## Search found 138 matches

- Mon Mar 27, 2017 7:05 pm
- Forum: Social Lounge
- Topic: BDMO Forum Mafia #1
- Replies:
**52** - Views:
**25346**

### Re: BDMO Forum Mafia #1

Yo, count me in.

- Mon Feb 27, 2017 5:33 pm
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**146** - Views:
**62902**

### Re: Geometry Marathon : Season 3

$\text{Solution to Problem 37:}$ Let $KT$ meet $\omega$ at $L'$ and let $XL'$ meet $\omega$ at $S'$. Since by angle chasing we get $TS'||KX$ and also $TS||KX$, we get that $S'$ coincides with $S$. So, $L'$ coincides with $L$. Also $KP||TR$. Now, $\angle TKP=\angle RTK=\angle RSL=\angle LRQ \Rightarr...

- Mon Feb 27, 2017 3:49 pm
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**146** - Views:
**62902**

### Re: Geometry Marathon : Season 3

$\text{Solution to Problem 35:}$ Let $D$ be the feet of perpendicular from $I$ on $BC$. And let $R$ be the circumradius of $\bigtriangleup ABC$. Obviously, $P,Q$ lie on $AC,AB$ respetively such that $BQ=BC=CP$.Also $IP=IB$ and $IQ=IC$. We will use the perpendicular lemma and show that $IP^2-IQ^2=OP^...

- Mon Feb 27, 2017 4:18 am
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**146** - Views:
**62902**

### Re: Geometry Marathon : Season 3

$\text{Problem 35:}$

Let $ABC$ be a triangle with incenter $I$ . Let $P$ and $Q$ denote the reflections of $B$ and $C$ across $CI$ and $BI$ respectively. Show that $PQ\perp OI$, where $O$ is the circumcenter of ABC.

Let $ABC$ be a triangle with incenter $I$ . Let $P$ and $Q$ denote the reflections of $B$ and $C$ across $CI$ and $BI$ respectively. Show that $PQ\perp OI$, where $O$ is the circumcenter of ABC.

- Mon Feb 27, 2017 4:13 am
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**146** - Views:
**62902**

### Re: Geometry Marathon : Season 3

$\text{Solution to Problem 34:}$ Let, $\bigtriangleup DEF$ be the intouch triangle. $M$ be the midpoint of $AI$ and $D_1$ be the antipode of $D$ with respect to the incircle. The Feuerbach point be $F_1$. $P,Q$ be the midpoints of the two arcs of the incircle created by the points $E,F$. $\text{Lemm...

- Sat Feb 25, 2017 12:53 am
- Forum: Social Lounge
- Topic: সেঞ্চুরি ! সেঞ্চুরি !! সেঞ্চুরি !!!
- Replies:
**10** - Views:
**7384**

### Re: সেঞ্চুরি ! সেঞ্চুরি !! সেঞ্চুরি !!!

Congo for 1000 posts bhai

- Wed Feb 22, 2017 7:54 pm
- Forum: Geometry
- Topic: For Raiyan
- Replies:
**2** - Views:
**1666**

### Re: For Raiyan

We use complex numbers. Set the incircle as the unit circle. Let $D$ be the touchpoint of $\omega$ with $BC$ and $Q$ be a point on $\omega$ such that $QEDF$ is harmonic.Now, since $(P,D;B,C)=-1$, we get that $PQ$ is tangent to $\omega$.Now, since $XY||BC$ and $XY$ is tangent to $\omega$, let at $Z$,...

- Tue Feb 21, 2017 11:09 pm
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**146** - Views:
**62902**

### Re: Geometry Marathon : Season 3

$\text{Problem 33:}$ Let $\bigtriangleup ABC$ be a scalene triangle with circumcenter $O$ and incenter $I$. Let $H, K, L$ be the feet of the altitudes of $\bigtriangleup ABC$ from the vertices $A, B, C$ respectively. Denote by $A_0, B_0, C_0$ the midpoints of these altitudes $AH, BK, CL$ respectivel...

- Tue Feb 21, 2017 11:01 pm
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**146** - Views:
**62902**

### Re: Geometry Marathon : Season 3

$\text{Solution of Problem 32:}$ Let, $D,E,F$ be the point of tangency of the incircle with $\bigtriangleup ABC$ at $BC,CA,AB$ respectively. $\bigtriangleup M_a,M_b,M_c$ be the midpoints of the arcs $BC,CA,AB$ opposite to $A,B,C$ respectively. Let the perpendicular to $AI$ at $I$ meet $AB,AC,BC$ at ...

- Tue Feb 21, 2017 2:56 am
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies:
**146** - Views:
**62902**

### Re: Geometry Marathon : Season 3

$\text{Problem 31:}$

Let $\bigtriangleup ABC$ be a triangle with incenter $I$. Prove that the $\text{Euler lines}$ of $\bigtriangleup AIB, \bigtriangleup BIC, \bigtriangleup CIA$ and $\bigtriangleup ABC$ are concurrent.

Let $\bigtriangleup ABC$ be a triangle with incenter $I$. Prove that the $\text{Euler lines}$ of $\bigtriangleup AIB, \bigtriangleup BIC, \bigtriangleup CIA$ and $\bigtriangleup ABC$ are concurrent.