Search found 110 matches

by Tahmid
Mon Feb 23, 2015 9:17 pm
Forum: Geometry
Topic: Italy TST 2000/2
Replies: 3
Views: 1586

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...
by Tahmid
Mon Feb 23, 2015 2:47 pm
Forum: Geometry
Topic: USAMO 2013/1
Replies: 3
Views: 1670

Re: USAMO 2013/1

tanmoy wrote:You did not state what is the point M is
oppsss sorry . didn't notice :? . now it is edited
by Tahmid
Mon Feb 23, 2015 2:38 pm
Forum: Geometry
Topic: Italy TST 2000/2
Replies: 3
Views: 1586

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 $\angle DFE$ .
by Tahmid
Mon Feb 23, 2015 2:26 pm
Forum: Combinatorics
Topic: Circular Table conference
Replies: 2
Views: 1251

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 :)
by Tahmid
Mon Feb 23, 2015 2:21 pm
Forum: Combinatorics
Topic: Circular Table conference
Replies: 2
Views: 1251

Re: Circular Table conference

answer is $5!/2!$
by Tahmid
Sun Feb 22, 2015 11:27 pm
Forum: Geometry
Topic: USAMO 2013/1
Replies: 3
Views: 1670

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\...
by Tahmid
Thu Feb 19, 2015 12:22 am
Forum: Combinatorics
Topic: Triangle in a 2n-Graph
Replies: 2
Views: 1405

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...
by Tahmid
Wed Feb 18, 2015 7:54 pm
Forum: Algebra
Topic: power of 2 or binomial?
Replies: 4
Views: 3137

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
by Tahmid
Sun Dec 21, 2014 2:03 am
Forum: Geometry
Topic: Side, Angle, and, (Ex)-Circle
Replies: 4
Views: 1982

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,$...
by Tahmid
Sun Oct 19, 2014 1:27 am
Forum: Geometry
Topic: Regional Mathematical Olympiad(India) 1994,P6
Replies: 10
Views: 3166

Re: Regional Mathematical Olympiad(India) 1994,P6

prya19970 wrote: So it's a rectangle. .
$\angle AMB$ is a part of $\angle KML$
so , $\angle KML$ must be greater than 90 .
so, $OKML$ is not a rectangle .