ $ ab \plus{} ac \plus{} bc \equal{} 3v^2,$ $ abc \equal{} w^3$ and $ u^2 \equal{} tv^2.$
Hence, $ \frac {(a \plus{} b \plus{} c)^2}{2(ab \plus{} bc \plus{} ca) \minus{} (a^2 \plus{} b^2 \plus{} c^2)} \plus{} \frac {2r}{R} \geq 4\Leftrightarrow$
$ \Leftrightarrow\frac {(a \plus{} b \plus{} c)^2}{2(ab \plus{} bc \plus{} ca) \minus{} (a^2 \plus{} b^2 \plus{} c^2)} \plus{} \frac {(a \plus{} b \minus{} c)(a \plus{} c \minus{} b)(b \plus{} c \minus{} a)}{abc} \geq 4\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a^5 \minus{} 3a^4b \minus{} 3a^4c \plus{} 2a^3b^2 \plus{} 2a^3c^2 \plus{} 9a^3bc \minus{} 8a^2b^2c)\geq0\Leftrightarrow$
$ \Leftrightarrow27u^5 \minus{} 72u^3v^2 \plus{} 48uv^4 \plus{} (13u^2 \minus{} 16v^2)w^3\geq0.$
If $ 13u^2 \minus{} 16v^2\geq0$ then your inequality is proved because $ 27u^5 \minus{} 72u^3v^2 \plus{} 48uv^4 \equal{} 3u(3u^2 \minus{} 4v^2)^2\geq0.$
If $ 13u^2 \minus{} 16v^2 < 0$ id est, $ 1\leq t < \frac {16}{13}$ then since, $ (a \minus{} b)^2(a \minus{} c)^2(b \minus{} c)^2\geq0$ gives
$ w^3\leq3uv^2 \minus{} 2u^3 \plus{} 2\sqrt {(u^2 \minus{} v^2)^3}$ we need to prove that
$ 27u^5 \minus{} 72u^3v^2 \plus{} 48uv^4\geq(16v^2 \minus{} 13u^2)\left(3uv^2 \minus{} 2u^3 \plus{} 2\sqrt {(u^2 \minus{} v^2)^3}\right).$
But $ 27u^5 \minus{} 72u^3v^2 \plus{} 48uv^4\geq(16v^2 \minus{} 13u^2)\left(3uv^2 \minus{} 2u^3 \plus{} 2\sqrt {(u^2 \minus{} v^2)^3}\right)\Leftrightarrow$
$ \Leftrightarrow u^5 \minus{} u^3v^2\geq2(16v^2 \minus{} 13u^2)\sqrt {(u^2 \minus{} v^2)^3}\Leftrightarrow t^3\geq4(16 \minus{} 13t)^2(t \minus{} 1)\Leftrightarrow$
$ \Leftrightarrow(15t \minus{} 16)^2(3t \minus{} 4)\leq0,$ which is true for $ 1\leq t < \frac {16}{13}.$ Done! :)[/quote]

It is really a complicated solution and of course it is not necessary for this one  :P
\[ \Leftrightarrow\frac {(a \plus{} b \plus{} c)^2}{2(ab \plus{} bc \plus{} ca) \minus{} (a^2 \plus{} b^2 \plus{} c^2)} \plus{} \frac {(a \plus{} b \minus{} c)(a \plus{} c \minus{} b)(b \plus{} c \minus{} a)}{abc} \geq 4\]
Let $ a\plus{}b\minus{}c\equal{}x,b\plus{}c\minus{}a\equal{}y,c\plus{}a\minus{}b\equal{}z$ the inequality is equivalent to:
\[ \frac{(x\plus{}y\plus{}z)^2}{xy\plus{}yz\plus{}zx}\plus{}\frac{8xyz}{(x\plus{}y)(y\plus{}z)(z\plus{}x)}\ge 4\]
\[ \iff \frac{x^2\plus{}y^2\plus{}z^2}{xy\plus{}yz\plus{}zx}\plus{}\frac{8xyz}{(x\plus{}y)(y\plus{}z)(z\plus{}x)}\ge 2\]
which is well-known  :wink:  SOS kills it easily ...

[quote="zaizai-hoang"][quote="arqady"][quote="tuandokim"]let a,b,c be the lengths of the sides of triangle ABC.and R is the  circumradius ang r is the inradius of triangle ABC
Prove this Inq
$ \frac {(a \plus{} b \plus{} c)^2}{2(ab \plus{} bc \plus{} ca) \minus{} (a^2 \plus{} b^2 \plus{} c^2)} \plus{} \frac {2r}{R} \geq 4$
[/quote]
Let $ a \plus{} b \plus{} c \equal{} 3u,$ $ ab \plus{} ac \plus{} bc \equal{} 3v^2,$ $ abc \equal{} w^3$ and $ u^2 \equal{} tv^2.$
Hence, $ \frac {(a \plus{} b \plus{} c)^2}{2(ab \plus{} bc \plus{} ca) \minus{} (a^2 \plus{} b^2 \plus{} c^2)} \plus{} \frac {2r}{R} \geq 4\Leftrightarrow$
$ \Leftrightarrow\frac {(a \plus{} b \plus{} c)^2}{2(ab \plus{} bc \plus{} ca) \minus{} (a^2 \plus{} b^2 \plus{} c^2)} \plus{} \frac {(a \plus{} b \minus{} c)(a \plus{} c \minus{} b)(b \plus{} c \minus{} a)}{abc} \geq 4\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a^5 \minus{} 3a^4b \minus{} 3a^4c \plus{} 2a^3b^2 \plus{} 2a^3c^2 \plus{} 9a^3bc \minus{} 8a^2b^2c)\geq0\Leftrightarrow$
$ \Leftrightarrow27u^5 \minus{} 72u^3v^2 \plus{} 48uv^4 \plus{} (13u^2 \minus{} 16v^2)w^3\geq0.$
If $ 13u^2 \minus{} 16v^2\geq0$ then your inequality is proved because $ 27u^5 \minus{} 72u^3v^2 \plus{} 48uv^4 \equal{} 3u(3u^2 \minus{} 4v^2)^2\geq0.$
If $ 13u^2 \minus{} 16v^2 < 0$ id est, $ 1\leq t < \frac {16}{13}$ then since, $ (a \minus{} b)^2(a \minus{} c)^2(b \minus{} c)^2\geq0$ gives
$ w^3\leq3uv^2 \minus{} 2u^3 \plus{} 2\sqrt {(u^2 \minus{} v^2)^3}$ we need to prove that
$ 27u^5 \minus{} 72u^3v^2 \plus{} 48uv^4\geq(16v^2 \minus{} 13u^2)\left(3uv^2 \minus{} 2u^3 \plus{} 2\sqrt {(u^2 \minus{} v^2)^3}\right).$
But $ 27u^5 \minus{} 72u^3v^2 \plus{} 48uv^4\geq(16v^2 \minus{} 13u^2)\left(3uv^2 \minus{} 2u^3 \plus{} 2\sqrt {(u^2 \minus{} v^2)^3}\right)\Leftrightarrow$
$ \Leftrightarrow u^5 \minus{} u^3v^2\geq2(16v^2 \minus{} 13u^2)\sqrt {(u^2 \minus{} v^2)^3}\Leftrightarrow t^3\geq4(16 \minus{} 13t)^2(t \minus{} 1)\Leftrightarrow$
$ \Leftrightarrow(15t \minus{} 16)^2(3t \minus{} 4)\leq0,$ which is true for $ 1\leq t < \frac {16}{13}.$ Done! :)[/quote]

It is really a complicated solution and of course it is not necessary for this one  :P
\[ \Leftrightarrow\frac {(a \plus{} b \plus{} c)^2}{2(ab \plus{} bc \plus{} ca) \minus{} (a^2 \plus{} b^2 \plus{} c^2)} \plus{} \frac {(a \plus{} b \minus{} c)(a \plus{} c \minus{} b)(b \plus{} c \minus{} a)}{abc} \geq 4
\]
Let $ a \plus{} b \minus{} c \equal{} x,b \plus{} c \minus{} a \equal{} y,c \plus{} a \minus{} b \equal{} z$ the inequality is equivalent to:
\[ \frac {(x \plus{} y \plus{} z)^2}{xy \plus{} yz \plus{} zx} \plus{} \frac {8xyz}{(x \plus{} y)(y \plus{} z)(z \plus{} x)}\ge 4
\]

\[ \iff \frac {x^2 \plus{} y^2 \plus{} z^2}{xy \plus{} yz \plus{} zx} \plus{} \frac {8xyz}{(x \plus{} y)(y \plus{} z)(z \plus{} x)}\ge 2
\]
which is well-known  :wink:  SOS kills it easily ...[/quote]
your solution is the same with my solution,zaiza hoang

[quote="arqady"][quote="zaizai-hoang"]

It is really a complicated solution and of course it is not necessary for this one  :P
\[ \Leftrightarrow\frac {(a \plus{} b \plus{} c)^2}{2(ab \plus{} bc \plus{} ca) \minus{} (a^2 \plus{} b^2 \plus{} c^2)} \plus{} \frac {(a \plus{} b \minus{} c)(a \plus{} c \minus{} b)(b \plus{} c \minus{} a)}{abc} \geq 4
\]
Let $ a \plus{} b \minus{} c \equal{} x,b \plus{} c \minus{} a \equal{} y,c \plus{} a \minus{} b \equal{} z$ the inequality is equivalent to:
\[ \frac {(x \plus{} y \plus{} z)^2}{xy \plus{} yz \plus{} zx} \plus{} \frac {8xyz}{(x \plus{} y)(y \plus{} z)(z \plus{} x)}\ge 4
\]

\[ \iff \frac {x^2 \plus{} y^2 \plus{} z^2}{xy \plus{} yz \plus{} zx} \plus{} \frac {8xyz}{(x \plus{} y)(y \plus{} z)(z \plus{} x)}\ge 2
\]
which is well-known  :wink:  SOS kills it easily ...[/quote]
zaizai-hoang, try to prove by SOS the following inequality.
For non-negative $ a,$ $ b$ and $ c$ prove that:
\[ \sum_{cyc}(a^5 \minus{} 3a^4b \minus{} 3a^4c \plus{} 2a^3b^2 \plus{} 2a^3c^2 \plus{} 9a^3bc \minus{} 8a^2b^2c)\geq0
\]
I don't sure that it's easily. :wink:[/quote]
What did you mean? I misunderstand  :maybe: I think you have known :
\[ \frac {x^2 \plus{} y^2 \plus{} z^2}{xy \plus{} yz \plus{} zx} \plus{} \frac {8xyz}{(x \plus{} y)(y \plus{} z)(z \plus{} x)}\ge 2
\]
which was apeared at least one time in Mathlink forum. Did you mean this one is not easy to solve by SOS method?! But I didn't numerator all of them, arqady  :wink:
The following is really easy
\[ \frac {x^2 \plus{} y^2 \plus{} z^2}{xy \plus{} yz \plus{} zx} \plus{} \frac {8xyz}{(x \plus{} y)(y \plus{} z)(z \plus{} x)}\ge 2
\]

\[ \iff \sum (x \minus{} y)^2\left(\frac {(x \plus{} y)(y \plus{} z)(z \plus{} x)}{xy \plus{} yz \plus{} zx} \minus{} 2z\right)\ge 0
\]

\[ \iff \sum (x \minus{} y)^2\left(x \plus{} y \minus{} z \minus{} \frac {xyz}{xy \plus{} yz \plus{} zx}\right)\ge 0
\]
Easy to see that if we assume $ x\ge y\ge z$ therefore $ S_z,S_y\ge 0$ and $ S_x \plus{} S_y \equal{} \frac {2z^2(x \plus{} y)}{xy \plus{} yz \plus{} zx}\ge 0$. We are done  :wink:
Generally, I said that:
\[ \frac {x^2 \plus{} y^2 \plus{} z^2}{xy \plus{} yz \plus{} zx} \plus{} \frac {8xyz}{(x \plus{} y)(y \plus{} z)(z \plus{} x)}\ge 2
\]
which was killed easily by SOS method. But I didn't say that your inequaliyt can be solved easily by SOS method:
\[ \sum_{cyc}(a^5 \minus{} 3a^4b \minus{} 3a^4c \plus{} 2a^3b^2 \plus{} 2a^3c^2 \plus{} 9a^3bc \minus{} 8a^2b^2c)\geq0
\]
In fact, your inequality is not easy with SOS method. Althought they are only one. That all  :D


given that abc is a triangle. n is a point inside the tringle. show that ab+ac >an+bn

It's placed in a wrong forum!
However, $ AB>AN\plus{}NB; AC>AN\plus{}NC$; by summing the 2 inequalities we get $ AB\plus{}AC>BN\plus{}CN\plus{}2\cdot AN>BN\plus{}CN$.

Best regards,
sunken rock


How many CDs which cost $ \$7.98$ each plus $ 4\%$ sales tax can Kylie buy with $ \$84$?

$ \frac{84}{7.98\cdot1.04}\equal{}10.121...$, so 10.
We can also see that 84:7.98 is near 10, and with the taxes, you know the answer is 10.

[quote="rightysportsfan1"]You can just do 84/7.48 and completely forget about the 4%. still works.

[code]sqrt2[/code][/quote]

If you ignore the $4%
$, it won't work...

It could work if it makes little difference, meaning you got lucky.

In this problem, it would work. In others, not so much. You just got lucky.


Perform the indicated operation and write your answer in a+bi form

[u]1+i  [/u]
2+4i

In general, if someone gives you $\frac{a+bi}{c+di}$ and wants it simplified, multiply by the conjugate of the denominator divided by itself, that is, $\frac{c-di}{c-di}$. This makes the denominator real, and you're pretty much done. It's the same idea as getting roots out of the denominator when you've got $\frac{\sqrt{a} + b}{\sqrt{c} + d}$ or something.

So I take it and multiply it by
[/u]2-4i  [u]
2+4i                       ???????????[/u]

[u]2-2i-4i to the 2[/u]
4-16i to the 2

Now what?

Well, just recall that $i$ is defined as $\sqrt{-1}$...

Looks good to me, though if you want it in $a + bi$ form I guess you should write $\frac{3}{10} - \frac{1}{10}i$  ;)

is i supose to be on the top or bottom of the fraction or to the side like you have it?

It's in the numerator. Generally, if you have a rational number times an irrational number, function, or some other symbol, it is conventional to put that symbol at the side rather than in the numerator, but they have the same meaning. Don't put the $i$ in the denominator -- $\frac{1}{i} = -i$.

Interesting - but not really calculus. I'm moving this and your other questions to another forum.


limit
logarithms
calculus
calculus computations

$H_{k}$ - k-th harmonic number; find
\[\lim_{n\to+\infty}\n\left( n(H_{2n}-H_{n}-\ln 2)-\frac{1}{4}\right)\]

[quote="boria"]$H_{k}$ - k-th harmonic number; find
\[\lim_{n\to+\infty}\n\left( n(H_{2n}-H_{n}-\ln 2)-\frac{1}{4}\right) \]
[/quote]

$\n\left( n(H_{2n}-H_{n}-\ln 2)-\frac{1}{4}\right=-1/4$
because $lim n(H_{2n}-H_{n}-\ln 2)$=0

sorry, i forgot one n
there should be :
\[\lim_{n\to+\infty}n\left( n(H_{2n}-H_{n}-\ln 2)-\frac{1}{4}\right) \]

[quote="Dr. Doe"]$ H_{n}\sim \ln n+\gamma+\frac{1}{2n}-\frac{1}{12n^{2}}+\frac{1}{120n^{4}}+O\left(\frac{1}{n^{6}}\right)$[/quote]

I don't understand from where this expansion come from. Could someone explain? Thanks.

If you want to know how to compute that expansion, then here's a method:
Apply Cesaro-Stolz to this limit: $ \lim_{n \to \infty}\frac{H_{n}-\ln n-\gamma}{\frac1{n}}$ (we can do it, since both the numerator and the denominator go to $ 0$).
Next, you can apply L'Hospital to get the desired result.
You can continue this way. However, the computations will get pretty nasty, so it's better to have a computer nearby :D

[Seeing into the future [b]EDIT[/b]]
Answer to Carcul's question: [b][i]YES[/i][/b]
[/[b]EDIT[/b]]

Thanks you so much Perfect_radio. One question though:

[quote="Perfect_radio"]You can continue this way.[/quote]

In the first "round", everythink ok. However, in the second "round" the limit to be taken is

$ \lim_{n\to\infty}\frac{H_{n}-\ln n-\frac{1}{2n}-\gamma}{\frac{1}{n^{2}}}$

right?


For what natural numbers, n, if any, is 2^(2n-1) + 1 prime?

Can you explain why?

Spoiler:

[hide] look at the powers of 2, mod 3. 2= 2(mod 3). 2*2= 4= 1 (mod 3). The product of a number congruent to 1(mod 3) and one congruent to 2(mod 3) is always equivalent to 2(mod 3)... thus the cycle continues. 2^n (mod 3) always equals 2 when n is odd (it's 1 when n is even.) Thus 2^n + 1 when n is odd is always divisible by 3.  The only case where this is prime is when the exponent is 1 and 2^n+1 is equal to 3. [/hide]

So what if we were to ask instead for which n can 2^n+1 possibly be prime? (That is, you can state that all exponents n except for one small class cannot possibly be prime.) What about 2^n-1?

do u want specific integers or just a group?

I was looking for a set of integers for which that expression might be prime.


abstract algebra
Ring Theory
Galois Theory
advanced fields
advanced fields unsolved

Let $ X$ be a connected $ k$-variety (i.e. $ X$ is a separated scheme of finite type over $ \text{Spec }k$ and the underlying topological space is connected). Show that if $ X(k) \neq \emptyset$ (i.e. there exists a morphism of schemes $ \text{Spec } k \to X$), then $ X$ is geometrically connected (i.e. the fibre product $ X \times_k \bar{k}$ is connected).

An idea concerning the affine case.. :roll: So let $ A$ be a finitely-generated $ k$-algebra without nontrivial idempotents and let $ {\mathfrak m}\subset A$ be a (maximal) ideal such that $ k\to A\stackrel{\pi}{\to} A/{\mathfrak m}$ is an isomorphism.Next, let $ K/k$ be an arbitary field extension, and suppose that $ x\in A\otimes_k K$ is an idempotent. Then $ (\pi\otimes\text{id}_K)(x)\in A/{\mathfrak m}\otimes_k K\cong K$ is an idempotent, thus it equals either $ 0$ or $ 1$. Wlog $ (\pi\otimes\text{id}_K)(x)\equal{}0$, so $ x\in{\mathfrak m}\otimes_k K$. But $ {\mathfrak m}_K : \equal{} {\mathfrak m}\otimes_k K$ is a maximal ideal in the Noetherian ring $ A\otimes_k K$, whence $ x\equal{}x^2\equal{}... \in \bigcap_{n\geq 0}{\mathfrak m}_K^n \equal{} \{0\}$.
What about the general case?

Use that the Galois group acts transitively on the connected components of $ X \times_k \bar{k}$, but leaves $ k$-rational points invariant.


[b]How do we find a point in the plane of a hexagon[/b] (can we???!!) [b]s.t. if we join this point to each of the vertices and draw perpendiculars to these lines[/b](joining the required point to vertices)[b] at the vertices, these perpendicular lines intersect to form a parrallel hexagon?[/b] i.e. a hexagon with opposite sides parallel.

Add::: Is it possible to construct a parallel (defines above) hexagon, s.t. opposite sides are unequal[b]???[/b][/b]

[quote="Rushil"][b]How do we find a point in the plane of a hexagon[/b] (can we???!!) [b]s.t. if we join this point to each of the vertices and draw perpendiculars to these lines[/b](joining the required point to vertices)[b] at the vertices, these perpendicular lines intersect to form a parrallel hexagon?[/b] i.e. a hexagon with opposite sides parallel.[/quote]

Did I understand you correctly that you have a hexagon ABCDEF and you want to find a point P such that the perpendiculars to the lines PA, PB, PC, PD, PE, PF through the points A, B, C, D, E, F enclose a hexagon whose opposite sides are parallel? If this is indeed your question, then it is trivial. In fact, consider the perpendicular to the line PA through the point A and the perpendicular to the line PD through the point D. These two perpendiculars must be parallel (since they are opposite sides of the enclosed hexagon), and thus the lines PA and PD must also be parallel to each other. But these lines have a common point (namely, the point P); thus, being parallel, they must coincide. Hence, the points A, P and D are collinear, i. e. the point P lies on the line AD. Similarly, the point P lies on the lines BE and CF. So your point P, if it exists, must lie on all three lines AD, BE and CF. Conversely, we can easily see that if the lines AD, BE and CF have a common point, then this common point is your point P (i. e. it satisfies the condition that the perpendiculars to the lines PA, PB, PC, PD, PE, PF through the points A, B, C, D, E, F enclose a hexagon whose opposite sides are parallel).

Hence, your point P exists if and only if the lines AD, BE and CF have a common point, and in this case it is this common point.

[quote="Rushil"]Add::: Is it possible to construct a parallel (defines above) hexagon, s.t. opposite sides are unequal[b]???[/b][/b][/quote]

Of course! Just draw three pairs of parallel lines; they enclose a parallel hexagon, and it would require very good luck to get opposite sides of this hexagon equal...

  darij

Sorry!!! I later realised its absolute triviality!!!! Basically, you can do this for only a special hexagon for generally, it is absolutely NOT necessary for the lines $AD$ , etc... to intersect...


Could someone prove the AM-GM inequality?  I think a proof would help me understand why the AM-GM inequality holds.

Wikipedia has a simple proof of the inequality: [url=http://en.wikipedia.org/wiki/AM-GM#Proof_by_Cauchy]see here[/url]

You'll get a better feel of how AM-GM works when you understand Jensen's Inequality and convexity, though.

Use the basic [a-b] the whole squere identity

AM-GM is a special case of Muirhead:

$ [1,0,\ldots,0]\ge[\frac1n,\frac1n,\ldots,\frac1n]$

[quote="batteredbutnotdefeated"]Try to prove the AM-GM inequality for 2 numbers(really obvious) and try to generalize. I think. :|[/quote]

It's simple to prove AM-GM for 2 numbers, but an example is not a proof.

How would you generalize it?

[hide="huge hint"] First use induction to prove that AM-GM is true as long as the number of variables is a power of 2. Next, suppose you know that AM-GM is true for variables, and you want to prove it for three variables. Generalize this method [/hide]

Yes, this was taken from ACoPS. :oops:

If you know calculus, you can use the method of Lagrange Multipliers to prove it too.


For any n exists $ a_n \in N$ so that $ (2\minus{}\sqrt{3})^n\equal{}\sqrt{a_n} \minus{} \sqrt{a_n\minus{}1}$

[hide="Solution"]Easy to see that exist $ x_n,y_n\in N$ so that $ (2\minus{}\sqrt{3})^n\equal{}x_n\minus{}y_n\sqrt{3}$ and $ (2\plus{}\sqrt{3})^n\equal{}x_n\plus{}y_n\sqrt{3}$ (By induction or by Newton formula)
Then $ (x_n\minus{}y_n\sqrt{3})(x_n\plus{}y_n\sqrt{3})\equal{}1\Leftrightarrow x_n^2\minus{}3y_n^2\equal{}1\Rightarrow (2\minus{}\sqrt{3})^n\equal{}\sqrt{x_n^2}\minus{}\sqrt{x_n^2\minus{}1}$[/hide]

you,ve posted solutions to all the problems I proposed  :D.And I have a question...do you have ebooks or electronic materials for math? If you want you can give me a yahoo messenger id(if you have one)  :blush: where we can talk about it


Hi,

I have used the "addtotoc" feature of pdflatex to include sections of a pdf file in the table of contents of my whole document. So far everything looks pretty good. However, the pdf document is not automatically numbered as it happens with the rest of the document. Therefore, if the pdf file inserted is big, it is not easy to find the section referenced in the table of contents.
A possible solution, I think, consist of numbering first the pdf file in such a way that the page numbers fit correctly within the whole document. But sometimes this is not possible and I guess there must be easier and more elegant ways to do that. Does anyone know how to do this ?

Thanks a lot

Have a look at the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=pdfpages]pdfpages package[/url] to see if it will do what you want.

Thanks stevem, I tried almost everything with this package, but I wasn't succesful. As each page is seen as an image and inserted within the main document, it seems to me that there is no way to numbers pages of the pdf file. Maybe I'm wrong ?

I'm not quite sure if you are creating a pdf file (call it foo.pdf) and then including it in a tex document. If you are creating foo.pdf then why aren't you including foo.tex instead? If, on the other hand, foo.pdf has already been created then I'm not sure how you can influence its numbering since you can't change foo.pdf.


Find the last two digits of $ \sum_{k\equal{}2}^{2008} k^2\binom{2008}{k}$.

[hide="hint(only look if you have tried this problem for at least 3 hours)"]creative committee forming argument[/hide]

$ \sum_{k \equal{} 0}^{n} k^2\binom{n}{k}\equal{}n(n\plus{}1)\cdot 2^{n\minus{}2}$.

um... neat!

Could you prove that, and then answer the problem?

Yes, but not factorials.
Also, calculus is discouraged in this forum.

All right. :lol:

$ \boxed{\sum_{k\equal{}0}^{n} \binom{n}{k}\equal{}2^n}$

$ \boxed{\sum_{k\equal{}0}^{n} k\binom{n}{k}\equal{}n\cdot 2^{n\minus{}1}}$

$ \sum_{k \equal{} 0}^{n} k^2\binom{n}{k} \equal{} \sum_{k\equal{}1}^n k\binom{n}{k}\plus{}\sum_{k\equal{}2}^{n }k(k\minus{}1)\binom{n}{k}$

$ \equal{}n\sum_{k\equal{}1}^{n}\binom{n\minus{}1}{k\minus{}1}\plus{}n(n\minus{}1)\sum_{k\equal{}2}^{n}\binom{n\minus{}2}{k\minus{}2}$

$ \equal{}n(1\plus{}1)^{n\minus{}1}\plus{}n(n\minus{}1)(1\plus{}1)^{n\minus{}2}$

$ \equal{}n(n \plus{} 1)\cdot 2^{n \minus{} 2}$.

uh-huh. nice one!

I actually extended this from a problem that asked you to evaluate the second box when $ n\equal{}2008$.

How do you do it with differentiation?  :huh:

$ (1 \plus{} x)^n \equal{} \sum _{k \equal{} 0}^n \binom{n}{k} x^k$

Differentiating both sides with respect to $ x$ gives $ n(1 \plus{} x)^{n \minus{} 1} \equal{} \sum_{k \equal{} 0}^n k\binom{n}{k} x^{k \minus{} 1}$,

Pllug $ x \equal{} 1$ into the both sides, we have $ \sum_{k \equal{} 0}^n k\binom{n}{k} \equal{} n\cdot 2^{n \minus{} 1}$.

And differentiating again, you get ..... :)


What part of 15 hours is 15 seconds? Express your answer as a
common fraction.

We can get rid of the 15's to make it easier. We just have what part of an hour is a second? There are 60 seconds in a minute and 60 minutes in an hour, so 60x60=3600 seconds in an hour, so 1/3600


what is the purpose of trying to make usamo? to represent the USA?

or just be happy and proud of how good you are  :D

[quote="PenguinIntegral"][quote="13375P34K43V312"]Well give us a reason why one should NOT try to qualify and we'll talk.[/quote]
You have provided no good reasons, and because nothing good comes out of it, it is a waste of my time.[/quote]

Elaborate?

[quote="Phelpedo"]A think he is trying to point out a logical fallacy. My answer would be...the same as 133fs;ldr12k4j's....why not? Qualification for the USAMO is really a pillar of achievement. Do you have to have some sort of monetary award for everything?[/quote]

Agreed. Also, it is one of the numerous tests required to make the IMO team.

USAMO in its best form is simply a contest to encourage mathematics in young students. It certainly encouraged me to get better and I can safely say that I probably wouldn't be where I am without it.

That being said, the main reason for trying to make USAMO is simply a proof that you've gotten better. It's an achievement marker; a goal to shoot for.

If you don't think of the USAMO as something to shoot for because there isn't an "incentive", as there is with other contests, you're approaching it with the wrong mindset.

At least, that's my two cents.

This is off-topic. Why don't you talk about whether qualifying for USAMO is a good idea in a different topic, not in one about whether there's going to be an expansion in 2007 >.<

[quote="chess64"][quote="MysticTerminator"]dude seriously note how ON TOP OF THE BALL mr joseph laurendi is proving to be

he is seriously ON TOP OF THE BALL

and making USAMO is pretty important

otherwise everyone's like

dude you are not on top of the ball[/quote]

agreed

you can obviously see that mister arnav tripathy is on top of the ball

us losers who aren't on top of the ball should really try to emulate his amazing on-top-of-the-ball-ness[/quote]

Yeah, I bet you like his On-Top-Of The Ballness

anyways:

Reasons to make USAMO:
Self-esteem( it is earned, so this will give you more self esteem
Confidence-

More goals in life:

Improvements  in math if you aren't that good:

bragging rights.

Reasons not to: It takes up a lot of your time.

You could do a lot of other subjects and other useful stuff in the time used to study for a non-school test. Time is money.

So what other reasons are there, and why should someone want to make USAMO?

I agree with everything about building confidence, USAMO being an achievement marker, and all.

But don't forget that taking the USAMO is just a ton of fun in and of itself!

[quote="now a ranger"]
Yeah, I bet you like his On-Top-Of The Ballness

anyways:

Reasons to make USAMO:
Self-esteem( it is earned, so this will give you more self esteem
Confidence-

More goals in life:

Improvements  in math if you aren't that good:

bragging rights.

Reasons not to: It takes up a lot of your time.

You could do a lot of other subjects and other useful stuff in the time used to study for a non-school test. Time is money.

So what other reasons are there, and why should someone want to make USAMO?[/quote]

It's not really the point. If you enjoy math and want a challenge, then it becomes a reward in and of itself. It's not about bragging rights or building self-confidence.

Um, sorry to kill the fun conversation we have here but I honestly do not see any reason to continue this topic. Locked.

There are already enough good reasons for "why" you should make USAMO or at least go for it. OK? Done.


if $z$ is a fifth root of one and $z \ne 1$, compute $z^{15}+z^{16}+...+z^{50}$.

[quote="Rzeszut"][hide="Solution"]Obviously
$z^{1}+\ldots+z^{50}= z^{0}+\ldots+z^{49}= \frac{z^{50}-1}{z-1}= 0$. [/hide][/quote]

Sorry, the question was wrong.  I edited it.

oh yeah... maye I should have lunch, instead of waiting until 3PM.


We know that the equation: ax^2+bx+c=0 (a,b,c are from Z, a>0) has two distinct solutions in (0,1), prove that a>=5, and for a=5 there is only one equation who verifies the conditions of the problems!

ee denote f(x)=ax^2+bx+c!
now we have the conditions f(0)>0, f(1)>0, delta >0, 0<-b/(2a)<1 which gives: c>0, sum(a)>0, b^2-4ac>0; 0<-b<2a.

from 0<-b<2a we have that b^2<4a^2 and because b^2>4ac we get that a>c. Because a,c are from Z we have that a>=c+1, c>=1 so we have a>=2.
from sum(a)>0 we get that sum(a)>=1 this means that -b<=a+c-1<=2a-2.
from b^2-4ac>0 we have b^2>=4ac+1, but from 0<-b<=2a-2 we get that b^2<=(2a-2)^2 so that 4a^2-8a+4>=4ac+1 or c<=a-2+3/(4a) which gives c<=a-2 so a>=3 and -b<=2a-3 which is 4ac+1<=b^2<=4a^2-12a+9 which leads us to c<=a-3+2/a. we have that a>=3, c natural which gives c<=a-3 so what we get is a>=4 and -b<=2a-4 from which we have 4ac+1<=b^2<=4a^2-16a+16 so c<=a-4+15/4a. because a<=4 we get that c<=a-4 which gives a>=5!

it's a bit complicated, but I think it's good! :D

oops I nearly forgot! :D
for the 2nd part of the problem we have that a=5 and from 1<=c<=a-4 so c=1.we know that b^2>=4ac+1 so we get that b^2>=21 so -b>=sqrt(21). From the fact that -b<=a+c-1 we get that -b<=5 and so we get that sqrt(21)<=-b<=5 which gives b=-5
SO the eqs is: 5x^2-5x+1=0.

tschuess!

Sorry for the super long bump, but could someone please write this in latex? It's kind of hard to read. Sorry.


Rachel and Stephanie earn $5.15/hour. Rachel works 13 hours each week and Stephanie works 20 hours per week. Stephanie does not get any paid vacation time. How long a vacation would Stephanie have to take to make the same amount of money as Rachel in one year?

[hide]
20 hours - 13 hours = 7 hours
7*52=$\framebox{364}$ hours = 15 days and 4 hours[/hide]


A train travels a certain journey and is sopposed to arrive at midday.Whenits average speed is 40km/h,it arrives at 1p.m.When its average speed is 48km/h,itarrives at 11a.m.What is the lenght of the journey.

[hide]Answer:  $ 480 km$

Solution:

The way i did it was $ D\equal{}RT$, we know the rate of each equation and we will asign a variable to the time it would take for one to arrive at midday.  so we have two different equations $ D\equal{}40(x \plus{} 1)$ and $ D\equal{}48(x \minus{} 1)$  Because they are both equal to $ D$ we can say that $ 40(x \plus{} 1) \equal{} 48(x \minus{} 1)$ solving for x yeilds $ x \equal{} 11$ and inputing $ x \equal{} 11$ into the equation $ 48(x \minus{} 1) \equal{} D$ gives us $ 48(10) \equal{} D$.  Therefore $ D\equal{}480 km$.[/hide]


I think that this must be proved:
 ((1/3)*\sum (xy) 4 ) 3 \leq ((1/3)* \sum x 3) 8   for any x,y,z.

And my opinion is that this makes the problem even more difficult. A have a three lines solution for this problem, but I will let you think.

Exactly, this inequality and Schur inequality solve the problem.

Harazi in which way do you apply Shur to :

 sum x^6y^6 + 2sum x^3y^3

to solve this beautiful inequality?

Thank you very much.

GM A = Gazeta Mathematica A; there is also a Gazeta Mathematica B.
Am I right?

Oh, I was so stupid, in fact the inequality sum x^6*y^6+2 sum x^3*y^3<=9 is not always true. I apologise and I give my solution, based on AM-GM and Schur inequality:
     We have sum x^4*y^4=sum xy*(x^3*y^3)<=sum (x^3+y^3+1)*(x^3*y^3)/3. So we have to prove that if a+b+c=3, then sum ab(a+b+1)<=9. Put a=3u, b=3v and c=3w. Now we have to prove that 4sum uv<=9uvw+1, which is true by Schur (it reduces via some computations to sum u(u-v)(u-w)>=0). Sorry again for my naivity.

It is a beautiful and direct solution.
Thank you very much Harazi.

I apologise for my misteak. I really though that Sum x^6y^6 + 2 Sum x^3.y^3 = (Sum x^3.y^3)^2 !!! The ineq  Sum x^6y^6 + 2 Sum x^3.y^3 <= 3, where x^3 + y^3 + z^3 = 3 in fact not true for all x, y, z, for example x^3 = y^3 = 3/2, z^3 = 0.

Harazi solution is nice and quite useful in methodology. Thank you.

Namdung

If i don't wrong, it is an inequality from Vasc'book  :)

[quote="chien than"][quote="harazi"]A very nice inequality from GM A:
  Prove that for all x,y,z>0 with x^3+y^3+z^3=3 we have x^4y^4+y^4z^4+z^4x^4<=3.
   V Cartoaje.[/quote]
By AM-GM we have:
$ \frac {1 \plus{} x^3 \plus{} y^3}{3} \geq xy$=>$ \frac {4 \minus{} z^3}{3} \geq xy$
=>$ x^4y^4 \leq \frac {4x^3y^3 \minus{} x^3y^3z^3}{3}$
Similar,then we have:
$ x^4y^4 \plus{} y^4z^4 \plus{} z^4x^4 \leq \frac {4\sum_{cyc}x^3y^3 \minus{} 3x^3y^3z^3}{3}$
Then we can prove that:
$ \frac {4 \sum_{cyc} x^3y^3 \minus{} 3x^3y^3z^3}{3} \leq 3$
or $ 4(x^3 \plus{} y^3 \plus{} z^3)(x^3y^3 \plus{} y^3z^3 \plus{} z^3x^3) \minus{} 9x^3y^3z^3 \leq (x^3 \plus{} y^3 \plus{} z^3)^3$(Schur)
Done![/quote]
It's  known harazi's proof. :wink:


inequalities
geometry
circumcircle
geometry solved

Let $a,b,c$ be the sides of a triangle and $m_{a},m_{b},m_{c}$ be the lengths of the medians. Show that

$\frac{b^{2}+c^{2}}{m_{a}}+\frac{a^{2}+c^{2}}{m_{b}}+\frac{a^{2}+b^{2}}{m_{c}}\geq \frac{8}{3}(m_{a}+m_{b}+m_{c})$

note that $b^{2}+c^{2}\ge 4m_{a}R\Rightarrow \sum\frac{b^{2}+c^{2}}{m_{a}}\ge 12R\ge\frac{8}{3}\sum m_{a}$

But I can't prove both left and right inequalities  :(

Hey, what about $b^{2}+c^{2}\geq 4m_{a}R$?

I wrote clearly, please read carefully all my post!

ok that's a stupid mistake sorry for it!
We can do that problem by other way:
Change into triangle with three side $m_{a},m_{b},m_{c}$ and their median $\frac{3}{4}a,\frac{3}{4}b,\frac{3}{4}c$ we have
$\sum\frac{4a^{2}+b^{2}+c^{2}}{3a}\ge 2(a+b+c)$
$\Leftrightarrow \sum\frac{b^{2}+c^{2}}{a}\ge 2(a+b+c)$ prove it quite easily!!
Sorry for all  :)


Hi again,

First sorry for my insisting about that bug but I would prefer to keep that pseudo if possible  :D.
Could an admin reinitialize My Favorites (like really clean up all that field from the database) ? Surely if this is reinitialized the bug should disappear, since it seems that there is an error in the syntax of the My Favorites module only. I do not care about its current content but I use this section a lot.

Thanks,

Julien.

Hi Julien.  It should be cleaned in a few minutes. Strangely enough, I coudn't recreate the bug. It was something very weird ...

Ok, i've partially identified and solved the problem. Topic closed.


I was thinking about putting together a collection of pictures of AoPS members! So if you could please post an image of yourself or PM it to me, i'll add it to the collection!  :D

see my gallery
there's one of Ubemaya (the chess one), and I'm putting up one of me by myself (except it looks really effed up) now.

I'm in kstans album once. It's a bad picture, but I'm too lazy to make another one even though I have a camera in my computer :rotfl:

Bump for a good thread. I'd post a picture myself, but I don't have one at the moment  :P

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=140822

Entitled "Post a Picture of Yourself" or something like that.

Only one pic in the two threads combined.

I summon the $\lceil x\rceil$ creature, who performs the ceiling function on everyone's HP.

randomdragoon: 24
probability1.01: 12
junggi: 3
bubka: 37
iamagenius: 93
1=2: 65
lingomaniac88: 32
daermon: 73
ars: 61
Treething: 90

Dear oh dear!
That is the silliest thing I've seen, mixing up threads! :rotfl:

somebody should totally put that in their sig

about pictures, i dont currently have any pictures of myself on my computer.  i could put one up of my brother though :D


$S_{n}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}$
does  $\lim_{n\to\infty}\S_{n}=\infty$? or it has a limit?
and why it doing cool S?  :P

Because you wrote \S instead of just S.  This question is on the forum probably a half-dozen times or more -- try a search for the words "harmonic series."

$k\in N^*\Longrightarrow e>\left(1+\frac 1k\right)^k\Longrightarrow \ln (k+1)-\ln k<\frac 1k\Longrightarrow$
$\sum^n_{k=1}\left[\ln (k+1)-\ln k\right]<\sum^n_{k=1}\frac 1k\Longrightarrow \ln (n+1)<\sum^n_{k=1}\frac 1k\Longrightarrow \lim_{n\to\infty}\sum^n_{k=1}\frac 1k =\infty\ .$

Allow me to explain why the harmonic series totals up to infinity:

$\sum_{k=1}^{\infty}\frac1k = \frac11+\frac12+\frac13+\frac14+\frac15+\frac16+\frac17+\frac18+\cdots$
$\frac11+\frac12+\frac13+\frac14+\frac15+\frac16+\frac17+\frac18+\cdots > 1 + \frac12 + \left( \frac14+\frac14 \right) + \left( \frac18+\frac18+\frac18+\frac18 \right) + \cdots = 1 + \frac12 + \frac12 + \frac12 + \cdots$

This is the easiest way there is, I guess...

yea the series $\sum \frac{1}{n}$ is a p-series with $p=1$, and therefore it diverges and the sum goes to infinity i believe....

You can always integrate it to infinity (if youre asking about this for calc BC)


Euler
calculus
geometry
3D geometry
sphere
tetrahedron
geometry theorems

let help me "caculus radius of    euler second   in space    :blush: thanks

Maybe you want compute the radius of the first sphere of twelve points in a orthocentric tetrahedron?


quadratics
algebra
polynomial
modular arithmetic

Can the discriminant of a quadratic polynomial with integer coefficients be a) $ 2007$? b) $ 2008$?

Discriminant: $ b^2\minus{}4ac\equal{}k$.

$ b^2\equal{}k\plus{}4ac$
$ b^2 \equiv k \pmod{4}$

Discriminant cannot be $ 2007$ for $ 2007 \equiv 3$, which is not a quadratic residue $ \pmod{4}$.

Discriminant can be $ 2008$. Let $ a\equal{}1$. Go find some numbers that match $ b^2\equal{}4\cdot(502\plus{}c)$.

$ (a,b,c)\equal{}(1,2,\minus{}501)$...

Of course you know that you should be giving an example...In this case, I feel like it is easier to just state the example instead of saying that one exists...


Who has already posted for Round 3 and received an adjusted score?

I protested. My score for #3 went from 1 to 3.

I protested on #3 and #4; my #3 went up 1 point and I haven't heard back in regards to #4 yet. However, I didn't receive an email regarding either of these - did you experience a similar issue?

I have not recieved a response
edit: score went up 1, but no response

I protested #5, and my score was bumped from a 3 to a 5.

Check your spam filters, junk mail folders, etc.; all protests have a email message sent out at the time the protest is reviewed.  If your score changed, you should have received an email.

I protested on #3 and got my score adjusted from 4 to 5:

The grader deducted 1 point because (s)he said I didn't consider the case with at least 4 primes dividing a,b,c. However on page 11, I wrote "If at least 4 primes divide a,b,c, then each is at least 2x3x5x7=210, the product of the four smallest primes. Then a+b+c>=630." (Note I explained on page 9 that a,b,c must have the same prime factors.)

However I protested on #5 (score: 4) and didn't get a score change:

The grader claimed that I did not handle the case t=0. (Here t is the index for a sequence I defined. The sequence k_1, k_2,... contains values of k that work in my restatement of the problem.) However on page 19, I specifically let the 0th term in the sequence be k_0=0, to start the sequence off. This is right below the start of the third line on the page, which says c=b^{k_t+1}/a. I deliberately excluded t=0 it in the line above because I did not want to imply k_0 is a value of k that works, but rather defined it so that it can be used to find a first value k_1 that works, using a procedure I explained later on the page.
---
I still don't really understand why I got a point taken off. Since the procedure for finding the first term of my sequence was the same for any other term, I didn't treat it separately and thought explicitly defining k_0 would be enough.

If anyone could read my solution and tell me what is wrong, I would greatly appreciate it.

I protested #3 Round 2 and went from a 0 to a 3 :maybe:

I originally got 55353 = 21, protested twice to get 55555 = 25.


ABCD is a quadrilateral and P, Q are midpoints of CD, AB respectively. Let AP, DQ meet at X and BP, CQ meet at Y. Prove that area(ADX)+area(BCY)=area(PXQY).

This should be moved to the high school section. Let C', P', D' be feet of the normals from C, P, D to the line AB. Since $CC' \parallel PP' \parallel DD'$ and P is the midpoint of CD, it follows that PP' is the average of CC' and DD', or $PP' = \frac{CC'+DD'}{2}.$ Since Q is the midpoint of AB,

$|\triangle ADQ|+|\triangle QCB| = \frac{DD' \cdot AQ+CC' \cdot QB}{2}=$

$= \frac{AQ (CC'+DD')}{2}= \frac{AB \cdot PP'}{2}= |\triangle APB|$

Now subtract areas $|\triangle AXQ|+|\triangle QYB|$ from both sides of the equation.


Find all possibilities for real numbers $ a$ and $ b$ such that the polynomial $ x^2\plus{}ax\plus{}b$ has values
between $ \minus{}2$ and $ 2$ inclusive, for all $ x$ with $ 0\le x\le 4$.

[quote="mdk"]Find all possibilities for real numbers $ a$ and $ b$ such that the polynomial $ x^2 \plus{} ax \plus{} b$ has values
between $ \minus{} 2$ and $ 2$ inclusive, for all $ x$ with $ 0\le x\le 4$.[/quote]

The only solution for $ (a,b)$ is $ (\minus{}4,2)$.

We can rewrite the polynomial into $ y\equal{}(x\minus{}m)^2\plus{}n$ and we see that $ m$ and $ n$ does not affect the shape of the graph.


geometry
combinatorial geometry
combinatorics proposed
combinatorics

$ n$ points are given in the plane such that all pairwise distances are at least 1. prove that there are fewer than $ 3n$ pairs of points with distances exactly equal to 1.

A structure $ S$ is the set of ordered pairs $ (e\in E(S),v\in V(S)$ where $ V(S)$ is the set of all points(vertices) on a plane separated by distance $ \geq 1$. $ V(S))$ is the set of all unordered pairs of points separated by distance 1. A structure $ S$ is said to be convex if
(i) $ \exists x\in V(S)$ so that $ (x,i) \in E(S)\quad \forall \in V(S)\backslash x$
(ii) The angle $ ixj < \pi^c\quad \forall (i,j)|(i,j)$ is a unordered pair with $ i,j\in V(S)\backslash x$.
(iii) $ \forall i\in V(S)\backslash x \quad \exists j\in V(S)\backslash x$ so that $ (i,j)\in E(S)$
Let $ x$ be called the convex vertex.

Clearly, there does not exist a convex structure with 5 or more vertices. There exist convex structures with less than 5 vertices.

We assume that there exists a convex hull that contains the structure $ S$. $ S$ has atleast one convex substructure (A convex substructure provides bending to a structure. If there did not exist a convex substructure then the structure cannot be contained within the convex hull). Pick a convex vertex of a convex sub structure of the structure S, delete the vertex and the edges incident on it. Continue the process for the resulting structure. At maximum, we delete 3 edges per vertex.

here's my solution.

assume that there do exist $ 3n$ pairs of points with distance 1. in the list of these $ 3n$ pairs, $ 6n$ points appear, and as there are only $ n$ points, it follows that each point occurs exactly 6 times or some point occurs 7 times. but the latter is impossible as then there would be two points at distance less than 1. so each point occurs in exactly 6 pairs. moreover, the 6 points with which a point is paired must surround it in the form of a regular hexagon. (or else again two would be at a distance < 1.) this implies that the $ n$ points must be arranged in a honeycomb pattern. but the boundary points of such an arrangement cannot occur in 6 pairs -- a contradiction.


need help with a mixture problem fast... thanks...

in 20 oz of one alloy, there are 6 oz of copper, 4 oz of zinc and 10 oz of lead. in 20 oz of a second alloy, there are 12 oz of copper, 5 oz of zinc and 3 oz of lead, while in 20 oz of a third alloy, there are 8 oz of copper, 6 oz of zinc, and 6 oz of lead. how many ounces of each alloy should be combined to make a new alloy containing 34 oz of copper, 17 oz of zinc and 19 oz of lead?

thanks so much.

Just convert this into a system of linear equations. Let $a$ be the number of ounces of the first one, $b$ the second and $c$ the third.

Then $\frac{6a}{20}+\frac{12b}{20}+\frac{8c}{20}=34$, or $3a+6b+4c=340$, $\frac{4a}{20}+\frac{5b}{20}+\frac{6c}{20}=17$, or $4a+5b+6c=340$, and $\frac{10a}{20}+\frac{3b}{20}+\frac{6c}{20}=19$, or $10a+3b+6c=380$.

Solve these simultaneously and you get $(a,b,c)=(20,40,10)$.

[quote="Megus"][b]md07918[/b] this is more proper place for your topics - and DON'T name topics like "need help fast" - it's only discouraging people[/quote]

Not to mention that it looks like you're desperately looking for help with your homework.  In general, forum participants are more happy to discuss problem solving methods than helping students answer homework problems directly.


geometry
3D geometry
sphere
geometry unsolved

Four distinct points A, B, C and M are given on a sphere, none of which is opposite to any other. Prove that if the great circles AM and BM are orthogonal to great circles BC and AC, respectively, then also the great circle CM is orthogonal to great circle AB.

In other words, the altitudes are concurrent in a spherical triangle.

Let $O$ be the center of the sphere. We know that $OM$ is contained in the unique plane through $OA$ orthogonal to $(OBC)$, and also in the unique plane through $OB$ orthogonal to $(OCA)$, and we want to prove that it is also contained in the plane through $OC$ which is orthogonal to $(OAB)$. The problem is now this:

Let $\alpha,\beta,\gamma$ be three planes through a point $O$, and let $a=\beta\cap\gamma,b=\gamma\cap\alpha,c=\alpha\cap\beta$. Prove that the planes $\pi_\alpha,\pi_\beta,\pi_\gamma$, passing through $a,b,c$ and orthogonal to $\alpha,\beta,\gamma$ respectively, share a point different from $O$ (in other words, they share a line).

Pick $A\in a$, and then $B\in b,C\in c$ such that $AB\perp c,AC\perp b$. It's easy to see that $BC\perp a$ (you can prove this just like you prove the concurrence of the altitudes in a triangle by using scalar products), which means that $\pi_\alpha,\pi_\beta,\pi_\gamma$ cut the triangle $ABC$ along its altitudes. Since they are concurrent, the concurrence point is a point different from $O$ belonging to all $\pi$'s.


What is the SC Nationals Team?

I know it's:

Phillip Cross
Harry.... someting
.............................
............................

Anyway, go Crayton!

Are you on the team?  Could you also post what school the people who you know are on the team are from or are both of those from Crayton? I'm in high school but I always like to find out what's going on with MC.

Phillip Cross from Crayton,
and David Richman from Dent

Congrats to SC. It sounds like they did just as well as we did (we also got two in the top 57). From what I've heard, you guys have really improved since last year. Come to think of it, so have we! Hopefully we'll both improve just as much at the ARML.

yeah. go carolina ARML teams. wanna place bets on what carolina team does better. i'm going with sc B team (i'll be on the team more than likely)  :P

[quote="joml88"]Results have been posted for Mathcounts Nats: [url=http://www.mathcounts.org/webarticles/articlefiles/536-2005_Natl_Comp_Rankings.pdf]click here[/url].

Looks like the SC team did really well compared to its usual results:

38th Individual - David Richman
43rd Individual - Rono Srimani

and as a team SC got 16th!  Awsome job guys!

Congrats also to our friends up north who got 10th as a team.[/quote]

70th individual - Phillip Cross

Hand's contestant was the only SC Mathlete not placing in the Top 100.  This is the first year SC ever had two people in the Top 50.  Also, we tied our record, 16th place.  Next year, with me on the Nats team (maybe ;) ) we are aiming for... say... number 1!

How do you find these records??? Not that we need them--I know that we [b]WON[/b] once and got the third place individual (I think the same year). I saw that we won in the mathcounts yearbook and Harold Reiter told me that his daughter got third. But how did you figure it out for SC?


Each integer is coloured either red, blue, green or white. Let $x$ and $y$ be odd integers such that $|x| \neq |y|$. Show that there exists two natural numbers of the same colour whose difference is one of $x, y, x + y, x - y$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=131878#p131878]IMO 1999 Shortlist, Problem C6[/url]


hello, do you want to solve it for $ x$?
Sonnhard.

[hide]$ \dfrac{x}{2\plus{}\frac13}\equal{}\frac{9}{7} \implies x(\frac37)\equal{}\frac97 \implies x\equal{}\boxed{3}$.[/hide]

Uh, what else would you want to do?

I can't see any tricks, so...
[hide="solution"]$ \frac{x}{2\plus{}\frac{1}{3}}\equal{}1\plus{}\frac{2}{7}$
$ x\equal{}\frac{9}{7}\cdot\frac{7}{3}$
$ x\equal{}\boxed{3}$[/hide]


Let [b][i]p>5[/i][/b] be a prime number. Let
[b][i]A={p-n 2 : (n is natural) , n 2 <p}.[/i][/b]PROVE that there exist two distinct number [b][i]x,y[/i][/b] in [b][i]A[/i][/b] such that [b][i]y[/i][/b] is devisible by [b][i]x[/i][/b] and [b][i]x isn't 1[/i][/b].

 :alien:

This one is from Balkan MO 1996.
I think it is already been solved here (even if I didn't search). You may find a solution on Kalva's site (I presume).

Pierre.

Bingo!
http://www.kalva.demon.co.uk/balkan/bsoln/bsol962.html

Pierre.

I solved the problem and I just want the others see and solve it,
not see the solution!

 :D  :D

you have to send problem that you haven't solved.
$S_hA_cH_iS_m$
halite$?^?$

So you have posted it in the wrong section. ;)  You should have posted it in the 'proposed and own problems section'...

Pierre.


Solve this
$ log_3(2x\plus{}2)\equal{}log_5(x\plus{}1)\plus{}1$

hello, write $ \frac{\ln(2x\plus{}2)}{\ln(3)}\equal{}\frac{\ln(2)\plus{}\ln(x\plus{}1)}{ln(3)}$ and
substitute $ t\equal{}\ln(x\plus{}1)$ so you have to solve:
$ \frac{\ln(2)\plus{}t}{\ln(3)}\equal{}\frac{t}{\ln(5)}\plus{}1$ with respect to $ t$.
Sonnhard.


It's really an easy inequality,but the way to do may be intresting.I wonder
can it be more stricter?:
(x+y)(y+z)(z+x) \geq 8xyz+(1+a)x(y-z) 2 +(1-a)y(x-z) 2 +(1-a)z(x-y) 2  where 0<a<1;x,y,z>0 and x is max or min{x,y,z} :D

In fact it reduces in the following way:
   x+y)(y+z)(z+x)-8xyz= \sum x(y-z) 2 , so it becomes (simplify with a) z(x-y) 2 +y(x-z) 2  \geq x(y-z) 2 . If we take a=x-y, b=x-z, it becomes ab(2x-a-b) \geq 0, or (y+z)(x-y)(x-z) \geq 0, true.

Conragulation :) !
You have gone the right way.But what about my question about the stronger
inequality.Can any one find this one?I wait for your answer! :D

What do you mean by stronger inequality? One that also is related with that one, or what?

Sorry about my vague request!!I  just want another form about:
(x+y)(y+z)(z+x) \geq 8xyz
But it can become the equivalent form like:
ax(y-z) 2 +by(x-z) 2 +cz(x-y) 2  \geq 0 where a,b,c can less than 0. :?.I really think my inequality is the only one.

Is he asking for the best possible weighted coefficient?

Yes,you are right,3x.lich!!! :) .Sorry,because my English is really bad so I don't know how to say my ideas. :(


Find the remainder upon dividing \[ x^{81}+x^{49}+x^{25}+x^{9}+x^{1}  \] by $x^3-x$.

[hide]
Note that $x^3-x=x(x+1)(x-1)$.
Let $P(x)=x^{81}+x^{49}+x^{25}+x^9+x$.

Then let $P(x)= x(x+1)(x-1) Q(x) +ax^2+bx+c$, where the remainder is $ax^2+bx+c$.

$P(1)=5=a+b+c$
$P(0)=0=c$
$P(-1)=-5=a-b+c$

Solving the system, we obtain $(a,b,c)=(0,5,0)$.

Therefore, the remainder is $0x^2+5x+0=\boxed{5x}$
[/hide]

Oops, I thought the remainder would be degree 80 instead of 2 :oops:


[quote="SplashD"][quote="bpms"]Can we discuss test A?[/quote]

Seeing as how people applying later can still choose to apply with Test A, the answer's probably no.[/quote]
Oh yeah, I so forgot.

This can be deleted then.
EDIT: Actually, can I make a poll for scores?

it would be illogical if not
that's what i concluded
though i don't in any way know almost anything about this camp
other than visiting their site
when they say tests they have dates

[quote="junggi"][quote="SplashD"][quote="bpms"]Can we discuss test A?[/quote]

Seeing as how people applying later can still choose to apply with Test A, the answer's probably no.[/quote]
But I thought even those people had to submit test A by yesterday?[/quote]
But you can still use test A for the other application times.

[quote="bpms"][/quote][quote="junggi"][quote="SplashD"][quote="bpms"]Can we discuss test A?[/quote]

Seeing as how people applying later can still choose to apply with Test A, the answer's probably no.[/quote]
But I thought even those people had to submit test A by yesterday?[/quote][quote="bpms"]
But you can still use test A for the other application times.[/quote]
But still, we(people doing test A) already submitted our solutions.

[quote="junggi"][/quote][quote="bpms"][/quote][quote="junggi"][quote="SplashD"][quote="bpms"]Can we discuss test A?[/quote]

Seeing as how people applying later can still choose to apply with Test A, the answer's probably no.[/quote]
But I thought even those people had to submit test A by yesterday?[/quote][quote="bpms"]
But you can still use test A for the other application times.[/quote][quote="junggi"]
But still, we(people doing test A) already submitted our solutions.[/quote]
But other people can still submit solutions to test A during the other times, so we can't discuss it on the forum.

[quote="bpms"][/quote][quote="junggi"][/quote][quote="bpms"][/quote][quote="junggi"][quote="SplashD"][quote="bpms"]Can we discuss test A?[/quote]

Seeing as how people applying later can still choose to apply with Test A, the answer's probably no.[/quote]
But I thought even those people had to submit test A by yesterday?[/quote][quote="bpms"]
But you can still use test A for the other application times.[/quote][quote="junggi"]
But still, we(people doing test A) already submitted our solutions.[/quote][quote="bpms"]
But other people can still submit solutions to test A during the other times, so we can't discuss it on the forum.[/quote]
uh I meant to say "we(people not doing early applications)." Which means non-early app people already submitted the test, I think....It says on the website
[quote="awesomemath website"]Test A (Feb. 2-12)[/quote]
which means test A is all due on the 12th, I believed. Otherwise, it just wouldn't be fair to the early-apps people to give them only 10 days, and others more than 2 months to do it.

Well early registration gets a 300 dollar discount, so they can be unfair...
http://www.awesomemath.org/application.shtml

[quote="chess64"]Dude bpms stop being stupid. Look at the "Required Admissions Test" under the "Regular Registration" column. See the "Test A (Feb. 2 - 12)"? That means test A has to be submitted before Feb 12. :wink:[/quote]
:ewpu:
Okay, I guess I was being stupid.

As it looks like you've figured out, anyone using Test A to apply must have submitted it already. Applicants may apply during the regular application period using Test A, but they would have had to submit their solutions by February 12.

Test A will be open for discussion soon. I ask you to hold off until Friday so we can make sure we've got them all. Beginning Friday, feel free to discuss any of the questions in this sub-forum or anywhere else on AoPS.

Thanks!
-Zac


How many of the natural numbers from 1 to 800, inclusive,
contain the digit 6 at least twice? (The numbers 266 and 663
are two natural numbers that contain the digit 6 at least twice,
but 430 or 16 are not.)

We have 5 cases to look at.

Case 1:

66_

$ 1 \times 1 \times 8 \equal{} 8$ choices

Case 2:

6_6

$ 1 \times 8 \times 1 \equal{} 8$ choices

Case 3:

_66

$ 8 \times 1 \times 1 \equal{} 8$ choices

Case 4:

666

$ 1$ choice

Case 5:

66

$ 1$ choice

$ 8 \plus{} 8 \plus{} 8 \plus{} 1 \plus{} 1 \equal{} \boxed{26}$ choices

actually, here's what i think

we could have 66_
in which case the last digit could be one of 9 others, not including the six

we could have 6_6
in which case the middle digit could be one of 9 others, not including the six

we could have _66
in which case the first digit could be one of 8 others, because it cannot be zero and we're not including the six

we could have 666
in which case there is only one possibility

or we could have 66
in which case there in only one possibility

so, we have $ 9 \plus{} 9 \plus{} 8 \plus{} 1 \plus{} 1 \equal{} 28$ possibilities

[quote="isabella2296"]Yes, that seem right, except I know for a fact that the answer is $ 26$...

EDIT: Hang on...

Case 1: 66_
The last digit can be $ 0 \sim 9$, not including a 6, so we have 9 choices.

Case 2: 6_6
The last digit can be $ 0\sim 9$, not including a 6, so we have 9 choices.

Case 3: _66
The last digit can be $ 0\sim 7$, not including a 0 and a 6, so we have 6 choices.

Case 4: 666
Just one choice.

Case 5: 66
Just one choice.

$ 9 \plus{} 9 \plus{} 6 \plus{} 1 \plus{} 1 \equal{} \boxed{26}$[/quote]

In your Case 3, the first digit can also be 8 or 9.

No it can't. The problem says "natural numbers from 1 to 800 inclusive".

oh right
stupid me
so, it would be from 1-7 without counting the six, would be 6 choices, so izzy right about that


The sum of a 15-term arithmetic series with first term 7  is $ \minus{}210$.  What is the common difference?

Write the first 15 terms of the sequence is 7, 7+d, 7+2d,..., 7+14d

Using Gauss' trick of writing the same sequence backwards underneath, we see that we're going to have 15 pairs each adding up to 14+14d.  However that's twice as much as the sum we want, so we'll have to divide through by 2 in the end:

$ \frac {15(14 \plus{} 14d)}{2} \equal{} \minus{} 210$
$ 15(7 \plus{} 7d) \equal{} \minus{} 210$
$ 15(1 \plus{} d) \equal{} \minus{} 30$
$ 1 \plus{} d \equal{} \minus{} 2$
$ d \equal{} \minus{} 3$


geometry
perimeter
trigonometry
trig identities
Law of Cosines

the sides of a triangle are consecutive odd numbers. The largest angle is 120 degrees. Find the perimeter of the triangle.

[hide]
Let the sides be $ x, x\plus{}2,$ and $ x\plus{}4$. Clearly, the $ x\plus{}4$ side is across the 120 degree angle.
Now we can use Law of Cosines and get:
$ (x\plus{}4)^2\equal{}x^2\plus{}(x\plus{}2)^2\minus{}2x(x\plus{}2)\cos(120)$
$ x^2\plus{}8x\plus{}16\equal{}2x^2\plus{}4x\plus{}4\minus{}x(x\plus{}2)2(\minus{}\frac{1}{2})$
$ 0\equal{}x^2\minus{}4x\minus{}12\plus{}x(x\plus{}2)$
$ 2x^2\minus{}2x\minus{}12\equal{}0$
$ x^2\minus{}x\minus{}6\equal{}0$
$ (x\minus{}3)(x\plus{}2)\equal{}0$
So $ x\equal{}3$ and the perimeter is $ 3\plus{}5\plus{}7\equal{}\boxed{15}$
[/hide]


Let $ k$ be a positive integer. What is the largest power of $ 3$ that divides $ 10^{k}-1$?

Um... I think that $ 10^{k}-1$ is 0 (mod 9)...
That might help :wink:


Let $ a$ be a positive integer. Set $ n \equal{} 2^{2a \plus{} 1} \minus{} 1$ and let $ \zeta \equal{} \exp\left(\frac{2 \pi i}{n}\right)$.

Let $ S$ be the smallest subset of $ \mathbb{C}$ such that

-if $ s \in S$, then $ s^2 \in S$, and

-for all $ k \in \left\{1,2,\,\cdots,2^a\right\}$, $ \zeta^k \in S$.

Show that $ S$ and $ \bar{S}$ (the set of conjugates of elements of $ S$) are disjoint.

Hint: consider the binary representation of positive integers.

$ S \equal{}${ $ r^{2^0} , r^{2^1} , r^{2^2} ... r^{2^{a\plus{}1}}$}
where $ r \equal{} \exp\left(\frac{2 \pi i}{n}\right)$

S conjugate are all $ r \equal{} \exp\left(\frac{\minus{}i 2 \pi 2^k }{n}\right)$
so if S and S conj have a common element , then $ \frac{2^k \plus{} 2^s}{n}$ = some positive integer
but n is greater than the numerator

is this right , or did i make a mistake  :maybe:


If n is an integer, how many different values of n make the inequality -7 < 3n <or= to 15 a true statement?

Choices:

(a) 9

(b) 8

(c) 7

(d) 5

I don't just want the answer.  I want to know how to find it should I ever face another similar question.

$3n$ is our function and it lies within the range of -6 to 15.
So we have two cases, one for the minimum, one for the max.
$min:$
$3n =-6 \Rightarrow n =-2$
Thus $-2 \le n$

$max:$
$3n = 15 \Rightarrow n = 5$
Thus $n \le 5$

Which leaves us with $-2 \le n \le 5$ and thus there are $8$ values for $n$

The book's answer is 8, which is choice b.

How do I get 8?

I edited my post I forgot the question was asking for how many numbers n could be and not the range specifically :o


Let $ f(x)$ be a polynomial of second degree the roots of which are contained in the interval $ [\minus{}1,\plus{}1]$ and let there be a point $ x_0\in [\minus{}1.\plus{}1]$ such that $ |f(x_0)|\equal{}1$. Prove that for every $ \alpha \in [0,1]$, there exists a $ \zeta \in [\minus{}1,\plus{}1]$ such that $ |f'(\zeta)|\equal{}\alpha$ and that this statement is not true if $ \alpha>1$.

If $f(x)=a(x-b)(x-c)$ (we can assume wlog that $a \ge 0$ because $|f|$ and $|f'|$ do not change when $f$ is replaced by $-f$), then introduce $m = \tfrac{b+c}{2}$ and consider two intervals $[-1,m]$ and $[m,1]$.  The longer of these two is of length at most $2$ and has `elevation change' of at least $1$ (either $f(m) \le -1$ and $f(\pm 1) \ge 0$ or $f(m) \le 0$ and $f(\pm 1) \ge 1$).  Shifting vertically and horizontally appropriately (so that $(m, f(m))$ becomes the origin) we get that $g(x) = ax^2$ and $g(2) \ge 1$, implying $a \ge \tfrac{1}{4}$ and $g'(x) = 2ax$ taking values from $0$ to $4a \ge 1$. The example when $|f'|$ is bounded by $1$ is $f(x) = \tfrac{1}{4} (x+1)^2$.  Indeed, $f(1) = 1$ and $f'(x) = \tfrac{x+1}{2} \in [0, 1]$.


ABC triangle.
(-a+b+c)(-b+c+a)(-c+a+b) \leq abc

This is well - known and not difficult.

Note that a + b - c > 0, a - b + c > 0 en -a + b + c > 0.
Then we have
(a + b - c)(a - b + c) = a2 - (b - c)2 \leq a2
(-a + b + c)(a + b - c) = b2 - (a - c)2 \leq b2
(-a + b + c)(a - b + c) = c2 - (a - b)2 \leq c2
Now multiplying these three inequalities en taking square roots gives the result.

This inequality is even true for positive a, b, c which are not the sides of a triangle.

nother thing we could do is consider a=y+z, b=x+z and c=x+y, since a, b, c are the sides of a triangle. Then we have to prove 8xyz<=(x+y)(y+z)(z+x), which is just AM-GM. However your soln is better, since it works for any positive numbers, not just sides of a triangle.

Yes I know it's an easy problem but I wanted to know how many solutions will be posted.


geometry
circumcircle
search
parallelogram
geometry proposed
U sviti matematyky

[quote="jayme"]Dear Discredit,
the problem is a dirct application of "the midcircle theorem".
Sincerely
Jean-Louis[/quote]

A search in mathlinks of "the midcircle theorem" reveals only this topic!

So, what is it, dear Jean-Louis?

Dear Discredit,
I don't remenber where I saw this term. The midcircle theorem is the following:
let 1 and 2 be two secant circles, A and B their points of intersection, La a line passing throuh A and intersecting 1 and 2 at M and N, Lb a line passing throuh B and intersecting 1 and 2 at M' and N',
I, J the midpoints of MN, M'N'.
Then A, B, I and J are concyclic.

If we can proof  O1PO2O is a parallelogram then the problem is done,but now I have no idea about that  :(

[quote="discredit"]The diagonals $ AC,BD$ of convex quadrilateral $ ABCD$ cut at $ P$. The circumcircles of triangles $ ABP,DCP$ cut at $ M$ distinct from $ P$. The circumcircles of triangles $ ADP,BCP$ cut at $ N$ distinct from $ P$. Perpendiculars to $ AC,BD$ passing through midpoints of $ AC,BD$ respectively meet at $ O$.

Prove that $ M,O,P,N$ are concyclic.[/quote]
[hide="Solution"]
Let the midpoints of $ AC$ and $ BD$ be $ X$ and $ Y$ respectively.  Since $ APND$ and $ BPNC$ are cyclic, we have that $ \angle DBN \equal{} \angle PBN \equal{} \angle PCN \equal{} \angle ACN$.  Similarly, $ \angle BDN \equal{} \angle CAN$, so $ \triangle CAN\sim \triangle BDN$.  Since $ X$ and $ Y$ are the respective midpoints of $ AC$ and $ BD$, we have that $ \triangle BYN\sim \triangle CXN$, so $ \angle CXN \equal{} \angle BYN$.  Thus, $ PYNX$ is cyclic.  Similarly is $ PMXY$.  Furthermore, $ \angle OYP \plus{} \angle OXP \equal{} 90 \plus{} 90 \equal{} 180$, so $ PXOY$ is cyclic.  Thus, $ N$, $ O$, and $ M$ all lie on the circumcircle of $ \triangle PXY$, so $ PNOM$ is cyclic. [/hide]


Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that
\[ \sum a^2\sqrt{a^2\plus{}3bc} \ge \sum ab\sqrt{2(a^2\plus{}b^2)}.\]
Equality holds if and only if $ (a,b,c) \sim (1,1,1), (1,1,0).$

My solution is ugly so I post it here to receive a nicer solution.

[quote=can_hang2007]Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Prove that
\[ \sum a^2\sqrt{a^2\plus{}3bc} \ge \sum ab\sqrt{2(a^2\plus{}b^2)}.\]
[/quote]
The is a proof by SOS, but it's very ugly.


Determine all continuous functions $f: \mathbb R \to \mathbb R$ such that
\[ f(x^2f(x) \plus{} f(y)) \equal{} (f(x))^3 \plus{} y\]
holds for all $x,y \in \mathbb R$.

Plugging in $ x\equal{}0$, we have $ f(f(y))\equal{}f(0)^3\plus{}y$, so $ f$ is bijective. Let $ r$ be the real number such that $ f(r)\equal{}0$. $ x\equal{}y\equal{}r$ gives $ f(f(r))\equal{}r$. But we also have $ f(f(r))\equal{}f(0)^3\plus{}r$, so $ f(0)\equal{}0$. Thus, we have $ f(f(y))\equal{}y$.
Now let $ x\equal{}1$, giving $ f(f(y)\plus{}1)\equal{}f(1)^3\plus{}y$. Plugging in $ y\equal{}0$ gives $ f(1)\equal{}f(1)^3$, or $ f(1)\equal{}0,1$. We cannot have $ f(1)\equal{}0$, so $ f(1)\equal{}1$.
Since $ f$ is continuous and injective, it must be monotonous, and since $ f(0)\equal{}0, f(1)\equal{}1$, $ f$ must be monotonically increasing.
Suppose there were numbers $ a\neq b$, with $ f(a)\equal{}b,f(b)\equal{}a$. (WLOG, $ a<b$.) Then we would have $ f(a)>f(b)$, contradiction.
Thus, we must have $ f(x)\equal{}x$ for all $ x$.

[quote="Brut3Forc3"]Plugging in $ x \equal{} 0$, we have $ f(f(y)) \equal{} f(0)^3 \plus{} y$, so $ f$ is bijective. Let $ r$ be the real number such that $ f(r) \equal{} 0$. $ x \equal{} y \equal{} r$ gives $ f(f(r)) \equal{} r$. But we also have $ f(f(r)) \equal{} f(0)^3 \plus{} r$, so $ f(0) \equal{} 0$. Thus, we have $ f(f(y)) \equal{} y$.
Now let $ x \equal{} 1$, giving $ f(f(y) \plus{} 1) \equal{} f(1)^3 \plus{} y$. Plugging in $ y \equal{} 0$ gives $ f(1) \equal{} f(1)^3$, or $ f(1) \equal{} 0,1$. We cannot have $ f(1) \equal{} 0$, so $ f(1) \equal{} 1$.
Since $ f$ is continuous and injective, it must be monotonous, and since $ f(0) \equal{} 0, f(1) \equal{} 1$, $ f$ must be monotonically increasing.
Suppose there were numbers $ a\neq b$, with $ f(a) \equal{} b,f(b) \equal{} a$. (WLOG, $ a < b$.) Then we would have $ f(a) > f(b)$, contradiction.
Thus, we must have $ f(x) \equal{} x$ for all $ x$.[/quote]

Nice demo, with a little mistake on the third line : Setting $ x\equal{}1$ gives $ f(f(1)\plus{}f(y))\equal{}f(1)^3\plus{}y$ and not $ f(f(y) \plus{} 1) \equal{} f(1)^3 \plus{} y$.

But this mistake as no consequence : setting then $ y\equal{}0$ as you suggested gives $ f(f(1))\equal{}f(1)^3$ (and not $ f(1) \equal{} f(1)^3$) and since $ f(f(1))\equal{}1$ you get immediately the same conclusion $ f(1)\equal{}1$

Congrats.

$ f(x^{2}.f(x) + f(y)) = (f(x))^{3} + y$.
if  $ f(y_{1}) = f(y_{2})$  $ \implies$  $ y_{1} = y_{2}$  $ \implies$ $ f$  is injective.
$ y = k - (f(x))^{3}$ $ \implies$  $ f$ is surjective.
$ f(u) = 0$  $ \implies$  $ f(f(y)) = y$,
$ f((f(y)) = f(0)^{3} + y = y$  $ \implies$  $ f(0) = 0$ $ \implies$  $ f(x^{2}.f(x)) = f(x)^{3}$
$ f(f(1)) = 1$ and $ f(f(1)) = f(1)^{3}$  $ \implies$  $ f(1) = 1$.
$ f$  is continous and ,$ f$ is injective $ \implies$  $ f$ is increasing function.$ (f(1)>f(0))$
$ f(f(x)) = x$  $ \implies$  $ f(x) = a$  $ \implies$ $ f(a) = x$
if $ a > x$ $ \implies$  $ f(x) > f(a)$ contradiction!
if $ a < x$  $ \implies$  $ f(x) < f(a)$  contradiction!
$ \implies$  only $ a = x$  $ \implies$  $ f(x) = x$ $ \Box$


function
logarithms
algebra proposed
algebra

Let f(n)=[1/{lnn}] for all positive integers n., where {},[] are the fractional and integer part and ln x is the natural logarithm of x. Prove that f is surjective. Is it true that for any positive integer k we can find an infinit set M of natural numbers such that for any n in M we have f(n)=f(n+1)=...=f(n+k)?

Let f(a) = b, then we have 1/(b+1) < {ln a} \leq  1/b.
So c + 1/(b+1) < ln a \leq c + 1/b for some positive integer c.
e^(c + 1/(b+1)) < a  \leq e^(c + 1/b).
If we let c be large enough we have e^(c + 1/(b+1)) + 1 < e^(c + 1/b), so for each positive integer b' we can find a positive integer a' such that f(a') = b'.

Using e^(c + 1/(b+1)) < a  \leq e^(c + 1/b) again, if we let c be (really) large enough we can have e^(c + 1/(b+1)) + k + 1 < e^(c + 1/b).  Then f([e^(c + 1/(b+1)) + 1]) = f([e^(c + 1/(b+1)) + 2]) = ... = f([e^(c + 1/(b+1)) + k + 1]) -- where [x] is the floor of x.

It's easy to see that [e^(c' + 1/(b+1)) + 1], where c'  > c is a positive integer can be an element of M for all c'.


Consider the sequence $ (a_n)_{n \ge 1}$ of real numbers and the sequences $ b_n\equal{}a_1\plus{}a_2\plus{}...\plus{}a_n, c_n\equal{}b_1\plus{}b_2\plus{}...\plus{}b_n \forall n \ge 1.$

$ a)$ If $ a_n\equal{}n\plus{}1, \forall n \ge 1$ then compute $ c_2$ and $ c_{2009}.$

$ b)$ If $ c_n\equal{}n^3, \forall n \ge 1$ then compute $ a_2$ and $ a_{2009}.$


function
real analysis
real analysis unsolved

find all function $f: \mathbb{R}\rightarrow\mathbb{R}$ such that there exist for this function atleast one interval with this property that for every point in this interval we have \[ f'(x)=f(x)+\frac{1}{(1-x)^2} \]

Multiplting by $e^{-x}$ we get $\left(e^{-x}f(x)\right)=\frac{e^{-x}}{(1-x)^2}$ Integrating we find the function. This is the solution, but there are many detailes to fix. For example for x=1 the equation could not hold. etc.


i'm a young enthusiast and wud like to ask all sorts of questions i get related to biology here..........i don't yet have proper biology in school......so i hope u people will help me quench my curiosity,won't ya??? :)
i welcome people like me to write on ......any silly but sensible question is welcome!!please feel free!!!! here goes my first question....

[u]question1[/u]
why do people have different coloured eyeballs? does this make any difference to their vision or perception?

About the first part of the question...

People have different colored eyeballs because of genes from their mom and dad. Say, that a child's mom has brown eyes and a dad has, let's say, blue eyes. If blue is recessive and brown is dominant, then there is a 3/4 prob. of the child having brown eyes and a 1/4 of the child having blue. There is also incomplete dominance, which is a mix of the two colors. PM me if you have any further questions.

 :)

Also, scientists are not sure whether color affects strength of the eye or perception, chances of blindness etc. They still have not reached a conclusion.

thankyou someperson01 and ragnarok!!!
[u]question2[/u]

why do tortoises have such a long lifespan?

There are many factors that affect its longetivity of life- one that is most noticable is it's shell. The shell gives an immense amount of protection- Many animals would not dare trying to crack it. It's size also gives it a good advantage, too (Mostly in the Giant Tortoise)
Another reason is that its young usually don't survive when it gives birth to them, many predators often feed on them.
In a species the most important thing is to continue it and reproduce. Since it is so hard for tortoises to have a child successfully live and become an adult, they have adapted themselves to live for a long time so that they can give birth more times. This way, there is more of a chance to reproduce.

please do answer....i have a long list of questions...... :(

[quote="firebolt"]thanks ragnarok!

now my next question........!!!!!

[u]question 3[u]

do animals have feelings i.e emotions like us?   do they feel other's pains......i know dogs dolphins do but do the other animals also hav such perceptions? is it true that reptiles have no emotions? also, does the extent of emotion expression determine the brain develpoment?

 :) [/u][/quote]
I don't know much about this topic, but ah well.
(Try googling things, too)
Scientists are not sure whether or not some animals have emotions, however many are beginning to believe they do. Animals may or may not have emotions- however they do have ways of expressing them. One that comes to mind is lizards- they bite each other's necks when they show love, as in a hug or a kiss. Dogs wag their tails when they are happy, cats purr when they feel comforted- It simply makes sense that if animals can express their emotions through actions, then they must have emotions themselves.

well ur answer does make sense...but what about the level of their brain development? does the expression of emotions mean they are showing mental maturity in perceptions?

anyways...thanks for the answer :lol: .....here goes my next question...

[u]question 4[/u]

we all know that there are different blood groups.......there are  compatible and non-compatible blood groups right? now when a mosquito bites two such persons consecutively whose blood groups are non-compatible.......doesn't blood agglutination take place within the mosquito itself? i hope ya get my question?!!  :|

please do answer!!!! :)

I think mosquitos have anticoagulants in their saliva, so this wouldn't be an issue.

What I have to add on:

Q1: The colored part of the eye, or iris, is dependent upon the chemical melanin. Melanin is responsible for  hair color, pigmented neural regions such as the locus ceruleus, skin color, certain parts of the adrenal gland, and more.

As someperson01 stated, genes coming from parents attribute to which color melanin determines.

WHERE MELANIN IS FOUND
Dermal melanin, which factors to skin color, is produced by melanocytes located in the epidermis of the skin. Melanocytes, can also be found, in the uvea, middle layer of the eye. Those uvea-positioned melanocytes factor to iris color. In a square millimeter of skin, you'll find from 1,000 to 2,00 melanocytes.

[img]http://upload.wikimedia.org/wikipedia/commons/d/dd/Illu_skin02.jpg[/img]

Certain types of melanin: eumelanin, neuromelanin, pheomelanin

Eumelanin: Consists of two types, black eumelanin and brown eumelanin. The most of all biological melanin.
Neuromelanin: As you can tell from the "Neuro", this melanin contributes to pigmented neural regions.
Pheomelanin: Determines hair color and skin color. Pheomelanin contributes to pink and red hues, being found mostly in red hair.

Q2: I would read this article: [url]http://www.slate.com/id/2138560/fr/rss/[/url] explaining some reasons towards why tortoises live long life spans. If you're having trouble with the url, I have copied the text of the article.

[quote]A 255-year-old giant tortoise named Adwaitya died at the Calcutta Zoo on Wednesday. The animal had been brought to India from the Seychelles Islands in the mid-18th century as a gift to the British colonial ruler Robert Clive. Why do giant tortoises live for such a long time?

So they can reproduce more effectively. Long life spans provide an evolutionary advantage for certain types of animals. It makes sense to stick around if you live in an unpredictable or harsh environment where it's hard to reproduce on a regular basis. (Desert animals, for example, tend to get quite old before they die.) You'd also want to have a long life if you could only give birth infrequently for some other reason, or if you spent a lot of time caring for each of your offspring.

Every species has its own "life history," or schedule according to which it passes on its genes to the next generation. Some animals invest their time and energy in having lots of babies while they're young; others use their resources to live longer. An animal might live fast if there are lots of predators around—if you're going to die young, you might as well get the baby-making over with as early as possible. (This strategy works only in a predictable environment, where you know there will be enough food around to feed all your babies.) The long-lived animals, on the other hand, have often evolved some way to protect themselves from predators.

That's why armored beasts—like turtles, armadillos, and beetles—tend to stick around for a while compared to similarly sized animals. Flying animals also seem to have an advantage; bats last much longer than rodents, perhaps because they can escape predators more easily. Very poisonous animals can last a long time, as well—any strong defense against predation will do the trick.

The giant tortoise has an armored shell for protection, but it also has a couple of other features that correlate with long life. As an island species, it has enjoyed natural protection from predators over the course of its history. It's also very large, and larger animals tend to live longer than small ones. (There are plenty of exceptions to the bigger-lives-longer rule—take the little bat, for example. Large body size also goes along with living on an island; click here for an Explainer on island gigantism and dwarfism.)

All of these factors interact over the course of an animal's evolutionary history, so you can't say one trait causes another. We can't conclude that a tortoise lives to 255 because it has a shell or because it's big; nor can we say that the tortoise evolved a shell or its large size because it has such a long life. Other types of animals—especially those with bigger brains—seem to have developed long life spans for different reasons entirely: In general, the more social the animal, the longer it lives. Social insects have more longevity than the lonesome varieties: Some flies last only a couple of days, while a termite queen can survive for 30 years.

Got a question about today's news? Ask the Explainer.

Explainer thanks David Reznick of the University of California at Riverside[/quote]

Q3: My best guess, would result in agreeing that animals expressing emotions is a sign of mental maturity. If you wish to learn more, search the indexes of neurology textbooks for the words, "paleopallium", "limbic system", or "old mammalian brain"

Q4: It wouldn't affect the mosquito in any negative aspect, such as if one would give type A blood to one with type B blood. Due to the fact, that the mosquito drinks the blood making it pass through the alimentary tract, and not into the circulatory system.

The type A blood would have already reacted with the hydrochloric acid of the stomach, before the type B blood even passed down the esophagus. If in the case that a trace of type A blood reacted with B, agglutination would only take place in the stomach.

The very same follows, not only for mosquitos, but for all organic life forms with a mouth. If an A blood patient, drank B blood, nothing would happen. Of course, a pyschiatrist would contact him on his "blood-drinking" dillemma.  :rotfl:

If you were to inject the type B blood into the a major vein of that A blood patient, then some great problems will arise.


ratio
number theory unsolved
number theory

Show that for every natural number $k \geq 2$ it is possible to find natural number $n$ which satisfies:

$\frac{n}{\pi (n)}=k$

I solved it some very long time ago. Give me some time to find it on the forum.

In fact, I still remember the magnificient solution. Since $\frac{\pi(mk)}{m}$ goes to 0 as $m$ goes to infinity and since $\pi(k)\geq 2$, there is a maximal $m$ such that $\pi(mk)\geq m$. Suppose that $\pi(mk)>m$ then $\pi((m+1)k)=\pi(mk+k)\geq \pi(mk)\geq m+1$, contradicting the maximality of $m$. Thus $\pi(mk)=m$ and it's done.

Thanks a lot Harazi for this beatiful solution. I realized that I might posted exactly the same problem some months ago - if this is the case excuse my ultra-short memory  :blush:


inequalities
trigonometry
geometry
vector
complex numbers
triangle inequality
inequalities proposed

Let $ ABC$ be an acute triangle. Let $ M$ be a point inside the triangle $ ABC$. Let $ x$, $ y$, $ z$ be nonnegative real numbers. Prove that:

$ x\cdot MB\cdot MC \plus{} y\cdot MC\cdot MA \plus{} z\cdot MA\cdot MB$
$ \geq 2\cdot \text{min}\left\{\frac {x}{\sin A},\frac {y}{\sin B},\frac {z}{\sin C}\right\}\cdot\left(\text{area of triangle }ABC\right)$.

When does the inequality become an equality?

is nobody interested in my problem?

It took me 3 months to get the first solution. So I am glad if someone finds a solution in the next week.

Using the substitutions
 $x=\frac{p}{bc}$,  $y=\frac{q}{ca}$,  $z=\frac{r}{ab}$
and
  $P=\frac{MB.MC}{bc}$,  $Q=\frac{MC.MA}{ca}$,  $R=\frac{MA.MB}{ab}$,
the inequality becomes as follows

   $pP+qQ+rR\geq min(p,q,r)$.

Using now the known inequality $P+Q+R\geq1$, it suffices to show that

$\frac{pP+qQ+rR}{P+Q+R}\geq min(p,q,r)$.

But this one is clearly true. ;)

[quote="Vasc"]Using the substitutions
 $x=\frac{p}{bc}$,  $y=\frac{q}{ca}$,  $z=\frac{r}{ab}$
and
  $P=\frac{MB.MC}{bc}$,  $Q=\frac{MC.MA}{ca}$,  $R=\frac{MA.MB}{ab}$,
the inequality becomes as follows

   $pP+qQ+rR\geq min(p,q,r)$.

Using now the known inequality[b] [u][i]$P+Q+R\geq1$[/i][/u][/b], it suffices to show that

$\frac{pP+qQ+rR}{P+Q+R}\geq min(p,q,r)$.

But this one is clearly true. ;)[/quote]

Yeah, you are right. The key is to prove that inequality, but P +Q + R >= 1 is the known inequality, is not it?  :( Does it have a name?

I read a lot of books and have never known that the inequality was found before. I found it and two different proofs in 1997  :(

You can find a first proof in the Geometry topic: "again circles and this time some radii inequalities ".

It is based on complex numbers or vectors, as you prefer.

Can anyone post also a simple trigonometric solution ?

I am very interested in it.

Thank you very much.

My first solution uses a formula containing MA, MB, MC, area of ABC, a,b,c. It is very complicated.

My second solution is simpler:

Let D, E, F be three points in MA, MB, MC respectively such that: MA * MD = MB * ME = MC * MF.

The inequality now becomes :

 MD^2 * EF  + ME ^2 * DF + MF^2 * DE >= DE * EF * FD

(Thanx manlio for the correction)

Now with the triangle DEF, using vector to find the minimal of the left hand side. It happens when M is the intersection of 3 bisec-angle of D, E, F angles and in fact it is equal to DE * EF *FD.

To prove in the algebraic way the inequality
$P+Q+R\geq1$,
I propose to multiply the inequality by

$\frac{abc}{MA.MB.MC}(a.MA+b.MB+c.MC )$.

I wait a such solution.

For Manlio and Nickolas, especially.
After multiplying, we obtain
$a^2+b^2+c^2+\frac{bc}{MB.MC}(MB^2+MC^2-a^2)+...\geq 0$,
then
$a^2+b^2+c^2+2bc.cosBMC+2ca.cosCMA+2ab.cosAMB\geq 0$,
that is
$(a+b.cosAMB+c.cosCMA)^2+(b.sinAMB-c.sinCMA)^2\geq 0$.

Dear Vasc,
thank you very much for your very nice solution.

Darij , has provided solution in geometry inequality forum, a proof of this of vectors in complex , here goes the proof
 [hide]We have
[b]Lemma 1.[/b]
$\frac{MB\cdot MC}{bc} + \frac{MC\cdot MA}{ca} + \frac{MA\cdot MB}{ab} \geq 1$.

[b]Lemma 2.[/b] For any three positive numbers $x, y, z $, if $yz+zx+xy \geq 1$, then $x+y+z \geq \sqrt3$.

From these lemmas, Theorem 1 quickly follows, since Lemma 2 yields $yz+zx+xy \geq 1$, where $x=\frac{MA}{a}$, $y=\frac{MB}{b}$ and $z=\frac{MC}{c}$, and thus by Lemma 1 we have $x+y+z \geq \sqrt3$, so that $\frac{MA}{a} + \frac{MB}{b} + \frac{MC}{c} \geq \sqrt3$ .
Remains to proves Lemmas 1. Well, Lemma 2 is almost evident, since $yz+zx+xy \geq 1$ implies $x^2+y^2+z^2 \geq 1$ (because of the famous inequality $x^2+y^2+z^2 \geq yz+zx+xy$), and thus

$\left(x+y+z\right)^2 = x^2+y^2+z^2 + 2\cdot\left(yz+zx+xy\right) \geq 1 + 2\cdot 1 = 3$

and hence $x+y+z\geq \sqrt3$.

So it remains to show Lemma 1. The most easy way to do it is to work in the complex plane, letting M be the origin and assigning to the points A, B, C the affixes u, v, w. Then,

$\frac{MB\cdot MC}{bc} = \frac{MB\cdot MC}{CA\cdot AB} = \frac{\left|v-0\right|\cdot \left|w-0\right|}{\left|w-u\right|\cdot\left|u-v\right|} = \frac{\left|v\right|\cdot \left|w\right|}{\left|w-u\right|\cdot\left|v-u\right|} = \left|\frac{vw}{\left(w-u\right)\left(v-u\right)}\right|$.

Similarly,

$\frac{MC\cdot MA}{ca} = \left|\frac{wu}{\left(u-v\right)\left(w-v\right)}\right|$;
$\frac{MA\cdot MB}{ab} = \left|\frac{uv}{\left(v-w\right)\left(u-w\right)}\right|$.

Thus, by the triangle inequality,

$\frac{MB\cdot MC}{bc} + \frac{MC\cdot MA}{ca} + \frac{MA\cdot MB}{ab}$
$= \left|\frac{vw}{\left(w-u\right)\left(v-u\right)}\right| + \left|\frac{wu}{\left(u-v\right)\left(w-v\right)}\right| + \left|\frac{uv}{\left(v-w\right)\left(u-w\right)}\right|$
$\geq \left|\frac{vw}{\left(w-u\right)\left(v-u\right)} + \frac{wu}{\left(u-v\right)\left(w-v\right)} + \frac{uv}{\left(v-w\right)\left(u-w\right)}\right|$.

But since

$\frac{vw}{\left(w-u\right)\left(v-u\right)} + \frac{wu}{\left(u-v\right)\left(w-v\right)} + \frac{uv}{\left(v-w\right)\left(u-w\right)} = 1$,

we thus get

$\frac{MB\cdot MC}{bc} + \frac{MC\cdot MA}{ca} + \frac{MA\cdot MB}{ab} \geq \left|1\right| = 1$,[/hide]
[hide="note"]I've copied the posts of darij from the forum mentioned but the link I'm still hunting for , If I happened to come across the link again I shall post the actual proof and the link too ....[/hide]


*points at the letters* I found them!

[hide="Seriously"]word, letter

A word contains several letters.
A letter (as in one you write) contains several words.[/hide]

Several maps are in the US, and the US is on several maps.

I [i]think[/i] "US" is more of an abbreviation...

[quote="The Zuton Force"]Fine... several maps are in [i]India[/i], and [i]India[/i] is in several maps[/quote]

But no map can contain several [i]Indias[/i] :P

Better change it to country:
A country contains several maps
A map contains several countries

coffee is in people and people are into coffee

Does that count?

Would these work?

[hide="A"]Bubble, containing several B's[/hide]

[hide="B"]Aardvark, containing several A's [/hide]

[quote="shaggy75"]Would these work?[hide="A"]Bubble, containing several B's[/hide][hide="B"]Aardvark, containing several A's [/hide][/quote]What?

[quote="randomdragoon"]A and B are words.

B is contains several A's.
A is contains  several B's.

Find A and B.[/quote]

B, aardvark, contains the letter 'a' several times

A, bubble, contains several 'b's

The object B contains several of the object A.
The object A contains several of the object B.

As far as I know, B's don't contain very many aardvarks.

people are in several memories [e.g. people are remembered]
memories are in several people [e.g. people remember]

ok that didn't work out too well

ahem let's get mathematical shall we?  :D

in $ R^n$ nontrivial closed spheres contain uncountably many open spheres and vice versa, so:
A - nontrivial closed spheres in $ R^n$
B - open spheres in $ R^n$

(just to be safe, we consider the euclidean metric throughout, though many others hould work just as well)

topology pwns!


Show:

\[ \frac{1}{9} \pi  \left(\sqrt{3}\minus{}3 \tan\left[\frac{1}{6} (\pi \minus{}2 a \pi )\right]\right)\equal{}
\frac{\pi \sin(\pi a/3)}{3\sin(\pi/3)\sin\big[\pi/3(a\plus{}1)\big]}\]


number theory
prime numbers
number theory unsolved

Does there exist two natural numbers such that the products of their divisors are equal?

No, if $a$ and $b$ are two positive integers such that $f(a)=f(b)$, where $f(n)=\prod_{d|n}d$. Then $\{p\; \text{is prime}|p|a\}=\{p\; \text{is prime}|p|b\}$. Put $a=p_{1}^{x_{1}}\cdot ... \cdot p_{k}^{x_{k}}$ and $b=p_{1}^{y_{1}}\cdot ...\cdot p_{k}^{y_{k}}$, where $p_{i}$ are prime numbers and $x_{i},y_{i}$ are positive integers. From $f(a)=f(b)$ we have $\frac{x_{1}}{y_{1}}=...=\frac{x_{k}}{y_{k}}$, and $x_{i}=y_{i}$. Done!

Does not.
Let $n=\prod_{i}p_{i}^{k_{i}}$, then $D(n)=\prod_{d|n}d=\prod_{i}p_{i}^{k_{i}\tau(n)/2}$.
Therefore $D(n)=D(m)$ equavalent to $m=\prod_{i}p_{i}^{m_{i}}, k_{i}=rt_{i},m_{i}=qt_{i}$ and $r\prod_{i}(1+rt_{i})=f(r)=f(q)=q\prod_{i}(1+qt_{i})$.
Because $t_{i}\ge 1$ can not be $f(r)=f(q)$, when $r\not =q$.


complex analysis
complex analysis unsolved

prove that $\sum_{n=1}^{\infty} \frac{1}{n^{2p}}=\frac{\pi}{2}Res_0(\frac{cos(\pi z)}{z^{2p}sin(\pi z)})$

It follows from the well-known formula $ctgz=\frac{1}{z}+\sum_{n=1}^{\infty} \frac{2z}{z^2-n^2 \pi^2}$ that
$\pi z ctg(\pi z)=1-2\sum_{m=1}^{\infty} \frac{z^2}{m^2-z^2}$ but for $|z|<1$ we have $\frac{z^2}{m^2-z^2}=\sum_{n=1}^\infty \left( \frac{x^2}{m^2} \right) ^n$, which means that $\frac{z^2}{m^2-z^2}=\sum_{n=1}^{\infty} \zeta_{2n} z^{2n}$, so for $|z|<1$ we actually have
$\pi z ctg(\pi z)=1-2\sum_{n=1}^{\infty} \zeta_{2n}z^{2n}$. To this end we have just to notice that $Res_{z=0} \frac{cos(\pi z)}{z^{2p}sin(\pi z)}=Res_{z=0} \frac{\pi z ctg(\pi z)}{z^{2p+1} \pi}$ is equal to the coefficient in the $2p$'th power of $z^{2p}$ in the expansion of it's numerator $\pi z ctg(\pi z)=1-2\sum_{n=1}^{\infty} \zeta_{2n}z^{2n}$ divided by $\pi$, which implies the problem assertion


Let $p(x)=x^5+x$ and $q(x)=x^5+x^2$, Find al pairs $(w,z)\in \mathbb{C}\times\mathbb{C}$, $w\not=z$ for which $p(w)=p(z),q(w)=q(z)$.

From $p(w)-q(w) = p(z)-q(z)$ we have
$$ w-w^2 = z-z^2 \quad \Longrightarrow \quad (w-z)(w+z-1) = 0 \quad \Longrightarrow \quad w = 1-z $$
$$ p(w) = p(z) \quad \Longrightarrow \quad w^5 + w = z^5 + z \quad \Longrightarrow \quad (w-z)(w^4+w^3z+w^2z^2+wz^3+z^4+1) = 0 $$
$$ w^4+w^3z+w^2z^2+wz^3+z^4+1 = 0 \; ........(1) $$
Substitute $w = 1-z$ to $(1)$, we get
$$ (z^2-z+1)(z^2-z+2) = 0 $$
$$ z_{1,2} = \frac{1}{2}\pm\frac{i\sqrt{3}}{2} \quad \textrm{or} \quad z_{3,4} = \frac{1}{2}\pm\frac{i\sqrt{7}}{2} $$
The answers are
$$ (w,z) = \left(\frac{1}{2}+\frac{i\sqrt{3}}{2}, \frac{1}{2}-\frac{i\sqrt{3}}{2}\right), \left(\frac{1}{2}-\frac{i\sqrt{3}}{2}, \frac{1}{2}+\frac{i\sqrt{3}}{2}\right), \left(\frac{1}{2}+\frac{i\sqrt{7}}{2}, \frac{1}{2}-\frac{i\sqrt{7}}{2}\right), \left(\frac{1}{2}-\frac{i\sqrt{7}}{2}, \frac{1}{2}+\frac{i\sqrt{7}}{2}\right). $$
Note: It is clear to verify that these solutions satisfy for the condition of the problem.

first we have $w^{5}+w=z^5+z$ and $w^5+w^2=z^5+z^2 \implies z-w=z^2-w^2 \implies z+w=1$\\

we have $w^2+z^2=1-2wz$ and $w^4+z^4=1+2w^2z^2-4wz$\\

So,alo we have $w^4+z^4+wz(w^2+z^2+wz)=-1\implies wz=1,2$\\

so we get: $w=(\varepsilon ,\varepsilon^2)$ and $z=(1-\varepsilon , 1-\varepsilon^2)$ , where $\varepsilon^3=1$\\

and also $w=\frac{1 \pm \sqrt{7}i}{2}$ , so $z=\frac{1 \mp \sqrt{7}i}{2}$\\

$\square$


As you can see, I've posted a couple of problems so far, because I need to get better at combinatorics, anyone have any suggestions of any websites or books I should read? I've done the AOPS book on Combinatorics but it goes a little fast and some of the problems at the end of the chapter are hard :o  :(  :?

maybe take the online class from this website, intermediate counting or something
that's supposed to be focused on probability and combinatorics

I found several PDF's about combinatorics on different websites while preparing for Mathcounts. Try googling "combinatorics" or "counting and probability" or something like that.

Where can I get excellent problems on combinatorics??/


geometry
perimeter
symmetry
rotation
inequalities
search
triangle inequality

Seniors (Grades 11 and up)

[b]International Mathematics
TOURNAMENT OF THE TOWNS[/b]

[b]O-Level Paper, Spring 2004[/b]

1. [4] Segments AB, BC and CD of the broken ABCD are equal and are tangent to a circle with centre at point O. Prove that the point of contact of this circle with BC, the point O and the intersection point of AC and BD are collinear.

2. [4] A positive integer a > 1 is given (in decimal notation). We copy it twice and obtain a number b = [u]aa[/u] which happened to be a multiple of a2. Find all possible values of b/a2.

(e.g. a = 47, b = 4747)

3. [4] Perimeter of a convex quadrilateral is 2004 and one of its diagonals is 1001. Can another diagonal be 1? 2? 1001?

4. [5] Arithmetical progression a1, a2, a3, a4, ... contains a12, a22 and a32 at some positions. Prove that all terms of this progression are integers.

5. [5] Two 10-digits integers are called neighbours if they differ in exactly one digit (for example, integers 1234567890 and 1234067890 are neighbours). Find the maximal number of elements in the set of 10-digit integers with no two integers being neighbours.

Now, why can't I just say [hide]"by symmetry" for the first one?  Stupid rigor .[/hide]

Number 2:

[hide]a2 | aa = b

a | 10000...001, where there is exactly one more digit on the right than on the left.  This means that whatever the number 10...01/a is, it can only have one digit.  The only 1-digit number that could possibly divide 10...01 is 7, since it would have to be odd, it couldn't be 1, and it clearly isn't 3, 5, or 9.  Thus, the only possibility is 7.  And this is conveniently also equal to b/a2.  All done .[/hide]

Number 4:

[hide]

So that I don't have to use <sub>, I'll call the first three numbers in the sequence a, b, and c

The common difference is b-a = c-b, so b^2 - a^2 is a multiple of b-a, thus a+b is an integer, and similarly c^2 - b^2 is a multiple of c-b, so b+c is an integer. c = 2b-a, so:

a+b and 3b-a are integers.

Adding these, 4b is an integer, and eliminating b, 4a is an integer.

Notice that we must also have a^2 - a is a multiple of b-a.

Firstly, if a is c/2, where c is odd, then b must also be of the same form, d/2, where d is odd (to make a+b integer). Now (a^2-a)/(b-a) = (c^2 - 2c)/(2d - 2c), which is not an integer since the numerator is odd.

Otherwise, a = c/4 and b = d/4 where c and d are both odd. Again, (a^2 - a)/(b-a) = (c^2 - 4c)/(4d - 4c), but the numerator is still odd.

So a must be an integer, and so must b, and so all terms are.

[/hide]

Number 3:

We have 3 separate problems.

First, the other diagonal is 2:

[hide]This is possible.  To see why, start with the two diagonals perpendicual to each other at their centers (giving us a rhombus).  It has side length :sqrt:(1 + (1001/2)2), so the perimeter is :sqrt:(16 + 20022), which is clearly less than 2004.  Now slide the diagonal of length 2 along the longer diagonal, keeping it always perpendicular at its own center.  When the center of the short diagonal is exactly at an endpoint of the long diagonal (which wouldn't give us a convex quadrilateral, since 3 vertices would lie on a straight line, although it doesn't matter) the two longer sides are each slightly greater than 1001 while the two shorter sides (which together make up the short diagonal) have combined length 2, making the total perimeter strictly greater than 2004.  Since this was a continuous process, there was some point in between where the perimeter was exactly 2004.  This is our quadrilateral.[/hide]

For the other diagonal of length 1001:

[hide]This is possible.  Set the two diagonals perpendicular to each other at their centers.  This gives us a perimeter of 2002 :rt:2 > 2004.  Rotate one of the diagonals continuously until the two diagonals overlap, when we have perimeter 2002 < 2004.  Again, this is a continuous process so at some point in between we must have a quadrilateral with perimeter exactly 2004.[/hide]

For the other diagonal of length 1:

[hide]This is not possible.  To see why, draw any quadrilateral with both diagonals, and apply the triangle inequality to each of the 4 small triangles formed, with the side of the quadrilateral as the lone side of each inequality.  Add the four inequalities to see that Perimeter < 2*(Sum of Diagonals) for any quadrilateral.  This means that the sum of the diagonals of our quadrilateral must be strictly greater than 1002 -- thus, the other diagonal cannot be 1.

This analysis leaves open the possibility that the other diagonal could have any length d, 1 < d < 1003 (apply the triangle inequality in a different way to get the upper bound), although it does not prove that such a quadrilateral exists for all such d.  (Even though I'm pretty sure they do in fact have to exist.)[/hide]

Man, I should know number 5 right away -- I bet a quick internet search for "error correcting codes" will find exactly what I need.

Oh, tripleM, re: your solution to number 4:

when you use c later on in the proof, that isn't supposed to be the same c you used to avoid using a3, is it?