geometry
analytic geometry
ratio
power of a point
radical axis
geometry proposed

$P,Q$ are two points in the plane of $ \triangle ABC.$ Let $ \triangle P_aP_bP_c$ and $ \triangle Q_aQ_bQ_c$ be the cevian triangles of $ P,Q$ WRT $ \triangle ABC.$ Sidelines $ P_bP_c$ and $ Q_bQ_c$ meet at $ A'$ and define cyclically the points $ B',C'.$ Triangle $\triangle A'B'C'$ is known as the 'side triangle' of $ \triangle P_aP_bP_c$ and $ \triangle Q_aQ_bQ_c.$ Show that $ \triangle A'B'C'$ and $ \triangle ABC$ are perspective and the perspector $ J$ is the intersection of the perspectrices of $ \triangle P_aP_bP_c$ and $ \triangle Q_aQ_bQ_c$ WRT $ \triangle ABC.$

Dear Luis and Mathlinkers,
this situation has been initiated by Schroeter.
You will find a synthetic proof on my website: http://perso.orange.fr/jl.ayme  vol. 2  Les deux points de Schroeter.

for the median and orthic triangle of a triangle see p. 4
for a generalization p. 6
for the position of the perspector p. 14

In my paper, I haven't studied the case of the orthocenter and Gergonne point.
Sincerely
Jean-Louis

Thanks for the references dear Jean-Louis, I found that proof very nice. Mine uses barycentric coordinates in order to identify J for particular situations.

Let $(u_1: v_1: w_1)$ and $( u_2: v_2: w_2)$ be the barycentric coordinates of $P,Q$

$ P_a \ (0: v_1: w_1) \ , \ P_b \ (u_1: 0: w_1) \  , \ P_c \ (u_1: v_1: 0)$

$ Q_a \ (0: v_2: w_2) \ , \ Q_b \ (u_2: 0: w_2) \ , \ Q_c \ (u_2: v_2: 0)$

Perspectrices $ p , q$ of $ \triangle P_aP_bP_c$ and $ \triangle Q_aQ_bQ_c$ WRT $ \triangle ABC$ are the trilinear polars of  $ P$ and $ Q.$ Namely

$ p \equiv \frac {x}{u_1} \plus{} \frac {y}{v_1} \plus{} \frac {z}{w_1} \equal{} 0 \ \ , \ \ q \equiv \frac {x}{u_2} \plus{} \frac {y}{v_2} \plus{} \frac {z}{w_2} \equal{} 0$

$p \cap q \equiv  J \ (u_1u_2(w_1v_2 \minus{} v_1w_2): v_1v_2(w_1u_2 \minus{} u_1w_2): w_1w_2(v_1u_2 \minus{} u_1v_2))$

Lines $ P_bP_c$ and $Q_bQ_c$ intersect at $A'$ with coordinates

$ A' \ (u_1u_2(v_1w_2 \minus{} w_1v_2): v_1v_2(w_1u_2 \minus{} u_1w_2): w_1w_2(v_1u_2 \minus{} u_1v_2))$

By cyclic exchange of notation, we get the coordinates of $B',C'$

$ B' \ ( u_1u_2(w_1v_2 \minus{} v_1w_2): v_1v_2(u_1w_2 \minus{} w_1u_2): w_1w_2(v_1u_2 \minus{} u_1v_2))$

$ C' \ ( u_1u_2(w_1v_2 \minus{} v_1w_2): v_1v_2(w_1u_2 \minus{} u_1w_2): w_1w_2( u_1v_2 \minus{} v_1u_2))$

Therefore, $ \triangle ABC$ and $ \triangle A'B'C'$ are perspective through

$ J' \ (u_1u_2(w_1v_2 \minus{} v_1w_2): v_1v_2(w_1u_2 \minus{} u_1w_2): w_1w_2(v_1u_2 \minus{} u_1v_2))  \Longrightarrow J\equiv J'$

Lemma (well known): Let the cross ratio of 4 collinear points $ A, B, C, D$ be harmonic, i.e., $ \frac {\overline{AC}}{\overline{AD}} \cdot \frac {\overline{BD}}{\overline{BC}} \equal{} \minus{} 1.$ Then $ \frac {\overline{CA}}{\overline{CM}} \equal{} \frac {\overline{CD}}{\overline{CB}}$ $ \Longleftrightarrow$ $ M$ is midpoint of $ AB.$

For example: $ (M, m), (N, n)$ are circles with diameters $ AB \equal{} 2m, MD \equal{} 2n,$ respectively. Their radical axis cuts their center line at $ C,$ because then $ \frac {\overline{AC}}{\overline{AD}} \cdot \frac {\overline{BD}}{\overline{BC}} \equal{} \frac {m \plus{} \frac {m^2}{2n}}{m \plus{} 2n} \cdot \frac {2n \minus{} m}{\frac {m^2}{2n} \minus{} m} \equal{} \minus{} 1$ $ \Longleftrightarrow$ powers of $ C$ to $ (M), (N)$ are equal, $ \overline{CA} \cdot \overline{CB} \equal{} \overline{CM} \cdot \overline{CD}$ $ \Longleftrightarrow$ $ \frac {\overline{CA}}{\overline{CM}} \equal{} \frac {\overline{CD}}{\overline{CB}}.$
_____________________________

Let $ X_a \equiv P_bP_c \cap BC,$ $ X_b, X_c$ defined cyclically, and let $ Y_a \equiv Q_bQ_c \cap BC,$ $ Y_b, Y_c$ defined cyclically. $ X_aX_bX_c$ is perspetrix of $ \triangle P_aP_bP_c, \triangle ABC$ and $ Y_aY_bY_c$ is perspetrix of $ \triangle Q_aQ_bQ_c, \triangle ABC.$ Project one perspetrix to infinity, say $ Y_aY_bY_c,$ and mark all projected points with asterisks. This makes projected $ \triangle Q_a^*Q_b^*Q_c^*$ medial triangle of the projected $ \triangle A^*B^*C^*$ and $ Q^*$ its centroid. Now drop the asterisks in the projected notation.

$ Q_bQ_c \parallel BC$ cuts $ P_bP_c$ at $ A'.$ Lines $ PC, P_bX_a, AB, X_bP_a$ concurrent at $ P_c$ form a harmonic pencil. $ Q_b$ is midpoint of $ AC$ $ \Longrightarrow$ (by the lemma) $ \frac {\overline{P_bC}}{\overline{P_bQ_b}} \equal{} \frac {\overline{P_bX_b}}{\overline{P_bA}}$ $ \Longrightarrow$ $ \triangle AA'Q_c \sim \triangle X_bX_aC$ are centrally similar with similarity center $ P_b$ $ \Longrightarrow$ $ AA' \parallel X_bX_a \equiv X_aX_bX_c.$ Similarly, $ BB' \parallel X_aX_bX_c,$ $ CC' \parallel X_aX_bX_c$ $ \Longrightarrow$ $ \triangle ABC, \triangle A'B'C'$ are perspective, their perspector is the infinite point of $ X_aX_bX_c,$ the intersection of $ X_aX_bX_c$ with the line at infinity, identical with $ Y_aY_bY_c.$


[quote="marin.bancos"]For $ a,b,c$ positive real numbers, prove that

$ \left.\frac {a^3 \plus{} b^3 \plus{} c^3}{ab \plus{} bc \plus{} ca}\cdot\left[\left(\frac {a}{b}\right)^6 \plus{} \left(\frac {b}{c}\right)^6 \plus{} \left(\frac {c}{a}\right)^6\right]\geq\frac {a^4}{b^3} \plus{} \frac {b^4}{c^3} \plus{} \frac {c^4}{a^3}\right.$[/quote]

From Cauchy shwarz inequality we have :

$ \left[\left(\frac {a}{b}\right)^6 \plus{} \left(\frac {b}{c}\right)^6 \plus{} \left(\frac {c}{a}\right)^6\right](a^2 \plus{} b^2 \plus{} c^2) \ge \left(\frac {a^4}{b^3} \plus{} \frac {b^4}{c^3} \plus{} \frac {c^4}{a^3}\right)^2$

Thus we need only to check that :

$ \left(a^3 \plus{} b^3 \plus{} c^3\right)\left(\frac {a^4}{b^3} \plus{} \frac {b^4}{c^3} \plus{} \frac {c^4}{a^3}\right) \ge (ab \plus{} bc \plus{} ca)(a^2 \plus{} b^2 \plus{} c^2)$

and From $ CS$ we have

$ \left(b^3 \plus{} c^3 \plus{} a^3\right)\left(\frac {a^4}{b^3} \plus{} \frac {b^4}{c^3} \plus{} \frac {c^4}{a^3}\right) \ge (a^2 \plus{} b^2 \plus{} c^2)^2 \ge (ab \plus{} bc \plus{} ca)(a^2 \plus{} b^2 \plus{} c^2)$

Very nice, [b]Abdek[/b]!
Using the same method, we can prove that

For $ a_1,a_2,\cdots,a_n\geq0$ and $ b_1,b_2,\cdots,b_n>0$, the following inequality holds

$ \left.\left(\frac{a_1}{b_1}\right)^6\plus{}\left(\frac{a_2}{b_2}\right)^6\plus{}\cdots\plus{}\left(\frac{a_n}{b_n}\right)^6\geq\frac{a_1^2\plus{}a_2^2\plus{}\cdots\plus{} a_n^2}{b_1^3\plus{}b_2^3\plus{}\cdots\plus{}b_n^3}\cdot\left(\frac{a_1^4}{b_1^3}\plus{}\frac{a_2^4}{b_2^3}\plus{}\cdots\plus{}\frac{a_n^4}{b_n^3}\right)\right.$

[quote="marin.bancos"]Very nice, [b]Abdek[/b]!
Using the same method, we can prove that

For $ a_1,a_2,\cdots,a_n\geq0$ and $ b_1,b_2,\cdots,b_n > 0$, the following inequality holds

$ \left.\left(\frac {a_1}{b_1}\right)^6 \plus{} \left(\frac {a_2}{b_2}\right)^6 \plus{} \cdots \plus{} \left(\frac {a_n}{b_n}\right)^6\geq\frac {a_1^2 \plus{} a_2^2 \plus{} \cdots \plus{} a_n^2}{b_1^3 \plus{} b_2^3 \plus{} \cdots \plus{} b_n^3}\cdot\left(\frac {a_1^4}{b_1^3} \plus{} \frac {a_2^4}{b_2^3} \plus{} \cdots \plus{} \frac {a_n^4}{b_n^3}\right)\right.$[/quote]

It's just the same methos Dear marin.bancos  :lol:

From $ CS$
$ \left[\left(\frac {a_1}{b_1}\right)^6 \plus{} \left(\frac {a_2}{b_2}\right)^6 \plus{} \cdots \plus{} \left(\frac {a_n}{b_n}\right)^6\right]\times \sum_{k \equal{} 1}^n a_k^2 \ge \left(\frac {a_1^4}{b_1^3} \plus{} \frac {a_2^4}{b_2^3} \plus{} \cdots \plus{} \frac {a_n^4}{b_n^3}\right)^2$

Thus it suffices to show that:

$ \left(\frac {a_1^4}{b_1^3} \plus{} \frac {a_2^4}{b_2^3} \plus{} \cdots \plus{} \frac {a_n^4}{b_n^3}\right)\times\sum_{k \equal{} 1}^n b_k^3 \ge \left[\sum_{k \equal{} 1}^n a_k^2\right]^2$

Which is Cauchy shwarz


modular arithmetic
number theory
prime factorization
AMC

For $ k>0$, let $ I_k\equal{}10\ldots 064$, where there are $ k$ zeros between the $ 1$ and the $ 6$.  Let $ N(k)$ be the number of factors of $ 2$ in the prime factorization of $ I_k$.  What is the maximum value of $ N(k)$?

$ \textbf{(A)}\ 6\qquad \textbf{(B)}\ 7\qquad \textbf{(C)}\ 8\qquad \textbf{(D)}\ 9\qquad \textbf{(E)}\ 10$

[hide="Click to reveal solution"]
Note that $ I_k \equal{} 10^{k\plus{}2} \plus{} 64 \equal{} 2^6(2^{k\minus{}4} \cdot 5^{k\plus{}2} \plus{} 1)$.  If $ 2^{k\minus{}4} \cdot 5^{k\plus{}2} \plus{} 1$ is an integer, it is odd.  Thus, the maximum value of $ N(k)$ is $ 6$, or $ \fbox{A}$.
[/hide]

[quote="erdos(II)"][hide="Click to reveal solution"]
Note that $ I_k \equal{} 10^{k \plus{} 2} \plus{} 64 \equal{} 2^6(2^{k \minus{} 4} \cdot 5^{k \plus{} 2} \plus{} 1)$.  If $ 2^{k \minus{} 4} \cdot 5^{k \plus{} 2} \plus{} 1$ is an integer, it is odd.  Thus, the maximum value of $ N(k)$ is $ 6$, or $ \fbox{A}$.
[/hide][/quote]

Wrong...

[hide]Unless the case where k = 4, where $ 5^6\plus{}1$ is even. In this case, 2 is the largest power of 2 that divides it, so the answer is 2^6*2 = 2^7, and the answer is b.[/hide]

no actually, note that $ (2^6\cdot 5^6 \plus{} 2^6) \equal{} 2^6(5^6 \plus{} 1)$ and $ 5^6 \plus{} 1$ has a multiple of 2, so the answer is $ \boxed{7}$ i have a rigorous proof, but i am too lazy to write it.

5^n mod 2 is always 1
5^n mod 4 is 1 also.

QED.

[hide="slightly more rigorous solution"]We can always write $ 100\ldots064$ as $ (2^n \cdot 5^n \plus{} 2^6)$ for $ n<6$, we can factor out only $ n$ factors of 2. For $ n\equal{}6$, we can factor out 6 factors of 2 and the remaining quantity is $ 2^6(5^6 \plus{}1)$ and since all power of 5 end in 25, we know that $ 5^6 \plus{} 1$ ends in 26. now $ 5^6 \plus{} 1$ has only 1 factor of 2, so the total number of factors of 2 for this case is 7. For $ n>6$, we can factor out 6 factors of 2, but the remaining quantity will be $ (2^k5^p \plus{} 1)$ which ends in 1, hence it has no factors of 2 and therefore the number of factors of 2 in this case is 6. therefore the maximum occurs when n=6 and the number of factors of 2 in that case is $ \boxed{7}$. [/hide]

This was the only one, I got wrong... I put A because I thought the divisibility rule for 128 wouldn't work for any number of zeros. :(

I must be the only one who didn't use $ 10^k\plus{}64$...
[hide]1000...64
5000...32
25000...16
125000...8
625000...4
...125000...2
...625000...1
That's six powers of 2, and you can get another one if the ...625 overlaps the 1. But ...626/2 is odd so you can't get any more powers of 2. Therefore the answer is $ 7$.[/hide]
The same general argument, but maybe a little bit faster to work out.

wait this problem was slightly tricky but I have already been similarly tricked before so I didn't get tricked this time

also, cognos - your solution is slightly too long :O

[hide]
edit: I replaced $ k\plus{}2$ with $ k$ so it's easier (i.e. $ I_k \equal{} 10^k \plus{} 2^6$ now)

We want the max number of 2s that go into $ 10^k \plus{} 2^6$.
If $ k > 6$, $ 10^k \plus{} 2^6 \equal{} 2^6(10^{k \minus{} 6}*5^6 \plus{} 1)$, but $ 10^{k \minus{} 6}*5^6 \plus{} 1$ is odd; hence only 6 2's go into it.
If $ k < 6$, $ 10^k \plus{} 2^6 \equal{} 2^k(5^k \plus{} 2^{6 \minus{} k})$, but $ 5^k \plus{} 2^{6 \minus{} k}$ is odd; hence only k<6 2's go into it.
If $ k \equal{} 6$, $ 10^k \plus{} 2^6 \equal{} 2^6(5^6 \plus{} 1)$. $ 5^6 \plus{} 1 \equiv 1 \plus{} 1 \equal{} 2 \pmod{4}$ (because 4 doesn't divide $ 5^6 \plus{} 1$); hence only 1 2 goes into $ 5^6 \plus{} 1$, so 7 2's go into $ 10^6 \plus{} 2^6$, and 7 is the maximum.
[/hide]

so pretty much, the max power of 2 that goes into $ 5^6 \plus{} 1$ is $ 2^1$ because $ 5^6 \plus{} 1 \equiv 2 \pmod{4}$ (superwizard - why'd you do the mod 2 thing first o.O)

Sorry for the year-old revive, but I had a burning question that has troubled me for quite a while and was wondering if anyone would be kind enough to answer it.

The reason why I am posting under this question is because my question does relate to this problem. Well, anyways, here it is...

When you have a number, say 10064, I know that you can do the "Factor-tree" where you take the lowest prime that divides that number (in this case 2) and try to factor the number that results from the division of 10064 by 2.

But what happens if you do as follows?
10064 = 10000 + 64  = $ 2^4 \cdot (5^4 \plus{} 2^2)$.
Then if I find that $ 5^4 \plus{} 2^2$ has no factor of 2 in it, can I conclude that the number of factors of two in 10064 is 4 (because of the initial 2^4)?

Likewise in this problem, we concluded that we can factor out the $ 2^6$ and we found one more factor in the $ (5^6 \plus{} 1)$ term, that there was a factor of $ 2^7$ and thus 7 factors of 2 in 1000064.

Thanks sooo much!

[quote="redsky"]Sorry for the year-old revive, but I had a burning question that has troubled me for quite a while and was wondering if anyone would be kind enough to answer it.

The reason why I am posting under this question is because my question does relate to this problem. Well, anyways, here it is...[/quote] Reviving an old thread is perfectly fine if you have a question like this. There's no need to be apologetic.  :)
[quote]When you have a number, say 10064, I know that you can do the "Factor-tree" where you take the lowest prime that divides that number (in this case 2) and try to factor the number that results from the division of 10064 by 2.

But what happens if you do as follows?
10064 = 10000 + 64  = $ 2^4 \cdot (5^4 \plus{} 2^2)$.
Then if I find that $ 5^4 \plus{} 2^2$ has no factor of 2 in it, can I conclude that the number of factors of two in 10064 is 4 (because of the initial 2^4)?[/quote]Yes, you factored it right, and you can conclude that. There are only 4 factors of 2 in 10064. I think the thing you have wrong is that 10...064 does not mean 10064. It means 10...064, with any number of 0s in between.

[quote="cognos599"][hide="slightly more rigorous solution"]We can always write $ 100\ldots064$ as $ (2^n \cdot 5^n \plus{} 2^6)$ for $ n < 6$, we can factor out only $ n$ factors of 2. For $ n \equal{} 6$, we can factor out 6 factors of 2 and the remaining quantity is $ 2^6(5^6 \plus{} 1)$ and since all power of 5 end in 25, we know that $ 5^6 \plus{} 1$ ends in 26. now $ 5^6 \plus{} 1$ has only 1 factor of 2, so the total number of factors of 2 for this case is 7. For $ n > 6$, we can factor out 6 factors of 2, but the remaining quantity will be $ (2^k5^p \plus{} 1)$ which ends in 1, hence it has no factors of 2 and therefore the number of factors of 2 in this case is 6. therefore the maximum occurs when n=6 and the number of factors of 2 in that case is $ \boxed{7}$. [/hide][/quote]

To clarify the n cognos is using is n = k + 2.

Also AMC 10 #25, this should be added to "Source".

[hide=Solution]
Note that $I_k={2^{k+2} \cdot 5^{k+2}+2^6}$. Then casework on whether $k+2 >6, =6,$ or $<6$. The maximum occurs when $k+2=6$ and notice that $5^6+1 \equiv 0 \pmod 2$, but not $0 \pmod 4$, so the answer is $6+1=\boxed{\textbf{(B)}\ 7 }$

[b]Solution:[/b] The problem statement is the definition of $\nu_p$. Notice that $10...064$ is just $X = 10^{k+2} + 64$. We have that $\nu_{p}(X) = 6$ when $\nu_{p}(10^{k+2}) \ne 6$. However, when it does equal $6$ (strictly when $k=4$), we have that $\nu_{p}(X) = \boxed{7}$, so we are done.


algebra
polynomial
quadratics
AMC
USA(J)MO
USAMO
algebra unsolved

We have polynomials P,Q,R,S and:
$ P(x^5) \plus{} x \cdot Q(x^5) \plus{} x^{2} \cdot R(x^5) \equal{} (x^4\plus{}x^3\plus{}x^2\plus{}x\plus{}1) \cdot S(x)$

Prove that: $ P(1)\equal{}0$

Notice that the quadratic $ P(1)\plus{}xQ(1)\plus{}x^2R(1)\equal{}0$ for 4 different values of $ x$, that is the 4 fifth roots of unity excluding 1.  Thus, it is $ 0$ for all $ x$.  Thus, $ P(1)\equal{}0$.

Nice.  Without Lagrange's theorem, we can still sum up the four equations we get which immediately gives $ 4P(1) \equal{} 0$.

[b]cosinator[/b], excellent :)
[b]t0rajir0u[/b], could you explain your mind?

[b]t0rajir0u[/b] plugged in to the equation the 4 fifth roots of unity excluding 1 and then summed the 4 equations.

Hm, but if the roots of $ K(x)\equal{}x^4\plus{}x^3\plus{}x^2\plus{}x\plus{}1$ are q,w,e,r, so by Viete's formulas: $ (q\plus{}w\plus{}e\plus{}r)\equal{}\minus{}1$ and $ (q^2\plus{}w^2\plus{}e^2\plus{}r^2)\equal{}(q\plus{}w\plus{}e\plus{}r)^2\minus{}2(qw\plus{}qe\plus{}qr\plus{}we\plus{}wr\plus{}er)\equal{}\minus{}1$, we get: $ 4P(1)\minus{}Q(1)\minus{}R(1)\equal{}0$ and I don't see solution that way.

oh right

then I don't know what t0rajir0u did.

Let $ f(x) \equal{} x^4 \plus{} x^3 \plus{} x^2 \plus{} x \plus{} 1$

We have $ x^5 \equiv 1 \bmod {f(x)}$

Hence $ P(x^5) \equiv P(1) \bmod {f(x)}$
Thus, $ P(1) \plus{} x Q(1) \plus{} x^2 Q(1) \equal{} 0$ for all x $ \Rightarrow P(1) \equal{} Q(1) \equal{} R(1) \equal{} 0$

My apologies.  (I am trying to use as few field properties as possible.)  If in addition we substitute $ x \equal{} 1$ then we actually get $ 5P(1) \equal{} 5S(1)$.

Or, to save [b]t0rajir0u[/b]'s first idea, one can also use the linear combination $ \zeta \times (i) \plus{} \zeta^2 \times (ii) \plus{} \zeta^3 \times (iii) \plus{} \zeta^4 \times (iv)$ with the obvious numbering of the equations and where $ \zeta \equal{} e^{\frac {2 i \pi}5},$ which yields $ Q(1) \plus{} R(1) \equal{} \minus{} P(1)$ and together with the first one $ Q(1) \plus{} R(1) \equal{} 4P(1)$ implies that $ P(1) \equal{} 0.$

Now, of course, [i]fifth[/i] roots of unity aren't important, and one can actually prove that if one takes $ n$ polynomials such that
$ P_0(x^n) \plus{} xP_1(x^n) \plus{} \ldots \plus{} x^{n \minus{} 2} P_{n \minus{} 2}(x^n) \equal{} (x^{n \minus{} 1} \plus{} x^{n \minus{} 2} \plus{} \ldots \plus{} 1)P_{n \minus{} 1}(x),$
then $ P_0(1) \equal{} P_1(1) \equal{} \ldots \equal{} P_{n \minus{} 2}(1) \equal{} 0,$ using exactly the same kind of arguments.

Actually, this problem comes from the USAMO 1976, and I see that it is already discussed :
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=89664]here : Harder poly.,[/url]
[url=http://www.artofproblemsolving.com/Wiki/index.php/1976_USAMO_Problems/Problem_5]and here : AoPS Wiki, USAMO 1976, #5.[/url]


why are a number and the sum of its digits congruent in mod 9?

indeedably so, it is basic, and obvious as well!

Because $ 10^k\equiv 1\pmod{9}$ for all integers $ k\ge 0.$

A decimal number whose digits are $ d_nd_{n\minus{}1}d_{n\minus{}1}\dots d_1d_0$ is the integer $ N\equal{}\sum_{k\equal{}0}^n10^kd_k.$ But since $ 10^k\equiv 1\pmod{9}$ for each $ k,$ we can reduce this to $ N\equiv\sum_{k\equal{}0}^nd_k\pmod{9}.$

Now: can you explain why $ N\equiv\sum_{k\equal{}0}^n(\minus{}1)^kd_k\pmod{11}?$


number theory
greatest common divisor
algorithm

[quote="gmac525"]Assume each letter in the following formula represents a unique digit: (for example, ABCD could be the number 1234)

NOMAD x 4 = DAMON

What is the number represented by the digits DAMON?[/quote]

[hide]Well $N$ is either 1 or 2 but since NOMAD x 4 is even then N is even, so $N=2$

Then because $4D \equiv 2\mod 10$ then $D\equiv 3 \mod 5$ and  D=8 or 9 so $D=8$

So then since NOMAD x 4 = DAMON and $D=8$, $O<3$ but $N=2$ so $O=1$

Then if you multiply 21MA8 x 4 = 8AM12, $4A+3 \equiv 1 \mod 10$ so $A\equiv 2 \mod 5$ but $N=2$ so $A=7$

And then $4M+3\equiv M\mod 10$ so $M+1\equiv 0\mod 10$ so  $M=9$

So then DAMON = $\boxed{87912}$[/hide]

I apologize for reviving an old thread, but if

DAMON X 7 = MONDAY,

Compute Y.

DAMON X 7 = MONDAY

$(DA)\cdot 7000+(MON)\cdot 7 = (MON)\cdot 1000+(DA)\cdot 10+Y$

$Y = (DA)\cdot 6990-(MON)\cdot 993 \ \ (1)$

[code]6990 =  7 * 993 + 39
 993 = 25 * 39  + 18
  39 =  2 * 18  + 3
  18 =  6 * 3[/code]

$\mbox{gcd}(6990,993) = 3 \mid Y \Rightarrow  Y=3,6 \text{ or }9$

From the above algorithm we get $3 = 51 \cdot 6990-359 \cdot 993 \ \ (2)$

So $Y=3$ is a solution of $(1)$.

Indeed $51329\cdot 7 = 359513$

[hide="other solutions"]On the other hand, every pair in the form
$\text{DA}= 51+331t$
$\text{MON}=-359-2330t$

satisfies $(1)$ but then $DA$ is not a 2-digit number

Also, $Y$ could not be $6$ or $9$. For example, if we multiply the equation $(2)$ by two, we get
$6 = 102 \cdot 6990-718 \cdot 993$
again 102 is not a 2-digit number[/hide]


$ a_{1}\equal{} 1$
$ a_{n\plus{}1}\equal{} a_{n}\plus{}\frac{1}{\sqrt{a_{n}}}$
Find all c such that
$ lim\frac{a_{n}^{c}}{n}$ exist and different from 0
General problem for b (instate 2)

I have method to find $ \frac{3}{2}$ and it can solve the general problem but can you post it?

It is your method but I use Newton represent in series form to find out c.


topology
advanced fields
advanced fields unsolved

My friend is doing some basic topology problem sets and he was asked to prove that if $A_{n}$ is a sequence of nonempty closed sets in $\mathbb{R}$ so that $A_{n+1}\subset A_{n}$, then $\cup_{n=0}^\infty A_{n}$ is nonempty.

Is this true in arbitrary metric spaces? What about spaces without even a metric? (I am more interested in the latter :) ) Sorry if they're dumb questions, I never studied topology and I don't have much intuition for it yet...

Of course you meant intersection. The crucial feature of $\mathbb{R}$ here is its completeness. The sets $(\sqrt{2}-1/n,\sqrt{2}+1/n)$ are closed in $\mathbb{Q}$ and nested, but have empty intersection.

I also think that your friend may have a problem with $A_{n}=[n,\infty)$.  :wink:

What you really need here is compactness. If any of these sets is compact, the statement is true and easy, even in a general topological space. If not, it's likely to fail as in mlok's examples.

Yes, in topological spaces one probably has to assume compactness. In metric spaces it is not necessary: a nested ($A_{n+1}\subset A_{n}$) sequence of shrinking ($\mathrm{diam}(A_{n})\to0$) closed subsets of a complete space has a common point.

Interestingly, the assumption $\mathrm{diam}(A_{n})\to0$ cannot be replaced with $\mathrm{diam}(A_{n})<\infty$. The sets must be very small to have a nonempty intersection.  :o


geometry
3D geometry
tetrahedron
geometry unsolved

A tetrahedron ABCD with $\angle BAD+\angle CAD+\angle CAB=\angle ABD+\angle ABC+\angle DBC=180^o$ . Prove that $CD\geq AB$

See http://www.mathlinks.ro/Forum/viewtopic.php?t=43654 .

  darij

Thanks darij . One thing I dont understand bout his solution , if $A,B,B'$ are collinear , then how come $ADB$ and $ADB'$ congruence ? I think this only happens if $\angle DAB=\angle DAB' = 90^o$ . Isn't it ?  :?


group theory
abstract algebra
linear algebra
matrix
superior algebra
superior algebra unsolved

Here is a really beautiful problem:
   G is a closed subgroup in $SL_2(R)$ such that $ G=-G$ and the trace of every element of G, different from $I,-I$ is strictly greater than 2 in absolute value. Then $I$ is an isolated point in G.

I think that the problem is not written correctly , because you can consider
$ G $ consisting of all diagonal matrices
$ \begin{pmatrix}
  \lambda & 0 \\
  0 & 1/\lambda
\end{pmatrix}
$ , when $ \lambda \in \mathbb{R} $, $ \lambda \neq 0 $.
Then the group satisfies all the conditions of the problem,
but $ I $ is not isolated.
May be you must say that $ G $ is countable ?

Well... this is the problem he gave me and even more, I see not only it is not correct, but my poor solution is wrong. Sorry.  :blush:  :blush:

In fact,I think one can show easily that if $ G $ is arbitrary closed countable subgroup
of $ SL(2,\mathbb{R}) $ then it is discrete.
( it is also true for any Lie group instead of $ SL(2,\mathbb{R} $ ).

I don't know how easy it is for you, but look what I have done:
  Suppose that $A_k$ are in G and converge towards $I$. Let $B_k=\sum_{i\geq 1}{-\frac{(I-A_k)^i}{i}}$, which converge to 0 and verifies $e^{B_k}=A_k$. Consider $A$ a limit point of the sequence $ B_k/{||B_k||}$ where $||$ is a norm. Then it is not difficult to see that $ e^{xA}$ is in G for any real x. From now on it's easy if the group is countable. IS there ay normal approach? I found this solution after 2 hours of battle.

I am not sure, but I think the following proof is ok:

Suppose $ G $ is a Lie group. Then it is of course locally compact, so at each
point there exists some open neighborhood of it, such that it's closure is compact.
Now suppose that $ G $ is not discrete and closed . Then $ 1 \in G $ is
an accumulation point of $ G $. Therefore every element of $ G $ is an accumulation
point of $ G $. Therefore every open neighborhood, which has a non-empty intersection
with $ G $, contains infinite number of points in $ G $, therefore it contains
$ 2 $ different points in $ G $.
Now take $ U $ as an open neighborhood of $ 1 \in G $, such that it's closure is compact.
Of course , $ U $ and $ G $ intersect, so there exist two distinct points
$ x,y \in G \cap U $. Take neighborhood $ x \in V',y \in V''$ of $ x,y $,
such that $ V' \cap V'' = \emptyset $ and $ \overline{V'} , \overline{V''} \subset U $.
Now take one of $ V' , V'' $ and do the same thing for each one of them:
take, for example , $ V' $, we know that $ V' \cap G \neq \emptyset $, so
$ V' \cap G $ contains two different points $ z,w $, then take neighborhoods
$ z \in W' , w \in W'' $, which do not intersect and their closures are contained
in $ V' $.
So we can do it, and we will obtain a tree of open subsets in $ U $, such that
every subset in this tree has two children - two open subsets, which do not intersect
and their closures lie in our subset.
Now suppose we go "down" in our tree in some way. We obtain a sequence
of open subsets $ V_{1} \supset V_{2} \supset \ldots $, such that $ V_{n} $
contains closure of $ V_{n+1} $. Also we know that $ V_{n} \cap G \neq \emptyset $.
Therefore we have a decreasing sequence of closed sets $ \overline{V_{1}} \cap G
\supset \overline{V_{2}} \cap G \supset \ldots $, which are nonempty and all contained
in $ \overline{U} $. Therefore because of compactness of $ \overline{U} $, their intersection
is nonempty. Also because $ V_{n} $ contains closure of $ V_{n+1} $, we get that
it's nonempty intersection is equal to $ G \cap \bigcap_{n=1}^{\infty} V_{n} $.
So we get that for every "way down" in our tree, the intersection of neighborhoods
in this "way" has nonempty intersection with $ G $, and if we pick an element
in this intersection, then for every such "way" we get an element of $ G $, and
it is clear that for different ways we get different elements of $ G $.
But number of such ways is uncountable, so we get that $ G $ is uncountable.

I hope we are solving the same problem  :)  .

The teacher who gave me the problem said that in fact the problem initialy posted is true if we add the condition that the subgroup is not commutative.  ;)

I think I can solve the initial problem, with this extra assumption (the fact that the group is not abelian).

Being a closed subgroup of the Lie group $Sl_{2}(\mathbb R),\ G$ is actually a Lie subgroup (a result due to Cartan). If we assume that it's not discrete, then its dimension is $1,\ 2$, or $3$. It's Lie algebra $\frak g$ is a Lie subalgebra of the $3$-dimensional Lie algebra $\frak{sl}_{2}(\mathbb R)$ consisting of real $2\times 2$ matrices with zero trace.

If $\dim G=3$, then $\frak g=\frak{sl}_{2}(\mathbb R)$. Since the exponential map $\exp: \frak g\to G$ maps a neighborhood of $0$ in $\frak g$ homeomorphically onto a neighborhood of $1$ in $G$, and a connected group is generated by arbitrarily small neighborhoods of the identity, it follows that $G=Sl_{2}(\mathbb R)$, contradiction.

Suppose now that $\dim G=2$. Every $2$-dimensional Lie algebra is solvable, so $\frak g$ can be made upper triangular through a similarity. There must be two different elements of $\frak g$ with the same diagonal entries (otherwise, it can easily be seen that $\frak g$ would be $1$-dimensional). Their difference $X$ is then a nilpotent element of $\frak g$, and $\exp X$ is an element of $G$ different from $1$, with both diagonal entries equal to $1$. This contradicts the requirement $|\mbox{tr}(\exp X)|>2$.

We're only left with the case $\dim G=1$. The component $G^{0}$ of the identity of $G$ must be the abelian group $\{\exp(tX)\ |\ t\in\mathbb R\}$ for some non-zero $X\in\frak g$. This group can, through a similarity, be turned into the group with elements $g(\lambda)=\begin{pmatrix}\lambda&0\\0&\frac 1\lambda\end{pmatrix},\ \lambda>0$. Note that $G^{0}$ is also normal in $G$ (in general, the identity component of a topological group is a closed normal subgroup).

Take a $g\in G\setminus G^{0}$. For every $\lambda$, the intersection between the similarity class of $g(\lambda)$ in $Gl_{2}(\mathbb R)$ and $G^{0}$ consists only of $g(\lambda),g\left(\frac 1\lambda\right)$, so $gg(\lambda)g^{-1}$ takes one of these two values. Write $g$ in matrix form, and notice that $gg(\lambda)g^{-1}= g\left(\frac 1\lambda\right)$ implies $\mbox{Tr}(g)=0$, which contradicts our hypotheses. This means that $g$ commutes with $g(\lambda)$ for any $\lambda$, which, in turn, implies that $G$ is abelian. We have a contradiction again.

P.S.

The condition $G=-G$ seems to play no role here :?.


This may be too picky, but the stats doesn't match with the leader board...

Sorry if I'm being annoying... :blush:

Not all participants and their parents have provided permission to list their names on the leaderboard.


What?
What Archimedes thing?
If you want to know who Archimedes is
[quote] what is the new archimedes (?) thing[/quote]
click[url=http://en.wikipedia.org/wiki/Archimedes#The_Golden_Crown]here[/url]


You can use Hospital rule or if you prefer another way just observe that $ (1\plus{}x)^a \sim 1\plus{}ax$ when $ x\to 0$. If you use the last arguments then the numerator of your fraction is ''something'' like $ (1\minus{}\frac{2008}{2009} x)(1\plus{}\frac{2009}{2010} x)\minus{}1$ and denominator is $ sin x \sim x$, hence our limit will be $ \frac{1}{2009\cdot 2010}$.

Why? leshik's solution is perfectly fine. I almost always use power series methods for such problems and that's exactly what leshik did.

In particular, we have the binomial series: $ (1\plus{}x)^a\equal{}1\plus{}ax\plus{}O(x^2)$ as $ x\to 0.$ That's what he used.


Find all contiuous functions $ f:R^+ \rightarrow R^+$ which satisfy($R^+$ doesnt cantain 0) :
                            $f(\frac{1}{f(xy)})=f(x)f(y)$

set f(1)=b.
putting xy=1 we see that f(x)f(1/x)=f(1/b)=b^2
now using this we get that : f(f(xy))f(x)f(y)=b^2
and finally putting y=1 in this equation we get f(f(x))=b/f(x)
hence  bf(xy)=f(x)f(y)
defining g(x)=f(x)/b we see that g(xy)=g(x)g(y) but sdince g(x) is continous we should have g(x)=x^a for some a.
and therefore f(x)=bx^a
now substituting this in the original equation yields that
f(x)=1/x or f(x)=1 for all x.


A triangle is such that at least two of the medians has a length less than 1. Can the area of the triangle be equal to $ ^8\log{(6)}$?

What is at the end of your question? Is it $ 8\ln 6$?

No, it's the logarithm of 6 in the base 8 (the 8-logarithm of 6). I don't really how to write that using LaTeX, but for simplicity, I mean the same as as $ ln(6)/ln(8)$.


geometry
circumcircle
trigonometry
geometric transformation
homothety
invariant
projective geometry

[quote="grupyorum"]$ w$ is the circumcircle of a triangle $ ABC$, $ D$ and $ E$ are arbitrary points on the sides AB and AC, respectively.(Different from vertices). Let the bisector of the angle $ BAC$ meet $ w$ at $ A'$. ($ A'$ is different from $ A$.) $ P$ and $ Q$ are the intersection points of the lines $ A'D$ and $ A'E$ with $ w$ , respectively. $ R$ and $ S$ are intersection of the circumcircles of the triangles $ APD$ and $ AQE$ with $ AA'$, respectively. Let $ DS$ and $ ER$ meet at $ X$. Show that $ X$ lies on a tangent line to $ w$ at $ A$.[/quote]
Such a nice problem. I have come out with a solution and I will redit this post and post my solution I have in mind later.

[quote="grupyorum"]$ w$ is the circumcircle of a triangle $ ABC$, $ D$ and $ E$ are arbitrary points on the sides AB and AC, respectively.(Different from vertices). Let the bisector of the angle $ BAC$ meet $ w$ at $ A'$. ($ A'$ is different from $ A$.) $ P$ and $ Q$ are the intersection points of the lines $ A'D$ and $ A'E$ with $ w$ , respectively. $ R$ and $ S$ are intersection of the circumcircles of the triangles $ APD$ and $ AQE$ with $ AA'$, respectively. Let $ DS$ and $ ER$ meet at $ X$. Show that $ X$ lies on a tangent line to $ w$ at $ A$.[/quote]

[geogebra]d4f1ef5b723ccd7e006925cc64eb1fca0f421842[/geogebra]

Let $ L,S',R'$ respective are the intersection points of the tangent line to  $ w$ at $ A$ and $ PQ$, $ AA'$ and $ BQ$, $ AA'$ and $ CP$ .

$ \widehat{AS'Q} \equal{} \frac {1}{2}.(m(\frown{AQ}) \plus{} m(\frown{BA'})) \equal{} \frac {1}{2}.[m(\frown{AQ}) \plus{} m(\frown{CA'})] \equal{} \widehat{AEQ}$$ \implies S' \equiv (AQE) \cap AA' \equiv S$

Then, similarly : $ R \equiv R'$

Use the Pascal's theorem for $ APCA'CA$ : $ \left\|\begin{array}{c}D \equiv AB \cap PA' \\
 \\
S \equiv AA' \cap BQ \\
 \\
\ L \equiv\ tangent\ line \ to\ w\ at\ A \ \cap PQ\end{array}\right\| \implies$ $ D, S, L$ are collinear.

Similarly : $ L, E, R$ are collinear.

Hence the intersection point of $ DS$ and $ RE$ lies on a tangent line to $ w$ at $ A$

$ \angle A'QB\equal{}\angle A'AB\equal{}\angle EAS\equal{}\angle EQS \Rightarrow Q,S,B$ are collinear.
$ \angle AES\equal{}\angle AQS\equal{}\angle ACB \Rightarrow ES//BC$. Similarly, $ DR//BC \Rightarrow DR//ES$.
$ \Rightarrow \frac{DR}{ES}\equal{}\frac{AD.\sin \angle DAR}{\sin \angle DRA}: \frac{AE.\sin \angle SAE}{\sin \angle ASE}\equal{}\frac{AD}{AE} (1)$
Let $ X'$ be the intersection of $ ER$ and the tangent of $ (O)$ at $ A, SE\cap AX\equal{}\{G\}, SE\cap AB\equal{}\{N\}$.
Applying Menelaus's theorem we get $ \frac{XR}{XE}\equal{}\frac{GS}{GE}.\frac{AR}{AS}\equal{}\frac{GA}{GE}.\frac{AD}{AN}$
$ \equal{}AD.\frac{GA}{GE.AN}\equal{}\frac{AD}{AE} (2)$
From $ (1)$ and $ (2)$ we obtain $ X',D,S$ are collinear or $ X'\equiv X \Rightarrow$ QED.

very nice geometry,, i think the turkey is coming with nice geometry [u]problems[/u] in the national olympiads recently...

Dear Mathlinkers,
1. Acording to Reim's theorem, it can be proven that DR // BC.
2. Let U, V be the meetpoints of AP' and BQ, AQ' and CP.
3. Acording to Pascal's theorem, UV //BC.
4. According to Desrgues theorem applied to PDR and QUV, P, Q and X are collinear.
5. Let Ta be the tangent to the circumcircle of ABC at A.
6. Acording to Pascal's theorem applied to Ta BQPA'A,  we are done.
Sincerely
Jean-Louis

Denote by $ C_1,C_2$ respectively the circumcircles of $ APD , AQE$. Let $ U , V$ be the centers of $ C_1,C_2$ respectively. It is easily verified that triangles $ DUR , SVE$ are similar and $ DR//SE$. Hence, it follows that $ X$ is the center of homethety that carries $ C_1$ into $ C_2$ and is also the center of inversion that carries  $ C_1$ into $ C_2$. Since $ A$ remains invariant under this inversion, if we are able to prove that $ P$ is carried into $ Q$, we'll be done.

We may proceed as follows. Let $ M , N$ be the feet of the perpendiculars from the centers to a common tangent as shown. It suffices to prove that $ PMNQ$ is cyclic.
We note that
$ \angle MPQ = \angle MPD + \angle A^'PQ = \frac{1}{2} \angle MUD + \angle A^'AQ = \frac{1}{2} \angle NVS + \angle A^'AQ = \angle NAS + \angle SAQ = \angle NAQ = 180^{\circ} - \angle MNQ$
where the third equality follows from similarity and the last one from tangency

[img]http://i48.tinypic.com/10mu7uu.jpg[/img]


Let $ a$, $ b$, and $ c$ denote the sides of a triangle and let $ 2s \equal{} a\plus{}b\plus{}c$. Show that
\[ 3\left(\sqrt{ bc(s\minus{}a)}\plus{}\sqrt{ ca(s\minus{}b)}\plus{}\sqrt{ ab(s\minus{}c)}\right)\geq 2s\left(\sqrt{s\minus{}a}\plus{}\sqrt{s\minus{}b}\plus{}\sqrt{s\minus{}c}\right).\]

[quote="Soarer"]let $x = s-a$ and so forth, it suffices to show $3\sum \sqrt{x(x+y)(x+z)}\ge 2(\sum x)(\sum \sqrt{x})$
After SOS, it is
$\sum \left(\frac{1}{\sqrt{x}+\sqrt{y}}-\frac{3\sqrt{z}}{(\sqrt{z+x}+\sqrt{z+y})^{2}}\right)(x-y)^{2}\ge 0$

this is true because $(\sqrt{z+x}+\sqrt{z+y})^{2}= z+x+z+y+2\sqrt{z^{2}+z(x+y)+xy}\ge 2\sqrt{zx}+2\sqrt{zy}+\sqrt{zx}+\sqrt{zy}= 3\sqrt{z}(\sqrt{x}+\sqrt{y})$[/quote]

Es ist gut, Soar!

  My curious!Can we solve this one with classic inequality? That would be great! :)

[quote="Soarer"]let $x = s-a$ and so forth, it suffices to show $3\sum \sqrt{x(x+y)(x+z)}\ge 2(\sum x)(\sum \sqrt{x})$
After SOS, it is
$\sum \left(\frac{1}{\sqrt{x}+\sqrt{y}}-\frac{3\sqrt{z}}{(\sqrt{z+x}+\sqrt{z+y})^{2}}\right)(x-y)^{2}\ge 0$

this is true because $(\sqrt{z+x}+\sqrt{z+y})^{2}= z+x+z+y+2\sqrt{z^{2}+z(x+y)+xy}\ge 2\sqrt{zx}+2\sqrt{zy}+\sqrt{zx}+\sqrt{zy}= 3\sqrt{z}(\sqrt{x}+\sqrt{y})$[/quote]

What's SOS?

[quote="sunchips"]

What's SOS?[/quote]
Sum of squares

[quote="Soarer"]let $x = s-a$ and so forth, it suffices to show $3\sum \sqrt{x(x+y)(x+z)}\ge 2(\sum x)(\sum \sqrt{x})$
After SOS, it is
$\sum \left(\frac{1}{\sqrt{x}+\sqrt{y}}-\frac{3\sqrt{z}}{(\sqrt{z+x}+\sqrt{z+y})^{2}}\right)(x-y)^{2}\ge 0$

this is true because $(\sqrt{z+x}+\sqrt{z+y})^{2}= z+x+z+y+2\sqrt{z^{2}+z(x+y)+xy}\ge 2\sqrt{zx}+2\sqrt{zy}+\sqrt{zx}+\sqrt{zy}= 3\sqrt{z}(\sqrt{x}+\sqrt{y})$[/quote]

Wow, it's nice.  :D SOS is that useful?

Hmm... it will be nice if somebody write a very detailed note on the
applications of SOS method with various examples!

[quote="Colaci"][quote="ideahitme"]Wow, it's nice.  :D SOS is that useful?

Hmm... it will be nice if somebody write a very detailed note on the
applications of SOS method with various examples![/quote]

hunghtn finished it a long time ago.  :lol:
link: http://www.mathlinks.ro/Forum/viewtopic.php?t=80127[/quote]

Thanks! It's also nice  :)


1. Why does my textbook say that if you round 2.5250 to three significant figures, you do not round the 2 in the hundredths place up since it is even?

According to my textbook, 2.5250 to three significant figures is 2.52 while 2.5350 rounds up to 2.54. Why does the parity matter or why did they make that rule?

2. Round 99.999 to four significant figures.

3. Complete the following calculation and round off the answers to the correct number of significant figures. $ 16 m \times 2.82m \times 0.05 m$..

Wait, for that last one would you have to round to only one significant figure since 0.05 only has one significant figure?

Sigfigs= useless in life.  Decimal places are a better marker of precision.

The parity rule for rounding up 5's based on the previous digit being even or odd is to "average" an otherwise systematic error.  Say, if you got 200 measurements all having .x5 in it.  If you rounded up each time, you would have more error than if you did their parity.

And you are right on number 3.  You use one sigfig.

Sig figs only make sense when you are measuring some units of something. They are lame. It's much more accurate to just use decimals (but that cannot be measured).

What's the difference between decimals and sigfigs?

Also, for rounding 99.999 to 4 sig figs, do we get 100.0? Do we kill the two other decimal places since 100.0 already has 4 sigfigs?

A measurement of .005 is obviously more useful than a measurement of .10 even though it has less sigfigs.

Yes.  The easiest way to use sigfigs is to make everything scientific notation so that all sigfigs appear as digits.  Thus, 99.999 becomes 100.0, or $ 1.000*10^2$.

[quote="ThetaPi"][b]Problem.[/b] Why do we write $ 3 \text{ significant figures}$ and not $ 3.00 \text{ significant figures}$? [For those of you who do Physics]

This is just a joke; but the problem is worth pondering on.[/quote]
Because we can measure the amount of significant figures exactly (i.e., we don't have to estimate it). :P

Hmm, I'll leave this here for now but if the talk goes further, I'll have to move it to either Chemistry or Physics Forum.

It's not a spam but this isn't really a "math" as in classroom math goes. It's more of science thing so that's why.  :)  Sig. figs is OK after you got sick and tired of them.  :rotfl:  :rotfl:  :rotfl:


In the expansion of $ (a \plus{} b)^n$ there are $ n \plus{} 1$ dissimilar terms. The number of dissimilar terms in the expansion of $ (a \plus{} b \plus{} c)^{10}$ is:

$ \textbf{(A)}\ 11\qquad
\textbf{(B)}\ 33\qquad
\textbf{(C)}\ 55\qquad
\textbf{(D)}\ 66\qquad
\textbf{(E)}\ 132$

$ (a\plus{}b\plus{}c)^{10}\equal{}[(a\plus{}b)\plus{}c]^{10}$

The expansion simply has (a+b) and c combined in terms with their exponents adding to 10. The first term in the expansion is $ (a\plus{}b)^{10}$ so it has 11 dissimilar terms. Following that, the exponent in the (a+b) term keeps going down by 1, so the subsequent terms yield out 10, 9, 8, ....1 dissimilar terms. Therefore the total number of dissimilar terms in the expansion of $ (a\plus{}b\plus{}c)^{10}$ is $ 1\plus{}2\plus{}3\plus{}...\plus{}11\equal{}\frac{11(12)}{2}\equal{}\boxed{66}$ terms.

Answer: D

Balls and urns is the appropriate generalization here.  That's why most variants on this problem put constraints on the terms to make it nontrivial.


What is the total number of square units in the shaded regions of the 3 x 4 grid of unit squares?  Express your answer as a common fraction.

[hide]If we set one of the height of one of the smaller triangle to $x$ we can make the proportion $\frac {x}{1} = \frac {3}{4}$ and solving we get $x = \frac {3}{4}$. The area of the two small triangles therefore is $\frac {3}{4}$.[/hide]

[hide]I split the diagram in half, and realized that the diagram was not in proportion (For all those people who depend on the diagram).  The line goes 1.5 units up in the center, but we need it to be at .25.  There, it is half the height at the middle, so .75.
.5(.75 x 1)=3/8
3/8 x 2 = 3/4[/hide]


Harvard
college
MIT
Yale
Stanford
CMU
Carnegie Mellon

How important is it to visit a college, with respect to one's application?

Does the college really care whether you come or not? Frankly, if they don't mind either way, then I don't want to spend time going there because I already have a sense of what it is like. I probably won't alter where I want to go just because the campus is somewhat unsightly or the students are somewhat not friendly.

What do people think?

Def. visit before you apply.  Once you get the application you will have NO idea what to write if you have not been there.  "Yes I want to go to harvard, I mean, it's harvard!" does not count.  Esp.  if you have no excuse, like you live w/in reasonable driving distance from the school.  They want to admit people who are interested in the school, so if it sounds like you want to go there and will enroll if they accept you - then you are more likely to get in.  Plus, a lot of schools keep track of if you visit and put that in your file.  I got into 7 out of the 12 schools I applied to.  Out of the 4 I was rejected from, I only had visited 2! So, that's not really proof, but do not make the same mistake as me.

[quote="JS1527"]I got into 7 out of the 12 schools I applied to.  Out of the 4 I was rejected from, I only had visited 2! So, that's not really proof, but do not make the same mistake as me.[/quote]
To make those statistics even somewhat valuable, you'll have to tell us how many of the schools you visited that later admitted you.

I think visiting is important, both for you and the college.

For you to ask yourself, do I REALLY want to go here? Does it hold up to the hype?

For some places, they want to know if you visit. I remember going to MIT and at the info session, they actually pass around cards for you to fill out and return, and the admissions officer SAYS "We want to make sure that you've been here." <--- that's a pretty blunt statement.

[quote]you got into seven out of twelve. four rejected you.... so.......... that other school... what happened to it?[/quote]

Hehe I forgot people on AoPS could add  :)  To answer your question, I was waitlisted.

[quote]To make those statistics even somewhat valuable, you'll have to tell us how many of the schools you visited that later admitted you.[/quote]

Um, let's see, out of top 50 schools:

*Brown - I have been there a million times, my dad is a professor there - rejected
*Yale - visited - rejected
*Cornell - didn't visit - rejected (I really wish I had visited, but I was so busy it never worked out, from what I had read I really liked it)
*NYU - didn't visit - rejected

*Vanderbilt - visited and I did a lot of other stuff like meet w/professor and stay w/a student who went there - waitlisted

*Rensselaer - never visited (I also wish I had visited here)- accepted and they gave me 1/4 scholarship

*Carnegie Mellon - I visited and I met w/a professor there - accepted

Out of not top 50 schools:

*Boston University - visited - accepted + $44,000 scholarship
*University of Rhode Island - I have been there a million times, I take classes there also - accepted + full scholarship
*WPI - visited - accepted + some scholarship money
*University of Vermont - never visited - accepted w/some scholarship
*Rochester Institute of Technology - never visited - accepted + 1/4 scholarship

Um, so yes that's all the info.  Visiting places helps, whether my info shows it or not.  For other info, I have a legacy at MIT and Boston University, I have all As in school, and my scores are about 700 on average.

I found that my college visits were not very helpful.  A visit really only shows you one or two sides of a school.  Often the tour guide was a theater major who knew nothing about the math and science departments, so I was left with an undeservedly negative impression of a school.  Sitting in on classes, despite what many "experts" say, is generally not useful because any individual lecture is only one of thousands that take place during the year and is unlikely to accurately represent the academic program as a whole.

In terms of getting in to colleges, it really varies by the school.  Stanford, for example, says that "demonstrated interest" plays no role in admissions decisions.  I received a "Likely letter"--a letter that a college sends to its top applicants informing them of their intention to admit them at a later date--from an Ivy League school which I had not visited.  It doesn't hurt to visit, just in case, but I don't think it helps you get into the most competitive schools.  It might help with less competitive admissions, though.

Your advice would be more useful if all of the colleges you listed were on an equal footing.  In particular, the ones you were rejected from were all Ivy League schools - I would say that visiting would probably not tip that one way or another.

Haha, thanks.

Look at it this way, visiting schools can't hurt you.  Maybe it does not give you an edge in terms of admission, but it's not like you are going to enroll there w/o visiting, so you might as well.

[quote="JS1527"]Look at it this way, visiting schools can't hurt you.
[/quote]

This is very true.  Also notable is that many of the 'richer' schools are willing to pay for you to visit them.  Unfortunately, many of the visiting dates end up conflicting, but it's still a nice perk.


Hello,

After a poll made in the [url=http://www.snarkianos.com/]Snark Mailing List[/url] few years ago, there is a list of [url=http://www.snarkianos.com/conocidos.html]The Most Famous Puzzles Ever[/url]. The list, as well as the whole site is in Spanish, but I will be posting a few of them a day. Here it is the first one.

A shepherd has to go across a river with a fox, a goat and a cabbage. He has a small boat and he can only carry one of the three things with him on each trip across the river. If the fox and the goat are alone, the fox would eat it. If the goat is left alone with the cabbage, it would eat it. How can the shepherd proceed?

I hope to see answer to these ones  ;)

[quote="djimenez"][quote="sky9073"]I remember this from.... fourth grade, I think. I loved it.
[hide]Take goat over, come back.
Take cabbage over, bring back goat.
Take wolf over, come back and take goat over again.[/hide][/quote]

Hehe, yes, it is not that hard, but... I said they were The Most Famous Puzzles Ever, not the The Hardest Puzzles Ever  :P  ;)[/quote]I think I first saw this in 1st or 2nd grade.  Took me a good 30-40 minutes to solve :lol:


So?  Final Four?

(I know it's really late for this topic, but....)
(and no cheating, tell us who your Final Four really were, don't just pick the highest seed left in the region).

Here's mine:
Duke
[color=red]Kansas[/color]
Uconn
Villanova

Also, go USC in the NIT!  We just beat Florida St. by a point.  One point.

so this is like the SC fun and games topic huh?

duke (winning over uconn in finals)
gonzaga
Uconn
florida

and i heard clemson was eliminated from the NIT. :(

[quote="furious"]
and i heard clemson was eliminated from the NIT. :([/quote]

YES.  GO USC!  Clemson lost to Louisville, 74-68 (I think). :P  :rotfl:  :rotfl:   Tough luck, but I could have called that one any day.  I'm surprised they even gave Clemson an NIT bid.  Oh well.

My entire Final Four was completely wrong. I had like all the number 1 seeds making it so that's not good. But I think now everyone's rooting for George Mason since they're the typical underdogs.

Yeah.  George Mason was one of my upsets (in the first round).  But I killed them off long before the Final Four.  Of course, now I have to root for them or UCLA.

Also: USC won the NIT semi by 15 or so.  So now we play Michigan tommorrow.

texas (b/c I didn't want to pick duke. I didn't really think texas could make it either, though, so I was all like well w/e)

ucla (yay!)

uconn (like most everybody else, I suppose)

bc (who *should* have beaten villanova. I don't know how they would have done against fla)

[u][b][size=200]USC wins NIT (again)!!!!!!![/size] [/b][/u] I am just making a prediction, considering the game is not over.  But USC is up 67-54.

[quote="Iversonfan2005"][quote="MysticTerminator"]b/c I didn't want to pick duke.[/quote]

Arnav,How can I say this any simpler? Duke>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>UNC

jk, UNC makes great NBA players :D[/quote]

exactly which is why duke (probably the second most popular pick to win it all) yet again failed to make it out of the sweet 16.

i know everyone around here is just sooo excited that florida beat george mason. yay

Oh yeah.  I was soooo happy *sarcasm dripping.*  But it's over now.  The championship will be UCLA and Florida.  I think Florida was luck that GM beat UConn, and Florida SHOULD NOT have gotten this far.  So I'm calling for UCLA to pull off the upset.  By a lot.

Hah...
But seriously, my picks were Duke, Florida, UNC, and UCLA, with Duke and UNC in the final (ugh..., almost had the final two).


geometry
geometric transformation
reflection

I got this problem a long time ago...
but could never solve it =[
"Five years ago, I planted a four-foot-tall tree in my front yard. It has grown the same amount each year, and last year it was twice as tall as the end of the first year. How tall is the tree now?

[hide] Let the amount that the tree grows each year (in ft) be defined as x.

Therefore we have the first expression, 4+5x, telling us (in ft) how tall the tree is now.

Last year (4+4x) the tree is twice the height it was after the first year (4+x)
so set up the following equation--

4+4x=2(4+x)

solve...
4+4x=8+2x
2x=4
x=2

plug that into our first expression...4+5(2) solve and you get [b]14[/b]

EDIT: check it--4+4(2) is twice as 4+2. [/hide]

Please don't name the topics "I neeeed helppppppp!" The name should reflect the problem.  :)

[/hide]  Last year the height of the tree is 4*2=8
So the every the tree grown up (8-4):4=1
Now the height of the tree is 8+1=9

 :oops: if you have any interesting or hard broblems,please post it to me.

The end of the first year the tree was $4+x$, and now it is $4+5x$, so the amount it grows each year is $4+4x=2(4+x)=2$, therefore now it is $4+5(2)$, which is 14 :|

:)  That was easy

[hide]

Since the tree started with 4 feet and 5 years has passed with same amount increasing every year we can write the equation of $4+5x$.
If last year's tree was double the end of first year's tree then the equation would be
$4+4x = 8+2x$ subtract those numbers and we get $2x=4$ which ends up in $x=2$.
Substitute 2 in the first equation $4+5(2)$ and we get $14$[/hide]


[hide="HINT"]

$ sin\beta$ has + sign if $ 0< \beta < \pi$
$ sin\beta$ has - sign if $ \pi<\beta<2\pi$

$ cos\beta$ has + sign if $ \frac{3\pi}{2}<\beta<\frac{\pi}{2}$

$ cos\beta$ has - sign if $ \frac{\pi}{2}<\beta<\frac{3\pi}{2}$

the rest is easy[/hide]

[quote="cgyao15"]Seriously salahcool, stop posting random problems, and this is WAYYYYYYYYYY BEYOND middle schooll math TRIG?[/quote]

Unless you can use a calculator. :)

BASIC BASIC trig is allowed.  This DOES fall under BASIC BASIC trig.

a. sin 50°     +

b. cos 100°   +

c. tan 200°    +

d. csc 300°    +

Maybe use the definitions of $ \cos$ and $ \sin$ as the coordinates of points on the unit circle, and the $ \tan \equal{} \frac{\sin}{\cos}$, $ \cot \equal{} \frac{1}{\tan}$, $ \sec \equal{} \frac{1}{\cos}$ and $ \csc \equal{} \frac{1}{\sin}$, or draw the graphs?

[quote="salahcool"]What is the sign (+ or - ?) of:

a. sin 50°

b. cos 100°

c. tan 200°

d. csc 300°

Do these without calculator so that you have a better idea what is going on.

thanks
salah[/quote]

[hide]Sin is positive from 0 to 180 degrees
Cos is negative from 90 to 270 degrees
tan is positive from 0 to 90, and 180 to 270 degrees
csc of 300 degrees= 1/sin 300, and has the same sign. sin of 300 is negative.[/hide]

It's better to think about this with quadrants, and just which funtions are positive.

Quadrant I - All

Quadrant II - Sine

Quadrant III - Tangent

Quadrant IV - Cosine

I remember it this way because a friend (Alan) taught it to me and to remember it as Alan Smokes Tobacco Constantly.  :D

[quote="modularmarc101"]
Quadrant I - All

Quadrant II - Sine

Quadrant III - Tangent

Quadrant IV - Cosine
[/quote]

Learned it as All Students Take Calculus. It's not very difficult to figure out if you think about the unit circle. If the center is (0,0) then anything with a positive X value has a positive cosine, and anything with a positive y value has a positive sine value. Tangent is just either both positive, or both negative.

[quote="AdithyaGanesh"][quote="salahcool"]What is the sign (+ or - ?) of:

a. sin 50°

b. cos 100°

c. tan 200°

d. csc 300°

Do these without calculator so that you have a better idea what is going on.

thanks
salah[/quote]

a. Positive, because all trig functions are positive in quadrant I.
b. Negative, because only sin functions are positive in quadrant II.
c. Positive, because all tan are positive in quadrant III.
d. Negative, because only cos functions are positive in quadrant IV.  (Cosecant=1/sin[theta])[/quote]

Please don't revive threads that are nearly a month old, especially when you merely repost the correct solution with nothing new to add.


This has been posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=28344 and subsequently locked. It is problem A.358. at http://www.komal.hu/verseny/2004-11/mat.e.shtml and it is still not a year old, but I think no harm can be done since the deadline is over. So let me post a solution:

Since abc = 1, the inequality in question,

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{3}{a+b+c}\geq2\left( \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\cdot\frac{1}{a^2+b^2+c^2}$,

rewrites as

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{3}{a+b+c}\geq 2\left( \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\right) \cdot \frac{abc}{a^{2}+b^{2}+c^{2}}$;

but this is true, since straightforward expanding and bunching yields

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{3}{a+b+c}-2\left( \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\right) \cdot \frac{abc}{a^{2}+b^{2}+c^{2}}$
$=\frac{bc\left(b+c\right)\left(b-c\right)^2+ca\left(c+a\right)\left(c-a\right)^2+ab\left(a+b\right)\left(a-b\right)^2}{abc\left(a+b+c\right)\left(a^2+b^2+c^2\right)}\geq 0$.

So the inequality is solved. This was indeed pretty easy, and I guess this is the two-lines solution using identities Harazi meant in http://www.mathlinks.ro/Forum/viewtopic.php?t=28344 .

  Darij

Very short indeed, Darij. :)

But I don't see the problem with posting problems that are "less than a year old"? I mean, when manlio posted that, it was end of February 2005, and the contest ended in November 2004? We also post problems from olympiads the day after the contest right? Since the deadline is the end of the contest...

I would like to know more about this, since I've also posted problems here with deadline something like end of May, also only some months ago. (which seems again totally harmless, since the deadline is passed for months...)


function
Support
advanced fields
advanced fields unsolved

Let $ Y$ be a quasi-affine variety of dimension $ \ge 2$ and let $ f$ be a regular function on $ Y\minus{}\{P\}$. If $ P$ is a normal point then the regular function can be extended to all of $ Y$.

Use the long exact sequence for cohomology with supports and local cohomology and Serre's criterion for normal. Alternatively, you could use the criterion with normality and height-1-prime ideals.

:huh:
this exercise is at the page ~20... so it should be solvable with very very basic stuff.
What's the criterion with height 1 prime ideals?


inequalities
geometry
geometric transformation
reflection
trigonometry
inequalities proposed

Let ABC be an acute-triangle. Prove that
$ \frac {1 \minus{} \cos A}{1 \plus{} \cos A} \plus{} \frac {1 \minus{} \cos B}{1 \plus{} \cos B} \plus{} \frac {1 \minus{} \cos C}{1 \plus{} \cos C}\le (\frac {1}{\cos A}\minus{}1)(\frac {1}{\cos B}\minus{}1)(\frac {1}{\cos C}\minus{}1)$

this problem is mine  :D  i can give you a funny lemma/hint for a couple of days...

[hide="{hint}"] if $ ABC$ is an acute angle, then for any real numbers $ x,y,z$ we have that $ (x\minus{}y)^2(\tan A\tan B\minus{}2)\plus{}(y\minus{}z)^2(\tan B\tan C\minus{}2)\plus{}(z\minus{}x)^2(\tan C\tan A\minus{}2)\geq 0$[/hide]

[quote="jerry_smith"]Let ABC be an acute-triangle. Prove that
$ \frac {1 \minus{} \cos A}{1 \plus{} \cos A} \plus{} \frac {1 \minus{} \cos B}{1 \plus{} \cos B} \plus{} \frac {1 \minus{} \cos C}{1 \plus{} \cos C}\le (\frac {1}{\cos A} \minus{} 1)(\frac {1}{\cos B} \minus{} 1)(\frac {1}{\cos C} \minus{} 1)$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=250853
Very nice your inequality, campos!


I am very experienced in java, but I am new to C++. I am trying to use strings, and I get the error: "invalid conversion from `const char*' to `char'". This is my program:

[code]#include <iostream>
#include <string>
using namespace std;

int main()
{
	string a;
	a = "the first letter of the alphabet.";
	cout << "a is ";
	cout << a << endl;
	return 0;
}
[/code]

The error occurs at the declaration of a ("string a;"). Can anyone help me??

Strangely, I erased and retyped it, and now it runs without error....  :huh:

What compiler do you use? I know for me, Eclipse acts really weird at times where I'll have to restart and then it works perfectly fine.

Yeah, I'm also using Eclipse. And I also have to restart at times to make it work correctly. Binaries act weird in eclipse sometimes.

Hm, so I keep getting the error "Binaries not found." I know its because in my project, there is no "binaries" folder. Some programs, when I build it, a binary folder appears. but sometimes it doesn't appear and eclipse says "Build error occurred, build is stopped." Does anyone know why?

[quote="ImUnderATree"]Eh, Eclipse can mess with you like that. Try Bloodshed, I find it's much more reliable:
 http://www.bloodshed.net/devcpp.html

Hope it helps! C++ can be a pain...[/quote]

This is the best compiler. It works perfectly and rarely bugs.

yes, i have started using dev on my new computer... thank you. :)


Hello there,

For $ \log_{7}((xy)^{8}/x^{3})$, is it possible to simplify it to $ \log_{7}(x^{5}y^{8})$?

Similarly, with $ \log_{10}(5^{2}x^{3}/y^{z})$, is it possible to simplify it to $ \log_{10}(25 x^{3}/y^{z})$

My algebra book did not show the simplified answers so I am unsure as to whether I can simplify in both cases.

Thank you.

Thank you very much for your help, Lider!


let $ (u_n)$ be sequence defined by:
$ u_n\equal{}(1\plus{} \frac{1}{n}) ^{n\plus{}k}$,  $ n \in N^*,k \in R$
determine all $ k$ such that:
a) $ u_n$ be decreasing
b)$ u_n$ be increasing

no one?  :oops:  who have nice solution for this?

I saw this problem here: http://www.mathlinks.ro/viewtopic.php?t=303201

No one has solved it there, either.


function
limit
topology
integration
real analysis
real analysis unsolved

$f$ is a real-valued continuous function defined on product space $E \times F$, ($E$, $F$ both metric space).
$x_{k}\to x$ in $E$, $y_{n}\to y$ in $F$. We know $f(x_{k},y)$ converges to a finite number. And for each $y_{n}$,
 $f(x_{k},y_{n})$ also converges to a finite number.

Prove:
$f(x,y)=\lim_{n\to \infty}(\lim_{k\to \infty}f(x_{k},y_{n}))=\lim_{k\to \infty}(\lim_{n\to \infty}f(x_{k},y_{n}))$

If the limit exist, does not matter the way of x or y going to some value, it will converge

does not make sense a funtion that converge to some value but if you change the order ir converge to another value

i think this can be found on google or in some calc book

we can make the following

make y dependent of x. or x dependent of y

and evaluate the limit

it need to converge to a unique value does not making matter the dependence of x or y

in a math languege $x= f(y)$ or $y = f(x)$ for all continuosly and smooth etc.... function f

this is justa idea

[quote="Thales418"]If the limit exist, does not matter the way of x or y going to some value, it will converge

does not make sense a funtion that converge to some value but if you change the order ir converge to another value[/quote]

Let $f(x, y) = \frac{x}{x^{2}+y^{2}}$.

$\lim_{x\to0}\lim_{y\to0}f(x, y) = \lim_{x\to0}\frac{1}{x}= \infty$.

$\lim_{y\to0}\lim_{x\to0}f(x, y) = \lim_{y\to0}0 = 0$.

There are, of course, many more such examples.  Perhaps [i]you[/i] should spend some time with a calc book and stop telling other people to do so.  This question, of course, asks us to prove that this pathalogical behavior cannot occur under certain conditions.  In the future, rather that merely asserting things with no proof, it would be nice if you would write complete, comprehensible solutions.

JBL was responding to Thales418, not to kalazealot. The original post contains the words "is a real-valued continuous function defined on product space."

So now the issue is just what those words mean. There is a standard product topology; in the case of the product of two metric spaces, that can be realized as a metric space (in more than one way in terms of the exact definition of the metric, but all of the possibilities give rise to equivalent metrics.)

Note that according to this product metric, the function in JBL's example is not continuous at the origin.

So I'll throw it back to you kalazealot: what exactly did you mean by those quoted words? Do you have a definition of what you mean by a "continuous function defined on product space"? And can you then use that definition to prove what you want?

I wanted to abstract my question but apparently I didn't do a good job.

This is what I was trying to solve:

"weak convergence" problem in $L^{p}$
Suppose $f_{n}$ is in $L^{p}(E)$ ($E$ could have infinite measure).
Suppose $f_{n}$ goes to $f$ a.e. and $\parallel f_{n}\parallel \leq  M$ for some fixed $M$.
Show $f_{n}$ goes to $f$ weakly. ($1<p<\infty$)

And I found the following hints:

The dual of $L^{p}$ is $L^{q}$. (1/p+1/q=1). so have to show that
$\int (f_{n}-f)g \to 0$.

Hint of the proof:
1. Using the Fatou lemma, $\parallel f\parallel _{p}\leq  M$
It is then sufficient to consider $f=0$.
2. The linear functionals $F_{n}: L^{q}\to  R$, $F_{n}(g) = \int f_{n}g$
are unformly bounded ($\parallel F_{n}\parallel \leq M$) using Holder.
3. span{$\chi_{A}: \mu(A)<\infty$} is dense in $L^{q}$, so it is sufficient
to consider $g = \chi_{A}$ (the characteristic function)
4. In order to prove $\int \chi_{A}f_{n}\to  0$, use the Egoroff theorem
and Holder again (for the constant 1 functiom).

My question is why step2 implies it is enough to consider characteristic function only in step3?
If simple integrable functions $g_{k}\to g$ in $L^{q}$. By step4, we can show
$\lim_{n\to \infty}\int g_{k}f_{n}= 0$ for all $k$, but it is not obvious this implies $\int g f_{n}\to  0$.

So at first I think this is a double limit of double series, for:
notice that $\lim_{k\to \infty}\int g_{k}f_{n}=\int g f_{n}$ because $f_{n}$ is a continuous linear functional on $L^{q}$.  So if we can show the following we are done:

$\lim_{n\to \infty}(\lim_{k\to \infty}\int g_{k}f_{n})=\lim_{k\to \infty}(\lim_{n\to \infty}\int g_{k}f_{n})$ (RHS=$\lim_{k\to \infty}0=0$).

Later I realised I complicated things too much: just notice that $(f,g) \mapsto \int gf$ is a continuous function defined on $L^{p}\times L^{q}$,
so we of course can switch the order the the double limit.

But still I feel like I missted something, maybe step2 implies we can switch order of double limit without knowing that $(f,g) \mapsto \int gf$ is a continuous function defined on $L^{p}\times L^{q}$, I remember somewhere in Rudin's "mathematical analysis" mentioned we can't stwich order of double limit as counter-example of "uniform convergence". what's the connection here?


geometry
3D geometry
topology
algebra
function
domain
invariant

This may be a pretty stupid exercise, but I can't figure it out. Consider $ A \equal{} \{z \in \mathbb{C}^3 \, : \, z_1z_2 \equal{} z_3^2 \}$. When removing $ 0$ we get a complex submanifold of dimension 2. How can one show that topological structure around $ 0$ is not that of an open set in $ \mathbb{R}^4$?

it doesn't sound that stupid..
$ A$ should be a cone over the 2-fold cover of $ S^3$ branched over the hopf link: i've been trying a bit to figure out what such an object is, but my (in)ability to imagine 3-manifolds stopped me (some googling tells it's a projective space). now you "just" would have to show that you can't embed $ \mathbb{RP}^3$ in $ \mathbb{R}^4$, which doesn't seem obvious (and stiefel-whitney machinery doesn't help, since orientable 3-manifolds are parallelizable; see below, though).

EDIT: in fact, Hirsch-Smale theorem that tells you whether or not you can embed $ \mathbb{RP}^n$ in $ \mathbb{R}^{n \plus{} d}$: the condition is that $ d$ is greater than or equal to the number of 1s in the binary expansion of $ n$ (!!!). so, in our case, you can't embed $ \mathbb{RP}^3$ in $ \mathbb{R}^4$.
moreover, i think you can build the double branched cover by quotienting $ \mathbb{RP}^3$ by the map $ \tau: (x_0: x_1: x_2: x_3)\mapsto( \minus{} x_0: \minus{} x_1: x_2: x_3)$: this branches over 2 circles in the projective space. i wasn't able to see what it covers, though.. any idea (following this path or brand new)?

I also don't see how this could be a "stupid exercise", but I feel there ought to be an easier way to do this.

Kalle, do you really need to work in the topological domain? I guess if you wanted to show that a neighborhood of $ 0$ is not [i]diffeomorphic[/i] to $ \mathbb{R}^4$, even if "diffeomorphic" just means $ C^1$, that should be fairly straightforward: you would determine the dimension of the tangent space and observe that it is invariant under diffeomorphisms. The Jacobian matrix is all $ 0$ at the origin, so the tangent space has dimension $ 6$ which is obviously wrong for a $ 4$ dimensional neighborhood in the [i]smooth[/i] domain.

The topological domain seems harder and I'm not sure how to approach it without doing something along the lines proposed by ma_go: identifying the actual structure of the neighborhood and appealing to some machinery to show that it's not homeomorphic to an open set in $ \mathbb{R}^4$.

randomgraph: I  think the purpose is to demonstrate that analytic sets can be complicated at singular points.

I think by not giving the whole exercise I probably missed some hints. The first part is to show that that $ A \setminus \{0\}$ is a connected complex manifold. I think a machinery solution to the original problem is that $ (z,w) \to (z^2, w^2, zw)$ gives a two fold covering map from $ \mathbb{C}^2$ to $ A$ except at $ 0$. But this contradicts that $ A$ is locally Euclidean because if it were then $ (A\setminus\{0\})\cap U$ should be simply connected for some open set $ U$.

I'm not sure if this is similar to your suggestion ma_go, I don't know nearly enough algebraic topology to follow.

uhm.. nice ;)
actually i was thinking it the other way (which is more complicated): if you restrict yourself to the unitary sphere $ S^3\subset\mathbb{C}^2$, you see that $ A\cap S^5$ is $ \mathbb{RP}^3$ (you're just quotienting $ S^3$ by the antipodal map, since you identify $ (z,w)$ with $ (\minus{}z,\minus{}w)$ when you map them to $ A$).
and now there's a classical result telling you that you can't embed $ \mathbb{RP}^3$ in $ \mathbb{R}^4$ (maybe not hirsch-smale, maybe i'm not so sure about the statement right now).


geometry
geometric transformation
homothety
ratio

A circle with center $ O$ and its chord $ AB$ is given. $ C$ is an arbitrary point on the circle.

i) Prove that the locus of the centroid of $ \triangle ABC$ is a circle.

ii) Find the radius of the loci circle in terms of $ OA \equal{} R$.

iii) Find the distance from $ O$ to the center of the loci circle in terms of $ OA \equal{} R$ and $ AB \equal{} x$.

Call G the centroid of  ABC 'cause of MG=1/3MC , G Moves on the homothetic circumference to (O) in the homothety with center M the midpoint of AB and ratio 1/3. the radius of this circle is 1/3R and its center K lies on MO such that  MK=1/3MO

[hide]Let $ M$ be the midpoint of $ AB$.

$ AM \equal{} MB \equal{} x/2$. Let $ C'C''$ be a chord passing through $ M$.

Let $ G'$ be the centroid of $ \triangle AC'B$. Let $ G''$ be the centroid of $ \triangle AC''B$.

$ G'M\cdot G''M \equal{} \frac {C'M\cdot C''M}{9} \equal{} \frac {\frac {x^2}{4}}{9}$ which is fixed. So locus of $ G'$ is a circle.

Another approach is drawing parallel line from $ G$ to the line $ OC$. Let this line intersect $ OM$ at $ O'$.
$ GO' \equal{} R/3$ and $ O'O \equal{} \frac23 \cdot OM$ are all fixed. $ OM^2 \equal{} R^2 \minus{} x^2/4$. Then $ OO' \equal{} \frac {\sqrt {4R^2 \minus{} x^2}}{3}$.[/hide]


limit
real analysis
real analysis unsolved


induction
number theory unsolved
number theory

Let be given a positive integer $ m >=9$ ( $ m$ is not a prime number).
Write $ m=\prod_{i=1}^n {p_i}^{\alpha_i}$ with $ p_i$ prime ,$ \alpha_i \in N$
Define $ f(m)=1+\sum_{i=1}^n \alpha_i.p_i$
Prove that $ f(m)  \leq  m-2$

I think this problem is almost identical to problem6 from the 1991 VietNam TST.
Darkmaster!
You can see the solution at page 94,"10000 primary problems:sequence and limit",author:Phan Huy Khai.Nice problem!  :P

[quote="pigfly"]
You can see the solution at page 94,"10000 primary problems:sequence and limit",author:Phan Huy Khai.Nice problem!  :P[/quote]

I don't have this book, could you post the solution ?
Thx

The proof is long    .I can't post it here :( .Moubinool,I'm sorry :(

sorry pigfly i think all the problem ,which were posted there, not for the solution
we think( just my opinion) we come here to discuss and find the nice of the problem
good look ;)

Dear Master,
I think if your's problem was solved and you find the nice solution then you post your's problem in Solved problem section or Proposed and Own problem section. ;)

Dear Master,
Maybe what u think is not what I think. I post this problem with only reason: its solution. When I can't solve a problem, I post it to Mathlinks for a proof, this is what I want.
I think we must thank pigfly because of his help, instead of "Bumming".

[quote="Moubinool"][quote="pigfly"]

I don't have this book, could you post the solution ?
Thx[/quote][/quote]
You can use induction on $ n$.  :wink:


An explorer sees a bear 100 meters due south of him.  The bear walks 100 meters due east while the explorer stands still.  The explorer then points his gun due south, fires, and kills the bear.
What color was the bear?

This should be in Puzzles & Brainteasers.

[quote="BlissIntelligence"]Is it Black, because at the time of the day the bear may have been thought of moving east due to its shadow.[/quote]
Good guess, but no, the bear actually moved east.

Yes bliss, you are correct.

This was on a geometry Mu Alpha Theta test a few years back.

[quote="BlissIntelligence"]White because its a polar bear?[/quote]
Yep  :lol:
The only way for the explorer to always shoot south is for him to be at the north pole.

black since polar bears are really black with white fur

Lol doesn't matter cause i answered both.


1. Find the equation of a line that touches the curve $ y\equal{}x^4 \minus{} 8x^2 \plus{} 2x \plus{} 17$ at two distinct points.

2. Determine the value of $ a$ and $ b$ so that the lines $ y \equal{} 2x\minus{}2$ and $ y\equal{}2x\plus{}2$ are tangents to the curve $ y\equal{}x^3 \plus{} ax \plus{}b$.

(For no.1 I meant touches, not intersects)

Thanks so much.  :)

1: We want $ x^4\minus{}8x^2\plus{}2x\plus{}17\minus{}L(x)$ to be a perfect square $ (x\minus{}a)^2(x\minus{}b)^2$. That means we need to complete $ x^4\minus{}8x^2$ to a square. Just at a glance, $ (x^2\minus{}4)^2\equal{}x^4\minus{}8x^2\plus{}16$, so $ L(x)\equal{}2x\plus{}1$ works. That line is tangent at $ x\equal{}2$ and $ x\equal{}\minus{}2$.


Is there anyone who has the instructions for the LaTeX stuff?
I only have the basics of it.
I need to know how to do absolute values etc..

Everything you need to know should probably be [url=http://www.artofproblemsolving.com/Wiki/index.php/Latex]here[/url].

Hi,

here you can find basic documents for LaTeX: [url=http://texblog.net/latex-link-archive/introduction-guide/]Introductions and guides to LaTeX[/url], here math specific introductions and tutorials: [url=http://texblog.net/latex-link-archive/mathematics/]Mathematics with LaTeX[/url].

Stefan

--
[url=http://texblog.net]TeXblog[/url]


Not again this one!!!  :bomb:   I have seldomly seen a more meaningless and idiotic inequality proposed in a competition, and that's not only because I haven't solved it. See http://www.mathlinks.ro/Forum/viewtopic.php?t=30865 for some solutions and you will understand why.

  Darij


Consider the $ \text{Haber Process}$ (ammonia synthesis):

$ \ce{N2}+\ce{H2}$ $ \rightleftharpoons$ $ \ce{NH3}$

Derive the condition under which $ \ce{N2}$ can be added (under constant pressure and temperature) such that the equilibrium shifts to the left.

I think that's the problem.

[hide="Hint"]

Well not really a hint. I did it using $ \Delta G = \Delta G^{\circ} + RT \ln{Q}$[/hide]

I have posted this problem before and no one answered. Actually, you don't need that expression: it is enough to work with $ Q_P$.

[hide="Answer"]The equilibrium shifts to the left when N2 is added only when $ x_{N_2} > \frac{1}{2}$.[/hide]

Which post was it?

Anyway, anyone else want to give it a try? :D


[quote="allrighty"]There are four urns. U1 can contain 6 balls, U2 can contain 4, U3 can contain 5 and U4 can contain 5.

There are 16 White balls and 4 Black balls in total. If these 20 balls are randomly placed into the four urns, what is the probability that no Black balls is placed in the U4 urn?[/quote]
[hide]We can just completely ignore urns U1 - U3, since they aren't important in this problem. The number of ways to place no black balls into the U4 urn is the number of ways to place all white balls into the urn, which is ${16 \choose 5}$. We divide this by the total number of possibilities, ${20 \choose 5}$, to get our answer,
$\frac{{16 \choose 5}}{{20 \choose 5}}=\frac{16 \times 15 \times 14 \times 13 \times 12}{20 \times 19 \times 18 \times 17 \times 16}=\boxed{\frac{91}{323}}$.[/hide]

[quote="nebula42"][hide]We can just completely ignore urns U1 - U3, since they aren't important in this problem. The number of ways to place no black balls into the U4 urn is the number of ways to place all white balls into the urn, which is ${16 \choose 5}$. We divide this by the total number of possibilities, ${20 \choose 5}$, to get our answer,
$\frac{{16 \choose 5}}{{20 \choose 5}}=\frac{16 \times 15 \times 14 \times 13 \times 12}{20 \times 19 \times 18 \times 17 \times 16}=\boxed{\frac{91}{323}}$.[/hide][/quote][hide="Equivalent"]Or the number of ways to place all black balls into the other urns, which is ${15 \choose 4}$, divided by the total number of ways to place the black balls, ${20 \choose 4}$:
$\frac{{15 \choose 4}}{{20 \choose 4}}=\frac{15 \times 14 \times 13 \times 12}{20 \times 19 \times 18 \times 17}$[/hide] :)


Is "Available http://..." for web references considered APA or MLA format or neither? If it is neither, then can you tell me what it is?

Also, how should I cite sources like MathWorld? Should I cite the homepage or do a bunch of references for each individual page i used?

I'd say use the MLA format whenever possible. Also, for MathWorld entries, there is information at the bottom of each page. It looks something like this:

Eric W. Weisstein. "Ackermann Function." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/AckermannFunction.html


geometry
inequalities
trigonometry
function
inequalities unsolved

Let $ABCD$ be a quadrilateral of sides $a,b,c,d$ inscribed in a circle of radius $R$ and circumscribed in a circle of radius $r$.  We have

$ 8R^2r^2 \geq abcd $

This one is easy. We assume that AB = a, BC = b, CD = c and DA = d.

At first, our quadrilateral ABCD is cyclic. Thus, according to [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15973]the Brahmagupta formula[/url], the area F of the quadrilateral ABCD is given by $F=\sqrt{\left( s-a\right) \left( s-b\right) \left( s-c\right) \left( s-d\right)}$, where $s=\frac{a+b+c+d}{2}$ is the semiperimeter of the quadrilateral ABCD. But since the quadrilateral ABCD also has an incircle, we have a + c = b + d, and thus $s=\frac{a+b+c+d}{2}=\frac{\left(a+c\right)+\left(b+d\right)}{2}=a+c$ and s - a = c. Similarly, s - b = d, s - c = a and s - d = b.

Thus,

$F=\sqrt{\left( s-a\right) \left( s-b\right) \left( s-c\right) \left( s-d\right)} = \sqrt{cdab} = \sqrt{abcd}$.

Hence, $abcd=F^2$. Thus, the inequality we have to prove, $8R^2r^2 \geq abcd$, is equivalent to $8R^2r^2 \geq F^2$. After taking the square root, this becomes $2\sqrt2 Rr \geq F$.

Now, the area of an arbitrary polygon which has an incircle equals the semiperimeter of the polygon times the inradius. In our case, this means that the area F of the quadrilateral ABCD equals sr. Thus, our inequality $2\sqrt2 Rr \geq F$ simplifies to $2\sqrt2 Rr \geq sr$, i. e. to $2\sqrt2 R \geq s$. But $s=\frac{a+b+c+d}{2}$. Denoting x = < ACB, y = < BAC, z = < CAD, w = < DCA, we have a = AB = 2R sin < ACB = 2R sin x, and similarly b = 2R sin y, c = 2R sin z and d = 2R sin w. Thus,

$s=\frac{a+b+c+d}{2}=\frac{2R\sin x + 2R\sin y + 2R\sin z + 2R\sin w}{2} = R\left(\sin x+\sin y+\sin z+\sin w\right)$.

And hence, the inequality $2\sqrt2 R \geq s$ rewrites as

$2\sqrt2 R \geq R\left(\sin x+\sin y+\sin z+\sin w\right)$,

i. e.

$2\sqrt 2 \geq \sin x+\sin y+\sin z+\sin w$.

But this is easy to prove: We know that x + y + z + w = < ACB + < BAC + < CAD + < DCA = (< ACB + < DCA) + (< BAC + < CAD) = < DCB + < BAD = 180° (since the quadrilateral ABCD is cyclic), and since the sine function is concave in the interval [0°; 180°], Jensen yields

$\sin x+\sin y+\sin z+\sin w \leq 4\sin \frac{x+y+z+w}{4} = 4\sin \frac{180^{\circ}}{4} = 4\sin 45^{\circ} = 2\sqrt2$,

qed..

  Darij


AMC
AIME
number theory
relatively prime
AMC 10

This isn't all that difficult if you make one observation.

If three positive numbers $x$, $y$, and $z$ satisfy $xyz = 1$, $x + \frac{1}{z} = 5$, and $y + \frac{1}{x} = 29$. Then $z + \frac{1}{y} = \frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

Just one observation.

[hide="HINT"] Multiply the terms together! [/hide]

BTW, this was number 7.

[hide]Well, z = 1/xy

So x + xy = 5
and y + 1/x = 29

x(1+y) = 5
So, x = 5/(1+y)

y + (1+y)/ 5 = 29

5y + 1 + y = 145
6y = 144
y= 24

Substituting,
z = 5/24

So z + 1/y = 5/24 +24 = 5/24 + 576/24 =581/24.
So m +n = 581 +24 = [b]605[/b][/hide]

[i]Moderator Note: Accidentally clicked the "edit" button while attempting to quote. Tried to fix it up as best as possible. Apologies.[/i]

[hide="solution"]lets say $z + \frac{1}{y} = n$

First, add all the equations:

$x + y + z + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 34 + n$  - reference1.

Then we MULTIPLY all the equations:

$xyz + x + y + z + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} + \frac{1}{xyz} = 145n$

$\left(1 \right) + x + y + z + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} + \frac{1}{\left(1 \right)} = 145n$

substitute the reference1.

$1 + 34 + n + 1 = 145n$

$36 = 144n$, $n = \frac{36}{144} = \frac{1}{4}$

then $m+n = 1 + 4 = 5$[/hide]

that seems really weird and too easy of an answer to be true.

you added z with [b]y[/b], not with [b]1/y[/b]

[quote="xxreddevilzxx"][hide="solution"]Then we MULTIPLY all the equations:

$xyz + x + y + z + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} + \frac{1}{xyz} = 145n$[/hide][/quote][hide]Which equations are you multiplying? I assume \[\left(x+\frac{1}{z}\right)\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)=5\cdot29\cdot\frac{m}{n}.\]Wow, I was just about to say that I don't think that's what you would get when you expand but it's right. Nice solution! Just one little thing.. you use $n$ to represent both the fraction $\frac{m}{n}$ and the denominator of the fraction $\frac{m}{n}$. If you want to replace the fraction with a symbol for abbreviation, pick something else like $u$, I guess.[/hide]

this was rather easy #7 on AIME, now i know why this was on the Getting Started, not Intermediate....

I have bad memorries with #7's (I dont think i get that many #7's right :( )

I personally don't think this should have been moved. Maybe I am misinterpreting what is good for getting started?

i think that should have been too hard for getting started because for you and me and others who have had many practices, these problems become too easy because we have done them (i have personally seen this problem about 4 times). but for getting started students, they aren't so easy. After all, it is a number 7.

It shouldn't matter where the problem came from. But by sight, beginners should be able to do some basic manipulation of equations to get what is needed. I still stand by my opinion unless someone points out a specific step in solving this that a person capable of doing AMC 10 ish type problems can't perform.

Yeah, they should be able to do it. Especially if you have plenty of time on the real AIME. (around 12 min a prob)


[quote="FOURRIER"]Prove that for every $ a,b,c > 0$ this inequalitie holds

$ \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a}\ge a \plus{} b \plus{} c \plus{} 4\frac {(a \minus{} b)^2}{a \plus{} b \plus{} c}$

I could turn it to prove:
$ \frac {(a \minus{} b)^2}{c} \plus{} \frac {(b \minus{} c)^2}{a} \plus{} \frac {(c \minus{} a)^2}{a}\ge 4\frac {(a \minus{} b)^2}{a \plus{} b \plus{} c}$
but .... :lol:[/quote]
  \[ L.H.S. \geq \frac{(|a\minus{}b|\plus{}|b\minus{}c|\plus{}|c\minus{}a|)^2}{a\plus{}b\plus{}c} \geq \frac{(|a\minus{}b|\plus{}|b\minus{}c\plus{}c\minus{}a|)^2}{a\plus{}b\plus{}c} \equal{} \frac{4(a\minus{}b)^2}{a\plus{}b\plus{}c}\]

[quote="apollo"][quote="FOURRIER"]Prove that for every $ a,b,c > 0$ this inequalitie holds

$ \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a}\ge a \plus{} b \plus{} c \plus{} 4\frac {(a \minus{} b)^2}{a \plus{} b \plus{} c}$

I could turn it to prove:
$ \frac {(a \minus{} b)^2}{c} \plus{} \frac {(b \minus{} c)^2}{a} \plus{} \frac {(c \minus{} a)^2}{a}\ge 4\frac {(a \minus{} b)^2}{a \plus{} b \plus{} c}$
but .... :lol:[/quote]
\[ L.H.S. \geq \frac {(|a \minus{} b| \plus{} |b \minus{} c| \plus{} |c \minus{} a|)^2}{a \plus{} b \plus{} c} \geq \frac {(|a \minus{} b| \plus{} |b \minus{} c \plus{} c \minus{} a|)^2}{a \plus{} b \plus{} c} \equal{} \frac {4(a \minus{} b)^2}{a \plus{} b \plus{} c}
\]
[/quote]

It was a simple consequence of CS  :(  :oops:
Thanks  :wink:

Please also visit  this link.   (Problem 3.)

[url]http://www.mathlinks.ro/Forum/resources.php?c=1&cid=19&year=2005[/url]


[color=green]Find the multiplicative inverse of [87] in Z150.

Use this answer to find all x in Z150 so that [87]x = [11].[/color]
Thanks guys

87 isn't invertible in $\mathbb{Z}/150\mathbb{Z}$ because it's divisible by 3 (and 150 is). There are no solutions to $87x\equiv 11\mod 150$.


integration
real analysis
real analysis unsolved

find the general solution of the differential equation(regarding x as dependant variable and y as independant variable
)
     (x+y(e^y))y'=1

i have no idea how to do it, some help pls..

$(x+ye^y)\frac{dy}{dx}=1$
Write it as
$\frac{dx}{dy} - x = ye^y$
Then we can solve it directly (recall that the solution of $y' + Px = Q$ is given by $e^{-\int P dx}\left(C+\int Qe^{\int P\ dx}dx\right)$)
$x = e^{y}\left(C+\frac{1}{2}y^2\right)$

no idea how you get this answer, can u pls explain in depth..
thanks a lot, get it now, pretty ez and i am pretty stupid... :blush:


ratio
geometry
perimeter
combinatorics proposed
combinatorics

Here is a terrific (and quite difficult) problem from the Tournament of Towns. Although the solution is not at all complicated, this is, in my humble opinion, [b]the[/b] most difficult problem ever proposed on a high school competition:

[i]On the white plane there are finitely many black unit squares, possibly overlapping. Those squares form a black polygonal figure $P$ (which may consist of several 'branches' and may have 'holes'). Is it possible that the ratio between the perimeter and the area of this figure exceeds $30 000$? [/i]

There was not $30000$ but $100000$ in the original problem, but it doesn't matter --- anyway the answer is 'no'. I know a proof of the estimate $\frac{P}{S}<16000$ but it isn't the best estimate.

P.S. I have participated in that Tournament, but I was junior, and this problem is from senior-level.

*** Can you show me an example, where the ratio is greater than 10?

*** Suppose that all unit-squares must be axes-parallel. Can the ratio then
be bigger than 10?

Vess, can you post your solution? thanks!

I agree with test20, something in this problem makes it look trivially impossible to be much bigger than 4...

[quote="Pascual2005"]In the case Peter VDD says, when they are parallel to grid lines, then it is easy to see a "tree" is our best choise, and a tree of area $n$ has perimeter $2n+2$ so our expression in this case is indeed at most 4. I dont know what about the general problem but i think strips are the best choise, because if many squares overlap, taking out one of them increases our expression.[/quote]

Still, I remain at my case that you cannot put unit squares in the plane so that the total black surface (in units) is less than 1/4 of the unit perimeter around the squares (in units)...

This is the difficulty of the problem: it seems so clear intuitively, yet you don't know where to start in this problem: the configuration is so messy and difficult to 'take under control.' No-one has any doubts that the ratio cannot be arbitrarily large, but who can actually [b]prove[/b] it? (A side note: no, there are easy examples where the ration is $> 4$ (but not much greater)). Actually, I think (Myth, Remike, correct me if I'm wrong) that none of the participants in the Tournament of Towns managed to solve this problem (although IMO the version in the contest was much easier, because it contained one of the two key observations as a hint).

This problem is actually taken from a research paper on a (much) more general extremal problem:

Tamas Keleti, [i]A Covering Property of Some Classes of Sets in $\mathbb{R}^n$[/i], Acta Univ. Carol., Math. Phys. [b]39[/b] (1998) No. 1--2, pp. 111--118.

The problem of finding the supremum of the ration of the perimeter to the area of our figure is open.

I'll give you the two (quite simple) hints that will enable you to solve this problem. First, put uniformly in the plane disks of small (but strictly positive) radii in such a manner that every square contains some of these circles completely in its interior (for example, it suffices to put circles of radii $0.05$ centered at the knots of an infinite square grid with unit $0.5$). The squares then break into several groups with respect to the small circles: two squares are in the same group iff they contain the same circle in their interiors. Then, reduce the problem to bounding the ratio of the perimeter to the area in [i]one[/i] of the groups (that is, reduce the problem to the case in which all squares are 'hammered' to the plane with a nail of radius $0.05$). Use some very rough estimates to finish the proof.

Hope these hints suffice!  :P

--Vesselin

Vess, i think next time, you should post the complete problem, because this problem had 3 parts, and the first part included the nail idea, so it was a clue! and without this clue this problem is i think impossible to solve! Thanks!

Maybe i'm just stupid but i don't get it even with the clue. Would someone mind posting a complete solution? I would be very grateful!

Iura, i can give you a link [url]http://www.math.toronto.edu/oz/turgor/archives/TT2000F_SAsolutions.pdf[/url]

hmm... but...

if the squares all follow the grid lines (sides parallel to grid lines) then the thing is trivially no greater than 4 (easy to prove right?)

But, if you turn them in any way, and make them overlap, you cannot ever gain more than sqrt(2) by any finite iteration, by pythagoras theorem. So that limits the thing to 4*sqrt(2) right? <therefor the little word 'finite' in the wording>

Where does my proof fail? :| I know it can't be so easy, I just don't see where I fail.


probability geometry inequalities trigonometry vector Stanford college

Is anybody interested in Olympiads[Physics (NSEP),Chemistry(NSEC),and Astronomy(NSEA)]???

I am interested in astronomy (i can't clear any other :rotfl: )

Can anybody provide details about question paper,etc of NSEA
I am not able to find it on hbcse site,also there is question of 2008 INAO on JOI,but i am unable to open it(it shows general error)but probably no question of NSEA.

Anybody who has given it,please help.

You can get previous year both NSEA junior and senior paper at http://www.iapt.org.in
BTW u will give junior or senior this year?

where did u find INAO 2008 paper? on JOI???? waht is it?

hey skand ill mail the paper to u both junior and senior and ofc u and i both will be giving senior astronomy however i think ill be givin physics although i know i wont be able to clear it!

to be honest,i'll give all of them :rotfl:

its of no use wasting 2-3 hrs waiting for one exam to be over
at-least i'll get an idea about my preparation for JEE

P.S- I am giving senior this year

Edit:@fbaggins Thanks a lot  :)
JOI is another sute(forum) that was maintained by some of seniors[url]http://joigroup.freezoka.com/[/url]

i mean ill be givin physics and mathe matics......................
and astronomy !

hey u shud better give it next yr :D

it wud increases my chances this yr :(

you don't need to improve your chances this year

i know you'll get thru(if you had given it previous year,you would have easily got through prev year also) :)
but i am not so genius :(

i tried prev year ques of phy and chem

chem is full or organic(i don't know 'o' of organic)

phy is full of 12th topics(E&M)etc
i could only solve 10-12 questions in physics  :oops: (most of them from from first 15-20 questions
and i don't see any reason for expecting much increase in next 2 months

i only have hopes from astro(seems to be easier,i cud solve much more quesrions)

btw,what are cut-offs for these

but i am sure had ritu given these examinations in 11th,he'ld certainly have cleared ohysics and astronomy apart from biology[i have not taken name of chem as probability(in my opinion was 90% only if compared from probability of 100% in other subjects)and he had cleared RMO already] :P

my god ritu is a thorough genius  :D
i feel so low :(

no need to feel low

just see me and feel good :rotfl:

come on u urself are no less than others !

hey! i am in class IX and is giving RMO this year. cud u plz advise me on what 2 prepare!!

according to me just clear ur concepts to a higher level and try out a book on higher mathematics like challenges and thrills of pre college mathematics

[quote="skand"]to be honest,juniors like him (and you too) make me feel bad  :( (as ypu'ld be knowing,i am no match to seniors here)

i could do nothing except NTSE,RMO and dav talent search examinations

and when i came to know you were in 10th only i was stunned(but not sad as now i am not much interested in olympiads  :P )[/quote]
That's why we should forget our academic achievements and try to share what we know with each other for the benefit of all. And here comes him, look at the introduction of himself, would you agree that this is the most appropriate form of intro?

Hey akashnil, I told that because the 11th stds couldn't solve that problem. They had no clue to it.
Please misunderstand ok. And don't genrealize just by one such post.
Sorry.

[quote="skand"]dav talent search examinations
[/quote]

never heard abt them..wat r they???? :D  :(

they are nothing,just exams at inter dav lvl(our batch was first batch they were expanded from Mumbai region to other places)

Akashnil is the only one who is completely honest with his replies. He is blunt and to the face. I remember him posting somewhere, he stood 27th in class and didnt care about it. Many others who have posted in this thread (not to give names), indirectly were trying to convey how great they were. "I [u]only[/u] came first in this competition and that, etc. "

All i want to say is that dont brag about your achievements people. You may never know when your talents be taken away from you.

yes that tells me a lesson :P  :rotfl:  :P

count me in that category too
i think hell_ever is correct and honest too.

[quote="hell_ever"]Akashnil is the only one who is completely honest with his replies. He is blunt and to the face. I remember him posting somewhere, he stood 27th in class and didnt care about it. [b]Many others who have posted in this thread (not to give names), indirectly were trying to convey how great they were. "I [u]only[/u] came first in this competition and that, etc. "[/b]

All i want to say is that dont brag about your achievements people. You may never know when your talents be taken away from you.[/quote]

I quite disagree ,but anyways chodo yaar :D

How did u guys fare in RMO? Mine was horrible but i am in class IX, so was not expecting much!

my general feeling was it was much easier than last few years. :D

is it???
i found 2006's most simple..... this year i dont know what happende and couldnt do a single question fully

well truly ...........this years was too simple than i expected.................i expected i will not b abl to solv any prob.............
but still solved at least 1 fully... :P

arnav i am in the same boat
could u post some questions


function
calculus
derivative
vector
invariant
geometry
geometric transformation

Let $f$ be a real continuous function and $y$ be defined on $I$ interval of $R$ with real values solution of the differential equation $y'=f(y)$. Prove that $y$ is monotonic.

is this question correct? I don't understand a part related to y. What do we know about y besides it's a solution of this equation?

Thx,

I also find it strange, but there is no other mention about $y$ except it is a solution of the equation. I could not solve the problem, I posted it and nobody sent a counter-example. But did anyone think on it?

I came up with the following:

If the solution changes the character of the monotony then it has a zero derivative in such a point. That means the equation's phase vector field has a singular point there (=equilibrium, here it's important that the equation is autonomous, i.e. invariant relatively to translations along time axis, i.e. we can draw filed of phase velocities or directional field lines only on the y axis and it will be the same for all t). But this point will be a separate phase trajectory (equilibrium is a separate solution, i.e. point can't go out of the equilibrium if it was there at the start). Two phase trajectories of the good (= autonomous and with good right side) equation can't intersect.

Seems to be ok, right?

Here's what I did (I'm not entirely sure it's Ok):

Suppose you have a differentiable function $g:[a,b]$ s.t. $g(a)=g(b)=t$, and $g'(a)=g'(b)>0$ (or $<0$; the point is that they have the same non-zero value). Then $g(x)=t$ has at least one solution in $(a,b)$. This is pretty obvious.

Assume $y$ is not monotonic, and we can find, for example, $p<q<r$ s.t. $y$ reaches its maximum on $[p,r]$ in $q$, and both $y(p),y(r)$ are strictly less than this maximum. $y'$ is continuous and $y'(q)=0$, so there must be a maximal $a\in(p,q)$ where $y'$ takes a certain fixed positive value, which can be chosen s.t. $y(a)>y(r)$. Since $y(u)=y(v)\Rightarrow y'(u)=y'(v)$, it means that $a$ is also the maximal real in $(p,q)$ where $y$ takes the value $y(a)$.  Now let $b\in(q,r)$ be minimal in $(q,r)$ s.t. $y(b)=y(a)$. Such a $b$ does exist, because $y$ has the intermediate value property.

Since $y(a)=y(b)$, we get $y'(a)=y'(b)$ and this common value is positive, because that's how we chose $a$ in the first place. We can now apply the observation in the second paragraph with $y$ instead of $g$: we can find at least one $c\in(a,b)$ with $y(c)=y(a)=y(b)$, and this contradicts the maximality of $a$ in $(p,q)$ and the minimality of $b$ in $(q,r)$ for which $y$ takes the value $y(a)$.

I updated my post: now proof is more accurate.

Of course, autonomous differential equations form a quite special classification, and the qualitative description of solutions of such equations can be explained in classes at a rather low level.

Note one of the consequences of this: since $y(t)$ is monotone, then either it is unbounded or it tends to a limit as $t\to\infty.$ (If $f$ is differentiable, or, weaker than that, Lipschitz on bounded domains, then the existence-uniqueness theorem would imply the existence of a solution on $[a,\infty)$ in the case that the solution stays bounded.)

It's not hard to show this: if $\lim_{t\to\infty}y(t)=L$ for some solution $y$ of the differential equation, then $f(L)=0.$

Autonomous systems also yield significant qualitative analysis, but with a system in two or more dimensions, there's no result quite as simple-minded as the result of this problem.