On a isosceles triangle $\triangle ABC$ with $AB=BC$ let $K,M$ be the midpoints of $AB,AC$ respectivily. Let $(CKB)$ intersect $BM$ at $N \ne M$, the line through $N$ parallel to $AC$ intersects $(ABC)$ at $A_1,C_1$. Show that $\triangle A_1BC_1$ is equilateral.
Problem
Source: Bolivia Ibero TST 2021 Day 2 P2
Tags: geometry, Bolivia, TST, isosceles
DottedCaculator
02.09.2022 23:41
If $M$ is the midpoint of $BC$ then doesn't that mean $BM$ is $BC$ so $(CKB)$ intersects $BC$ at $B$ and $C$?
MathLuis
02.09.2022 23:42
DottedCaculator wrote: If $M$ is the midpoint of $BC$ then doesn't that mean $BM$ is $BC$ so $(CKB)$ intersects $BC$ at $B$ and $C$? it was AC, not BC :facepalm:
Captainscrubz
13.09.2024 14:34
No solution?
balllightning37
13.09.2024 18:00
Cute! Notice that since $\angle KBN=\angle NBC$, $NK=NC$. Also, by symmetry, $NC=NA$, so $N$ is in fact the circumcenter of $KAC$. Thus, $N$ lies on the perpendicular bisector of $AK$. Let $J$ be the midpoint of $AK$ and $O$ be the circumcenter of $ABC$. Since triangles $BJN$ and $BKO$ are similar, $\frac{BN}{BO}=\frac{BJ}{BK}=\frac32$. Thus, it is obvious by a simple phantom point argument that $A_1BC_1$ must be equilateral.
falantrng
13.09.2024 21:55
See here