VMO 2010-Circle geometry

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VMO 2010-Circle geometry

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#1 h Dec 1, 2010, 7:43 AM • 2 Y

Y by Adventure10, Mango247

In plane,let a circle $(O)$ and two fixed points $B,C$ lies in $(O)$
such that $BC$ not is the diameter.Consider a point $A$ varies in
$(O)$ such that $A\neq B,C$ and $AB\neq AC$ .Call $D$ and $E$
respective is intersect of $BC$ and internal and external bisector
of $\widehat{BAC}$ , $I$ is midpoint of $DE$ .The line that pass through
orthocenter of $\triangle ABC$

and perpendicular with $AI$ intersects $AD,AE$ respective at $M,N$ .

1/Prove that $MN$ pass through a fixed point

2/Determint the place of $A$ such that $S_{AMN}$ has maxium value

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#2 h Dec 1, 2010, 11:59 AM • 2 Y

Y by Adventure10, Mango247

(1) Let $\angle BAC=\alpha$ (asume that $\alpha < \frac{\pi}{2}$ ) and $BC=a$ , construct point $K$ , such that : $OK \perp BC$ and $OK=a|\cot \alpha|$ ，then we would prove that $MN$ always pass through point $K$ .It is famous that in a triangle $ABC$

with orthocenter $H$ , $AH=a|\cot \alpha|$ , so we only need to prove that $OKHA$ is a parallelogram . Obviously , $OK//AH$ ; $KH \perp AI$ ,to prove $OA// KH$ we only need to prove $OA \perp IA$ since $AD$ and $AE$ are angle bisectors , $AD \perp AE$ , we

only need to prove $\angle OAD=\angle IAE$ , with $I$ ,the midpoint and easy angle chasing ,we can prove it . Hence $MN$ pass

through $K$ ,the fixed point .

(2) To solve this one , we may need the following lemma : in triangle $ABC$ the angle bisector of $\angle BAC$ is also the

angle (internal or external) bisector of $\angle OAH$ where $O$ and $H$ are the ,circumcentre and the orthocenter of the triangle .

Easy angle chasing can prove this lemma . So by the lemma given above , we get that $HM=AH=NH= a|\cot \alpha|$ ,so $S_{AMN}$

can be written as $AH^2 \sin\theta$ where $\theta=\angle AHN$ ,when the area get its largest value , $\theta = \frac{\pi}{2}$ ,

and $\theta=\pi-2\beta-\alpha$ , so we get $\beta=\frac{\pi}{4}-\frac{\alpha}{2}$ . The maximum value of triangle $AMN$ is $a^2 \cot^2\alpha$ .The area gets the maximum value if and only if $\angle BAC$ is acute and $AO // BC$ .( $\angle ABC=\frac{\pi}{4}-\frac{\angle BAC } {2},\angle ACB=\frac{3 \pi}{4}-\frac{\angle BAC}{2}$ or $\angle ACB=\frac{\pi}{4}-\frac{\angle BAC }{2},\angle ABC=\frac{3 \pi} {4}-\frac{\angle BAC}{2}$ )

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#3 h Jun 25, 2016, 12:45 AM • 2 Y

Y by Adventure10, Mango247

there is little excess space

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#4 h Jun 1, 2019, 5:12 PM • 1 Y

Y by Adventure10

For the First part,
Let $AA'||BC$ and $A' \in \odot (ABC)$ and Let $G$ be the midpoint of $BC$
$-1=( B, C; G, \infty_{BC}) \overset{A'}{=} ( B, C ; GA' \cap \odot (ABC), A)$ $\implies$ $GA' \cap \odot (ABC)$ $\in$ $A-$ symmedian. Let $O$ be the circumcenter WRT $\Delta ABC$ . Let $\overset{\longrightarrow}{GO} \cap \odot (ABC)=M_{BC}$ . Let $M_{BC}D$ $\cap$ $\odot (ABC)$ $=$ $L$ . If $AD$ $\cap$ $\odot (ABC)$ $=$ $M_A$ $\implies$ $M_AGDL$ is cyclic $\implies$ $LD$ bisects $\angle ALG$ $\implies$ $GA'$ $\cap$ $\odot (ABC)$ $=$ $L$
$-1=(M_{BC}, M_A ; A , A') \overset{A}{=} ( E , D ; AA ~ \cap ~ BC , \infty_{BC}) \implies AA ~ \cap ~ BC =I$ Also, Since, $D$ is the orthocenter WRT $\Delta EM_{BC}M_A$ $\implies$ $E - L - M_A$
$\implies -1 = ( AG ~ \cap ~ \odot (ABC), L ; M_{BC} , M_A) \overset{L}{=} ( \infty_{BC}, LL ~ \cap ~ BC ; D ,E) \implies LL ~ \cap ~ AA ~\cap ~BC= I$ Let $H$ be the orthocenter WRT $\Delta ABC$ , Let $H'$ be the reflection of $H$ over $BC$ . Let $H'O$ $\cap$ $BC$ $=$ $X'$
$\angle HX'K=\angle H'CA=90^{\circ}+\angle C - \angle B=\angle IAH \implies AIH'X' \text{ is cyclic}$ $H$ is the orthocenter WRT $\Delta AIX'$ $\implies$ $HX' \cap M_AM_{BC}=F$ , hence, $F$ is the reflection of $O$ over $BC$ which is indeed fixed

This post has been edited 7 times. Last edited by AlastorMoody, Jun 1, 2019, 7:27 PM

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