Search found 461 matches
- Sat Aug 06, 2011 10:24 pm
- Forum: Combinatorics
- Topic: Good problem
- Replies: 1
- Views: 2066
Good problem
1.1. Determine the number of functions $f : \{1, 2, . . . , 1999\} \mapsto \{2000, 2001, 2002, 2003\}$ satisfying the condition that $f(1)+ f(2)+. . . + f(1999)$ is odd.
- Sat Aug 06, 2011 5:47 pm
- Forum: Divisional Math Olympiad
- Topic: Divisional (Jesoore) 2011- Sec - 10
- Replies: 7
- Views: 4420
Re: Divisional (Jesoore) 2011- Sec - 10
In solution of this problem doesn't need the value of MN. Then why did they give it?
- Thu Aug 04, 2011 11:59 am
- Forum: Geometry
- Topic: locus of point $D$
- Replies: 6
- Views: 3647
Re: locus of point $D$
You must count the degenerate cases
- Thu Aug 04, 2011 11:55 am
- Forum: Geometry
- Topic: locus of point $D$
- Replies: 6
- Views: 3647
Re: locus of point $D$
you don't mention to not count A,C.Hi, Hi, Hi
- Thu Aug 04, 2011 11:48 am
- Forum: Geometry
- Topic: locus of point $D$
- Replies: 6
- Views: 3647
Re: locus of point $D$
Upps. I forget to mentio short arc(AC)
- Thu Aug 04, 2011 11:35 am
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO longlist 1989 [reformed]
- Replies: 1
- Views: 2446
IMO longlist 1989 [reformed]
$0<a<1\ constant,\ f:R\mapsto R, \ f(0)=0,\ f(1)=1,\ f(\frac{x+y}{2})=(1-a)f(x) +$ $af(y) $ for $x\leq y$. if $f(x)$ is not equal$\ 0$ for all $x$, prove that $a = \frac{1}{2}$
- Thu Aug 04, 2011 10:58 am
- Forum: Geometry
- Topic: locus of point $D$
- Replies: 6
- Views: 3647
Re: locus of point $D$
$DA.BC + DC.AB = DB.AC$ [Ptolemy].So locus of $D$ is the circumcircle.
- Tue Aug 02, 2011 12:28 pm
- Forum: Combinatorics
- Topic: Count subsets of finite positive integers constrained by sum
- Replies: 3
- Views: 3432
- Mon Aug 01, 2011 3:21 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO shortlist 2010 G1
- Replies: 4
- Views: 3906
IMO shortlist 2010 G1
Let $ABC$ be an acute triangle with $D$;$E$;$F$ the feet of the altitudes lying on $BC$;$CA$;$AB$
respectively. One of the intersection points of the line $EF$ and the circumcircle is $P$: The
lines $BP$ and $DF$ meet at point $Q$: Prove that $AP = AQ$:
Proposed by United Kingdom
respectively. One of the intersection points of the line $EF$ and the circumcircle is $P$: The
lines $BP$ and $DF$ meet at point $Q$: Prove that $AP = AQ$:
Proposed by United Kingdom
- Mon Aug 01, 2011 12:13 am
- Forum: Geometry
- Topic: Canada 2002 problem:2(Good)
- Replies: 1
- Views: 2076
Canada 2002 problem:2(Good)
Let $\tau$ be a circle with radius $r$. Let $A$ and $B$ be distinct points on $\tau$ such that $AB < \sqrt{3}r$. Let the circle with center $B$ and radius $AB$ meet $\tau$ again at $C$. Let $P$ be the point inside $\tau$ such that triangle $ABP$ is equilateral. Finally, let line $CP$ meet $\tau$ aga...