AoPS - Problem

Source: VMO 2023 day 1 P4

Tags: geometry

a_507_bc

24.02.2023 18:47

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 .

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

$\text{Parts a. and b. are solved in separate images because they are not related to each other.}$