A point is on a fixed line

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A point is on a fixed line

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Source: Vietnam Mathematical Olympiad 2016, Day 1, Problem 3

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#1 h Jan 6, 2016, 1:17 PM • 3 Y

Y by Davi-8191, Adventure10, Mango247

Let $ABC$ be an acute triange with $B,C$ fixed. Let $D$ be the midpoint of $BC$ and $E,F$ be the foot of the perpendiculars from $D$ to $AB,AC$ , respectively.
a) Let $O$ be the circumcenter of triangle $ABC$ and $M=EF\cap AO, N=EF\cap BC$ . Prove that the circumcircle of triangle $AMN$ passes through a fixed point;
b) Assume that tangents of the circumcircle of triangle $AEF$ at $E,F$ are intersecting at $T$ . Prove that $T$ is on a fixed line.

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#2 h Jan 6, 2016, 6:21 PM • 2 Y

Y by Adventure10, Mango247

My solution.

Let $O$ be the circumcenter of $\odot(ABC)$ , $P\equiv AO\cap DE$ , $X\equiv AO\cap BC$ , $L$ be the midpoint of $EF$ .

(a). I'll claim that $D \in \odot(AMN)$ .

Note that $\angle EDB=90^\circ-\angle ABC=\angle OAC \implies P,D,C,A$ are concyclic $\implies \angle DEF=\angle DAF=\angle DPC \implies EF\parallel CP$ . Thus $\angle MND=\angle PCD=\angle PAD$ as desired.

(b). I'll cliam that $T,O,D$ are collinear.

At the beginning, see CGMO P1 here. CGMO P1 give us $DL\parallel AO \implies \angle LDE=\angle APE=90^\circ - \angle PAE=\angle ACB=\angle ODF \implies OD$ is symmedian wrt $\odot(AFDE)$ . The result then follows.

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This post has been edited 3 times. Last edited by Complex2Liu, Jan 6, 2016, 6:31 PM

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#3 h Jan 7, 2016, 8:00 AM • 2 Y

Y by Adventure10, Mango247

My solution for part b)

Let $DO\cap \odot (AEF)=X$ , then $(X,D;E,F)=A(\infty,D;B,C)=-1\implies XEDF$ is harmonic quadrilateral.

Hence $T,D,X$ collinear so $T$ lies on perpendicular bisector of $BC$ (which is fixed).

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#4 h Aug 15, 2016, 7:46 AM • 2 Y

Y by Adventure10, Mango247

My solution for a)

This post has been edited 2 times. Last edited by quangminhltv99, Aug 15, 2016, 2:41 PM

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#5 h Apr 26, 2019, 2:22 PM • 2 Y

Y by Adventure10, Mango247

Here is my solution for this problem
a) We have: ( $DA$ ; $DB$ ) $\equiv$ ( $DA$ ; $DE$ ) + ( $DE$ ; $DB$ ) $\equiv$ ( $FA$ ; $FE$ ) + $\dfrac{\pi}{2}$ $-$ ( $BC$ ; $BA$ ) $\equiv$ ( $FA$ ; $FE$ ) + ( $AO$ ; $AC$ ) $\equiv$ ( $NA$ ; $NM$ ) (mod $\pi$ )
So: $D$ $\in$ ( $AMN$ )
b) Let $S$ $\equiv$ $DO$ $\cap$ ( $AEF$ ) ( $S$ $\not \equiv$ $D$ ) then $AS$ $\parallel$ $BC$
We have: $\dfrac{SE}{SF}$ = $\dfrac{AD . \sin \widehat{SFE}}{AD . \sin \widehat{SEF}}$ = $\dfrac{\sin \widehat{SAE}}{\sin \widehat{FAS}}$ = $\dfrac{\sin \widehat{ABC}}{\sin \widehat{ACB}}$ = $\dfrac{DE}{BD}$ . $\dfrac{DC}{DF}$ = $\dfrac{DE}{DF}$
So: $DESF$ is harmonic quadrilateral
Hence: tangents at $E$ , $F$ of ( $AEF$ ) and $DS$ concurrent at $T$ or $T$ lies on perpendicular bisector of $BC$

This post has been edited 1 time. Last edited by khanhnx, Apr 26, 2019, 2:23 PM

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