SMC 2001 - Death to Circle theorems!

Questions 20 and 23 on this cost me 100% damn it, and after looking at them again, especially 23, I don't know what on earth I am doing...any help on how to do these two questions would be great, and any advice on circle theorems/Shapes inside circles would be greatly appreciated!

Q. 20

In diagram (attached below) AB, CB and XY are tangents to the circle with centre O, and <ABC = 48°

What is the size of <XOY? (66° )

Q. 23

An equilateral triangle is inscribed in a circle (other attachment). Another equilateral triangle is drawn in one of the segments so that the final diagram has a line of symmetry.

What is the ratio of the length of the side of the bigger triangle to the length of the side of the smaller triangle?
([√5 + 1] : 1 )

Thanks!

without loss of generality we can choose X to be vertically above Y.

let T be the tangent point for the line XY.

now draw in the two tangent points P (near X) and Q (near Y) for the sloping tangents.

i want to show that triangles OQY, OYT,OTX and OXP are congruent.

OQ = OT (radii). OY is common to triangle OQY and OYT. angle OQY = angle OTY = 90°
which is sufficient for congruency.
hence angle QOY = angle YOT = angle TOX = angle XOP...

angle POQ = 360° - 90° - 90° - 48°.

so angle YOT = 33° and so does angle TOX...so angle YOX = 66° as required

Reply 2

the bear

KAISER_MOLE

Q. 23

An equilateral triangle is inscribed in a circle (other attachment). Another equilateral triangle is drawn in one of the segments so that the final diagram has a line of symmetry.

What is the ratio of the length of the side of the bigger triangle to the length of the side of the smaller triangle?
([√5 + 1] : 1 )

Thanks!

if the big sides are each 2b and the small sides each 2a and the radius is r...

you can show that r = b/cos30° or 2b/√3

the line from the centre O to the middle point M of the lowest chord ST is made up of btan30° plus atan60° or (b/√3 + a√3)

now do pythagoras on triangle OSM:

r²= a² + ( b/√3 + a√3)² but this must also equal (2b/√3)²

so expanding and simplifying we get

4a² + 2ab +b²/3 = 4b²/3

now divide everything by a²

4 + 2b/a +b²/(3a²) = 4b²/(3a²)

call b/a some letter t

4 + 2t +t²/3 = 4t²/3

solve this quadratically to find t which is the ratio you require....