Let $H$ be the orthocentre of $ABC$ triangle.
Does the sum of disks of diameters $AH,BH,CH$ cover the whole triangle?
It seems so, each disk covers the part of triangle nearest to appropriate vertex, all disks meet "close to the centre of the triangle" and cover this region. But this is an informal observation, not a formal proof. How it can be proven?
Let the feet of the altitudes from $A$, $B$, and $C$ be $D$, $E$, and $F$, respectively.
Note that the quadrilaterals with vertices $AFHE$, $BFHD$, and $CEHD$ cover the entire triangle and are cyclic because the angles at $D$, $E$, and $F$ are right angles. The diameters of the circumcircles are the segments $AH$, $BH$, and $CH$, so the three discs will cover the entire triangle.