Let $ABC$ be an acute-angled triangle with circumcenter $O$ and let $M$ be a point on $AB$. The circumcircle of $AMO$ intersects $AC$ a second time on $K$ and the circumcircle of $BOM$ intersects $BC$ a second time on $N$.
Prove that $\left[MNK\right] \geq \frac{\left[ABC\right]}{4}$ and determine the equality case.
Nice problem
Claim1 : CKON is cyclic.
Proof : ∠KON = 360 - (180 - ∠A) - (180 - ∠B) = ∠A + ∠B = 180 - ∠C.
Claim2 : NKM and ABC are similar.
Proof : ∠MNK = ∠MNO + ∠KNO = ∠MBO + ∠KCO = ∠OAB + ∠OAC = ∠BAC. with same approach we can prove ∠MKN = ∠ABC and ∠KMN = ∠ACB.
Now note that if we prove MK/BC ≥ 1/2 then $\left[MNK\right] \geq \frac{\left[ABC\right]}{4}$.
Claim3 : MK/BC ≥ 1/2.
Proof : MK ≥ AO . Sin ∠A = BC/2 so MK/BC ≥ 1/2.
The equality case is when MK = AO . Sin ∠A which is when M,N and K are midpoints of AB,BC and CA.