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Source: Final Korean Mathematical Olympiad 2014, #4

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#1 h Mar 23, 2014, 5:48 AM • 1 Y

Y by Adventure10

Let $ABC$ be a isosceles triangle with $AC=BC$ . Let $D$ a point on a line $BA$ such that $A$ lies between $B, D$ . Let $O_1$ be the circumcircle of triangle $DAC$ . $O_1$ meets $BC$ at point $E$ . Let $F$ be the point on $BC$ such that $FD$ is tangent to circle $O_1$ , and let $O_2$ be the circumcircle of $DBF$ . Two circles $O_1 , O_2$ meet at point $G ( \ne D)$ . Let $O$ be the circumcenter of triangle $BEG$ . Prove that the line $FG$ is tangent to circle $O$ if and only if $DG \bot FO$ .

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#2 h Mar 25, 2014, 12:09 AM • 6 Y

Y by frill, eziz, rkm0959, mijail, Adventure10, Mango247

Obviously $DBE\sim CAB$ so $DE=DB$ , therefore $DO$ is the perpendicular bisector of $BE$ , say that it bisects $BE$ at $M$ . Since $\angle EBG=\angle GDF=\angle DEG$ , therefore $DE$ is tangent to $(O)\Rightarrow OG^2=OE^2=OM*OD\Rightarrow$ $OMG\sim OGD\Rightarrow \angle OGM=\angle ODG$

Since $\angle OMF=90$ , therefore $FG$ is tangent to $(O)\iff \angle OGF=90\iff O,M,G,F$ are concyclic $\iff \angle OFM=\angle MGO=\angle ODG\iff DF\perp OF$ . Done

Btw these problems are all about observations along with having deductions and equivalences in your mind. Of course what I've written here isn't my very first thought, it's been finalized after getting familiar with the diagram.

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#3 h Mar 25, 2017, 2:46 AM • 3 Y

Y by duongminh10, Adventure10, Mango247

$\measuredangle EBG = \measuredangle GDF = \measuredangle DEG$ and $\measuredangle GEC = \measuredangle DEC - \measuredangle DEG = \measuredangle CBA - \measuredangle EBG = \measuredangle GBD$ . Furthermore, $DB$ and $DE$ are tangents to $\odot (EGB)$ .

Let $X \equiv DG \cap \odot (EGB)$ and let $Y \equiv XB \cap GE$ .

If $FG$ is tangent to $\odot (EGB)$ . Pascal on $GGXBEE$ shows that $Y$ lies on $DF$ . Pascal on $XXBBEG$ shows that $XX \cap BE$ lies on $DY$ . Thus $FX$ is tangent to $\odot (EGB)$ . Furthermore $FG=FX \Longrightarrow GX \perp FO$ .

Similarly if $GX \perp FO$ . Pascal on $GGXBEE$ shows that $GG \cap BE$ lies on $DY$ . Pascal on $XXBBEG$ shows that $XX \cap BE$ lies on $DY$ . Thus $XX \cap GG \equiv F$ . Thus $FG$ is tangent to $(EGB)$ .

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#4 h Oct 11, 2020, 1:10 PM • 3 Y

Y by Mango247, Mango247, Mango247

CLAIM. $DE,DB$ are tangent to $(EGB)$ .
Proof.
This is just angle chasing. From spiral sim. lemma, $\triangle GDA\sim \triangle GFB$ .
$\angle CDG=\angle BEG$ and $\angle DCG=\angle GAB=\angle GDB+\angle DGA=\angle GFB+\angle FGB=\angle GBE$ Hence $\triangle GDC\sim\triangle GEB$ .
Therefore $\angle DEG=\angle DCG=\angle GBE$ . Hence $DE$ is tangent to $(EGB)$ . Moreover, $\angle DEB=\angle DEG+\angle GBE=\angle GAB+\angle CAG=\angle CAB=\angle EBD$ . Therefore $DE=DB$ and $DB$ is tangent to the circumcircle of $\triangle EGB$ as well. $\blacksquare$
Therefore, $EB$ is the polar of $D$ with respect to $(EGB)$ . By La hire's theorem, $D$ lies on the polar of $F$ with respect to the same circle. Hence $G$ lies on the polar of $F$ if and only if $DG\perp FO$ .

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