Stay ahead of learning milestones! Enroll in a class over the summer!

G
Topic
First Poster
Last Poster
Parallel lines in incircle configuration
GeorgeRP   4
N 14 minutes ago by VicKmath7
Source: Bulgaria IMO TST 2025 P1
Let $I$ be the incenter of triangle $\triangle ABC$. Let $H_A$, $H_B$, and $H_C$ be the orthocenters of triangles $\triangle BCI$, $\triangle ACI$, and $\triangle ABI$, respectively. Prove that the lines through $H_A$, $H_B$, and $H_C$, parallel to $AI$, $BI$, and $CI$, respectively, are concurrent.
4 replies
GeorgeRP
May 14, 2025
VicKmath7
14 minutes ago
Geometry
shactal   1
N 19 minutes ago by ricarlos
Two intersecting circles $C_1$ and $C_2$ have a common tangent that meets $C_1$ in $P$ and $C_2$ in $Q$. The two circles intersect at $M$ and $N$ where $N$ is closer to $PQ$ than $M$ . Line $PN$ meets circle $C_2$ a second time in $R$. Prove that $MQ$ bisects angle $\widehat{PMR}$.
1 reply
shactal
Today at 3:04 PM
ricarlos
19 minutes ago
No more topics!
Symmedian line
April   93
N May 18, 2025 by aidenkim119
Source: All Russian Olympiad - Problem 9.2, 10.2
Let be given a triangle $ ABC$ and its internal angle bisector $ BD$ $ (D\in BC)$. The line $ BD$ intersects the circumcircle $ \Omega$ of triangle $ ABC$ at $ B$ and $ E$. Circle $ \omega$ with diameter $ DE$ cuts $ \Omega$ again at $ F$. Prove that $ BF$ is the symmedian line of triangle $ ABC$.
93 replies
April
May 10, 2009
aidenkim119
May 18, 2025
Symmedian line
G H J
Source: All Russian Olympiad - Problem 9.2, 10.2
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
AshAuktober
1009 posts
#82
Y by
Perform a force-overlaid inversion (complete with the reflection about the angle bisector) about $B$. Let the resultant image of any object $X$ be $X'$. Note that $F', E'$ must lie on line $A'C'$, and also that $90^\circ = \angle DFE = \angle BFE - \angle BFD - \angle BE'F' - \angle BD'F = \angle E'F'D'$; and thus $D'F' \perp A'C'$; however, as $D'$ is the midpoint of arc $A'C'$, $F'$ is nothing but the midpoint of $A'C'$. Thus $BF$ is the reflection of the median about the angle bisector, i.e. the symmedian. $\square$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Aiden-1089
297 posts
#83
Y by
Let $E'$ be the antipode of $E$ in $(ABC)$. Since $\angle EBE' = 90 ^{\circ}$, we have $(A,C;D,BE' \cap AC)=-1$.
Also, $F$ lies on $DE'$, so $(A,C;F,B) \stackrel{E'}{=} (A,C;D,BE' \cap AC) = -1$, which implies that $BF$ is a symmedian of $\Delta ABC$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
MagicalToaster53
159 posts
#84
Y by
Let $\omega$ denote the circle with diameter $\overline{DE}$. Suppose $\omega$ meets $\overline{BC}$ at $M$ other than $D$. Then observe that $M$ is the midpoint of $\overline{BC}$ as $E$ bisects minor arc $\widehat{BC}$ and $\measuredangle DME = 90^{\circ}$.

Now consider the inversion at $A$ with radius $\sqrt{bc}$ followed by a reflection about $AD$, the angle bisector of $\angle BAC$. Then $D$ maps to $E$ under this inversion, and it evidently follows that the image of $M$ is $F$. Hence $AM$ and $AF$ are isogonal, as desired. $\blacksquare$

Remark
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Ywgh1
139 posts
#85
Y by
Russia 2009

Consider a $\sqrt{bc}$ inversion, we have that $D,E$ will be swapped, hence circle with diameter $DE$ fixed, $F$ will be sent to foot to $BC$ which is the mid of $BC$ hence $AF$ is Symedian
This post has been edited 2 times. Last edited by Ywgh1, Aug 24, 2024, 5:30 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Mathandski
772 posts
#86
Y by
$        $
Attachments:
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Anshu_Singh_Anahu
69 posts
#87
Y by
In the Following diagram

Define A function F(Z) = ±(BZ/CZ) , Now clearly BE = EC so F(E) = -1 aswell H is midpt of BC so F(H)= 1 also by angel bisector theorem F(D)= F(A) , by two circle intersecting at BC lemma F(H) × F(D) = F(E) × F(G) so ,F(G) = F(D) = F(A) , also F(G) × F(A) = F(M) , hence F(M) = (F(A))^2 and we are dn with AM as the symmedian cute~ :)

I don't have cared about the direct length but no problem ig
Attachments:
This post has been edited 1 time. Last edited by Anshu_Singh_Anahu, Dec 11, 2024, 8:37 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Anshu_Singh_Anahu
69 posts
#88
Y by
Above sol is too short ig it may be wrong check someone please :blush:
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
lelouchvigeo
183 posts
#89
Y by
Sol
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Ilikeminecraft
658 posts
#90
Y by
Note that the midpoint of $BC, M,$ passes through $(DE).$ Take $\sqrt{bc}$ inversion, and we get that $\omega$ maps to itself. Thus, $K \leftrightarrow M,$ so $AK$ and $AM$ are isogonal.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
TestX01
341 posts
#91
Y by
ma 29 wrote:
http://i524.photobucket.com/albums/cc322/khanhto-photo/untitled-108.jpg
borrowing points here. Redefine $F$ as on symmedian. RTP $F,D,P$ collinear. DDIT on $A'FDE$ means that $EF\cap AC$ and $DF\cap EA'$ are isogonal from existing isogonal pairs $BD,BE$ and $BF, BA'$. RTP now $EF, BP,AC$ concur clearly. This is as $-1=(B,F;A,C)$ is projected to $(D,EF\cap AC,A,C)$. But well-known by apollonian that $BP\cap AC$ is harmonic conj of $D$, so coincide with $EF\cap AC$, yay.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Avron
37 posts
#92
Y by
Let M, N be midpoints of BC and arc BAC, then $N,M,E$ are collinear and $ADMN$ is cyclic. Note that $\angle EFD=90=\angle EFN$ so $F,D,N$ are also collinear and we get:
\[
\angle FAD=\angle FAE = \angle FNE = \angle DNM=\angle DAM
\]and we're done since $AD$ is the angle bisector
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
bjump
1032 posts
#93
Y by
Today is the day I finally understand how force overlaid inversion can actually be useful.
Perform a Force-Overlaid inversion with respect to $A$. Note that $D \leftrightarrow E$, so the diameter of $(DE)$ is fixed, therefore $F$ is sent to the midpoint of $BC$ finishing.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
BS2012
1047 posts
#94
Y by
This problem is 2012 AIME II P15 $-$ 2024 AIME I P10 explanation

Also slightly off topic but force overlay inversions are also useful to prove that the config in 2024 aime p10 is in fact a symmedian
This post has been edited 1 time. Last edited by BS2012, Apr 29, 2025, 1:05 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
zuat.e
66 posts
#95
Y by
Consider the $\sqrt{bc}$ inversion around $A$, which sends $D$ to $E$ and the intersection point of the symmedian with $(ABC)$, $S$, to $M$, the midpoint of $BC$. Now ,as $\measuredangle DME=90º$ and $DMES$ is cyclic, it follows, that $\measuredangle ESD=90º$, hence $S\equiv F$, as desired.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
aidenkim119
34 posts
#96
Y by
let the midpoint of arc $ABC$ be $M$. Then obviously $M, D, F$ is co line.
Therefore $BA:BC = AD:DC = AF:FC$ by angle bisector, so square $BCFA$ is harmonic, so $BF$ is symmedian!
Z K Y
N Quick Reply
G
H
=
a