Search found 461 matches
- Mon Sep 05, 2011 10:36 am
- Forum: Number Theory
- Topic: 2003 USAMO[Nice problem]
- Replies: 3
- Views: 2938
Re: 2003 USAMO[Nice problem]
I have used induction. Let $x_{n}$ means a number with n digit. $n=1$ is clear with 5. Assume the existence for $n=m$ and let $x_{m}$ be the number. Now let $ P \in \{ 1, 3, 5, 7, 9 \}$ (NOTE: This set is a complete residue of mod 5). Now $P * 5^{m} * 2^{m} + x_{m}$ , a $m+1$ digit number is divisib...
- Thu Sep 01, 2011 9:53 am
- Forum: Geometry
- Topic: Just Beauty (Bulgaria 2002 : Problem 4)
- Replies: 25
- Views: 12417
Re: Just Beauty (Bulgaria 2002 : Problem 4)
I thought the problem would be solved in Euclidean way .I was so amused to solve the problem that it was more than beauty for me. I don't think it is a very good idea to start with complex and not with Euclidean (My personal comment, don't take it seriously :D ). I am posting my solution now as i th...
- Wed Aug 31, 2011 3:44 pm
- Forum: Geometry
- Topic: Just Beauty (Bulgaria 2002 : Problem 4)
- Replies: 25
- Views: 12417
Re: Just Beauty (Bulgaria 2002 : Problem 4)
Don't you try homothety? Seriously I'm expecting the solution with homothety. One more day and i will post my solution.
- Mon Aug 29, 2011 5:41 pm
- Forum: Geometry
- Topic: Just Beauty (Bulgaria 2002 : Problem 4)
- Replies: 25
- Views: 12417
Re: Just Beauty (Bulgaria 2002 : Problem 4)
I'm expecting rather beautiful solution Mahi. Your solution is also complex like the official solution. Don't you find out the short proof??
- Fri Aug 12, 2011 11:34 am
- Forum: Geometry
- Topic: Just Beauty (Bulgaria 2002 : Problem 4)
- Replies: 25
- Views: 12417
Just Beauty (Bulgaria 2002 : Problem 4)
Problem 4 Let $I$ be the incenter of non-equilateral triangle $ABC$, and let $T_{1}, T_{2}, T_{3}$ be the tangency points of the incircle with sides $BC, CA, AB$ respectively. Prove that the orthocenter of triangle $T_{1}T_{2}T_{3}$ lies on line $OI$, where $O$ is the circumcenter of triangle $ABC$....
- Fri Aug 12, 2011 11:11 am
- Forum: International Mathematical Olympiad (IMO)
- Topic: 39th IMO
- Replies: 3
- Views: 2897
Re: 39th IMO
now for the only if part,for the sake of contradiction we consider the existence of a cyclic quadrilateral $ABCD$ for which the area of $\triangle APB,\triangle CPD$ are not equal.but the 'if' process can be done here and proven $\triangle APB=\triangle CPD$,so a contradiction. Completely wrong. Yo...
- Wed Aug 10, 2011 7:18 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: 39th IMO
- Replies: 3
- Views: 2897
39th IMO
Problem 1. In the convex quadrilateral ABCD, the diagonals AC and BD are perpendicular and the opposite sides AB and DC are not parallel. Suppose that the point P, where the perpendicular bisectors of AB and DC meet, is inside ABCD . Prove that ABCD is a cyclic quadrilateral if and only if the trian...
- Tue Aug 09, 2011 5:50 pm
- Forum: Number Theory
- Topic: a congruence problem
- Replies: 10
- Views: 7222
Re: a congruence problem
Use Farmet's little theorem
- Mon Aug 08, 2011 11:24 pm
- Forum: Number Theory
- Topic: help please
- Replies: 7
- Views: 4191
Re: help please
For $5^{2n} + 3.2^{5n} - 2$ you get contradiction for $n = 2$. For $5^{2n} + 3.2^{5n - 2}$ $2^{5n}\equiv 32^{n}\equiv 4^{n}\ (mod \ 7 ) $ $g.c.d(2^2,7)=1$ So, $2^{5n-2}\equiv 4^{n-1}\ (mod \ 7 ) $ $5^{2n}\equiv 25^{n}\equiv 4^{n}\ (mod \ 7 ) $ and so, $5^{2n} + 3.2^{5n - 2} \equiv 4^{n-1} (3 + 4 ) \...
- Mon Aug 08, 2011 8:01 pm
- Forum: Number Theory
- Topic: Problem of Charles Hermite
- Replies: 1
- Views: 1974
Problem of Charles Hermite
Let $n$ be a positive integer. Prove that for any real number $x$, $\left \lfloor nx \right \rfloor \ = \ \sum_{n-1}^{i=0}\left ( \left \lfloor x + \frac{i}{n} \right \rfloor \right )$
Charles Hermite (1822-1901): French mathematician who did brilliant work in many
branches of mathematics.
Charles Hermite (1822-1901): French mathematician who did brilliant work in many
branches of mathematics.