Search found 79 matches
- Sat Sep 27, 2014 6:08 am
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Marathon
- Replies: 184
- Views: 115487
Re: IMO Marathon
$Y$ lies outside $ABC$, $X$ lies inside. As $AY$ is the perpendicular bisector of $EF, \angle YEF = \angle YFE = \angle DFE = 90-(C/2)$. For this reason, $\angle AYE = \angle AYF = ( \angle EYF)/2 = (180-2\angle YEF)/2 = \angle C/2$. Then $\angle DYE = \angle AYF+\angle AYE = \angle C,$ so $DYCE$ is...
- Fri Sep 26, 2014 6:48 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Marathon
- Replies: 184
- Views: 115487
Re: IMO Marathon
Denote the main equation by (M). Suppose f(0)=c. First we check for constant solutions: (M) $\Rightarrow$ c=c+y$^2$c-2c, taking y nonzero and y$\neq \pm\sqrt{2}$ implies c=0. Constant f=0 is a solution. Now assume f is nonconstant. (M)x=y=0 $\Rightarrow$ f(c)=-c…..(i). (M) y = 0 $\Rightarrow$ f(f(x)...
- Sun Sep 21, 2014 8:58 pm
- Forum: Number Theory
- Topic: Magic-Square preserving functional equation(Self-made)
- Replies: 4
- Views: 3765
Re: Magic-Square preserving functional equation(Self-made)
I have noted that f should be injective. This is easy to see: if f(a)=f(b); we construct the 'general magic square' (f(a),f(a),......($n^2-1$ times),f(b)); we get so is (a,a,a,.....($n^2-1$ times),b). So a=b. I have another idea, which may be interesting to try: that is, using a latin square is a ge...
- Sun Sep 21, 2014 6:25 pm
- Forum: Geometry
- Topic: Parallelism(self-made)
- Replies: 0
- Views: 1856
Parallelism(self-made)
Let $ABC$ a triangle with incircle touching $BC,CA,AB$ respectively at $X,Y,Z$. Suppose $D,E,F$ are respectively the feet of altitudes in $ABC$ from $A,B,C$. Similarly, define points $M,N,L$ to be feet of altitudes of triangle $XYZ$ from $X,Y,Z$. Finally, let pairs $DF,AC; DE,AB; MN,XY; ML,XZ$ inter...
- Fri Sep 19, 2014 5:43 am
- Forum: Number Theory
- Topic: Magic-Square preserving functional equation(Self-made)
- Replies: 4
- Views: 3765
Magic-Square preserving functional equation(Self-made)
Find all functions f:$\mathbb{N} \to \mathbb{N}$ that preserves 'general magic square's. That is, if a n*n square's entries are serially in standard written as ($a_1, a_2,........, a_{n^2}$); then ($a_1, a_2,........, a_{n^2}$)(positive integer entries) is a 'general magic square' if and only if so ...
- Wed Sep 10, 2014 9:20 pm
- Forum: Number Theory
- Topic: An honouring divisibility(self-made)
- Replies: 4
- Views: 3104
Re: An honouring divisibility(self-made)
hmm....... If,we give n freedom, then ?(edited problem)
- Wed Sep 10, 2014 8:44 pm
- Forum: Number Theory
- Topic: An honouring divisibility(self-made)
- Replies: 4
- Views: 3104
An honouring divisibility(self-made)
Prove that for all k$\in \mathbb N$, k > 1; they're exists n$\in \mathbb N$ , n < 2k such that
$\frac {\sigma(n)}{n} = \frac{k+1}{k}$. [Here $\sigma(n)$ denotes the sum of positive divisors of n; as usual.]
$\frac {\sigma(n)}{n} = \frac{k+1}{k}$. [Here $\sigma(n)$ denotes the sum of positive divisors of n; as usual.]
- Wed Sep 03, 2014 5:21 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Marathon
- Replies: 184
- Views: 115487
Re: IMO Marathon
My solution to problem 34 is as follows: The solution set is S=$\left \{ \forall r\in\mathbb{N}\cup\left \{ +\infty\ \right \}|\;f_r(n)= \begin{cases} & n, \ {if }\ n<r \\ & n+1, \ { if }\ n\geqslant r \end{cases} \right \}$ We start by defining $a_k$:= min{f(i)|$i\in\mathbb{N},i\geqslant k$}. For $...
- Thu Aug 28, 2014 11:17 pm
- Forum: Geometry
- Topic: Collinearity and elegance revisited(self-made)
- Replies: 0
- Views: 1870
Collinearity and elegance revisited(self-made)
Two triangles 123, 456 have the same incircle. Let the respective intersections of 26,35; 12,45; 13,46 are 7,8,9. Prove that 7,8,9 are collinear.
- Thu Aug 28, 2014 11:14 pm
- Forum: Geometry
- Topic: Elegant collinearity is in my intuition(Self-made)
- Replies: 1
- Views: 2357
Elegant collinearity is in my intuition(Self-made)
Let 123 be a triangle, in which medians from 1,2,3 respectively intersect it's circumcircle again at 4,5,6. The tangents to this circumcircle at 4,5,6 respectively intersects 23,31,12 at 7,8,9. Prove that 7,8,9 are collinear.