Summer is a great time to explore cool problems to keep your skills sharp!  Schedule a class today!

G
Topic
First Poster
Last Poster
Might be slightly generalizable
Rijul saini   5
N 39 minutes ago by guptaamitu1
Source: India IMOTC Day 3 Problem 1
Let $ABC$ be an acute angled triangle with orthocenter $H$ and $AB<AC$. Let $T(\ne B,C, H)$ be any other point on the arc $\stackrel{\LARGE\frown}{BHC}$ of the circumcircle of $BHC$ and let line $BT$ intersect line $AC$ at $E(\ne A)$ and let line $CT$ intersect line $AB$ at $F(\ne A)$. Let the circumcircles of $AEF$ and $ABC$ intersect again at $X$ ($\ne A$). Let the lines $XE,XF,XT$ intersect the circumcircle of $(ABC)$ again at $P,Q,R$ ($\ne X$). Prove that the lines $AR,BC,PQ$ concur.
5 replies
Rijul saini
Yesterday at 6:39 PM
guptaamitu1
39 minutes ago
Cute geometry
Rijul saini   5
N an hour ago by everythingpi3141592
Source: India IMOTC Practice Test 1 Problem 3
Let scalene $\triangle ABC$ have altitudes $BE, CF,$ circumcenter $O$ and orthocenter $H$. Let $R$ be a point on line $AO$. The points $P,Q$ are on lines $AB,AC$ respectively such that $RE \perp EP$ and $RF \perp FQ$. Prove that $PQ$ is perpendicular to $RH$.

Proposed by Rijul Saini
5 replies
Rijul saini
Yesterday at 6:51 PM
everythingpi3141592
an hour ago
No more topics!
sharky devil or no sharky devil
rafaello   1
N Jan 15, 2023 by Zena_B
Source: MODSMO 2019 September Advanced Contest P3
$ABC$ is a triangle with incentre $I$. The feet of the altitudes from $I$ to $BC, AC, AB$ are $D, E, F$ respectively, and the line through $D$ parallel to $AI$ intersects \(AB\) and \(AC\) at \(X\) and \(Y\) respectively. Prove that the circles with diameters $XF$ and $YE$ have a common point on the circumcircle of $ABC.$
1 reply
rafaello
Oct 27, 2021
Zena_B
Jan 15, 2023
sharky devil or no sharky devil
G H J
G H BBookmark kLocked kLocked NReply
Source: MODSMO 2019 September Advanced Contest P3
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
rafaello
1079 posts
#1
Y by
$ABC$ is a triangle with incentre $I$. The feet of the altitudes from $I$ to $BC, AC, AB$ are $D, E, F$ respectively, and the line through $D$ parallel to $AI$ intersects \(AB\) and \(AC\) at \(X\) and \(Y\) respectively. Prove that the circles with diameters $XF$ and $YE$ have a common point on the circumcircle of $ABC.$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Zena_B
3 posts
#2
Y by
Notice that $EF\perp AI, EF\perp DX $ Define $P=DX\cap EF $ and $ S=IP\cap (AIF) $ See that $P$ and $S$ swap under incircle inversion. Also, $(ABC)$ swaps with the nine-point circle of $DEF$. Hence, $S\in (ABC)$ (sharky-devil)
Now let $F'$ be the $F$-Antipode in $(AIF)$. Applying Pascal's Theorem on $SF'EFAI$ we get $X\in SF'$, since $EF'\perp EF, EF'\parallel AI\parallel XP$. Thus, we have $\angle XSF=90, S\in (XF) $. Similarly, $S\in (YE) $ so we are done.
Z K Y
N Quick Reply
G
H
=
a