Japanese Math Olympiad | A Very Nice Geometry Problem

[@ДмитрийИвашкевич-я8т]

tgx/tg10°=tg60°/tg20° tg3α=tgα*tg(60°-α)*tg(60°+α) tg60°=tg20°*tg40°*tg80° tgx=tg60°*tg10°/tg20° tgx=tg40°*tg10°*tg80°=tg40° x=40°

[@vcvartak7111]

But proof of tan(3a) in terms of tan(60-A) is needed?

[@ДмитрийИвашкевич-я8т]

@vcvartak7111 tgα=t tg3α=(3t-t^3)/(1-3t^2) tg3α=t(√3+t) (√3-t)/[(1-t√3) (1+t√3)]=t[(√3+t)/(1-t√3)]*[(√3-t)/(1+t√3)]=tgα*tg(60°+α)*tg(60°-α)

[@imetroangola17]

Verifiquei isso, porém, resolver isso, não é fácil!

[@ДмитрийИвашкевич-я8т]

​@imetroangola17Формула красивая, запоминается легко, может еще пригодится 🙂

[@RAG981]

I do not see what is wrong with tanx = rt3tan10/tan20, which gives x=40.

[@Z-eng0]

This was magnificent, absolutely genius. I'll never be able to know how people come up with such creative constructions, like, I can solve hard geometry puzzles, but I get stuck on this specific type. Can you tell me how your thinking process goes to know these type of problems, please

[@MathBooster]

It takes a lot of practice to get the intuition for solving geometric problems.

[@Z-eng0]

@MathBooster I know, I'm just saying that problem solving in general takes practice but it has strategies, for example, sometimes a good strategy in geometry when we see a midpoint, is constructing a circle centred at it, then constructing an extra radius touching 1 side then seeing where that goes. So I'm just asking what strategy you used here.

[@RealQinnMalloryu4]

{10°A+10°A+20°A+20°A60°C60°C}=180°AAAACC 10^10^20^20^60^60 1^1^2^2^6^6 1^2^3^3^3^3 1^2^1^1^1^3 23(AAAACC ➖ 3AAAACC+2).

[@와우-m1y]

asnwer=40 isit

[@dickroadnight]

(atan(tan(10° )/tan(20°)*tan(60°))*180/π‎ = 40

[@wasimahmad-t6c]

40