Still not interested. $SIGH$
btw,Happy Birthday.
Search found 217 matches
- Thu Jan 24, 2013 6:18 pm
- Forum: Secondary Level
- Topic: Solve this congruence equation
- Replies: 2
- Views: 2482
- Wed Jan 23, 2013 1:59 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Marathon
- Replies: 184
- Views: 114654
Re: IMO Marathon
I propose a new rule that anyone who posts a solution of a problem must certainly post another problem.Or we'll have to wait for someone's grace. Problem $\boxed{22}$:Let $ABCD$ be a parallelogram.A variable line $l$ passing through the point $A$ intersects the rays $BC$ and $CD$ at points $X$ and $...
- Wed Jan 23, 2013 1:29 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Marathon
- Replies: 184
- Views: 114654
Re: IMO Marathon
I Love Ratio. :mrgreen: $\frac{SC}{SB}=\frac{SA}{SC}$ $\Longrightarrow \frac{SP}{SB}=\frac{SA}{SP}$ $\triangle{SPB} \sim \triangle{SAP}$ and $\triangle{SCB} \sim \triangle{SAC}$ $\frac{AP}{BP}=\frac{AS}{PS}=\frac{AS}{CS}=\frac{AC}{BC}$ from alternate segment theorem, $\frac{MK}{MP}=\frac{AC}{AP}$ an...
- Wed Jan 23, 2013 10:03 am
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Marathon
- Replies: 184
- Views: 114654
Re: IMO Marathon
I wish every geometry problems were as easy and as beautiful as this.
- Wed Jan 23, 2013 9:56 am
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Marathon
- Replies: 184
- Views: 114654
Re: IMO Marathon
Problem 21:Let $ABC$ be a triangle with $P$ in its interior(with $BC \neq AC$).The lines $AP,BP,CP$ meet $\odot{ABC}$ again at $K,L,M$.The tangent line at $C$ intersects $AB$ at $S$.Show that from $SC=SP$ it follows that $MK=ML$
Source: IMO 2010-Problem: 4
Source: IMO 2010-Problem: 4
- Tue Jan 22, 2013 10:13 pm
- Forum: Higher Secondary Level
- Topic: Secondary and Higher Secondary Marathon
- Replies: 128
- Views: 300965
Re: Secondary and Higher Secondary Marathon
Problem $\boxed{34}$:$ABCD$ is a cyclic quadrilateral.$E$ and $F$ are variable points on sides $AB$ and $CD$ respectively such that $\displaystyle \frac{AE}{BE}=\frac{CF}{DF}$. $\; P$ is a point on the segment $EF$ such that $\displaystyle \frac{EP}{PF}=\frac{AB}{CD}$.Show that $\displaystyle \frac{...
- Tue Jan 22, 2013 8:10 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Marathon
- Replies: 184
- Views: 114654
Re: IMO Marathon
Solution $\boxed {19}$:Same as Tahmid.
Solution $\boxed {20}$:Actually $APIE$ is concyclic .Which implies $\angle {A}$=$90$
Problem $19$ seemed harder to me than problem $20$.
Solution $\boxed {20}$:Actually $APIE$ is concyclic .Which implies $\angle {A}$=$90$
Problem $19$ seemed harder to me than problem $20$.
- Tue Jan 22, 2013 1:21 pm
- Forum: Higher Secondary Level
- Topic: Secondary and Higher Secondary Marathon
- Replies: 128
- Views: 300965
Re: Secondary and Higher Secondary Marathon
If we draw a circle centering at $A$ with radius $AC$ as $CD^2=AD.BD$ it holds that $D$ lies on the radical axis of the two circles.And $DK$ is the radical axis.And we now that the radical axis is perpendicular to the line joining the centres of the concerning circles.The result follows.
- Tue Jan 22, 2013 12:59 pm
- Forum: Higher Secondary Level
- Topic: Secondary and Higher Secondary Marathon
- Replies: 128
- Views: 300965
Re: Secondary and Higher Secondary Marathon
Well,Adib notice that $AK=AC$ he said.
- Mon Jan 21, 2013 12:07 am
- Forum: Higher Secondary Level
- Topic: Secondary and Higher Secondary Marathon
- Replies: 128
- Views: 300965
Re: Secondary and Higher Secondary Marathon
Well,this is problem 3 of geomtry problem set of BDMC 2012.