Correction: $G$ lies on $T_1$, not $T_2$.

Solution: $I$ be a point on line $AF$ such that it is the reflection of $D$ under $AB$. We get $HEBI$ is cyclic. So, $AH \times AI=AE \times AB=AF^2$ from which we can get the result.

Correction: $G$ lies on $T_1$, not $T_2$.

Solution: $I$ be a point on line $AF$ such that it is the reflection of $D$ under $AB$. We get $HEBI$ is cyclic. So, $AH \times AI=AE \times AB=AF^2$ from which we can get the result.

Solution: $I$ be a point on line $AF$ such that it is the reflection of $D$ under $AB$. We get $HEBI$ is cyclic. So, $AH \times AI=AE \times AB=AF^2$ from which we can get the result.

- Sun Oct 01, 2017 2:59 pm
- Forum: Geometry
- Topic: CGMO 2002/4
- Replies:
**2** - Views:
**61**

Problem 43: An equilateral pentagon $AMNPQ$ is inscribed in triangle $ABC$ such that $M\in\overline{AB}$, $Q\in\overline{AC}$, and $N,P\in\overline{BC}$. Let $S$ be the intersection of $ \overleftrightarrow{MN}$ and $ \overleftrightarrow{PQ}$. Denote by $\ell$ the angle bisector of $\angle MSQ$. Pr...

- Tue Aug 08, 2017 12:58 am
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**92** - Views:
**3894**

Solution to Problem 42: $a)$ $P$ is just the reflection of $F$ under $AC$ and $Q$ is the reflection of $E$ under $AB$. $AFDCP$ and $AEDBQ$ cyclic. The rest is trivial by taking homothety with centre $A$ and ratio 2. $b)$ Let $H_1$ be the reflection of $H$ under $A$. We can find that the perpendicul...

- Tue Aug 08, 2017 12:37 am
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**92** - Views:
**3894**

$\text{Problem 40:}$ Let the Nagel point of triangle $ABC$ be $N$. We draw lines from $B$ and $C$ to $N$ so that these lines intersect sides $AC$ and $AB$ in $D$ and $E$ respectively. $M$ and $T$ are midpoints of segments $BE$ and $CD$ respectively. $P$ is the second intersection point of circumcirc...

- Sat Apr 29, 2017 11:01 pm
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**92** - Views:
**3894**

$\text{Solution to Problem 39: }$ We do angle chase and get $\angle LCJ=\angle IBC$ and $\angle LBJ=\angle ICB$. And the we use $\text{trig ceva}$ w.r.t. $\bigtriangleup BJC$ and the points $I$ and $L$ and we get that $\frac{sin \angle CJL}{sin \angle BJL}=\frac {sin \angle CJI} {sin \angle BJI} \Ri...

- Sat Apr 29, 2017 10:45 pm
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**92** - Views:
**3894**

dshasan wrote:Define $H_BH_C$.

$H_B$ is the feet of perpendicular from $B$ to $AC$ nd figure $H_C$ urself.

Count me in

- Tue Apr 11, 2017 6:47 pm
- Forum: Social Lounge
- Topic: BDMO Forum Mafia #2
- Replies:
**27** - Views:
**1353**

I'm also suspecting epshita seeing her behaviour in tg nd the forum.

- Mon Apr 03, 2017 7:43 pm
- Forum: Social Lounge
- Topic: BDMO Forum Mafia #1
- Replies:
**50** - Views:
**2309**

Well, its online mafia and maybe the mafia wanted to kill the strong players first(rahul bhai and itti) so that he won't get into trouble. There's nothing else i can say :/. Btw, $1)$Can't say nothing more than what nahin and epshita said about dshasan being a mafia. $2)$Maybe, Nahin's not a mafia. ...

- Mon Apr 03, 2017 1:26 am
- Forum: Social Lounge
- Topic: BDMO Forum Mafia #1
- Replies:
**50** - Views:
**2309**

Same as Nahin, my tutorial xm will also start in 5 April. Usually will be able to give 10-12 hrs a week but not much during the xms. Nd ik basics of FE. Experience? almost 0 maybe.

- Sat Apr 01, 2017 3:26 pm
- Forum: National Math Camp
- Topic: The Gonit IshChool Project - Beta FE Class
- Replies:
**8** - Views:
**641**