## Search found 36 matches

Sat Sep 30, 2017 12:02 pm
Forum: Geometry
Topic: CGMO 2002/4
Replies: 2
Views: 8223

### CGMO 2002/4

Circles $T_1$ and $T_2$ intersect at two points $B$ and $C$, and $BC$ is the diameter of $T_1$. Construct a tangent line to circle $T_1$ at $C$ intersecting $T_2$ at another point $A$. Line $AB$ meets $T_1$ again at $E$and line $CE$ meets $T_2$ again at $F$. Let $H$ be an arbitrary point on seg...
Sat Jul 29, 2017 12:38 pm
Topic: IMO 2017 P4
Replies: 4
Views: 6975

We get $TA$ parallel to $KR$ because $\angle{ATS}=\angle{SJK}=\angle{SRK}$ We extend $KS$ to $P$ where $KS$ intersects $TA$ at $P$ Now, It's easy to prove that $TPRK$ is a rombus So, $\angle{TPK}=\angle{PKR}$ Again, $\angle{ARS}=\angle{SKR}$ So, $\angle{TPK}=\angle{ARS}$ So, $APSR$ is cyclic. $\angl... Sat Jul 29, 2017 12:29 pm Forum: International Mathematical Olympiad (IMO) Topic: IMO 2017 P4 Replies: 4 Views: 6975 ### IMO 2017 P4 Let R and S be dierent points on a circle Ω such that RS is not a diameter. Let be the tangent line to Ω at R. Point T is such that S is the midpoint of the line segment RT. Point J is chosen on the shorter arc RS of Ω so that the circumcircle Γ of triangle JST intersects at two distinct points. Let... Sat Jun 03, 2017 3:21 pm Forum: Geometry Topic: CGMO 2007/5 Replies: 1 Views: 7662 ### CGMO 2007/5 Point$D$lies inside triangle$ABC$such that $$\angle{DAC}=\angle{DCA}=30^o$$ and $$\angle{DBA}=60^o$$. Point$E$is the midpoint of segment$BC$. Point$F$lies on segment$AC$with$AF=2FC$. Prove that$DE$I$EF$. Fri Jun 02, 2017 3:04 pm Forum: International Mathematical Olympiad (IMO) Topic: IMO 2001 Problem1 Replies: 2 Views: 5432 ### Re: IMO 2001 Problem1 $$\angle{A}+\angle{COP} <90° \Rightarrow \angle{COP} <90°-\angle{COD}$$ $$\Rightarrow \angle{COP} <\angle{OCP} \Rightarrow CP<PO$$ is enough to prove.$\angle C -\angle B \geq 30°$as both$(\angle{C}-\angle{B})$and 30° are less than$90°, \begin{align*}&\sin(\angle{C}-\angle{B})\geq \sin30°\\ \... Fri Jun 02, 2017 3:01 pm Forum: International Mathematical Olympiad (IMO) Topic: IMO 2001 Problem1 Replies: 2 Views: 5432 ### IMO 2001 Problem1 Consider an acute angled triangle ABC. Let P be the foot of the altitude of triangle ABC issuing from the vertex A, and let O be the circumvented of triangle ABC. Assume that\angle{C}\geq\angle{B}+30°$$. Prove that$$\angle{A}+\angle{COP}<90°$$Fri Jun 02, 2017 1:59 pm Forum: Geometry Topic: IMO 2007 Problem 4 Replies: 4 Views: 8903 ### Re: IMO 2007 Problem 4 Let's draw RX perpendicular on PK which intersects thevcircumcirle of triangle ABC at M RX is parallel to BC So, BRMC is a trapizoid. So, BR=MC Again, BR=AR is known as CR in the angle bisector of angle ACB So, AR=MC and it makes ARCM cyclic trapizoid Then MR=b X... Thu Apr 27, 2017 4:31 pm Forum: National Math Olympiad (BdMO) Topic: BDMO 2017 National round Secondary 2 Replies: 2 Views: 1879 ### Re: BDMO 2017 National round Secondary 2 Here, AD is perpendicular on BC Hence, BE=CE We can show that$$\triangle$$BDE and$$\triangleCFE$are congruent. So,$FE=ED=OF=\frac{1}{3}OD=\frac{1}{3}OB$In the right angle triangle$OBE$,$OE^2+BE^2=OB^2$or,$4OF^2+5=9OF^2$or,$OF=1=DE$in rigth angle triangle$CED$,$CE^2+DE^2=CD^2\$ ...
Wed Apr 12, 2017 10:42 am
Forum: Social Lounge
Topic: BDMO Forum Mafia #2
Replies: 30
Views: 28496

### Re: BDMO Forum Mafia #2

I am also in
Sun Apr 02, 2017 2:31 pm
Forum: National Math Camp
Topic: The Gonit IshChool Project - Beta FE Class
Replies: 8
Views: 5831

### Re: The Gonit IshChool Project - Beta FE Class

I will be able to dedicate about 10 hours per week.
My experience is Adib vai's class.