<Qua-2026-820>, hint

triangles ABX and  YZX are similar, Let Y=t. [ABZY]=1/2+t/2, [XYZ]=2/3

Hence [XAB]/[XYZ]=1^2/t^2=(1/2+t/2+2/3)/(2/3). You get  t=2/3

<Qua-2026-820>, square m area

<South Africa MO 2012>

 

<Tri-2026-2330>, hint

Since AH=2R cosA, OD= RcosA, AH//OD, triangles POD and PAH are similar

By vectors OG=1/3(OA+OB+OC) . 

since  IOAI=IOPI, OP=-OA, OH=OA+OB+OC, OG'=OA+OP(-OA)+OH(OA+OB+OC)=1/3(OA+OB+OC). Hence G=G'

<Tri-2026-2330>, general, centroid

<South Africa MO 2012>

 

<Tri-2026-2329>, hint

Apply Cosine 2nd Law to triangles ABD and ACD with angles <ADB. <ACB

You can get  6.