KOMATH

<Cir-2025-791>, hint

Let <AOB=2z, <OAK=y, <OBK=x A circle inscribed in angle O touches A, B

Hence <OAK=<ABK=y <OBK=<BAK=x. Since LK//OA <KLB=<KMB=<AOB=2z, L,K,B,M are cyclic

<KLB=<KMB=2z=<AOB Thus A,O, M, B are cyclic.

OB cuts MK at N, <OMA=<OBA=x+y, <BAM=x=<OBK=<OMK,

<OAM=<LKM=y, <OML=<OMN-<LMN=(x+y-x)=y=<MKL. Hence OM tangent to circle w.

<Cir-2025-791>, tangent

<Qua-2025-789>, hint

Consider a rectangle ABCD , A(0,2), B(0,0), C(4,0), D(4,2)

CM: y=-(1/4)x+1, BN: y=-(2/3)x+2, Then Q(12/5, 2/5)

Slope of AQ ; (2-2/5)/(4-12/5)=1 Hence AQ is the angle bisector of <A (=90)

<Qua-2025-789>, parallelogram, bisector

<Polygon-2025-278>, hint

Consider a regular pentagon ABCDE inscribed in a circle O .

While C,D are fixed, pull A inside O to make <A be almost 180 (max angle among A,B,C,D,E)

then at least one of B, E are pushed out of O. Adjust <B smaller than <E.

At this moment <C, <D become bigger than 108, because B, E are outside O.

Thus <A is max and <B is min