Search found 110 matches
- Mon Feb 23, 2015 9:17 pm
- Forum: Geometry
- Topic: Italy TST 2000/2
- Replies: 3
- Views: 3168
Re: Italy TST 2000/2
i think my solution is too much boring :? let's apply Menelaus's and Menelaus's ; let , $CF\cap AB=N ; BF\cap AC=P ; BF\cap CM=R ; CF\cap BD=S ; CM\cap BD=T$ Now $\frac{BD}{DT}\frac{TF}{FA}\frac{AM}{MB}=1\Leftrightarrow \frac{AF}{FT}=\frac{BD}{DT}$ $\frac{AC}{CD}\frac{DS}{ST}\frac{TF}{FA}=1 \Leftrig...
- Mon Feb 23, 2015 2:47 pm
- Forum: Geometry
- Topic: USAMO 2013/1
- Replies: 3
- Views: 3062
Re: USAMO 2013/1
oppsss sorry . didn't notice . now it is editedtanmoy wrote:You did not state what is the point M is
- Mon Feb 23, 2015 2:38 pm
- Forum: Geometry
- Topic: Italy TST 2000/2
- Replies: 3
- Views: 3168
Italy TST 2000/2
Let $ABC$ be an isosceles right triangle and $M$ be the midpoint of its hypotenuse $AB$ . Points $D$ and $E$ are taken on the legs $AC$ and $BC$ respectively such that $AD=2DC$ and $BE=2EC$. Lines $AE$ and $DM$ intersect at $F$ . Show that $FC$ bisects the $\angle DFE$ .
- Mon Feb 23, 2015 2:26 pm
- Forum: Combinatorics
- Topic: Circular Table conference
- Replies: 2
- Views: 2672
Re: Circular Table conference
if you want to know how it works , read any books of combinatorics .
it could be "principles and techniques in combinatorics" or google it
it could be "principles and techniques in combinatorics" or google it
- Mon Feb 23, 2015 2:21 pm
- Forum: Combinatorics
- Topic: Circular Table conference
- Replies: 2
- Views: 2672
Re: Circular Table conference
answer is $5!/2!$
- Sun Feb 22, 2015 11:27 pm
- Forum: Geometry
- Topic: USAMO 2013/1
- Replies: 3
- Views: 3062
Re: USAMO 2013/1
let, $M=w_{A}\cap w_{B}\cap w_{C}$ $\frac{XY}{XZ}=\frac{sin\angle YMX \cdot YM}{sin\angle ZMX\cdot ZM}$ or, $\frac{XY}{XZ}=\frac{sin\angle YMX \cdot sin\angle YZM}{sin\angle ZMX\cdot sin\angle ZYM}$ now , $\angle YMX=180-\angle MYX-\angle MXY=\angle MRB-\angle MRP=\angle PRB=\angle PMB$ and let $XM\...
- Thu Feb 19, 2015 12:22 am
- Forum: Combinatorics
- Topic: Triangle in a 2n-Graph
- Replies: 2
- Views: 2869
Re: Triangle in a 2n-Graph
it is easy to check that it works for $n=2$ let it is true for $n=m$ ...... now for $n=m+1$ we have $2(m+1)=2m+2$ points and we need to connect $m^{2}+2m+2$ edges let there are $2m+2$ points in a plane such that no three are collinear .... now draw an area S such that $2m$ points are inside S. other...
- Wed Feb 18, 2015 7:54 pm
- Forum: Algebra
- Topic: power of 2 or binomial?
- Replies: 4
- Views: 6748
Re: power of 2 or binomial?
$\frac{2n}{n}=2 $
$\frac{2n-1}{n-1} > 2$
$\frac{2n-2}{n-2} > 2$
.
.
$\frac{2n-(n-1)}{n-(n-1)} > 2$
multiplying all ....we have ,
$\frac{2n(2n-1)(2n-2)......(n+1)}{n(n-1)(n-2)......1}>2^{n}$
or, $\binom{2n}{n}>2^{n}$
it works for all n>1 . for n=1 both of them are equal
$\frac{2n-1}{n-1} > 2$
$\frac{2n-2}{n-2} > 2$
.
.
$\frac{2n-(n-1)}{n-(n-1)} > 2$
multiplying all ....we have ,
$\frac{2n(2n-1)(2n-2)......(n+1)}{n(n-1)(n-2)......1}>2^{n}$
or, $\binom{2n}{n}>2^{n}$
it works for all n>1 . for n=1 both of them are equal
- Sun Dec 21, 2014 2:03 am
- Forum: Geometry
- Topic: Side, Angle, and, (Ex)-Circle
- Replies: 4
- Views: 4004
Re: Side, Angle, and, (Ex)-Circle
[1] Let $AD\cap OC=N$ For triangle $\Delta AOC$ ; $\frac{AP}{PO}\cdot \frac{ON}{NC}\cdot \frac{CM}{MA}=1$ [ceva] or, $\frac{AP}{PO}=\frac{NC}{NO}$ Now let $I$ be the incentre of triangle $\Delta ABC$ ...... $\therefore I\epsilon AP$ $(A,I,P,O) $ harmonic $\therefore \frac{AI}{IP}=\frac{AO}{PO}$ or,$...
- Sun Oct 19, 2014 1:27 am
- Forum: Geometry
- Topic: Regional Mathematical Olympiad(India) 1994,P6
- Replies: 10
- Views: 6771
Re: Regional Mathematical Olympiad(India) 1994,P6
$\angle AMB$ is a part of $\angle KML$prya19970 wrote: So it's a rectangle. .
so , $\angle KML$ must be greater than 90 .
so, $OKML$ is not a rectangle .