It's easy to see that $\angle (CBD) = \angle(DAC) $ so, by the law of sines in $ \triangle{ANM} $and in$ \triangle{BKM} $ we get that $MN=MK$.We show that $MK$ is perpendicular to $OC$. Let $\angle{BAC}$ be x.Then $\angle{MKC} $ is x and $\angle{OCB} $is 90- x, so OC is perpendicular to MK.The conclusion follows from the fact that the area of a quadrilateral is the product between the diagonals times the sin of the angle between them, over 2.

I hope someone will write official solution...

IstekOlympiadTeam wrote: I hope someone will write official solution... me too

What about the cutoff.

12-17 bronze, 18 -22 silver, 23-42 gold .

Popescu wrote: It's easy to see that $\angle (CBD) = \angle(DAC) $ so, by the law of sines in $ \triangle{ANM} $and in$ \triangle{BKM} $ we get that $MN=MK$.We show that $MK$ is perpendicular to $OC$. Let $\angle{BAC}$ be x.Then $\angle{MKC} $ is x and $\angle{OCB} $is 90- x, so OC is perpendicular to MK.The conclusion follows from the fact that the area of a quadrilateral is the product between the diagonals times the sin of the angle between them, over 2. I think your solution is right, but I don't understand why the conclusion follows from the quadrilateral formula. (Edit: Sorry, I misread your meaning, you're right. Very nice solution. ) When I encounter this problem, I conjecture that if $OD$ is perpendicular to $NM$, since if the above statement is right, the area is very easy to express. Now I will claim that $DO \perp NM$ indeed. Note that $\angle DNM=\angle DBA=\angle DCA$ and $\angle NDO=90^\circ-\angle DCA$, so $DO\perp NM$. Analogously $KM \perp OC$. Since $\angle DAC=\angle CBD=\angle CAK$, it amounts to $MK=MN$. I will use $\left[\quad\right]$ to denote the area. 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Complex2Liu wrote: Popescu wrote: It's easy to see that $\angle (CBD) = \angle(DAC) $ so, by the law of sines in $ \triangle{ANM} $and in$ \triangle{BKM} $ we get that $MN=MK$.We show that $MK$ is perpendicular to $OC$. Let $\angle{BAC}$ be x.Then $\angle{MKC} $ is x and $\angle{OCB} $is 90- x, so OC is perpendicular to MK.The conclusion follows from the fact that the area of a quadrilateral is the product between the diagonals times the sin of the angle between them, over 2. I think your solution is right, but I don't understand why the conclusion follows from the quadrilateral formula. (Edit: Sorry, I misread your meaning, you're right. Very nice solution. ) When I encounter this problem, I conjecture that if $OD$ is perpendicular to $NM$, since if the above statement is right, the area is very easy to express. Now I will claim that $DO \perp NM$ indeed. Note that $\angle DNM=\angle DBA=\angle DCA$ and $\angle NDO=90^\circ-\angle DCA$, so $DO\perp NM$. Analogously $KM \perp OC$. I dont understand that step why those angles are equal?? or why $\angle NDO=90^\circ-\angle DCA$??

sinus Law

the perpendicularity of $MN$ and $DO$ follows from the simple angle calculation. $\angle MAN=\angle MBC=\angle KAM$ $KM=MN$ The rest is easy.

\[\angle{CAD}=\angle{DBC} \to \widehat{MN}=\widehat{MK} \to MN=MK \]Since $OD$ and $OC$ are radii, then $OC=OD$ Let $l$ be the tangent line to $(ABCD)$ at point D and let $Z$ be on $l$, such that $\angle{ADZ}=\angle{ABD}$. Then $\angle{ADZ}=\angle{ABD}=\angle{MND}$(Since $ABMN$-cyclic).$\to MN || ZD \to OD \perp NM$. Simillarly, $OC \perp KM$. So $[NOMD]=[KOMC]$, since the diagonals are equal

Proposed by Sava Grozdev, Bulgaria.