Given is a triangle $ABC$ and let $D$ be the midpoint the major arc $BAC$ of its circumcircle. Let $M , N$ be the midpoints of $AB , AC$ and $J , E , F$ are the touchpoints of the incircle $(I)$ of $\triangle ABC$ with $BC, CA, AB$. The line $MN$ intersects $JE , JF$ at $K , H$ respectively; $IJ$ intersects the circle $(BIC)$ at $G$ and $DG$ intersects $(BIC)$ at $T$. a) Prove that $JA$ passes through the midpoint of $HK$ and is perpendicular to $IT$. b) Let $R, S$ respectively be the perpendicular projection of $D$ on $AB, AC$. Take the points $P, Q$ on $IF , IE$ respectively such that $KP$ and $HQ$ are both perpendicular to $MN$. Prove that the three lines $MP , NQ$ and $RS$ are concurrent .
Problem
Source: VMO 2023 day 1 P4
Tags: geometry
TheMathBob
24.02.2023 20:37
a_507_bc wrote: $JA$ intersects the circle $(BIC)$ at $G$ Is it correct? I think it must be $JA$ intersects the circle $(ABC)$ at $G$.
a_507_bc
24.02.2023 20:59
It is IJ, sorry for the typo
TheMathBob
24.02.2023 21:00
First part of a) $JA$ passes through the midpoint of $HK$ is just Iran Lemma. We know that $B, I, K$ are collinear and $\angle BKA = 90^\circ$, so $\angle AKH = 90^\circ - \angle HKB = 90^\circ - \frac{1}{2}\angle B = \angle FJB = \angle JHK$, so $AK$ and $HJ$ are parallel. And because $MN$ bisects $AJ$, then $AJ$ bisect $HK$. ($AHJK$ is a parallelogram).
VicKmath7
24.02.2023 22:59
Let $L$ be the midpoint of the minor arc $BC$ and let $IL \cap (BIC)=T'$ be the antipode of $T$ in $(BIC)$. Hence, we want $TT' \parallel AJ$. Note that there is a homothety at $I$ sending $\triangle AHK$ to $\triangle T'BC$, and since $AJ$ bisects $HK$, we need to prove that $TT'$ bisects $BC$. Indeed, $TBGC$ is harmonic (since $DB$ and $DC$ are tangents to $(BIC)$) and $T'G \parallel BC$, so projecting $(T, G, B, C)=-1$ from $T'$ onto $BC$ finishes.
Easy angle chasing yields $DR=DS$, and in particular, $RS \parallel AI$. By Simson line, $RS$ bisects $BC$. Note that $KPFM$ is cyclic, so $\angle KMP= \angle KFP = \angle KFI= \angle KHI$, so $MP \parallel CI$. Similarly, $NQ \parallel BI$, so homothety sending $\triangle ABC$ to the medial triangle finishes.
Paramizo_Dicrominique
25.02.2023 08:21
this geo easy for a VMO's
Noob_at_math_69_level
25.02.2023 10:22
Some good looking properties (May be c)) Prove that: $PQ \perp IZ$ with $Z$ midpoint $BC$ Prove that: $PQ$ bisects $IZ$. (Collected)
799786
25.02.2023 18:40
Way too many points to look at Part a is easy For part b, the lines concurrent at Spieker Center
799786
25.02.2023 18:44
Well, in recent years VMO geometry problems aren't good. Part a and b are independent of each other, and the committee just use old ideas to create these unsatisfying geo
Paramizo_Dicrominique
25.02.2023 18:47
wardtnt1234 wrote: Well, in recent years VMO geometry problems aren't good. Part a and b are independent of each other, and the committee just use old ideas to create these unsatisfying geo Yes, you are right. We just need 1 high quality and good problem in the test more than 2 independent problems but not very nice and complicated. There are still more issues about this but I will just say about the Geometry...
trinhquockhanh
10.07.2023 15:49
$\text{Parts a. and b. are solved in separate images because they are not related to each other.}$