## 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

$\text{Solution to Problem 37:}$ 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.
Mon Feb 27, 2017 4:13 am
Forum: Geometry
Topic: Geometry Marathon : Season 3
Replies: 146
Views: 62902