Search found 244 matches
- Fri Nov 16, 2012 1:10 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Marathon
- Replies: 184
- Views: 116628
Re: IMO Marathon
yap .. thanks .
- Fri Nov 16, 2012 8:06 am
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Marathon
- Replies: 184
- Views: 116628
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
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Marathon
- Replies: 184
- Views: 116628
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
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Marathon
- Replies: 184
- Views: 116628
Re: IMO Marathon
Why don't you post one A samrt one ?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.
- Sat Nov 10, 2012 8:28 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Marathon
- Replies: 184
- Views: 116628
Re: IMO Marathon
What if Q' lie on LQ ?SANZEED wrote: $LQ\parallel LQ'$, which is impossible.
- Sat Nov 10, 2012 8:07 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Marathon
- Replies: 184
- Views: 116628
Re: IMO Marathon
Edited some comma ,.. :S
- Sat Nov 10, 2012 6:48 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Marathon
- Replies: 184
- Views: 116628
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: 17629
Re: Active users for marathon
Count me Too ..
Test exam er kheta puRi XD
Test exam er kheta puRi XD
Re: Triangle
$Q,P$ are midpoints of $CD,CE$ respectively . (retio)Phlembac Adib Hasan wrote:নাদিম ভাই, $QD+PE=QC+PC$ কীভাবে হল? ঠিক বুঝতে পারলাম না।Nadim Ul Abrar wrote:Now $AC+BC=(QC+PC)+(QD+PB)=AB+(QD-AD+PE+BE)$ $
=AB+(QD+PE)=(QC+PC)+AB=2AB$
Re: Triangle
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 \...