## Search found 244 matches

Fri Nov 16, 2012 1:10 pm
Topic: IMO Marathon
Replies: 184
Views: 64072

### Re: IMO Marathon

yap .. thanks .
Fri Nov 16, 2012 8:06 am
Topic: IMO Marathon
Replies: 184
Views: 64072

### Re: IMO Marathon

Problem $9$ A circle intersects sides $BC,CA,AB$ of $\triangle ABC$ at two points for each side in the following order: $(D_1 , D_2 ), (E_1 , E_2 )$ and $(F_1 ,F_2)$. Line segments $D_1E_1$ and $D_2F_2$ intersect at point $L$, $E_1F_1$ and $E_2D_2$ intersectat point $M$, $F_1D_1$ and $F_2E_2$ inters...
Fri Nov 16, 2012 7:30 am
Topic: IMO Marathon
Replies: 184
Views: 64072

### Re: IMO Marathon

Problem $8$:Let $ABC$ be a triangle such that $AB \neq AC$. The internal bisector lines of the angles $ABC$ and $ACB$ meet the opposite sides of the triangle at points $B_0$ and $C_0$, respectively, and the circumcircle $ABC$ at points $B_1$ and $C_1$, respectively. Further, let $I$ be the incentre...
Sat Nov 10, 2012 9:30 pm
Topic: IMO Marathon
Replies: 184
Views: 64072

### Re: IMO Marathon

sourav das wrote:Come on guys, IMO standard problems won't be easy to crack (Doesn't mean ugly). They'll be smart and hidden. So please post a new one.
Why don't you post one A samrt one ?
Sat Nov 10, 2012 8:28 pm
Topic: IMO Marathon
Replies: 184
Views: 64072

### Re: IMO Marathon

SANZEED wrote: $LQ\parallel LQ'$, which is impossible.
What if Q' lie on LQ ?
Sat Nov 10, 2012 8:07 pm
Topic: IMO Marathon
Replies: 184
Views: 64072

### Re: IMO Marathon

Edited some comma ,.. :S
Sat Nov 10, 2012 6:48 pm
Topic: IMO Marathon
Replies: 184
Views: 64072

### Re: IMO Marathon

Waiting for confirmation It seems the problem statement is not true .We may disprove the statement . The condition $\angle KAP =90- \angle LBP$ says $\triangle KQP$ is right angled triangle with right angle $Q$ . Now draw a perpendicular $QX$ from $Q$ to $KL$ so that $KL\cap QX=X$ . Then the line t...
Tue Nov 06, 2012 8:07 am
Forum: News / Announcements
Topic: Active users for marathon
Replies: 23
Views: 9974

### Re: Active users for marathon

Count me Too ..
Test exam er kheta puRi XD
Thu Nov 01, 2012 8:37 pm
Forum: Geometry
Topic: Triangle
Replies: 13
Views: 6781

### Re: Triangle

Nadim Ul Abrar wrote:Now $AC+BC=(QC+PC)+(QD+PB)=AB+(QD-AD+PE+BE)$ $=AB+(QD+PE)=(QC+PC)+AB=2AB$
নাদিম ভাই, $QD+PE=QC+PC$ কীভাবে হল? ঠিক বুঝতে পারলাম না। $Q,P$ are midpoints of $CD,CE$ respectively . (retio)
Nadim Ul Abrar.PNG Draw a parellal line $l$ to $QP$ through the point $R$ so that $l \cap AC=D$,$l\cap AB=E$. its easy to prove that $Q,P$ are midpoints of $CD,CE$ respectively Using sine rule in $\triangle ADR$ and $\triangle BER$ \$\frac {AD}{BE}=\frac{\frac {AR}{sin \angle CDE}} {\frac {BR}{sin \...