Your construction inspired me to find a second method: In ∆BPQ, sin theta=x/BP In ∆BAC, sin theta=y/AB So, x/BP=y/(x+BP) => x.(x+BP)=y.BP But, x.(x+BP)=y.y from semicircle Hence, BP=y => y^2=x^2+x.y etc.
This is a math olympiad problem that shows how useful a golden angle really is!!! The angle theta is arcsin[(sqrt(5)-1)/2]. Also I wish that there was a mnemonic device for understanding HL congruence involving beta and the right angle. No lie. I shall use this for practice!!!
2:30 ∆PQB is Congruent to ∆APC, hence PC =PB If BQ=a and QC=b, then by the Similarity of ∆PQB and ∆CQP PQ/a=b/PQ PQ=√(ab) In ∆PQC, a²=b²+(√(ab))² (b/a)²+(b/a)-1=0 b/a= ½(√5 -1) Sinθ=QC/PC=b/a=½(√5 -1) θ≈38.17°
In this figure there are 5 similar right-angled triangles, 2 of which are congruent, and it is not difficult to show that AQ is the bisector of angle A. I would have expected that theta was a notable angle! But it is not so... I found the solution by setting the radius = 1 and then calculating the tangent of theta in two different ways: tan(theta)=y/2 PC = y*tan(theta) x = BQ*tan(theta) PC = BQ x = y*tan(theta)*tan(theta) tan²(theta) = x/y and then compare. We obtain that x = y³/4 to be substituted in: y² = x² + xy (which is also obtained with AC/PQ=AB/BP) tan(theta) = (√ 2√ 5 - 2)/2 theta = 38.17°
PQ=a...r=raggio..2rcosθ=a/sinθ..r=(a/2)/sinθcosθ...(a/sinθ):a=((a/sinθ)+a):2rtgθ...semplifico la a,risulta...1/sinθ=(1+1/sinθ)/(tgθ/sinθcosθ)..1/sinθ=(cosθ)^2(1+1/sinθ)... calcoli (sinθ)^2+sinθ-1=0..sinθ=(-1+√5)/2...θ=arcsinφ=38,17
