If A and B are real numbers, how many solutions to the equation  $ \; \; \left|\left\parallel{}x\minus{}1|\minus{}2\right|\minus{}3\right|\equal{}Ax\plus{}B \; \; \; ,\; \; x\in {\mathbb R}\; \; ?$

It is easy to see after drawing this graphic that there could be from $ 0$ to $ 6$ solutions depends on values of $ A$ and $ B$.

[quote="aktyw19"]If A and B are real numbers, how many solutions to the equation  $ \; \; \left|\left\parallel{}x \minus{} 1| \minus{} 2\right| \minus{} 3\right| \equal{} Ax \plus{} B \; \; \; ,\; \; x\in {\mathbb R}\; \; ?$[/quote]
This is a crazy exercise and I cant understand how a teacher / contest / book may ask for such a stupid subject :(

Let $ f(x)\equal{}\parallel{}|x\minus{}1|\minus{}2|\minus{}3|$ :
If $ x>1$ $ f(x)\equal{}\parallel{}x\minus{}3|\minus{}3|$
If $ x\le 1$ $ f(x)\equal{}\parallel{}\minus{}1\minus{}x|\minus{}3|$

If $ x>3$ $ f(x)\equal{}|x\minus{}6|$
If $ 1<x\le 3$ $ f(x)\equal{}|\minus{}x|\equal{}x$
If $ \minus{}1<x\le 1$ $ f(x)\equal{}|x\minus{}2|\equal{}2\minus{}x$
If $ x\le \minus{}1$ $ f(x)\equal{}|\minus{}4\minus{}x|$

If $ x>6$ $ f(x)\equal{}x\minus{}6$
If $ 6\ge x>3$ $ f(x)\equal{}6\minus{}x$
If $ 1<x\le 3$ $ f(x)\equal{}x$
If $ \minus{}1<x\le 1$ $ f(x)\equal{}2\minus{}x$
If $ \minus{}4<x\le \minus{}1$ $ f(x)\equal{}x\plus{}4$
If $ x\le \minus{}4$ $ f(x)\equal{}\minus{}4\minus{}x$

So the equation is :
either $ x>6$ and $ x\minus{}6\equal{}Ax\plus{}B$ $ \iff$ $ (A\minus{}1)X\equal{}\minus{}B\minus{}6$
either $ 6\ge x>3$ and $ 6\minus{}x\equal{}Ax\plus{}B$ $ \iff$ $ (A\plus{}1)x\equal{}6\minus{}B$
either $ 1<x\le 3$ and $ x\equal{}AX\plus{}B$ $ \iff$ $ (A\minus{}1)x\equal{}\minus{}B$
either $ \minus{}1<x\le 1$ and $ 2\minus{}x\equal{}AX\plus{}B$ $ \iff$ $ (A\plus{}1)x\equal{}2\minus{}B$
either $ \minus{}4<x\le \minus{}1$ and $ x\plus{}4\equal{}AX\plus{}B$ $ \iff$ $ (A\minus{}1)x\equal{}4\minus{}B$
either $ x\le \minus{}4$ and $ \minus{}4\minus{}x\equal{}AX\plus{}B$ $ \iff$ $ (A\plus{}1)x\equal{}\minus{}4\minus{}B$

1) $ A\equal{}1$. The equation becomes :
either $ x>6$ and $ 0\equal{}\minus{}B\minus{}6$ $ \iff$ $ B\equal{}\minus{}6$
either $ 6\ge x>3$ and $ 2x\equal{}6\minus{}B$ $ \iff$ $ x\equal{}3\minus{}\frac B2$ but this requires $ 6\ge 3\minus{}\frac B2>3$ $ \iff$ $ \minus{}6\le B<0$
either $ 1<x\le 3$ and $ 0\equal{}\minus{}B$ $ \iff$ $ B\equal{}0$
either $ \minus{}1<x\le 1$ and $ 2x\equal{}2\minus{}B$ $ \iff$ $ x\equal{}1\minus{}\frac B2$ but this requires $ \minus{}1<1\minus{}\frac B2\le 1$ $ \iff$ $ 4>B\ge 0$
either $ \minus{}4<x\le \minus{}1$ and $ 0\equal{}4\minus{}B$ $ \iff$ $ B\equal{}4$
either $ x\le \minus{}4$ and $ 2x\equal{}\minus{}4\minus{}B$ $ \iff$ $ x\equal{}\minus{}2\minus{}\frac B2$ but this requires $ \minus{}2\minus{}\frac B2\le \minus{}4$ $ \iff$ $ B\ge 4$

And so :
If $ B<\minus{}6$ : no solution
If $ B\equal{}\minus{}6$ : solution is $ x\in[6,\plus{}\infty)$
If $ \minus{}6<B<0$ : solution is $ x\equal{}3\minus{}\frac B2$
If $ B\equal{}0$ : solution is $ x\in[1,3]$
If $ 0<B<4$ : solution is $ x\equal{}1\minus{}\frac B2$
If $ B\equal{}4$ : solution is $ x\in[\minus{}4,\minus{}1]$
If $ 4<B$ : solution is $ x\equal{}\minus{}2\minus{}\frac B2$

2) $ A\equal{}\minus{}1$ The equation becomes :
either $ x>6$ and $ x\equal{}\frac B2\plus{}3$ but this requires $ \frac B2\plus{}3>6$ $ \iff$ $ B>6$
either $ 6\ge x>3$ and $ B\equal{}6$
either $ 1<x\le 3$ and $ x\equal{}\frac B2$ but this requires $ 1<\frac B2\le 3$ $ \iff$ $ 2<B<6$
either $ \minus{}1<x\le 1$ and $ B\equal{}2$
either $ \minus{}4<x\le \minus{}1$ and $ x\equal{}\frac B2\minus{}2$ but this requires $ \minus{}4<\frac B2\minus{}2\le \minus{}1$ $ \iff$ $ \minus{}4<B\le 2$
either $ x\le \minus{}4$ and $ B\equal{}\minus{}4$

And so :
If $ B<\minus{}4$ : no solution
If $ B\equal{}\minus{}4$ : solution is $ x\in(\minus{}\infty,\minus{}4]$
If $ \minus{}4<B<2$ : solution is $ x\equal{}\frac B2\minus{}2$
If $ B\equal{}2$ : solution is $ x\in[\minus{}1,\plus{}1]$
If $ 2<B<6$ : solution is $ x\equal{}\frac B2$
If $ B\equal{}6$ : solution is $ x\in[3,6]$
If $ 6<B$ : solution is $ x\equal{}\frac B2\plus{}3$

3) $ A\notin\{\minus{}1,\plus{}1\}$ The equation becomes :
either $ x>6$ and $ x\minus{}6\equal{}Ax\plus{}B$ $ \iff$ $ x\equal{}\frac{\minus{}B\minus{}6}{A\minus{}1}$ but this requires $ \frac{\minus{}B\minus{}6}{A\minus{}1}>6$ and so :
$ \minus{}B>6A>6$ gives a solution $ x\equal{}\frac{\minus{}B\minus{}6}{A\minus{}1}$
$ \minus{}B<6A<6$ gives a solution $ x\equal{}\frac{\minus{}B\minus{}6}{A\minus{}1}$

either $ 6\ge x>3$ and $ 6\minus{}x\equal{}Ax\plus{}B$ $ \iff$ $ x\equal{}\frac{6\minus{}B}{A\plus{}1}$ but this requires $ 6\ge \frac{6\minus{}B}{A\plus{}1}>3$ and so :
$ 6>3\minus{}3A>B\ge \minus{}6A$ gives a solution $ x\equal{}\frac{6\minus{}B}{A\plus{}1}$
$ 6<3\minus{}3A<B\le \minus{}6A$ gives a solution $ x\equal{}\frac{6\minus{}B}{A\plus{}1}$

either $ 1<x\le 3$ and $ x\equal{}AX\plus{}B$ $ \iff$ $ x\equal{}\frac{\minus{}B}{A\minus{}1}$ but this requires $ 1<\frac{\minus{}B}{A\minus{}1}\le 3$ and so :
$ 0<A\minus{}1<\minus{}B\le 3A\minus{}3$ gives a solution $ x\equal{}\frac{\minus{}B}{A\minus{}1}$
$ 0>A\minus{}1>\minus{}B\ge 3A\minus{}3$ gives a solution $ x\equal{}\frac{\minus{}B}{A\minus{}1}$

either $ \minus{}1<x\le 1$ and $ 2\minus{}x\equal{}AX\plus{}B$ $ \iff$ $ x\equal{}\frac{2\minus{}B}{A\plus{}1}$ but this requires $ \minus{}1<\frac{2\minus{}B}{A\plus{}1}\le 1$ and so :
$ \minus{}A\minus{}1<2\minus{}B\le A\plus{}1$ gives a solution $ x\equal{}\frac{2\minus{}B}{A\plus{}1}$
$ \minus{}A\minus{}1>2\minus{}B\ge A\plus{}1$ gives a solution $ x\equal{}\frac{2\minus{}B}{A\plus{}1}$

either $ \minus{}4<x\le \minus{}1$ and $ x\plus{}4\equal{}AX\plus{}B$ $ \iff$ $ x\equal{}\frac{4\minus{}B}{A\minus{}1}$ but this requires $ \minus{}4<\frac{4\minus{}B}{A\minus{}1}\le \minus{}1$ and so :
$ 4\minus{}4A<4\minus{}B\le 1\minus{}A$ gives a solution $ x\equal{}\frac{4\minus{}B}{A\minus{}1}$
$ 4\minus{}4A>4\minus{}B\ge 1\minus{}A$ gives a solution $ x\equal{}\frac{4\minus{}B}{A\minus{}1}$

either $ x\le \minus{}4$ and $ \minus{}4\minus{}x\equal{}AX\plus{}B$ $ \iff$ $ x\equal{}\frac{\minus{}4\minus{}B}{A\plus{}1}$ but this requires $ \frac{\minus{}4\minus{}B}{A\plus{}1}\le \minus{}4$ and so :
$ B\ge 4A>\minus{}4$ gives a solution $ x\equal{}\frac{\minus{}4\minus{}B}{A\plus{}1}$
$ B\le 4A<\minus{}4$ gives a solution $ x\equal{}\frac{\minus{}4\minus{}B}{A\plus{}1}$

4) Trial for a synthesis
=========================
Consider the plane of points $ (A,B)$ and the lines :
$ B\equal{}\minus{}6A$
$ B\equal{}3\minus{}3A$
$ B\equal{}1\minus{}A$
$ B\equal{}3\plus{}A$
$ B\equal{}4A$
$ A\equal{}1$
$ A\equal{}\plus{}1$

These seven lines split the plane in $ 24$ areas + some pecular segments and points giving :
Area 1 : $ A>1$ and $ B<\minus{}6A$
Area 2 : $ A>1$ and $ 3\minus{}3A>B\ge\minus{}6A$
Area 3 : $ A>1$ and $ 1\minus{}A>B\ge 3\minus{}3A$
Area 4 : $ A>1$ and $ A\plus{}3>B\ge 1\minus{}A$
Area 5 : $ A>1$ and $ 4A>B\ge A\plus{}3$
Area 6 : $ A>1$ and $ B\ge 4A$
Area 7 : $ 1>A>\minus{}1$ and $ B<4A$ and $ B<\minus{}6A$
Area 8 : $ 1>A>\minus{}1$ and $ B\ge\minus{}6A$ and $ B<4A$ and $ B<1\minus{}A$
Area 9 : $ 1>A>\minus{}1$ and $ B\ge\minus{}6A$ and $ B\ge 4A$ and $ B<1\minus{}A$
Area 10: $ 1>A>\minus{}1$ and $ B\ge 4A$ and $ B<\minus{}6A$ and $ B<1\minus{}A$
Area 11: $ 1>A>\minus{}1$ and $ B\ge 1\minus{}A$ and $ B\le 3\minus{}3A$ and $ B<4A$
Area 12: $ 1>A>\minus{}1$ and $ B>3\minus{}3A$ and $ B<4A$
Area 13: $ 1>A>\minus{}1$ and $ B\ge 4A$ and $ B\le 3\minus{}3A$ and $ B\le A\plus{}3$ and $ B\ge\minus{}6A$ and $ B\ge 1\minus{}A$
Area 14: $ 1>A>\minus{}1$ and $ B\ge 1\minus{}A$ and $ B<\minus{}6A$ and $ B\le A\plus{}3$
Area 15: $ 1>A>\minus{}1$ and $ B\ge 4A$ and $ B\le A\plus{}3$ and $ B>3\minus{}3A$
Area 16: $ 1>A>\minus{}1$ and $ B>A\plus{}3$ and $ B<\minus{}6A$
Area 17: $ 1>A>\minus{}1$ and $ B>A\plus{}3$ and $ B\ge\minus{}6A$ and $ B\le 3\minus{}3A$
Area 18: $ 1>A>\minus{}1$ and $ B>A\plus{}3$ and $ B>3\minus{}3A$
Area 19: $ \minus{}1>A$ and $ B\le 4A$
Area 20: $ \minus{}1>A$ and $ B>4A$ and $ B\le A\plus{}3$
Area 21: $ \minus{}1>A$ and $ B>A\plus{}3$ and $ B\le 1\minus{}A$
Area 22: $ \minus{}1>A$ and $ B>1\minus{}A$ and $ B\le 3\minus{}3A$
Area 23: $ \minus{}1>A$ and $ B>3\minus{}3A$ and $ B\le\minus{}6A$
Area 24: $ \minus{}1>A$ and $ B>\minus{}6A$

And we have the following result :
==================================
Area 1 : $ A>1$ and $ B<\minus{}6A$. One solution : $ \frac{\minus{}B\minus{}6}{A\minus{}1}$

Area 2 : $ A>1$ and $ 3\minus{}3A>B\ge\minus{}6A$. One solution : $ \frac{6\minus{}B}{A\plus{}1}$

Area 3 : $ A>1$ and $ 1\minus{}A>B\ge 3\minus{}3A$. One solution : $ \frac{\minus{}B}{A\minus{}1}$

Area 4 : $ A>1$ and $ A\plus{}3>B\ge 1\minus{}A$. One solution : $ \frac{2\minus{}B}{A\plus{}1}$

Area 5 : $ A>1$ and $ 4A>B\ge A\plus{}3$. One solution : $ \frac{4\minus{}B}{A\minus{}1}$

Area 6 : $ A>1$ and $ B\ge 4A$. One solution : $ \frac{\minus{}4\minus{}B}{A\plus{}1}$

Area 7 : $ 1>A>\minus{}1$ and $ B<4A$ and $ B<\minus{}6A$. No solution.

Area 8 : $ 1>A>\minus{}1$ and $ B\ge\minus{}6A$ and $ B<4A$ and $ B<1\minus{}A$. Two solutions : $ \frac{\minus{}B\minus{}6}{A\minus{}1},$ $ \frac{6\minus{}B}{A\plus{}1}$

Area 9 : $ 1>A>\minus{}1$ and $ B\ge\minus{}6A$ and $ B\ge 4A$ and $ B<1\minus{}A$. Four solutions : $ \frac{\minus{}B\minus{}6}{A\minus{}1},$ $ \frac{6\minus{}B}{A\plus{}1},$ $ \frac{4\minus{}B}{A\minus{}1},$ $ \frac{\minus{}4\minus{}B}{A\plus{}1}$

Area 10: $ 1>A>\minus{}1$ and $ B\ge 4A$ and $ B<\minus{}6A$ and $ B<1\minus{}A$. Two solutions : $ \frac{4\minus{}B}{A\minus{}1},$ $ \frac{\minus{}4\minus{}B}{A\plus{}1}$

Area 11: $ 1>A>\minus{}1$ and $ B\ge 1\minus{}A$ and $ B\le 3\minus{}3A$ and $ B<4A$. Four solutions : $ \frac{\minus{}B\minus{}6}{A\minus{}1},$ $ \frac{6\minus{}B}{A\plus{}1},$ $ \frac{\minus{}B}{A\minus{}1},$ $ \frac{2\minus{}B}{A\plus{}1}$

Area 12: $ 1>A>\minus{}1$ and $ B>3\minus{}3A$ and $ B<4A$. Two solutions : $ \frac{\minus{}B\minus{}6}{A\minus{}1},$ $ \frac{2\minus{}B}{A\plus{}1}$

Area 13: $ 1>A>\minus{}1$ and $ B\ge 4A$ and $ B\le 3\minus{}3A$ and $ B\le A\plus{}3$ and $ B\ge\minus{}6A$ and $ B\ge 1\minus{}A$. Six solutions : $ \frac{\minus{}B\minus{}6}{A\minus{}1},$ $ \frac{6\minus{}B}{A\plus{}1},$ $ \frac{\minus{}B}{A\minus{}1},$ $ \frac{2\minus{}B}{A\plus{}1},$ $ \frac{4\minus{}B}{A\minus{}1},$ $ \frac{\minus{}4\minus{}B}{A\plus{}1}$

Area 14: $ 1>A>\minus{}1$ and $ B\ge 1\minus{}A$ and $ B<\minus{}6A$ and $ B\le A\plus{}3$. Four solutions : $ \frac{\minus{}B}{A\minus{}1},$ $ \frac{2\minus{}B}{A\plus{}1},$ $ \frac{4\minus{}B}{A\minus{}1},$ $ \frac{\minus{}4\minus{}B}{A\plus{}1}$

Area 15: $ 1>A>\minus{}1$ and $ B\ge 4A$ and $ B\le A\plus{}3$ and $ B>3\minus{}3A$. Four solutions : $ \frac{\minus{}B\minus{}6}{A\minus{}1},$ $ \frac{2\minus{}B}{A\plus{}1},$ $ \frac{4\minus{}B}{A\minus{}1},$ $ \frac{\minus{}4\minus{}B}{A\plus{}1}$

Area 16: $ 1>A>\minus{}1$ and $ B>A\plus{}3$ and $ B<\minus{}6A$. Two solutions : $ \frac{\minus{}B}{A\minus{}1},$ $ \frac{\minus{}4\minus{}B}{A\plus{}1}$

Area 17: $ 1>A>\minus{}1$ and $ B>A\plus{}3$ and $ B\ge\minus{}6A$ and $ B\le 3\minus{}3A$. Four solutions : $ \frac{\minus{}B\minus{}6}{A\minus{}1},$ $ \frac{6\minus{}B}{A\plus{}1},$ $ \frac{\minus{}B}{A\minus{}1},$ $ \frac{\minus{}4\minus{}B}{A\plus{}1}$

Area 18: $ 1>A>\minus{}1$ and $ B>A\plus{}3$ and $ B>3\minus{}3A$. Two solutions : $ \frac{\minus{}B\minus{}6}{A\minus{}1},$ $ \frac{\minus{}4\minus{}B}{A\plus{}1}$

Area 19: $ \minus{}1>A$ and $ B\le 4A$ One solution : $ \frac{\minus{}4\minus{}B}{A\plus{}1}$

Area 20: $ \minus{}1>A$ and $ B>4A$ and $ B\le A\plus{}3$. One solution : $ \frac{4\minus{}B}{A\minus{}1}$

Area 21: $ \minus{}1>A$ and $ B>A\plus{}3$ and $ B\le 1\minus{}A$. One solution : $ \frac{2\minus{}B}{A\plus{}1}$

Area 22: $ \minus{}1>A$ and $ B>1\minus{}A$ and $ B\le 3\minus{}3A$. One solution : $ \frac{\minus{}B}{A\minus{}1}$

Area 23: $ \minus{}1>A$ and $ B>3\minus{}3A$ and $ B\le\minus{}6A$. One solution : $ \frac{6\minus{}B}{A\plus{}1}$

Area 24: $ \minus{}1>A$ and $ B>\minus{}6A$. One solution : $ \frac{\minus{}B\minus{}6}{A\minus{}1}$

Plus the specific segments and points :
If $ A\equal{}\minus{}1$ and $ B<\minus{}4$ : no solution
If $ A\equal{}\minus{}1$ and $ B\equal{}\minus{}4$ : infinitely many solutions $ x\in(\minus{}\infty,\minus{}4]$
If $ A\equal{}\minus{}1$ and $ \minus{}4<B<2$ : solution is $ x\equal{}\frac B2\minus{}2$
If $ A\equal{}\minus{}1$ and $ B\equal{}2$ : infinitely many solutions $ x\in[\minus{}1,\plus{}1]$
If $ A\equal{}\minus{}1$ and $ 2<B<6$ : solution is $ x\equal{}\frac B2$
If $ A\equal{}\minus{}1$ and $ B\equal{}6$ : infinitely many solutions $ x\in[3,6]$
If $ A\equal{}\minus{}1$ and $ 6<B$ : solution is $ x\equal{}\frac B2\plus{}3$

If $ A\equal{}1$ and $ B<\minus{}6$ : no solution
If $ A\equal{}1$ and $ B\equal{}\minus{}6$ : infinitely many solutions $ x\in[6,\plus{}\infty)$
If $ A\equal{}1$ and $ \minus{}6<B<0$ : solution is $ x\equal{}3\minus{}\frac B2$
If $ A\equal{}1$ and $ B\equal{}0$ : infinitely many solutions $ x\in[1,3]$
If $ A\equal{}1$ and $ 0<B<4$ : solution is $ x\equal{}1\minus{}\frac B2$
If $ A\equal{}1$ and $ B\equal{}4$ : infinitely many solutions $ x\in[\minus{}4,\minus{}1]$
If $ A\equal{}1$ and $ 4<B$ : solution is $ x\equal{}\minus{}2\minus{}\frac B2$

[hide="Examples"]
Area 1 : $ A\equal{}2$ and $ B\equal{}\minus{}14$ : $ \parallel{}|x\minus{}1|\minus{}2|\minus{}3|\equal{}2x\minus{}14$ and solution $ 8$

Area 2 : $ A\equal{}2$ and $ B\equal{}\minus{}9$ : $ \parallel{}|x\minus{}1|\minus{}2|\minus{}3|\equal{}2x\minus{}9$ and solution $ 5$

Area 3 : $ A\equal{}2$ and $ B\equal{}\minus{}2$ : $ \parallel{}|x\minus{}1|\minus{}2|\minus{}3|\equal{}2x\minus{}2$ and solution $ 2$

Area 4 : $ A\equal{}2$ and $ B\equal{}2$ : $ \parallel{}|x\minus{}1|\minus{}2|\minus{}3|\equal{}2x\plus{}2$ and solution $ 0$

Area 5 : $ A\equal{}2$ and $ B\equal{}6$ : $ \parallel{}|x\minus{}1|\minus{}2|\minus{}3|\equal{}2x\plus{}6$ and solution $ \minus{}2$

Area 6 : $ A\equal{}2$ and $ B\equal{}11$ : $ \parallel{}|x\minus{}1|\minus{}2|\minus{}3|\equal{}2x\plus{}11$ and solution $ \minus{}5$

Area 8 : $ A\equal{}\frac 12$ and $ B\equal{}\minus{}1$ : $ \parallel{}|x\minus{}1|\minus{}2|\minus{}3|\equal{}\frac x2\minus{}1$ and solutions $ 10,\frac{14}3$

Area 9 : $ A\equal{}\frac 16$ and $ B\equal{}\frac 34$ : $ \parallel{}|x\minus{}1|\minus{}2|\minus{}3|\equal{}\frac x6\plus{}\frac 34$ and solutions $ \frac{81}{10},\frac 92,\minus{}\frac{39}{10},\minus{}\frac{57}{14}$

Area 10: $ A\equal{}\minus{}\frac 12$ and $ B\equal{}\minus{}1$ : $ \parallel{}|x\minus{}1|\minus{}2|\minus{}3|\equal{}\frac \minus{}x2\minus{}1$ and solutions $ \minus{}6,\minus{}\frac{10}3$

Area 11: $ A\equal{}\frac 12$ and $ B\equal{}1$ : $ \parallel{}|x\minus{}1|\minus{}2|\minus{}3|\equal{}\frac x2\plus{}1$ and solutions $ 14,\frac{10}3,2,\frac 23$

Area 12: $ A\equal{}\frac 34$ and $ B\equal{}1$ : $ \parallel{}|x\minus{}1|\minus{}2|\minus{}3|\equal{}\frac {3x}4\plus{}1$ and solutions $ 28,\frac 47$

Area 13: $ A\equal{}0$ and $ B\equal{}2$ : $ \parallel{}|x\minus{}1|\minus{}2|\minus{}3|\equal{}2$ and solutions $ 8,4,2,0,\minus{}2,\minus{}6$

Area 14: $ A\equal{}\minus{}\frac 12$ and $ B\equal{}2$ : $ \parallel{}|x\minus{}1|\minus{}2|\minus{}3|\equal{}\minus{}\frac x2\plus{}2$ and solutions $ \frac 43,0,\minus{}\frac 43,\minus{}12$

Area 15: $ A\equal{}\frac 12$ and $ B\equal{}3$ : $ \parallel{}|x\minus{}1|\minus{}2|\minus{}3|\equal{}\frac x2\plus{}3$ and solutions $ 18,\minus{}\frac 23,\minus{}2,\minus{}\frac {14}3$

Area 16: $ A\equal{}\minus{}\frac 34$ and $ B\equal{}3$ : $ \parallel{}|x\minus{}1|\minus{}2|\minus{}3|\equal{}\minus{}\frac {3x}4\plus{}3$ and solutions $ \frac{12}7,\minus{}28$

Area 17: $ A\equal{}\minus{}\frac 12$ and $ B\equal{}4$ : $ \parallel{}|x\minus{}1|\minus{}2|\minus{}3|\equal{}\minus{}\frac x2\plus{}4$ and solutions $ \frac{20}3,4,\frac 83,\minus{}16$

Area 18: $ A\equal{}0$ and $ B\equal{}4$ : $ \parallel{}|x\minus{}1|\minus{}2|\minus{}3|\equal{}4$ and solutions $ 10,\minus{}8$

Area 19: $ A\equal{}\minus{}2$ and $ B\equal{}\minus{}10$ : $ \parallel{}|x\minus{}1|\minus{}2|\minus{}3|\equal{}\minus{}2x\minus{}10$ and solution $ \minus{}6$

Area 20: $ A\equal{}\minus{}2$ and $ B\equal{}\minus{}5$ : $ \parallel{}|x\minus{}1|\minus{}2|\minus{}3|\equal{}\minus{}2x\minus{}5$ and solution $ \minus{}3$

Area 21: $ A\equal{}\minus{}2$ and $ B\equal{}2$ : $ \parallel{}|x\minus{}1|\minus{}2|\minus{}3|\equal{}\minus{}2x\plus{}2$ and solution $ 0$

Area 22: $ A\equal{}\minus{}2$ and $ B\equal{}6$ : $ \parallel{}|x\minus{}1|\minus{}2|\minus{}3|\equal{}\minus{}2x\plus{}6$ and solution $ 2$

Area 23: $ A\equal{}\minus{}4$ and $ B\equal{}18$ : $ \parallel{}|x\minus{}1|\minus{}2|\minus{}3|\equal{}\minus{}4x\plus{}18$ and solution $ 4$

Area 24: $ A\equal{}\minus{}2$ and $ B\equal{}15$ : $ \parallel{}|x\minus{}1|\minus{}2|\minus{}3|\equal{}\minus{}2x\plus{}15$ and solution $ 7$
[/hide]

[quote="Yuriy Solovyov"]:o [b]pco[/b], i hope this is not autogenerated code  :rotfl: you are very patient[/quote]

Not at all, I really created this post, letter per letter, line per line, + examples in order to check errors.

I wonder where aktyw19 found this question  :mad:

[quote="aktyw19"][quote="pco"]

I wonder where aktyw19 found this question  :mad:[/quote]
what are you so angry?[/quote]

Because I dont think it is a real problem found in contest / book / class.
I think it is an invented problem which must not be posted in "algebra unsolved problems".

[quote="aktyw19"]After all, you do not have to answer[/quote]

You're right. But there is a space for invented problem for which poster does not know if a solution exists : "open problems" and I never spend time on these problems.

When you post in "unsolved" forum, you claim that this is a normal olympiad problem whose solution exists but you dont know it ...
So, in this forum, I generally spend time because I trust you and I think there is really a normal solution.

So, I'm angry because I think that some posters lie when posting in this forum because they dont know if a normal olympiad solution exist. It would be a polite attitude to post in the correct forum.

But, you're right, I may also just consider some posters are liers and just ignore their posts.

[i][b]aktyw19[/b] , I think you are absolutely wrong

   You must understand that your initial problem doesn't deserve to be post in [b]Olympiad box[/b] :mad:

I guess that you don't know the origin of your problem  :rotfl:

  Pco , I have advised you many times that with your ability , you shouldn't spent time with this kind of crazy problem . :lol: [/i]

YES....
We are really glad to have you here in AoPS [b]pco[/b].....
Thank you for your solutions......
And please dont stop giving us your precious ideas......... :wink:  :)  :P


geometry inradius trigonometry circumcircle

Each side of a regular dodecagon $ A_1A_2A_3...A_{12}$ has length $ 2$. What is the area of
the pentagon $ A_1A_2A_3A_4A_5$?

[hide]
Let the center be $ O$; the desired area is $ [A_1A_2A_3A_4A_5] \equal{} [OA_1A_2A_3A_4A_5] \minus{} [OA_1A_5] \equal{} 4[OA_1A_2] \minus{} [OA_1A_5]$. We see that the inradius is $ r \equal{} \frac {1}{\tan 15} \equal{} \frac {1}{\tan (45 \minus{} 30)} \equal{} 2 \plus{} \sqrt {3}$ and the circumradius is $ R \equal{} \frac {1}{\sin 15} \equal{} \sqrt {6} \plus{} \sqrt {2}$. Thus $ [OA_1A_2] \equal{} \frac {1}{2} \cdot 2r \equal{} 2 \plus{} \sqrt {3}$ and, since $ \triangle OA_1A_5$ is $ 30 \minus{} 60 \minus{} 90 \triangle$, $ [OA_1A_5] \equal{} 2\left(\frac {1}{2} \cdot \frac {R}{2} \cdot \frac {R\sqrt {3}}{2}\right) \equal{} \frac {R^2\sqrt {3}}{4} \equal{} 2\sqrt {3} \plus{} 3$. Therefore, the answer is $ 4(2 \plus{} \sqrt {3}) \minus{} (2\sqrt {3} \plus{} 3) \equal{} \boxed{5 \plus{} 2\sqrt {3}}$.

[/hide]


We define f(m,n) by this formular (m,n are positive integers):
  f(m,n) =  \sum( (-1)^(m-i) *(i,m)*(n,i*n) ) for i = 1 to m
  Here (u,v) is equal to the number of the subsets consisting of u elements
from the set v elements(u<=v)
       a) Prove that if n is a prime then k=f(m,n)/(n^m) is a non-negative integer.
       b) Evaluate f(m,m).
       c) Find the condition between m and n such that f(m,n)*f(n,m)=0.

This old one looked somewhat interesting -- I don't have time to hack at it now, but I thought I'd convert it to a legible form so others can play around with it:

$ f(m,n) \equal{}  \sum_{i \equal{} 1}^m (\minus{}1)^{m\minus{}i} \cdot \binom{m}{i} \cdot \binom{ni}{n}$

       a) Prove that if $ n$ is a prime then $ k\equal{}\frac{f(m,n)}{n^m}$ is a non-negative integer.
       b) Evaluate $ f(m,m)$.
       c) Find the condition between $ m$ and $ n$ such that $ f(m,n)\cdot f(n,m)\equal{}0$.


inequalities
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inequalities unsolved

Im curious if anyone knows the proof to the kantorovich inequlaity? Im not sure if it should go in here but i think its quite an interesting topic.

J.D.

Kantorovich ineq was posted and solved on ML before . Try to search for it !


It should be $\frac{1}{2}\binom{m}{n}$,but I can't prove.

I mean the coefficient in power series expansion,is it too difficult? :(

[quote="shencaili"]What's the coefficient of $x^{n}$ in $\frac{(1+x)^{2m+1}}{(2+x)^{n+1}}$[/quote]
hi,shencaili,ni hao .
x+2=t,then,

[quote="milin"]Hint:

$ \frac {1}{(x \plus{} 2)^{n \plus{} 1}} \equal{} \frac {1}{2^{n \plus{} 1}}\sum^\infty_{i \equal{} 0}( \minus{} 1)^i\binom{n \plus{} i}{i}\frac {x^i}{2^i}$

$ (x \plus{} 1)^{2m \plus{} 1} \equal{} \sum^{2m \plus{} 1}_{i \equal{} 0}\binom{2m \plus{} 1}{i}x^i$[/quote]

And then？ :maybe:

Coefficient is \[ a_n\equal{}\sum_{k\equal{}0}^n\binom{2m\plus{}1}{k}\binom{\minus{}n\minus{}1}{n\minus{}k}2^{k\minus{}1\minus{}2n}.\]
For example $ a_0\equal{}\frac{1}{2}, \ a_1\equal{}\frac{m}{2}\equal{}\frac{1}{2}\binom{m}{1}, \ a_2\equal{}\frac{m(m\minus{}1)}{4}\equal{}\frac 12 \binom{m}{2}$.
[quote="shencaili"]It should be $ \frac {1}{2}\binom{m}{n}$,but I can't prove.[/quote]
It may be true.

It should be $ \frac {1}{2}\binom{m}{n}$,but I can't prove. :(


I'm stuck in first step itself  :(  :ninja:

The first step looks like a reduction: L is very likely the alcohol. Does this help?

Can you pls cite a reference for the first step reduction? I have never come across it!  :maybe:

One more time, this would make more sense if at least molecular formulas were given: otherwise, there is more than one possible answer. Anyway, the reduction that I have in mind is a bimolecular reduction - coupling of two molecules to the pinacol.

We had used Mg Zn , to convert the ketones to pinacols.

Does Mg Hg do the same operation as Mg Zn?

[quote="hell_ever"]Surprisingly, the first reaction is given in our state board textboook (maharashtra)  :P[/quote]

In 12th grade state board text book?  :o

Then give the mechanism.  :)

Next L to M pinacol-pinacolone?
so i think M is 3-isopropyl-3,4-dimethyl-pentan-2-one(isopropyl grp migrating). correct?  :D

Yes, that should be correct: the most substituted alkyl group migrates preferantly.

not surprisingly, the only intermediate given is some complex with no hint about the mechanism.

What complex? And the mechanism of what?

I was answering Da vinci's question. The complex formed in the mechanism of the first reaction is (ketone is R-CO-R')


geometry
parallelogram
vector
trigonometry
circumcircle
geometric transformation
homothety

There are several solutions to the so called Napolean's Theorem. Most of them are based on complex number or some very weired transfomation geometry. I have discovered the following extension to the theorem, which I have not found in any documents/forums so far :

[b]"The Napolean triangle shares the same centroid with the original triangle."[/b]

I also have a very elegant plane-geometric proof for the above statement, which also establishes the Napolean's Theorem simultaneously. It requires a somewat involved construction which I am not able to present here right now. I wish to publish it in some print forum if I come to know that nobody has thought of it yet.
I want to have the members' response/feedback on the matter.

Let ABC be the subject triangle,  BCD, CAE and ABF be the isosceles triangles having the angles D, E and F of $ 120^\circ$ each. If M and N are midpoints of BC and EF respectively, our aim is to prove $ MN \parallel AD$ and $ MN \equal{} \frac {AD}{2}$, statement equivalent with the conclusion. Let's take D' simmetrical of D about BC.
It is classical by now that AFD'E is a parallelogram, hence N is the middle of AD' and, with M middle of DD', the problem is solved.

Best regards,
sunken rock

Thank you both for the beautiful problem and proof.

Thanks sunken rock for your solution. The simplicity is really nice.  However it presumes that 'AFDE is a parallelogram' theorem and the existence of the Napolean theorem. Also how 'MN || AD and half of it' leads to the conclusion is not very apparent.

this result has been found (i don't by who  but i found it in the book Geometry revisited by H.S.M.Coxeter,S.L.Greitzer):
the napolean triangles (the inner and the outer one) share the same centroid.
by what i see here we have :
the napolean triangles and the original triangle share the same centroid.
that's really nice (thx shanku!)

also it can be proved by using vectors
suppose that $ G$ is the centroid of $ ABC$ we must show that $ G$ is the centroid of $ O_aO_bO_c$ but we have :
$ 2\vec{GA}\equal{}(\vec{O_cA}\plus{}\vec{O_bA})\plus{}(\vec{GO_c}\plus{}\vec{GO_b})$
and similarly for the other sides which gives
$ 0\equal{}2(\vec{GA}\plus{}\vec{Gb}\plus{}\vec{GC})\equal{}(\vec{DO_a}\plus{}\vec{EO_b}\plus{}\vec{FO_c})\plus{}2(\vec{GO_c}\plus{}\vec{GO_b}\plus{}\vec{GO_a})$
so we must prove that :
$ \vec{DO_a}\plus{}\vec{EO_b}\plus{}\vec{FO_c}\equal{}0$
but we know that

$ \frac{DO_a}{\sin 180^\circ\minus{}\angle A}\equal{}\frac{EO_b}{\sin 180^\circ\minus{}\angle B}\equal{}\frac{FO_c}{\sin 180^\circ\minus{}\angle C}\equal{}\frac{2}{3}R . \sin 60^\circ$

where $ R$ is the circumradius of $ \Delta ABC$ and $ D,E,F$ are the vertices of the equilateral triangles which share a side with $ BC,CA,AB$ respectively.
so we are done!

My proof is as following : (I could not give the figure here, Could somebody provide the diagram pl.?)
-------------------------------------------------------
Let $ {BX} \perp{AD}$ and $ {CY} \perp{AF}$ and $ P, Q$ are midpoints of $ AB$ and $ AC$ respectively. Therefore, $ AD: AX \equal{} AF: AY \equal{} 2: 3$ implying $ {DF} \parallel{XY}$. We also have Δ$ MXP$ and Δ$ YMQ$ congruent, $ M$ being the midpoint of $ BC$, as $ XP \equal{} MQ$( = ½ $ AB ), PM \equal{} QY$( = ½ $ AC )$ and $ \angle XPM \equal{} \angle MQY ( \equal{} \angle BAC$ + 60º ). Therefore $ XM \equal{} MY$ and $ \angle XMY \equal{} \angle PMQ \plus{} \angle XMP \plus{} \angle QMY \equal{} \angle BAC \plus{} \angle QYM \plus{} \angle QMY \equal{} \angle BAC$ + 180º – ($ \angle BAC$ + 60º) = 120º.
A 2/3-scale homothety about A will take Δ$ XMY$ to Δ$ DOF$, where $ MO: OA$ = 1:2 implying $ O$ to be the centroid of Δ$ ABC$.
Therefore $ \angle ODF \equal{} \angle OFD$.
Since the centroid of Δ$ ABC$ is independent irrespective of the choice of D, E or F, it follows that $ \angle ODE \equal{} \angle OED$ and $ \angle OEF \equal{} \angle OFE$, and thus all the angles of Δ$ DEF$ are equal, i.e. Δ$ DEF$ is equilateral that shares the same centroid of the original triangle.
-----------------------------------------------------
Here, $ ABC$ is the original Triangle and $ EFD$ its Napolean Triangle, where $ DAB$, $ EBC$ and $ FCA$ are isoceles triangles with vertical angles D, E and F respectively of 120 deg.
 :lol:  :lol:  :D

See also a generalization [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=315336][color=blue]Here[/color][/url], which is coming form the past.

Kostas Vittas.

All postings are very recent ones after start of the present topic. :o


A regular pentagon with side length 300 cm has the same perimeter as a particular square. What is the area of the square? Express your answer to the nearest thousand square centimeters.

The perimeter of the square is 1500 so side length is 375.  Area is 140625, rounding gives 141000.


search
combinatorics solved
combinatorics

Let $M$ be the set of all sequences $a = (a_1,...,a_n)$ of fixed length $n$, $a_i \in \{-1,1\}$ for all 1,2,...,$n$. The product of two such sequences $a$ and $b = (b_1,b_2,...,b_n)$ is defined as $ab = (a_1b_1,a_2b_2,...,a_nb_n)$. For $a \in M$ and an arbitrary proper subset $T$ of $M$ let $aT = \{ab|b \in T\}$. Prove that for each proper subset $K$ of $M$ with $k$ elements there is a sequence $c = (c_1,c_2,...,c_n)$ from $M$ such that the intersection of $K$ and $cK$ contains at most $k^2/2^n$ sequences.

[i]Edited by Myth[/i]

I can't believe it!!!! Such easy problem has been lieing in Unsolved section for 9 months!

It is clear $ac=b$ iff $c=ab$. So consider all possible products $C=\{ab\mid a,b\in K\}$. This set consists of $k^2$ elements. There are $2^n$ binary sequences, so there exist $c$ s.t. $c$ can be expressed as $ab$ ($a,b\in K$) in no more than $\frac{k^2}{2^n}$ ways.

That's all.

Happy New Year!
 :lol:  :lol:  :lol:

Actually, I seem to remember it being posted by orl in another forum, and I think the problem was solved there. I remember a nice probabilistic approach.

I think it was a subforum of the "Problem Solving" section, but I don't have time to search right now.

Here was the previous topic:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14999[/url].


graph theory
linear algebra
number theory

Hi all,
Please answer the poll and post your comments.

I have the will to specialize in informatics so i'm supposed to be close to math in a way (algorithmic & networks:graph theory,combinatorics,linear algebra,number theory) ,your poll ignores this case ;)

I feel that you could vote Yes because your interests are also closely related to mathematics.
Please more people reply.

I'm not sure. However, I still have one year and a half to think about it...

Strictly math? No. However, whatever I do will undoubtably involve some level of math.

I need to find a job involving math, dvorak keyboards, and hot women.


Putting bad words, or being extremely aggresive, or puting sth too bad to another to insult him... or spamming a lot

[quote="whosthis2"]i just want to know rules[/quote]
Each specific forum has guidelines for you to follow (they're usually located at the top of the forum). It's best to read them before posting. Also, don't post anything that is useless, mean, flamey, etc. This forum is a math problem solving forum. Don't post something completely unrelated or inappropriate. If you do post spam, moderators or administrators will just delete it. Basically, all you need to do is to just try to use common sense. If you have a question, you can PM a moderator or an administrator and he/she will try to help you out. Thanks.

[quote="Treething"]being like me?[/quote]
Yeah, that pretty much sums it up.

I think this belongs in Questions, Suggestions, & Announcements forum.
Moved.
MATHCOUNTS >> Questions, Suggestions, & Announcements

[quote="whosthis2"]i just want to know rules[/quote]

Basically, you get banned for doing what you did last time.  Posting obscenities and things that, while many people don't get, can be extremely offensive to people who do understand you.

Um...just saying, levans said (I think) that that will happen, and just wait. Also, you should have not revived a 9 year old thread.

[quote=tarikjaber]I was banned, but i didn't mean to, it was my friend. I really want to play ftw, but I cant, so someone please help.[/quote]

We've answered your email query on this issue.


Determine the number of positive integers $ n$ such that $ \phi (n) \equal{} 8$
I got $ 5$. They are $ 15,16,20,24,30$. Are these the only?

Edit: Forgot 24


Ring Theory
calculus
integration
algebra
function
domain
superior algebra

Let $R$ be a $UFD$ and $S$ be a multiplicative subset of $R$. Prove that $S^{-1}R$ is a $UFD$ as well.

[quote="sissa"]Isn't this obvious from the description of $S^{-1}R$ as the set of $\frac{r}{s^{n}}$?[/quote]

Maybe it is if $S=\{s^{n}|\ n\in\mathbb N\}$.

In the other cases, we can use:
[b]Theorem:[/b] An integral domain $R$ is an UFD if and only if every nonzero prime ideal of $R$ contains a prime element.

To our problem: $R$ is UFD, so it is a domain, hence $S^{-1}R$ is also a domain.
Let $P$ be a nonzero prime ideal of $S^{-1}R$. The preimage of $P$ through the natural  injective (because we are working with domains) ring homomorphism $R\to S^{-1}R: r\to\frac{r}{1}$ is a prime ideal of $R$ that we call $Q$.

[[i]addendum[/i]: It is known that the preimage map described above gives a bijective correspondence between prime ideals of $S^{-1}R$ and prime ideals of $R$ not intersecting $S$.]

By the addendum, but it is anyway easy to prove, $Q$ is nonzero. By the hypothesis, $Q$ contains a prime element $q$ of $R$. Let $p=\frac{q}{1}\in P$.
First of all, since $Q$ does not intersect $S$ and $q$ is not invertible in $R$ because it is prime, $p$ is not invertible in $S^{-1}Q$.
Assume that it is not prime. Then there exist $a,b\in R$ and $s,t\in S$ such that $p|\frac{a}{s}\cdot\frac{b}{t}$ and $p$ divides neither of the factors. Then $\frac{ab}{st}=\frac{qx}{y}$ for some $x\in R$ and $y\in S$. It is equivalent to $aby=stxq$. Since $q$ is prime in $R$, it divides $a$, $b$ or $y$. $q$ does not divide $y$, otherwise $y$ is in $Q$ because $q$ is and we saw that $Q$ and $S$ are disjoint. So $q$ divides $a$ or $b$. This means that $p$ divides $\frac{a}{s}$ or $\frac{b}{t}$ which solves the problem.


This hardly qualifies as a combinatorial identity; it's just algebra.  $(1+x)(1-x) = (1-x^{2})$, and multiplication is commutative, and so forth.

I think you meant the combinatorial implication of this identity, which is something like

$\sum_{k=0}^{2i}(-1)^{k}{n \choose k}{n \choose (2i-k) }= (-1)^{i}{n \choose i}$


Let $s_1, s_2, s_3, \dots$ be an infinite, nonconstant sequence of rational numbers, meaning it is not the case that $s_1 = s_2 = s_3 = \dots.$  Suppose that $t_1, t_2, t_3, \dots$ is also an infinite, nonconstant sequence of rational numbers with the property that $(s_i - s_j)(t_i - t_j)$ is an integer for all $i$ and $j$.  Prove that there exists a rational number $r$ such that $(s_i - s_j)r$ and $(t_i - t_j)/r$ are integers for all $i$ and $j$.

Paul Christiano's solution:

[hide]
We can always re-normalize by translating all the elements in a sequence so that $ s_1 \equal{} t_1 \equal{} 0$. Now divide the second sequence by a proper amount and multiply the first sequence by a proper amount so that $ t_2 \equal{} 1$. Then $ s_2 \equal{} k$ for some integer $ k$ (since $ (s_2 \minus{} s_1)(t_2 \minus{} t_1)$ is an integer), and we have

$ (s_i \minus{} 1)(t_i \minus{} k) \equal{} s_it_i \plus{} k \minus{} (t_i \plus{} ks_i) \in \mathbb{Z}$

But $ s_it_i \equal{} (s_i \minus{} 0)(t_i \minus{} 0) \equal{} (s_i \minus{} s_1)(t_i \minus{} t_1)$ must be in $ \mathbb{Z}$, and certainly $ k \in \mathbb{Z}$, so $ t_i \plus{} ks_i \in \mathbb{Z}$. Now suppose $ ks_i \not\in \mathbb{Z}$, then $ s_i$ has some factor $ p$ in its denominator that isn't cancelled out by $ k$, and furthermore $ t_i$ certainly can't have that term in its denominator since $ s_it_i \in \mathbb{Z}$. But then $ t_i \plus{} ks_i$ can't be an integer. This is a contradiction, so $ ks_i \in \mathbb{Z}$, and consequently $ t_i \in \mathbb{Z}$.

So we have shown that we can normalize both sequences so that one sequence is always integers and the other has denominator divisible by some fixed integer $ k$. In particular, normalize the sequence of integers (i.e. the $ t_i$) so that its overall $ gcd$ is exactly $ 1$. We claim that this will cause the other sequence to consist of integers as well.

Indeed, suppose for the sake of contradiction that $ p$ divides the denominator of some $ s_i$. Then $ p$ does not divide some $ t_j$ (since their overall $ gcd$ is $ 1$). Then $ p$ must divide $ t_i$ since $ s_it_i$ must be an integer. So $ t_i \minus{} t_j$ is not divisible by $ p$. Similarly, $ p$ cannot divide the denominator of $ s_j$ since $ s_jt_j$ must be an integer, so $ s_i \minus{} s_j$ must have denominator divisible by $ p$. But then $ (s_i \minus{} s_j)(t_i \minus{} t_j)$ has denominator divisible by $ p$, and in particular it is not an integer. This is a contradiction, so actually all the $ s_i$ must be integers, so we are done.
[/hide]

:o The second half of his solution matched my first half, but my second half didn't match his first half due to me blanking out and being unable to continue further :( (So I ended up writing junk for the second half)

[hide="My (arrogantly claimed to be) elegant solution"]
We start out with only the sequence $ s_n$. Let all the terms $ s_i$ be representable as $ \frac {a_i}{b_i}$, where $ a_i$ and $ b_i$ are integers.

We can select $ t_1$ to be anything. Let all terms $ t_i$ be representable as $ \frac {c_i}{d_i}$. Same as above.

Net, we choose a $ t_2$ such that $ (t_2 \minus{} t_1)(s_2 \minus{} s_1)$ is an integer. We can declare $ r$, at this point, to be $ \frac {b_1b_2}{d_1d_2}$. This works because $ (s_2 \minus{} s_1) \equal{} \frac {a_1}{b_1} \minus{} \frac {a_2}{b_2}$, the common denominator of which is $ b_1b_2$, and $ (t_2 \minus{} t_1) \equal{} \frac {c_1}{d_1} \minus{} \frac {c_2}{d_2}$, the common denominator of which is $ d_1d_2$, so multiplying or dividing appropriately will always produce an integer.

We keep adding $ t_m$ so that $ (t_m \minus{} t_i)(s_m \minus{} s_i)$ is an integer for all $ 1 \le i < m$. Then $ r$ is $ \frac {b_1b_2b_3\ldots b_m}{d_1d_2d_3\ldots d_m}$.

As we keep on adding rational numbers in this way, r will always have a value.

QED, LOL
[/hide]

I probably got one point docked because in my written solution I wrote that $ s_1t_1$ had to be an integer, which it didn't.

[quote="sdkudrgn88"]No, we can't. What if none of the s_n and t_n are zero? The sequences are an infinite number of rational numbers, but they don't have to be all of them. For all we know, it's the harmonic series, or one of the terms are 1 while the others are all 2/7...the problem only said they couldn't all be the same, so it could oscillate between two, etc...[/quote]
We can translate it, though, because if you subtract or add anything to all of the terms of the sequence, it still preserves all of the differences.

Edit: Sorry, should have checked before posting.

This was a fun one.

Here's my approach:
[hide="Solution"]For convenience, shift indices to start at zero. As in JSteinhardt's post, translate to $ s_0\equal{}t_0\equal{}0$.
By setting $ j\equal{}0$ in the given property, we get that $ s_it_i$ is an integer for all $ i$. By setting $ i\neq j$, we get that $ s_it_i\plus{}s_jt_j\minus{}(s_it_j\plus{}s_jt_i)$ is an integer, and thus $ s_it_j\plus{}s_jt_i$ is an integer in this case.
Now, for all $ n,k$, we have that $ \sum_{\substack{i,j\le n\\ i\plus{}j\equal{}k}}s_it_j$ is an integer; we can pair off terms $ s_it_{k\minus{}i}$ to integer sums, and $ s_{k/2}t_{k/2}$ is an integer if it appears.

This convolution is the rule for polynomial multiplication, and thus $ (s_1x\plus{}s_2x^2\plus{}\cdots\plus{}s_nx^n)\cdot(t_1x\plus{}t_2x_2\plus{}\cdots\plus{}t_nx^n)$ is an integer polynomial for each $ n$. Suppose $ n$ is large enough that neither factor is the zero polynomial, and let $ d_n$ be the gcd of the product's coefficients. By Gauss' lemma, we can write this product as $ \left(r_ns_1x\plus{}r_ns_2x^2\plus{}\cdots\plus{}r_ns_nx^n\right)\cdot\left(\frac{t_1}{d_nr_n}x\plus{}\frac{t_2}{d_nr_n}x^2\plus{}\cdots\plus{}\frac{t_n}{d_nr_n}x^n\right)\cdot d_n$ such that all of the $ r_ns_i$ and $ \frac{t_j}{d_nr_n}$ are integers and $ gcd(r_ns_i)\equal{}gcd(\frac{t_j}{d_nr_n})\equal{}1$.
Now, note that the gcd of $ \{r_ms_1,\cdots,r_ms_n\}$ is a factor of the gcd of $ \{r_ms_1,\cdots,r_ms_m\}$ when $ n<m$, so that $ \frac{r_m}{r_n}$ is an integer. Similarly, $ \frac{dnr_n}{d_mr_m}$ is an integer when $ m>n$, and the $ d_n$ are a decreasing sequence. This decreasing sequence of positive integers must have a minimum $ d_N$, at which point $ d_n\equal{}d_N$ and $ r_n\equal{}r_N$ for all $ n\ge N$. Choose $ r\equal{}r_N$ and we are done.

Gauss' Lemma for formal power series.[/hide]

[quote="aznlord1337"]? What JSteinhardt means is that given any pair of infinite sequences which satisfies the property, you can obviously subtract or add the same amt from each term (they would cancel). Similarly you can say divide every element in s by k and divided all elements in t by k and it still has the property. Obviously this is reversible.[/quote]

Oops... I removed the comment from my own post after I realized what he meant.

[quote="Yongyi781"]But in the infinite case, $ r$ might either be (1) infinity, or (2) an irrational number. So proving it for "arbitrarily large $ n$" doesn't work, as I showed someone else who made this assumption.[/quote]

That means I didn't get any of the problems at all. Wahhh... well, there's always next year.

Wasn.t able to solve anything of this question.

Actually, to prove that $ r$ was indeed rational and not infinity or irrational, all you needed to show was that there are a finite number of prime factors between the denominators of $ s_i\minus{}s_j$ and $ t_i\minus{}t_j$ for all $ i$ and $ j$, i.e. there are not an infinite number of prime factors in the denominators of these two matrices.

I'm not sure if this is correct, but here's my approach:

[hide]
For integers $ i$, $ j$, and $ k$, we have that $ (s_i\minus{}s_j)(t_i\minus{}t_j)$, $ (s_j\minus{}s_k)(t_j\minus{}t_k)$, and $ (s_i\minus{}s_k)(t_i\minus{}t_k)$ are integers. Therefore, $ (s_i\minus{}s_j)(t_i\minus{}t_j)\plus{}(s_j\minus{}s_k)(t_j\minus{}t_k)\minus{}(s_i\minus{}s_k)(t_i\minus{}t_k)\equal{}(s_j\minus{}s_k)(t_j\minus{}t_i)\plus{}(s_j\minus{}s_i)(t_j\minus{}t_k)$ is also an integer.

Subtract $ (s_j\minus{}s_i)(t_j\minus{}t_i)$ from this (needless to say, another integer), and depending on the way the terms are grouped, we have both $ (s_i\minus{}s_k)(t_j\minus{}t_i)\plus{}(s_j\minus{}s_i)(t_j\minus{}t_k)$ and $ (s_j\minus{}s_k)(t_j\minus{}t_i)\plus{}(s_j\minus{}s_i)(t_i\minus{}t_k)$ as an integer.

Comparing $ (s_i\minus{}s_k)(t_j\minus{}t_i)\plus{}(s_j\minus{}s_i)(t_j\minus{}t_k)$ with $ (s_j\minus{}s_k)(t_j\minus{}t_i)\plus{}(s_j\minus{}s_i)(t_j\minus{}t_k)$, we can see that $ (s_i\minus{}s_k)(t_j\minus{}t_i)$ and $ (s_j\minus{}s_k)(t_j\minus{}t_i)$ must have the same fractional part in order for both expressions to be integers. Now let $ t_j\minus{}t_i\equal{}\frac{x}{y}$, where $ x$ an $ y$ are relatively prime integers (we can do this because $ t_i$ and $ t_j$ are rational). So now we have that $ \frac{s_ix}{y}\minus{}\frac{s_kx}{y}$ and $ \frac{s_jx}{y}\minus{}\frac{s_kx}{y}$ have the same fractional part. We can get rid of the common term $ \frac{s_kx}{y}$, leaving us with $ \frac{s_ix}{y}$ and $ \frac{s_jx}{y}$ having the same fractional part. Therefore, $ s_ix\cong s_jx \mod{y}$. Since $ x$ is relatively prime to $ y$, we can divide it out, leaving $ s_i\cong s_j \mod{y}$. This means that $ s_i\minus{}s_j\equal{}ny$, for some integer $ n$. Since both $ n$ and $ y$ are integers, $ s_i\minus{}s_j$ must be an integer. Similar reasoning using $ (s_j\minus{}s_k)(t_j\minus{}t_i)\plus{}(s_j\minus{}s_i)(t_i\minus{}t_k)$ shows that $ t_i\minus{}t_j$ must also be an integer.

So if we chose $ r\equal{}1$, $ (s_i\minus{}s_j)r$ and $ (t_i\minus{}t_j)/r$ will be integers.
[/hide]

I feel like I'm not getting something right.

By the way, the solution I posted was by Paul Christiano.

Indeed, $ r\equal{}1$ is not supposed to be the number you're getting...

To maru:
Yes, it's wrong.
[quote="maru"]Since $ x$ is relatively prime to $ y$, we can divide it out, leaving $ s_i\cong s_j \mod{y}$.[/quote]No, you can't. This only works if you have beforehand that $ s_i$ and $ s_j$ are integers.
Furthermore, that $ \frac{s_ix}{y}$ and $ \frac{s_jx}{y}$ have the same fractional part is simply that $ (s_i\minus{}s_j)\frac{x}{y}\equal{}(s_i\minus{}s_j)(t_i\minus{}t_j)$ is an integer, something you knew from the start. Your post represents no progress on the problem.

^ Yeah that looks fine.  It was pretty simple for a #6 I think.  Here's my phrasing of more or less the same thing.

Like everyone else, shift $s_1 = t_1 = 0$, and WLOG assume $s_2 \neq 0$. Call a [b]scaling[/b] an operation where we multiply $\{s_n\}$ by some rational amount and divide $\{t_n\}$ by the same amount. Then perform a scaling so that $s_2 = 1$.  Now observe that for any $i$, $j$ we have $s_it_i \in \mathbb Z$ and also $s_it_j + s_jt_i \in \mathbb Z$.  In particular, $t_2$ is an integer, say $t$.

Note that if $s_3 = \frac ab \neq 0$ for $(a,b) = 1$ and $t_3 = \frac ba \cdot r$ for some $r \in \mathbb Z$ then $t \cdot \frac ab  + r \cdot \frac ba$, implying $b \mid t$. That therefore means all denominators in $\{s_n\}$ divide $t$. Thus if we scale $\{s_n\}$ by the LCM of all their denominators, we arrive at a situation where $s_n \in \mathbb Z$ for all $n$, and $\{s_n\}$ has no common divisor.

Now we claim that all the $\{t_n\}$ must be integers, which then completes the problem.  Assume not, and that for some $k$, $t_k = \frac ab \neq 0$ with $(a,b) = 1$ and $p \mid b$ for a prime $p$. Then $p \mid s_k$, but we can also select an $s_\ell$ not divisible by $p$, meaning $t_\ell$ does not have $p$ in its denominator. Now $s_\ell t_k + s_k t_\ell$ is an integer, but looking at the $p$-adic evaluation this is clearly impossible.

The condition only depends on the difference of the elements of the sequence, so we can assume $s_1 = t_1 = 0$. Also, there is some $x$ so that $s_x$ is non-zero. Then multiplying all of the $s_i$ by $\frac{1}{s_x}$ and multiplying all of the $t_i$ by $s_x$, so we assume that $s_x = 1$, for some $x$. Note that because $s_1 = t_1 = 0$, the condition easily tells us that $s_i t_i$ is an integer for all $i$. Note that $-(s_i t_j + s_j t_i) = (s_i - s_j)(t_i - t_j) - s_i t_i - s_j t_j$, which tells us that $s_i t_j + s_j t_i$ is an integer for all $i$ and $j$. But note that $s_i t_j s_j t_i$ is also an integer. Thus, $s_i t_j$ and $s_j t_i$ are both rational roots of some monic quadratic polynomial with integer coefficients, which tells us that they are algebraic integers which means they must be integers themselves (this also follows from the rational root theorem). Now, taking $i = u$, we get that $t_j$ is an integer for all $j$. Let $d$ be the greatest common denominator of all of the $t_i$. Then divide all of the $t_i$ by $d$ and multiply all of the $s_j$ by $d$. Clearly all of the $t_i$ are still integers, but actually we have that all of the $s_j$ are all integers now as well. To see this, note that we can express $d$ as an integer linear combination of all of the $t_i$ by bezout's lemma, but then $ds_i$ is clearly an integer because $s_i t_j$ is an integer for all $i$ and $j$, so we are done.

A purly p-adic solution also exist.I am on the phone now I will edit the post to show my solution.

If rationals $a$ and $b$ satisfy $a+b\in \mathbb{Z}$ and $ab \in \mathbb{Z}$ then clearly both $a$ and $b$ are integers, this is a quick application of Rational Root Theorem. $(\bigstar)$

Let $r$ be an integer such that $M=r(t_2-t_1)$, $N=r(s_2-s_1)$ and $C=r(s_1t_1-s_2t_2)$ are integers.
$$(s_i-s_1)(t_i-t_1) \in \mathbb{Z} \ \ \text{and} \ \ (s_i-s_2)(t_i-t_2) \in \mathbb{Z} $$ $$\implies s_i(t_2-t_1)+t_i(s_2-s_1)+s_1t_1-s_2t_2 \in \mathbb{Z} \implies Ms_i+Nt_i+C \in \mathbb{Z}$$ Mind that $MN(s_i-s_j)(t_i-t_j) \in \mathbb{Z}$ for any $i$ and $j$.$$Ms_i+Nt_i+C \in \mathbb{Z} \ \ \text{and} \ \ Ms_j+Nt_j+C \in \mathbb{Z} \implies $$ $$\implies M(s_i-s_j)N(t_i-t_j) \in \mathbb{Z} \ \ \text{and} \ \ M(s_i-s_j)+N(t_i-t_j) \in \mathbb{Z} \stackrel{(\bigstar)}{\Longrightarrow} M(s_i-s_j) \in \mathbb{Z}$$
And the conclusion easily follows.

We begin with an important technical lemma.

[u]Lemma:[/u] Suppose
\begin{align*}
\nu_p(a)+\nu_p(b)&\ge 0 \\
\nu_p(c)+\nu_p(d)&\ge 0 \\
\nu_p(a\pm c)+\nu_p(b\pm d)&\ge 0
\end{align*}
for any of the four possible choices of sign for the last inequality. Then, we have
\[\nu_p(a)+\nu_p(d),\nu_p(b)+\nu_p(c)\ge 0.\]

[u]Proof:[/u] By symmetry, it suffices to show that $\nu_p(a)+\nu_p(d)\ge 0$. Suppose for sake of contradiction that we had $\nu_p(a)+\nu_p(d)<0$. Then, $\nu_p(c)\ge-\nu_p(d)>\nu_p(a)$, so
\[\nu_p(a+c)=\nu_p(a).\]
Also, $\nu_p(b)\ge-\nu_p(a)>\nu_p(d)$, so
\[\nu_p(b+d)=\nu_p(d).\]
Thus,
\[\nu_p(a+c)+\nu_p(b+d)=\nu_p(a)+\nu_p(d)<0,\]
which is the desired contradiction. $\blacksquare$

From this, we derive the following extraordinary claims.

[u]Claim (Weak):[/u]
For any $i,j,k$, we have
\[\nu_p(s_i-s_j)+\nu_p(t_j-t_k)\ge 0.\]

[u]Proof:[/u] Apply the lemma with $a=s_i-s_j$, $c=s_j-s_k$, $b=t_i-t_j$, and $d=t_j-t_k$, with the signs for the third inequality chosen to be $+,+$. $\blacksquare$

[u]Claim:[/u] For any $i,j,k,\ell$, we have
\[\nu_p(s_i-s_j)+\nu_p(t_k-t_\ell)\ge 0.\]

[u]Proof:[/u] Apply the previous claim and the lemma with $a=s_i-s_j$, $c=s_i-s_\ell$, $b=t_i-t_k$, and $d=t_k-t_\ell$, with the signs for the third inequality chosen to be $-,+$. $\blacksquare$

The problem is all but destroyed at this point. The final claim tells us that $(s_i-s_j)(t_k-t_\ell)$ is an integer for all $i,j,k,\ell$. For convenience, we'll rephrase the problem as having two sequences $(S_i)$ and $(T_j)$ such that $S_iT_j\in\mathbb{Z}$ for all $i$ and $j$, with our goal being to show that there exists a rational number $r$ such that $S_ir$ and $T_j/r$ are always integers.

Note that for any $i$ and fixed $j$, there exists an integer $k$ such that $S_i=k/T_j$. Thus, the number of prime that can be in the denominator of any $S_i$ is finite, and same with $T_j$. Let $P$ be the (finite) set of all primes in the denominators of all $T_i$ and $S_j$. Furthermore, for any $p\in P$, the equation $S_i=k/T_j$ guarantees that $\nu_p(S_i)$ is bounded below. Let $r_p$ be this minimum value. Then, $\nu_p(T_j)\ge -r_p$, as $S_iT_j$ is an integer where $i$ is such that $\nu_p(S_i)=r_p$.

Now, let $r=\prod_{p\in P}p^{-r_p}$. We see that for all $p\in P$, $\nu_p(S_ir)\ge 0$ and $\nu_p(T_j/r)\ge 0$. Again, since only primes in $P$ appear in the denominators of $S_i$ and $T_j$, this forces $S_ir$ and $T_j/r$ to always be integers, as desired.

First we note a easy fact. If rationals $a,b$, satisfy $a+b , ab \in \mathbb Z$, then $a,b \in \mathbb Z$.

Proof : Define $f(x)= x^2-(a+b)x+ab)$. On one hand, $f(x)\in \mathbb Z[X]$, so $a,b \in \overline {\mathbb Z}$. Also $a,b \in \mathbb Q$. This gives $a,b \in \mathbb Q \cap \overline {\mathbb Z}\equiv \mathbb Z$, so the lemma holds.

We shift such that $s_1=t_1=0$. Also scaling the $s_i,t_i$'s doesnt change the truth of the statement, so WLOG $s_2=1$.

Next note that $s_it_i\in \mathbb Z$ and $s_it_j+s_jt_i \in \mathbb Z$ for all $i,j \in \mathbb N$. Using the fact on $s_it_j$ and $s_jt_i$, we get $s_it_j\in \mathbb Z$, for all $i,j$. This directly gives that $t_j$ are integers for all $j$. Note that $t_i\cdot s_j$ is a integer. This gives that the denominator of the $s_i$'s divides $t_j$ for all $i,j$. So divide the $t_i$ by their gcd and multiply the $s_i$ by the same and we are happy !

Solution with lots of help from @above.

First we can subtract some rational from all the $s_i$ such that $s_1=0$, and similarly $t_1 = 0$. Moreover by multiplying each $s_i$ appropriately we can assume $s_n=1$ for some $n$. It follows that $s_it_i$ is an integer for all $i$, and as a result $s_it_j + s_jt_i$ is also an integer. But also $s_is_jt_it_j$ is an integer, so we can conclude that $s_it_j$ is an integer for all pairs $(i,j)$. (Here's a $\nu_p$ argument to see this: let $a = \nu_p(s_it_j)$ and $b = \nu_p(s_jt_i)$. Note that $\min(a,b) \geq 0$ is true unless $a=b$; but then $a+b \geq 0$ implies $a,b \geq 0$ anyways.)

Let $e$ be the minimal $\nu_p$ of all the numbers in $\{s_i\}$: we will show that $\nu_p(t_i) \geq e$ for all $i$. But note that $\nu_p(s_it_j) \geq 0$ for all pairs $(i,j)$, done.

For rational numbers $a,b$, we write $a\equiv b \pmod{k}$ if $k \mid a-b$, where $k$ is an integer. This replaces the conventional modulo meaning with all those inverses.
We begin with the following claim:

[b]Claim 1:[/b] For any prime $p$, the quantity $\nu_p(t_i-t_j)$ across all $i,j$ is bounded below.
[b]Proof:[/b] I will prove that $\nu_p(t_i)$ across all $i$ is bounded below. Suppose otherwise. Then, there exist infinitely many positive integers $k>1$ such that $\nu_p(t_k)<\nu_p(t_i)$ for all $i<k$. Call such indices "mountainous". For every mountainous index $k$, we have $\nu_p(t_k-t_i)=\nu_p(t_k)$ for all $i<k$. It follows that for such $k$, we have $\nu_p(s_k-s_i)\geq -\nu_p(s_k)$ for all $i<k$. Then, for all $i,j<k$,
$$\nu_p(s_i-s_j)=\nu_p((s_k-s_j)-(s_k-s_i))\geq -\nu_p(s_k).$$
But since $-\nu_p(s_k)$ is unbounded above as $k$ grows large, it follows that $\nu_p(s_i-s_j)$ isn't bounded, so $s_i-s_j=0$. This implies $(s_n)$ is a constant sequence, which violates the condition. Thus $\nu_p(t_i)$ is bounded below, so $\nu_p(t_i-t_j)\geq \min\{\nu_p(t_i),\nu_p(t_j)\}$ is bounded below. $\blacksquare$

Of course, this claim also applies to $\nu_p(s_i-s_j)$.

We now prove the following claim, the purpose of which is to avoid issues with the fact that multiplying infinitely many rationals doesn't necessarily give a rational number. We will use the fact that if $q$ is rational and nonzero, then there are finitely many primes $p$ with $\nu_p(q) \neq 0$ [hide=(proof)]
We can write $q=\tfrac{a}{b}$, where $a,b$ are integers and $\gcd(a,b)=1$. Then if a prime $p$ satisfies $\nu_p(q) \neq 0$, we must have $p \mid a$ or $p \mid b$.
[/hide].

[b]Claim 2:[/b] There are finitely many primes $p$ such that the minimum of $\nu_p(t_i-t_j)$ over all $i,j$ is nonzero.
[b]Proof:[/b] Suppose otherwise. Then there are either infinitely many primes $p$ such that $\min\{\nu_p(t_i-t_j)\}$ is positive or infinitely many $p$ such that the quantity is negative. However, the former is impossible, since it would then follow that there are infinitely many $p$ such that $\nu_p(t_i-t_j)$ is positive for all $i,j$, so $t_i-t_j=0$ for all $i,j$ and $(t_n)$ is constant: contradiction. Thus we would need infinitely many $p$ with $\min\{\nu_p(t_i-t_j)\}<0$. As $\nu_p(t_i-t_j) \geq \min\{\nu_p(t_i),\nu_p(t_j)\}$, we would then have infinitely many $p$ with $\min\{\nu_p(t_i)\}<0$. By the first claim, for such $p$ $\min\{\nu_p(t_i)\}$ is bounded below, so let $\min\{\nu_p(t_i)\}$ be $f(p)<0$. Also let $S(p)$ denote the set of indices $i$ such that $\nu_p(t_i)=f(p)$. Pick some $k \in S(p)$. Then for all $i \neq j \not \in S(p)$, we have
$$\nu_p(t_k-t_i)=\nu_p(t_k-t_j)=f(p),$$
so $\nu_p(s_k-s_i)=\nu_p(s_k-s_j)\geq -f(p)$. It follows that $\nu_p(s_i-s_j)\geq -f(p)$ for all $i,j \not \in S(p)$, so $\nu_p(s_i-s_j) \geq -f(p)>0$ where $i$ and $j$ are not both in $S(p)$. If there exist $i \neq j$ such that $i,j$ are not both in $S(p)$ for infinitely many primes $p$, then $\nu_p(s_i-s_j)$ is positive for infinitely many primes $p$, which is impossible. Hence for every pair $(i,j)$ with $i \neq j$, there are finitely many primes $p$ such that $i,j$ are not both in $S(p)$, and thus infinitely many primes $p$ such that $i,j$ are both in $S(p)$. But this implies that there are infinitely many primes $p$ such that $i \in S(p)$ for all $i$, which means $\nu_p(t_i)=f(p)<0$ for infinitely many $p$. This is, however, impossible, as $t_i$ is rational. $\blacksquare$

Observe that multiplying every element of $(s_n)$ by some rational $q$ and dividing every element of $(t_n)$ by $q$ changes nothing. Now, let $S$ be the set of all primes $p$ such that $\min\{\nu_p(t_i-t_j)\}$ over all $i,j$ is nonzero, which is finite by claim 2. Let
$$q=\prod_{p \in S} \min\{\nu_p(t_i-t_j)\},$$
which is rational as $S$ is finite. Then, multiply all elements of $(s_n)$ by $q$ and divide all elements of $(t_n)$ by $q$. We can also shift $(s_n)$ and $(t_n)$ such that $s_1=t_1=0$. After this transformation, I claim $r=1$ works. Since $\min\{\nu_p(t_i-t_j)\}=0$ for all $p$, it follows that $(t_i-t_j)/r=t_i-t_j$ has nonnegative $\nu_p$ and is therefore an integer. Now let $p$ be an arbitrary prime and let $\min\{\nu_p(s_i-s_j)\}=k$. If $k \geq 0$, then $r=1$ obviously works. Suppose $k<0$. Pick $i \neq j$ with $\nu_p(s_i-s_j)=k$, so $\nu_p(t_i-t_j)\geq -k$. Now for every $n \neq i,j$, we have
$$\nu_p(s_i-s_j)=\nu_p((s_n-s_j)-(s_n-s_i))\geq \min\{\nu_p(s_n-s_i),\nu_p(s_n-s_j)\}.$$
Since $\nu_p(s_n-s_i)\geq k$ and likewise for $s_j$, this means that either $\nu_p(s_n-s_i)=k$ or $\nu_p(s_n-s_j)=k$. If $\nu_p(s_n-s_i)=k$, then $\nu_p(t_n-t_i)\geq -k$. This also means $\nu_p(t_n-t_j)\geq -k$, otherwise $\nu_p((t_n-t_j)-(t_n-t_i))<-k$, contradicting the fact that it's also at least $-k$. Hence we have
$$t_n \equiv t_i \equiv t_j \pmod{p^{-k}}$$
for all $n \neq i,j$. But this means
$$t_a \equiv t_b \pmod{p^{-k}} \implies p^{-k} \mid t_a-t_b,$$
so $\nu_p(t_i-t_j)\geq -k>0$ for all $a,b$, which contradicts $\min\{\nu_p(t_i-t_j)\}=0$. Thus $\min\{\nu_p(s_i-s_j)\}\geq 0$, so $(s_i-s_j)r=s_i-s_j$ is an integer for all $i,j$, as desired. $\blacksquare$

Scale both sequences so that $s_1=t_1=0$. This implies that for every $i$ $s_it_i$ is an integer.

Take any prime $p$. If for every index $v_p(s_i)\geq 0$ and $v_p(t_i)\geq 0$ we can put $v_p(r)=0$ and we're good $mod p$. Otherwise assume that WLOG there exists an index $a$ such that $v_p(s_q)<0$ and is minimal. Denote by $A$ the set of indices for which we have $v_p(s_{a\in A})=v_p(s_q)$; and by $B$ the set of indices for which $v_p(s_{b\in B})>v_p(s_q)$.Clearly both of these sets are non-empty. Now take any two indices $a\in A$ and $b\in B$. We know that $v_p(s_a-s_b)=v_p(s_q)$, so we need to have $v_p(t_a-t_b)\geq -v_p(s_q)$.

[b]Lemma:[/b]$v_p(t_x-t_y)=v_p(t_x-t_z-(t_y-t_z))\geq min(v_p(t_x-t_z);v_p(t_y-t_z))$.

If $a_1;a_2\in A$ and $b\in B$, then we get that $$v_p(t_{a_1}-t_{a_2})\geq min(v_p(t_{a_1}-t_b);v_p(t_{a_2}-t_b))=-v_p(s_q)$$ Analogically, if $b_1;b_2\in B$ and $a\in A$, then we get that $$v_p(t_{b_1}-t_{b_2})\geq min(v_p(t_{b_1}-t_a);v_p(t_{b_2}-t_a))=-v_p(s_q)$$ This tells us that $v_p(t_i-t_j)\geq -v_p(s_q)$ and $v_p(s_i-s_j)\geq v_p(s_q)$, thus putting $v_p(r)=-v_p(s_q)$ fixes everything $mod $ $p$.

If there is an index $i$ with $v_p(t_i)<0$, then we proceed analogically to the above.

[b]Claim 1:[/b] $v_p(s_i-s_j)$ is bounded below for all $p$, across all $i, j$; the same is true for $\{t_i\}$.
[i]Proof:[/i] If its unbounded below, then take terms $(s_i, s_j, s_k)$ such that $v_p(s_i-s_j) = -a$ and $v_p(t_i-t_j) = b$ where $b \ge a > 0$, and $v_p(s_j-s_k) \le -69420 b$. Clearly, $v_p(t_j-t_k) \ge 69420b$.

Yet, $$v_p(s_i-s_k) = v_p((s_i-s_j) + (s_j - s_k)) = \min(v_p(s_i-s_j), v_p(s_j - s_k)) \le -69420b$$ $$v_p(t_i-t_k) = v_p((t_i-t_j) + (t_j - t_k)) = \min(v_p(t_i-t_j), v_p(t_j - t_k)) = b,$$ yielding $v_p((s_i-s_k)(t_i-t_k)) < 0$, contradiction. Same result for $\{t_i\}$. $\blacksquare$

[b]Claim 2:[/b] Finitely many primes appear in the factorization of denominators of $s_i-s_j$ through all $i, j$; same for $\{t_i\}$.
[i]Proof:[/i] Assume not. WLOG $t_1 \neq t_2$ (otherwise we can permute the two sequences correspondingly). Take any (of infinitely many) primes $p$ not in the factorization of the denominator of $s_1-s_2$ (note that the following argument works even if this is $0$). Let $s_k$ be the first term such that the factorization of the denominator of $s_k-s_1$ contains $p$. It follows that $s_k-s_2$'s denominator factorization also contains $p$.  Thus, $p \mid t_k - t_1$ and $p \mid t_k - t_2$, meaning $p \mid t_1 - t_2$. Since there are infinitely such primes $p$, and $t_1-t_2$ is nonzero, contradiction. $\blacksquare$

[b]Claim 3:[/b] Let $S$ be the minimum positive integer such that $S(s_i-s_j)$ is an integer for all $i, j$. This is finite due to the previous claims. Then, all the numerators of $t_i-t_j$ for all $i, j$ are divisible by $S$. Define $T$ similarly and a similar result holds.
[i]Proof: [/i] Assume FTSOC that there exists $s_i, s_j, t_a, t_b$ and prime $p$ such that $v_p(s_i-s_j) = -\alpha$ and $v_p(t_a-t_b) = \beta$ with $\alpha > 0$ and $\beta < \alpha$. Then, at least one of $v_p(s_a - s_i)$ and $v_p(s_a - s_j)$ is not greater than $-\alpha$. WLOG $v_p(s_a - s_i) \le -\alpha$. Then $v_p(t_a-t_i) \ge \alpha$. Since $v_p(t_a-t_b) = \beta < \alpha$, we have $v_p(t_i-t_b) = \beta$.

Meanwhile, $v_p(s_a-s_b) \ge -\beta$. As $v_p(s_a - s_i) \le -\alpha$, we have $v_p(s_i-s_b) \le -\alpha$. Thus, $v_p((t_i-t_b)(s_i-s_b)) \le \beta - \alpha < 0$, contradiction. $\blacksquare$.

It follows that setting $r= \frac{S}{T}$ works. We are done.

[hide=wrong soln]Lol, this was locked from the past 2 days so couldn't post back then :P

[hide=Page 1][img]https://i.imgur.com/oCHv4wr.jpg[/img][/hide]

handwritten solution images attached[/hide] :noo:

@below, you are totally correct, sorry for the flawed solution :(

[quote=kamatadu]Lol, this was locked from the past 2 days so couldn't post back then :P

[hide=Page 1][img]https://i.imgur.com/oCHv4wr.jpg[/img][/hide]

handwritten solution images attached[/quote]

There's a problem in this solution,
If $S_{\text{min}}=v_p(S_i-S_j)$ then it need not be true that,
$t_{\text{min}}=v_p(t_i-t_j)$.
You cannot write that,
$v_p((S_i-S_j)(t_i-t_j))=S_{\text{min}}+t_{\text{min}}$

My point is if power of $p$ in the sequence $S_i$ be minimum for $S_i-S_j$,then $t_{\text{min}}$ might be equal to $v_p(t_a-t_b)$ where $a,b \not = i,j$.

You need to fix this , nice idea by the way

Replacing $s_{i}$ to $s_{i} - a$ satisfies problem condition, so WLOG we can assume $s_{0} = t_{0} = 0$. Let $p$ be a prime such that there exists $n$ such that $\nu_{p}(s_n) < 0$. Then since $s_{i}t_{i}$ is integer, so $\nu_p(t_{i}) > -\nu_{p}(s_{i})$. If $j$ is positive integer such that $\nu_{p}(t_{j}) < \nu_{p}(t_{i})$, then $\nu_{p}(t_{i} - t_{j}) = \nu_{p}(t_{j})$. Thus it forces $\nu_{p}(s_{j}) = \nu_{p}(s_{i})$. Then for all $n$ either $\nu_{p}(s_{n}) = \nu_{p}(s_{i})$ or $\nu_{p}(t_{n}) = \nu_{p}(t_{i})$. Thus $p$ cannot be large. Similarly, if $\nu_{p}(t_{n}) < 0$ for some $n$, then $p$ cannot be large. Let $p$ be a prime such that $\nu_{p}(s_{n}) < 0$ for some $n$.

Claim:

The sequence $(\nu_{p}(s_{n}))$ is bounded below.
Proof:

Assume the contrary, for large $N$, $\nu_{p}(s_n) \le -N$ for some $n$. Then $\nu_{p}(s_{n} - s_{1}) = -N$ and $\nu_{p}(t_{n} - t_{1}) = \nu_{p}(t_{1})$. So $-N + \nu_{p}(t_{1}) \ge 0$, a contradiction. $\blacksquare$

Let $-M = \inf \nu_{p}(s_n)$. Then we'll show that for all $i, j$, $\nu_{p}(t_i - t_j) \ge M$. Assume the contrary, there is some $i, j$ such that $\nu_{p}(t_i - t_j) < M$. Thus $\min(\nu_{p}(t_i), \nu_{p}(t_j)) \le \nu_p(t_i - t_j) < M$. WLOG, assume $\nu_p(t_i) = \min(\nu_{p}(t_i), \nu_{p}(t_j))$. Then $\nu_p(t_i) < M$. Take $k$ such that $\nu_p(s_k) = -M$. Then $\nu_p(t_k) \le M$. Thus $\nu_p(t_i - t_k) = \nu_p(t_i) < M$. Thus $\nu_p(t_i - t_k) = \nu_p(t_i) < M$. Thus $\nu_p(s_i - s_k) > -M$, so $\nu_p(s_i) = \nu_p(s_k) = -M$. Therefore $\nu_p(t_i) + \nu_p(s_i) < 0$, a contradiction.

Therefore taking $r$ such that $\nu_p(r) = -M$ would works.(Since there are finitely many $p$). This completes proof. $\blacksquare$

Consider each prime $p$ seperately.\\

Claim 1: For each prime $p$, $v_p(s_i-s_j)$ is bounded below. This is clearly true as otherwise we can pick $(t_1-t_2)(s_i-s_j)$ with $v_p(s_i-s_j)$ arbitrarily low to make it not an integer.\\

Claim 2: There can only be finitely many primes that appear in the denominator of any difference $s_i-s_j$ or $t_i-t_j$ at all. Suppose not. Then, at least one of the difference sets will contain infinitely many primes. Due to this, the numerator of any difference in the other sequence must be divisible by all of these infinitely many primes, contradiction.\\

Thus, consider any prime $p$. Suppose that the minimal $v_p$ of the difference of two terms in $s$ is $k$. By the above claim, $k$ exists, and since the sequence is nonconstant, $k$ is finite. This means that the difference between any two terms of $t$ has to have a $v_p$ of at least $-k$, otherwise the product of that and the one with $v_p$ of $k$ would not be an integer. Thus, if we pick a factor of $p^{-k}$ for $r$, the product of any difference of two terms of $s$ with $r$ would have a nonnegative $v_p$, since we have at least $k$ minus $k$, and for $t$, we would have at least $-k$ plus $k$. Repeating this for all primes $p$ that appear in some denominator gives us a valid value of $r$, since there are only finitely many "concerning" primes.

The “finite difference” or “FD” operation on a sequence $a_1$, $a_2$, $\dots$, is defined to be the sequence $a_2-a_1$, $a_3-a_2$, $\dots$, $a_{i+1}-a_i$, $\dots$

Let $a_1$, $a_2$, $\dots$, $a_n$ and $b_1$, $b_2$, $\dots$, $b_n$ be any sequence such that $(a_i-a_j)(b_i-b_j)$ is always an integer for any pair of natural numbers $(i,j)$. If two sequences satisfy this, we will call them “appley”.

[color=#9a00ff][b]Apple Claim -  [/b][/color]
I claim that for the sequences $\{c_i\}$ and $\{d_i\}$, where $c_i=a_{i+1}-a_i$ and $d_i=b_{i+1}-b_i$, we must have that
\[(c_p-c_q)(d_p-d_q),\]
is an integer for any pair of natural numbers $(p,q)$. In other words, $c_i$ and $d_i$ must be appley.

First, note that $c_kd_k$ for any natural number $k$ must always be an integer, because we have that
\[c_kd_k=(a_{k+1}-a_k)(b_{k+1}-b_k),\]
which must be an integer by our assumptions about the two sequences $a$ and $b$. Now, subtraction $c_qd_q$ and $c_pd_p$ from $(c_p-c_q)(d_p-d_q)$, all that is left to prove is proving that $c_pd_q+c_qd_p$ must also be an integer.

We will prove this by strong induction. First, we assume that if $|p-q|\leq k$, then $c_pd_q+c_qd_p$ is an integer. Now, we will prove that if $|p-q|=k$, then $c_pd_q+c_qd_p$ must still be an integer.

First, note that $c_i+c_{i+1}+\dots+c_{j-1}=a_j-a_i$. Similarly, $d_i+d_{i+1}+\dots+d_{j-1}=b_j-b_i$. Therefore, note that if we have $r$ for some integer $r$; we have that
\[(c_r+c_{r+1}+\dots+c_{r+k})(d_r+d_{r+1}+\dots+d_{r+k})=(a_{r+k+1}-a_r)(b_{r+k+1}-b_r),\]
must be an integer, since $(a_i-a_j)(b_i-b_j)$ is always an integer for any pair of natural numbers $(i,j)$. However, we can also decompose the sum
\[(c_r+c_{r+1}+\dots+c_{r+k})(d_r+d_{r+1}+\dots+d_{r+k}),\]
into a sum of many $c_pd_q+c_qd_p$ for $|p-q|\leq k$ and $c_rd_{r+k}+c_{r+k}d_r$. Since the large sum must be an integer and all of the other $c_pd_q+c_qd_p$’s are integers, we conclude that $c_rd_{r+k}+c_{r+k}d_r$ for any natural number $r$, proving our inductive claim.

As for our base case, notice that as proved above, we have that $c_pd_p+c_pd_p$ (for $p=q$) must always be an integer since $c_pd_p$ must be an integer.

Therefore, if two sequences $a$ and $b$ satisfy that $(a_i-a_j)(b_i-b_j)$, their sequences of their differences $a_2-a_1$, $a_3-a_2$, etc. and similarly for the other one; also satisfy the same condition.

Now let us consider the problem conditions $s_i$ and $t_i$. Let us take the first $n$ numbers of the sequences as smaller sequences, namely
\[s_1, s_2, \dots, s_n,\]
and
\[t_1, t_2, \dots, t_n.\]
Now allow us to take finite differences of these until one of them converges to a sequence of constant numbers (it will not be a constant sequence of $0$). It is well known that for all finite sequences, they will eventually converge to a constant sequence of length at least $2$ with the constant not being equal to $0$. WLOG, let’s assume that $t_i$ converges first. Let the constant it converges to be called $a$, and assume that this sequence is $k$ numbers long. Note that $s_i$ can converge similarly to a constant, but it does not necessarily have to; so let us consider the finite difference sequence of $s_i$ after applying the finite difference operation the same number of times to be $x_1, x_2, \dots, x_k$. I will now prove that $a(s_i-s_j)$ and $(t_i-t_j)/a$ are always integers using the following two claims.

[color=#9a00ff][b]Bapple Claim - [/b][/color]
Let $m_i$ be the sequence “before” $x_i$ (meaning that in the chain of finite difference operations on $s_i$, applying finite differences on $m_i$ resulted directly in $x_i$). Then I claim that $a(m_i-m_j)$ is always an integer.

Note that since $x_i$ and the constant sequence $a$ are appley together by the [color=#9a00ff][b]Apple Claim[/b][/color], we have that $ax_i$ must always be an integer for any $1\leq i\leq k$. Additionally, note that
\[m_i=m_1+(x_1+x_2+\dots+x_{i-1}),\]
which gives that (WLOG let $j\leq i$)
\[m_j-m_i=x_{j-1}+x_{j-2}+\dots+x_{i},\]
and multiplying this by $a$ gives a sum of $ax_c$’s for integer $c$’s, meaning that $a(m_i-m_j)$ is always an integer.

[color=#9a00ff][b]Catpple Claim - [/b][/color]
Let $n_i$ be the sequence “before” the constant sequence consisting of $a$’s. Then I claim that $(n_i-n_j)/a$ is always an integer.

Note that since the finite differences of $n_i$ are a constant sequence, $n_i$ is an arithmetic sequence with common difference $a$. Therefore, any two numbers in $n_i$ differ by an integer multiple of $a$, meaning that $(n_i-n_j)/a$ is always an integer.

[color=#9a00ff][b]Doneapple Claim - [/b][/color]
For any sequence $e_1$, $e_2$, $\dots$ and any constant $c$, if $c(e_i-e_j)$ is always an integer, then for any sequence $b_1$, $b_2$, $\dots$ that is “before” the $e_i$ sequence, we have that $c(b_i-b_j)$ is always an integer.

This can be proved siimlarly to the [color=#9a00ff][b]Bapple Claim[/b][/color] by taking sums and then differences.

Finally, combining the [color=#9a00ff][b]B[/b][/color], [color=#9a00ff][b]C[/b][/color], and [color=#9a00ff][b]D[/b][/color] claims and inducting backwards from the $x_i$ and $a$ constant sequence, we get that $a(s_i-s_j)$ and $(t_i-t_j)/a$ must be integers. We continue taking larger and larger $n$, which in the end yields that the infinite sequence satisfies the same conditions, finishing the problem.


A box of $ 100$ gumdrops is guaranteed to have at least $ 8$ of each color and no more than $ 12$ of any one color. What is the least number of colors that can be used to meet these requirements?

Dividing $ 100$ by $ 12$, we get $ 8\frac{1}{3}$. Thus the least number of colors that fulfill the requirements is $ \boxed{9}$.

We look for maximum values close to $ 100$. $ 12\times{8}\equal{}96$ leaves us with $ 4$ left over, and that won't work. $ 12\times{7}\equal{}84$ leaves $ 16$, and that works if I split it into $ 2$ groups of $ 8$. So the answer is $ \boxed{9}$.


calculus
integration
geometry
logarithms
vector
calculus computations

Let $ S(t)$ be the area of the traingle $ OAB$ with $ O(0,\ 0,\ 0),\ A(2,\ 2,\ 1),\ B(t,\ 1,\ 1 \plus{} t)$.
Evaluate $ \int_1^ e S(t)^2\ln t\ dt$.

$ S(t)\equal{}\frac{3}{2}\sqrt{1\plus{}t^{2}}$

Hence,

$ \int_{1}^{e}S^{2}(t)\ln tdt\equal{}\frac{9}{4}\int_{1}^{e}\left( 1\plus{}t^{2}\right) \ln tdt\equal{}\allowbreak \frac{1}{2}\left( e^{3}\plus{}5\right)$

I got $ \frac {e^3 \plus{} 5}{2}$ for the answer.

Indochina, you out-typed me;- My solution used the general formula of the area of triangle in three dimensions.

no , indochina  is  right ,  even i am getting the same answer as him , akech u kindly recheck

hehe ,  u retyped  ,  anyways  now  u r correct    :D  :lol:

Vector Product?

Japanese high school students use $ \boxed{\triangle{OAB}\equal{}\frac{1}{2}\sqrt{|\overrightarrow{OA}|^2|\overrightarrow{OB}|^2\minus{}(\overrightarrow{OA}\cdot\overrightarrow{OB})^2}}$

I was using the fact that Area of $ \Delta OAB\equal{}\frac{1}{2}\left| \overrightarrow{OA}\times \overrightarrow{OB}\right|$

The formula you have quoted is actually a variant saying the same thing.

we  can  even use determinants  .

or  we can  also find the  lengths of the sides  and use heron`s formula.


limit
real analysis
real analysis unsolved

$ g(x) \equal{} \lim_{t\minus{}>0}{\frac{f(x\plus{}t) \minus{} f(x\minus{}t)}{t}}$
and $ g(x)$ exists for every point $ x \in (a,b)$


Given an equilateral triangle, all points on its sides are colored in one of two given colors. Prove that the is a right-angled triangle such that its three vertices are in the same color and on the sides of the equilateral triangle.
[i]Alhaji Akbar, Jakarta[/i]

assume the contrary -- that there is no monochromatic right triangle. label the equilateral triangle $ ABC$. trisect the sides of the triangle to get 6 points that form a regular hexagon. if any two opposite vertices of this hexagon are colored the same color, then the other four vertices must all be of the opposite color, which give us a right triangle of this color. so opposite vertices must be of opposite colors. it's easy to see that no matter how we color the vertices of the hexagon so that opposite vertices are oppositely colored, two adjacent vertices which lie on a side of the triangle must have opposite colors. call these two points $ P, Q$ and assume they lie on side $ AB$, with $ P$ red and $ Q$ blue. let $ P'\in BC$ be $ P$'s opposite vertex in the hexagon, and $ Q'\in AC$ $ Q$'s opposite. then $ P, Q'$ are both red and form a right triangle with $ A$, while $ Q, P'$ are both blue and form a right triangle with $ A$ as well. thus $ A$ can't be colored red or blue - a contradiction.


algebra
difference of squares
special factorizations

Find all pairs of positive integers $ (x,y)$ such that $ x^{6}=y^{2}+127$

[hide="..."]We just might need to use the difference of squares factorization.

$ (x^{3}-y)(x^{3}+y)=127$ and $ 127$ is prime...[/hide]

Continuing Altheman's hint...
[hide]That would mean $ (x^{3}+y) = 127$ and $ (x^{3}-y) = 1$, yielding $ y = 63$ and $ x = 4$ after you solve the system of equations.

For positive integer solutions, I think thats the only solution; I might be wrong, though.[/hide]

Yes that is correct. Since $ x$ and $ y$ are positive then $ x^{3}+y$ is positive. But since $ 127$ is positive then $ x^{3}-y$ is positive. But  $ x^{3}+y > x^{3}-y$. This forces $ x^{3}+y = 127$ and $ x^{3}-y = 1$.


How many integers from 1 through 9999, inclusive, do not
contain any of the digits 2, 3, 4 or 5?

For integers 1000-9999, there are 6^4 or 1,296 numbers that do not contain those digits.
For integers 100-999, there are 6^3 or 216 numbers that do not contain those digits.
For integers 10-99, there are 6^2 or 36 numbers that do not contain those digits.
For integers 1-9, there are 6^1 or 6 numbers that do not contain those digits.

Add 'em up, and you should get [b]1,554[/b], or at least, I did.

I hope you learned sufficient information from my [color=red]200th post[/color] on how to solve this problem.

wh00ps...

my bad.

thank you for correcting my 200th post.


algebra
polynomial
search
number theory solved
number theory

show that the polynomial $x^p-x+c$
is irreducible when $p$ is a prime and c is relatively prime to $p$

If it were reducible, then it would be reducible in $\mathbb F_p[x]$ as well, where $\mathbb F_p$ is the field with $p$ elements. Let $\alpha$ be a root of our polynomial $f$ in the algebraic closure of $\mathbb F_p$. Then the roots of $f$ are $\alpha,\alpha+1,\ldots,\alpha+p-1$. Notice that any non-empty subset of $\{\alpha+i\ |\ i=\overline{0,p-1}\}$ has a sum not belonging to $\mathbb F_p$, so we have a contradiction.

Posted before: http://www.mathlinks.ro/Forum/viewtopic.php?t=68010 .
Use search button please.


limit
function
calculus
calculus computations

How Can I do a graphic for?
$%Error. "displaysty" is a bad command.
{ \lim_{x\to3}{2x+1}}$

Be the definition of limit

$\forall\epsilon >0 \exists\delta>0/0<|x-a|<\delta \Rightarrow |f(x)-L|<\epsilon$

What happens if i change delta for epsilon?
$\forall \delta>0 \exists\epsilon>0/0<|x-a|<\delta \Rightarrow |f(x)-L|<\epsilon$

In this case
${ \lim_{x\to1}{2x}=2}$
${ \lim_{x\to1}{2x}=3}$
Is this a contradiction?

[quote="Rushil"]what is $n$ ?????[/quote]

sorry is $x$

The $ \epsilon - \delta $ definiton can only allow you to prove and test your limit, not find it (unless you can think of the limit yourself and test it)

By the way, the answer is $7$ . SInce $ 2x+1 $ is continous, the limit at any point equals the value of the function at that point(definiton)...

An interesting thing comes up. If this particular thing follows from the definiton of continuity that the limit equals the value of the fn, how do we prove the fn. is continous non-graphically???
Basically I am asking for the proof of the above fact!!

[quote="felipesa"][quote="Rushil"]The $ \epsilon - \delta $ definiton can only allow you to prove and test your limit, not find it (unless you can think of the limit yourself and test it)

By the way, the answer is 47$ . SInce 4 2x+1 $ is continous, the limit at any point equals the value of the function at that point(definiton)...

An interesting thing comes up. If this particular thing follows from the definiton of continuity that the limit equals the value of the fn, how do we prove the fn. is continous non-graphically???
Basically I am asking for the proof of the above fact!![/quote]

Can you show how did you find 47?[/quote]

It was a typo the answer is 7


quadratics
algebra
algebra proposed
Combinatorial games
game

Two players play a game as follows. The first player chooses two non-zero integers A and B. The second player forms a quadratic with A, B and 1998 as coefficients (in any order). The first player wins iff the equation has two distinct rational roots. Show that the first player can always win.

[quote="Rust"]A=B=-1998/2.[/quote]

The equation $-{1998\over 2}x^2+1998x-{1998\over 2}=0$ doesn't have [u]distinct[/u] roots (check the text of the problem).

Well, when I try to solve this problem, my first idea is to set 1 as a root of the quadratic equation. It makes life easier on this problem :P
And so, the problem is to assure that 1 is not a double root!

a quadratic equation that has rational roots has rational roots: we can notice that the two roots are in the form $frac{m}{n}$ for $(m,n)$ respectively divisor of the last and the first terms' coefficent in $ax^2+bx+c.$

[quote="M4RI0"]We can make six equations, but we'll have only 3 discriminants, and they must be greater than 0, so the strategy is the next:

The first player chooses a positive number A, and then a negative B such that:
$B^2>4(A)(1998)$
$B^2-4(A)(1998)>0$

Then: $A^2-4(b)(1998)>0$ is obvius because B es negative, and the all expression is positive

Also: $1998^2-4(A)(B)>0$ is true

So, the first player always has a winning strategy[/quote]

Don't forget that the discriminants must be full squares, in order for solutions to be rational.

[quote="M4RI0"]You're wrong they should be squares if the solutions were integers.

If the solutions are rationals, it doesn't matter if the discriminants are squares or not.[/quote]

Now you'll tell me that the solutions to e.g. $x^2+x-1=0$ are rationals? Try harder, please :)

[quote=jin]Just let $A+B+1998=0$,$A\neq B,A,B\neq0$ Then 1 must be one of the root and easy to see the other one is not 1.[/quote]

Not necessarily. If the first player give $A=1998$ and $B=-3996$, then the second can create the equation $1998x^2-3996x+1996=0$ which has only solution and win the game.

But, if we have $A+B+1998=0$,$A\neq B,A,B\neq0$, $(A,B) \neq (1998,-3996)$ and $(-3996,1998)$, then the first player can always win the game.


Okay, so I'll need 16 players to start.  Games which we will be playing include Mafia, Russian Roulette, Elect The Executioner.  I WILL BE TAKING SUGGESTIONS ON GAMES TO PLAY!

me will be replacement, or player if signups die off.
the "ernie" challenge? seriously?

EDIT: "beat the puppet" now?? seriously??? DO NOT KILL INNOCENT UNHURTFUL THINGYS!!!!!!

[quote="Chompy"]the "ernie" challenge? seriously?[/quote]

Well I'm sorry, I think I'll just change the game's name to [i]Beat The Mod![/i] if you really want me too. :P

So you are playing, ernie?  Also I say you completely stole the idea for the game name from Teh Meewhee Challenge.   :P

In other news, I join.

Well, I might be playing in [i]some[/i] challenges.  You just don't know which ones. :P

master of puppets i'm pulling your strings.

i join

The message is too small. Please make the message longer before submitting.

[quote="Bugi"]Yeah, like 11.10 xD

I think only one didn't throw egg...[/quote]

Who???

wait, so you are abandoning this? ok, can i be the replacement mod?

I propose that we just let this thread die. GAME OVER!

okay, since i was the first to point out that we need a new mod, i get to be one. so deadline is today.
uhh... can anybody tell me who that one person who did not throw was? cause I will make it as if the egg was thrown at king (since that was my aim)
i guess this can be renamed to "Sour Cream And Onion Revolution"

very well. by quitting, i have stored all the eggs thrown at you. if at some random point any one of them explodes, I die.

I also quit. Not enough activity...
I hopes mathelete dies :D

mathelete, you can be the replacement mod, you just need to change the title.

Hmm. In light of the situation, I [b]quit[/b].

aah! too many eggs! i cant see from under a pile! i cant tell who is still left! And they smell!

Start a new thread, lots of spam here.
And I didn't throw :P :P :P

i just joined a dying game...i [b]quit[/b]

Okay, you guys don't all have to quit.  The mod already quit, so you don't have to quit yourselves.  I quit for you all.


function
integration
limit
calculus
derivative
real analysis
real analysis unsolved

Assume $\phi: \mathbb R\to\mathbb R$ is a differentiable function such that ($L$ and $K$ are two real numbers):

$\int_1^{\infty}\frac{\phi(x)}{x^2}dx=L$   and $\phi'(x)\leq K$, $\forall x\geq 0$

Then  $\lim_{x\to+\infty}\frac{\phi(x)}{x}=0$.

One can take $K=1$:
suppose that $\frac{\phi(x)}{x}$ does not tends towards $0$. WLOG, then there exists a sequence of real numbers $x_n>1$ such that $x_{n}$ tends towards $+\infty$ and $\frac{\phi(x_n)}{x_n} > a$, where $a>0$ is a certain real.

1/first one show that the function stays positive during a long time after $x_n$

Take such a real number $x_n$:
Let $y$ the smallest real number such that $y>x_n$ and $\phi(y)=0$, and $y=+ \infty$ if it does not exists.
$\frac{\phi(x)}{x} > \frac{\phi(x)}{y}$ for all $x \in [x_n; y]$.
But $\frac{\phi(x_n)}{y}>\frac{ax_n}{y}$, and because $| \phi'(x) | < 1$,  $\frac{\phi(x)}{x}>\frac{\phi(x)}{y}>0$ for all $x \in [x_n;(1+a)x_n)[$.
This is why $y \geq (1+a)x_n$

2/from here one can obtain a contradiction:

$\frac{\phi(x)}{x} > \frac{\phi(x)}{(1+a)x_n}$ for all $x \in [x_n; (1+a)x_n]$.
But $\frac{\phi(x_n)}{(1+a)x_n}>\frac{a}{1+a}$, and because $| \phi'(x) | < 1$,  $\frac{\phi(x)}{(1+a)x_n}>\frac{a}{2(1+a)}$ for all $x \in [x_n;(1+C)x_n]$ where $C=\frac{a}{2}$   .

This is why $\int_{x_n}^{(1+C)x_n} \frac{\phi(x)}{x^2}dx>\frac{1}{(1+C)x_n}\int_{x_n}^{(1+C)x_n} \frac{\phi(x)}{x}dx>\frac{1} {(1+C)x_n} C x_n \frac{a} {2(1+a)} = E > 0$, where $E$ is a constant independant of $x_n$.
From this we obtain that $\int_{1}^{+\infty} \frac{\phi(x)}{x^2}dx$ does not exist, contradiction.

it is not entirely correct. you use $|\phi'(x)|\leq 1$ yet only $\phi'(x)\leq 1$ is given, and that actually makes a difference.

also, you say at some point that $\frac{\phi(x)}{(1+a)x_n}>\frac{a}{2(1+a)}$, which is incorrect since $x_n$ itself is large.

actually it is almost correct:

*$\frac{\phi(x_n)}{(1+a)x_n}>\frac{a}{1+a}$. And because the derivative of $x\to \frac{\phi(x)}{(1+a)x_n}$ is superior to $\frac{-1}{(1+a)x_n}$ if we assume $|\phi'(x)|<1$, then $\frac{\phi(x)}{(1+a)x_n}>\frac{a}{2(1+a)}$ for all $x \in [x_n; (1+\frac{a}{2})x_n]$

*but I just used the fact that $\phi'(x)>-1$. The same idea works : it suffices to work on $[(1-\frac{a}{2})x_n;x_n]$ instead of  $[x_n; (1+\frac{a}{2})x_n]$.

this is indeed the sort of analysis that is needed here.
but you need more than that under the assumption that $\phi'$ is bounded from above (below). as it is the argument doesnt stand. you can convince yourself of that.


number theory
prime numbers
number theory unsolved

find all [b]p,q,r[/b] in prime numbers such that  [b]p <= q <= r [/b]and [b]pq+r,pq+r 2, qr+p,[/b][b]qr+p 2,pr+q,pr+q 2[/b] all of them is prime.

This problem comes from LIII Polish Math Olimpiad (2002).

All I can see is that p must be 2 because, pq+r must be prime and since it is prime it must be odd (because even if you put in the smallest possible values for p, q, and r in it is 2*2+2=6).  If you assume that p is not 2 then, p, q, and r must be odd. pq would be odd and pq+r would be even(and greater than 2) and thus not be prime.


I am Nam, from to VietNamese. I am a student class 9.

 Cho a,b>= 0 CMR a^2/b + b^2/a  >=  SQRT[2(a^2 + b^2)]

     Good luck!.

If I can understand any of that non-LaTeX mess, perhaps I could move it to the appropriate forum.  :P

I ran it through a Vietnamese dictionary, and I think the problem is basically
Solve for $ a$ and $ b$ if $ \frac{a^2}{b}\plus{}\frac{b^2}{a}\geq \sqrt{2(a^2\plus{}b^2)}$

[color=darkblue]This is the solution for your problem:[/color]
$ \bullet$ $ \frac {a^2}{b} \plus{} \frac {b^2}{a}\ge \sqrt {2(a^2 \plus{} b^2})$

$ \Longleftrightarrow \frac {a^4}{b^2} \plus{} \frac {b^4}{a^2} \plus{} 2ab\ge2a^2 \plus{} 2b^2$

$ \Longleftrightarrow\frac {a^4}{b^2} \plus{} \frac {b^4}{a^2}\ge2(a^2 \minus{} ab \plus{} b^2)$

$ \Longleftrightarrow(a \plus{} b)(\frac {a^4}{b^2} \plus{} \frac {b^4}{a^2})\ge2(a^3 \plus{} b^3)$

$ \Longleftrightarrow(\frac {a^5}{b^2} \plus{} \frac {b^5}{a^2} \minus{} a^3 \minus{} b^3) \plus{} (\frac {a^4}{b} \plus{} \frac {b^4}{a} \minus{} a^3 \minus{} b^3)\ge0$

$ \bullet$ [color=darkblue]But [/color]$ \forall a, b \in R$ we have: $ (a^5 \minus{} b^5)(a^2 \minus{} b^2)\ge0$

$ \Longleftrightarrow\frac {a^5}{b^2} \plus{} \frac {b^5}{a^2} \minus{} a^3 \minus{} b^3\ge0$[color=darkblue], (1)[/color]

$ \bullet$ [color=darkblue]Same[/color] $ \forall a, b \in R$ $ (a^4 \minus{} b^4)(a \minus{} b)\ge0$

$ \Longleftrightarrow\frac {a^4}{b} \plus{} \frac {b^4}{a} \minus{} a^3 \minus{} b^3\ge0$[color=darkblue], (2)[/color]

$ (1) \plus{} (2)$ [color=darkblue]we get [/color]$ (\frac {a^5}{b^2} \plus{} \frac {b^5}{a^2} \minus{} a^3 \minus{} b^3) \plus{} (\frac {a^4}{b} \plus{} \frac {b^4}{a} \minus{} a^3 \minus{} b^3)\ge0$
[color=darkblue][u]Note[/u]: You must post your problem by English at this forum. If you want post by Vietnamese you can post at this link: [/color] http://www.mathlinks.ro/index.php?f=284

Yea i agree.

1. I have absolutely no clue how to speak Vietnamese

I'm sorry, but can you please explain why for all real numbers a and b (unless I misunderstood the symbolism),
[code](a^5-b^5)(a^2-b^2)\ge0[/code]
Can't you just set a to be -2 and b to be -1?

EDIT: oops, sorry, i'm still trying to figure out this latex code stuff

I sorry because I can't speak english well Thanks you very much!

[color=darkblue]Another solution for your problem:

Inequality $ \Longleftrightarrow$ $ a^6 \minus{} 2b^2a^4 \plus{} 2b^3a^3 \minus{} 2b^4a^2 \plus{} b^6 \equal{} (a \minus{} b)^2(a^4 \plus{} 2ba^3 \plus{} b^2a^2 \plus{} 2b^3a \plus{} b^4) \ge 0$[/color]

[quote="thanhnam2902"][color=darkblue]This problem is:
Let $ \boxed{a > ,b > 0}$. Prove that $ \boxed{\frac {a^2}{b} \plus{} \frac {b^2}{a}\ge\sqrt {2(a^2 \plus{} b^2)}}$[/color][/quote]

$ 2(a^3\plus{}b^3)^2\ge (a^2\plus{}b^2)^3\ge 4a^2b^2(a^2\plus{}b^2)\implies \left(\frac{a^2}{b}\plus{}\frac{b^2}{a}\right)^2\ge 2(a^2\plus{}b^2)$

hi this is onther solution i may call it the EASIER
using chebchev we have :
$ a^2/b \plus{}b^2/a >\equal{} 1/2(a^2\plus{}b^2)(1/a\plus{}1/b)$
so we have to prouve that $ 1/4(a^2\plus{}b^2)(1/a\plus{}1/b)^2>\equal{}2$ ==> $ (a^2\plus{}b^2)(a\plus{}b)^2>\equal{}8(ab)^2$
and we know that $ a^2\plus{}b^2>\equal{}2ab$ and $ (a\plus{}b)^2>\equal{}4ab$   :ninja:

[quote="thanhnam2902"][color=darkblue]This problem is:
Let $ \boxed{a > ,b > 0}$. Prove that $ \boxed{\frac {a^2}{b} \plus{} \frac {b^2}{a}\ge\sqrt {2(a^2 \plus{} b^2)}}$[/color][/quote]
Vamathletes clearly stated to solve for a and b  :D

I think thanham's version of this problem is right as he's from vietnam. :D

[quote="ckck"]Vamathletes clearly stated to solve for a and b  :D[/quote]

If so, then the solution is $ a > 0,b > 0$. :P

This problem was asked to be proven obviously.


geometry
circumcircle
congruent triangles
cyclic quadrilateral
geometry proposed

I hope this problem is not easy. ;)
  Let $ABC$ be an acute triangle and $M$ be the midpoint of the arc $BC$ which doesn't contain the point $A$. Let $D$ and $E$ be the feet of the $A$ bisector and $A$ altitude of $ABC$. The circle $(\pi)$ that is tangent to $OM$ at $M$ and passes through $D$ intersects the line $OD$ at a point $F$ which is different from $D$. Let $EF$ cut the arc $BC$ which doesn't contain the point $A$ at $K$.
   Prove that the circumcircle of $ODM$ is tangent to $OK$ at $O$.

This one is for Yetti:   after several pages of tightly packed trigo he can convince
     most of us that anything is possible in Q. mechanics and E. geometry.

     Maj.  Pestich.

Are you sure it's true, treegoner? :? It's pretty clear from my sketch that's it's not generally true in the case $\angle B=\frac \pi 2$ (I know you said the triangle was acute-angled, but it should also work for right triangles by continuity).

[quote="prowler"]I didn't find contradiction, but anyway :
$\angle OAM = \angle EAM = \angle AMO = \angle FOM$
since $OM$ is parallel to $AE$ and is middle perpendicular to $BC$.

$\angle AMO = \angle OFM$  from condition of tangency.

Hence $FM =OM = AM = R$, triangles $AOM$ and $FMO$ are congruent.

So $AOFM$ trapezoid with equal sides and $A,E,F$ are collinear.

We have $K$ is intersection of height $AE$ and circle.
$\angle KOM = 2\angle EAM $ and $\angle DOK =\angle DMO$ c.t.d.[/quote]
Why $\angle AMO=\angle FOM$ ;)

[quote="xeroxia"]in my skecth, it looks true[/quote]
Could you post your sketch?

It [b]looks[/b] Ok ;). Experiment with it more. Like I said above, take $\angle B=\frac\pi 2$, to make the picture simpler, and drag the points around like crazy :). You'll see that there are positions where it doesn't look Ok at all. It looked Ok to me too at first.

Sorry, i think problem is wrong. Assuming $\bigtriangleup ABC$ as acute, i get $\angle KOL_1 = 70$, where $L_1$ is the circumcenter of $\bigtriangleup ODM$.

I have no clue what happens to my mind these day.
 The question is :
    Prove that the line $OK$ is tangent to the circumcircle of $ODM$ if and only if $\measuredangle {BAC} = 60^0$, here $O$ is the circumcenter of $ABC$
  Very sorry ! :oops:

By the way, it can be true to any triangle, but here I want to make sure that $EF$ intersects the circle $(O)$

[quote="prowler"]Sorry for stupid mistake mecrazywong.[/quote]
Nevermind ;)
But your solution works for the correct version now :)

Yep, because $OB=OC=BM=CM= R$ and $OD=DM$ :D

What about the inverse case, prowler ?  :D
 If $KO$ touches the circumcircle of $ODM$ then $\measuredangle{BAC} = 60^0$ ?

[quote="treegoner"]Noone has any idea ?
  I will post the solution next week.  ;)[/quote]
Next week I still have my school exam...

Here is my solution for the other direction:
We will show that A,E,K,F are collinear.
Suppose the contrary, that is, line FKE intersects the circle at another point $A'\neq A$. Note that F is the inverse of D, so A',O,D,K are concyclic. This implies $\angle EAD=\angle DMO=\angle KOD=\angle KA'D$. Thus A,A',D,E are concyclic. We get $\angle AA'D=90^\circ$. Now we obtain that the other intersection of A'D, X is the antipode of A.
$\angle OXD=\angle OA'D=\angle OKD$, together with common side OD and OX=OK, $\triangle ODK\cong\triangle ODX\Rightarrow\angle KOD=\angle XOD$, but $\angle KOD=\angle KA'D=\angle EA'D=\angle EAD=\angle XAD$. Thus $\angle XOD=\angle XAD$, this yields A,O,D are collinear and hence ABC is isosceles. Now clearly contradiction occurs.

Therefore we have shown A,E,K,F are collinear. Then $\angle MAO=\angle DMO=\angle OFM$, so A,O,M,F are concyclic. Now it's easy to see that OD=OM and the conclusion follows immediately.

There should be a much simpler proof... :(

Are you sure that the two triangles $ODK = ODX$ ? (As you draw a circle $(O)$, and a point $D$. Your two points $K$ and $X$  lie on the two symmetric arcs of circles constructed on the segment $OD$ which may cut your circle $(O)$ at 4 different points. This yields that not always it is true. Notice that you make a mistake in the SAS condition for the two congruent triangles  :?. Or I misunderstand you ? )
 Anyway, it is a very nice approach  ;)

Be frank, I really didn't notice that :oops:
Anyway, this can be corrected easily:
We know that OK=OX, $\angle OKD=\angle OXD$. This implies $\angle ODK=\angle ODX$ or $\angle ODK+\angle ODX=180^\circ$.
For the former case, it is exactly the same as what I have proved.
For the latter case, it follows that K,D,X are collinear. We sub-divide it into 2 cases:
(1)K,X lies on different side of BC. Then A,D,E,K are concyclic and thus K,E,M are collinear and clearly $F=EK\cap OD\in (EK)$(is it the right notation for line segment :? )
(2)KX coincides with BC. Then B or C is right angle and we get that $A=60^\circ$ easily by the concyclicity of A,O,D,E(BorC).

I hope this is correct now :)

Are you sure that $K, D,X$ are collinear in the latter case? Be careful with your figure. You can not 100% believe in it, right ?
 But you are getting very, very close to your success if you reconsider the point $A'$ in a more general way. ;)

So I screw up again :(
$\angle ODK+\angle ODX=180^\circ$. Then if K and X lie on different sides of OD, we have K,D,X are collinear.
If K and X lie on the same side of OD, then O,D,K,X are concyclic. This means that K,X,F are collinear or K=X.
If K=X, then O,A',D,X are concyclic, which is obviously absurd(well, actually not really if $O\in A'X$ :P )
If K,X,F are collinear, then K=X(done already) or K=A'. Recall that K and A' are the points of intersection of line FK with the circle. This can only occur if FK is tangent to the circle. But since $K=FE\cap(O)$, $E=K\Rightarrow E=BorC$.

[quote="treegoner"]But you are getting very, very close to your success if you reconsider the point  in a more general way. [/quote]
Indeed, define A' again to be the intersection point of DX with the circle. Then we obtain A',E,M are collinear and M=K. Now since F lies on A'K=A'M and OD, we see that F lies the circle. Contradiction.(note that it's possible for F=M, which I neglect)

I strongly feel that all the cases I dealt with can actually be done by a strong lemma or transformation-since they are all very similar.

[quote="treegoner"]That is what I want to say, mecrazywong. Congratulations !
 By the way, your proof is pretty shorter than mine. I will post it later.[/quote]
So finally solved :blush:  :)
Anyway, Treegoner, how do you come up with so many nice problems?

I guess with some luck, mecrazywong :D
 As promise, I give my solution to the inverse part of the problem. Here are some nice facts that are true to any triangle (some are discovered accidentally by me ;) )
   1) $OD.OF = R^2$, and therefore $F$ lies on the circumcircle of $OBC$.
   2) Call $A'$ the second intersection of $AE$ with $(O)$. Call $V$ the second intersection of the circumcircles of $OA'M$ and $OBC$. Then $OV, BC, MA'$ are concurrent at $T$ (this is obvious as they are the radical axises of three circles $(O)$, the circumcircles of $OBC$ and $OA'M$)
   3) This point $V$ lies on the circle $(\pi)$ defined in the first post [hide="Hint"] use some angle chasing [/hide]
   4) $V$ also lies on the line $EF$ [hide="Hint"] By calling $P$ the second intersection of the circle with diameter $AD$ and $(O)$, we have $P, E, M$ are collinear. Now, you can either use the property of cyclic quadrilateral or consider the three radical axises of $(\pi)$, $(O)$, $(OBC)$ [/hide].
   5) $OV.OT = R^2$ (in fact this is true for any point $V$ lying on the circumcircle of $OBC$, and $T$ is the intersection of $OV$).
   6) $TKDO$  is cyclic [hide="Hint"] from 4 and 5 [/hide]
   From that, you can prove that $\measuredangle{KTC} = \measuredangle{A'TC}$, which yields that the lines $TK$ and $TA'$ are the same, or $A' = K$. Then $M$ is the circumcenter of $FBC$ whose circumcircle passes through $O$, which implies that  $\measuredangle{A} = 60^0$.
   That is how I developed this problem. At first, I made a mistake when I asked you the original question as my sketch was not good. Therefore, later I elaborated the condition so that it is right.