Japanese Math Olympiad | A Very Nice Geometry Problem

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[@georgexomeritakis2793]

This is a great puzzle. You can solve it faster focusing on the isosceles triangle APC you constructed in the beginning. Just draw the bisector AM of angle PAC with M being the center point of base PC. Now notice that CM=CO. Suppose N is the point that AM crosses BO. Now the quadrilateral CONM comprises of two isosceles triangles OCM and ONM, sharing common base OM. Now CN is the bisector of angle OCM, so angle OCN = 80/2 = 40.

[@marcociot]

Why is CM equal to CO?

[@marcociot]

Oh, I see!
60-30-triangle: so OC is half of BC and CP = CB.

[@mihaipuiu6231]

Nice demonstration.Thank you very much.

[@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 🙂

[@MataniMath]

​@@ДмитрийИвашкевич-я8тnice solution and nice proven

[@epsom2024]

tanx*tan20°=tan10°*tan60°
tan(3θ)=tan(60°+θ)*tan(60°-θ)*tanθ
θ=20°=> tan60°=tan80°*tan40°*tan20°=> tan10°*tan60°=tan40°*tan20°=> x=40°

[@forest-c1m]

I couldn’t solve this problem.
Thank you for your explanation.

[@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.

[@RAG981]

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

[@MataniMath]

@RAG981 In ∆ADO, OD = AO.tan10 & in ∆ODC, OD = OC.tan x. In ∆ABO, OB = AO.tan20 & in ∆BOC, OB = OC.tan60 = OC.sqrt3.
So BD = AO(tan20+tan10) and
BD = OC(tan x + ✓3), then
AO(tan20 + tan10) = OC(tan x + ✓3). Since AO = OC.tan x /tan10, then (OC.tan x / tan10)(tan10+tan20) = OC(tan x + ✓3) ------> tan x = ✓3.tan10/tan20, tan x = tan40

[@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).

[@wasimahmad-t6c]

40

[@축복-l1l]

asnwer=40 isit

[@dickroadnight]

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

[@Klaatù74]

Questa é Geometria, non Matematica.

[@joegillian6781]

激ムズ！