Search found 217 matches

by zadid xcalibured
Thu Jan 24, 2013 6:18 pm
Forum: Secondary Level
Topic: Solve this congruence equation
Replies: 2
Views: 1548

Re: Solve this congruence equation

Still not interested. $SIGH$
btw,Happy Birthday.
by zadid xcalibured
Wed Jan 23, 2013 1:59 pm
Forum: International Mathematical Olympiad (IMO)
Topic: IMO Marathon
Replies: 184
Views: 63440

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 $...
by zadid xcalibured
Wed Jan 23, 2013 1:29 pm
Forum: International Mathematical Olympiad (IMO)
Topic: IMO Marathon
Replies: 184
Views: 63440

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...
by zadid xcalibured
Wed Jan 23, 2013 10:03 am
Forum: International Mathematical Olympiad (IMO)
Topic: IMO Marathon
Replies: 184
Views: 63440

Re: IMO Marathon

I wish every geometry problems were as easy and as beautiful as this. :D
by zadid xcalibured
Wed Jan 23, 2013 9:56 am
Forum: International Mathematical Olympiad (IMO)
Topic: IMO Marathon
Replies: 184
Views: 63440

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
by zadid xcalibured
Tue Jan 22, 2013 10:13 pm
Forum: Higher Secondary Level
Topic: Secondary and Higher Secondary Marathon
Replies: 127
Views: 58282

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{...
by zadid xcalibured
Tue Jan 22, 2013 8:10 pm
Forum: International Mathematical Olympiad (IMO)
Topic: IMO Marathon
Replies: 184
Views: 63440

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 $20$.
by zadid xcalibured
Tue Jan 22, 2013 1:21 pm
Forum: Higher Secondary Level
Topic: Secondary and Higher Secondary Marathon
Replies: 127
Views: 58282

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.
by zadid xcalibured
Tue Jan 22, 2013 12:59 pm
Forum: Higher Secondary Level
Topic: Secondary and Higher Secondary Marathon
Replies: 127
Views: 58282

Re: Secondary and Higher Secondary Marathon

Well,Adib notice that $AK=AC$ he said.
by zadid xcalibured
Mon Jan 21, 2013 12:07 am
Forum: Higher Secondary Level
Topic: Secondary and Higher Secondary Marathon
Replies: 127
Views: 58282

Re: Secondary and Higher Secondary Marathon

Well,this is problem 3 of geomtry problem set of BDMC 2012.