Japan Math Olympiad | A Very Nice Geometry Problem 

OB =√2R. ∆ABC is Isosceles, hence, BP= 2a. If OP= b, then b=2a-√2R In the ∆OPD, R²=b²+a² This gives, 5a²-4√2Ra+R²=0 Cosθ= a/R = 1/(2√2-√3)

Your method and solution are so intresting!! Wish you the best 🙂

Nice Video , Keep on going . I sugest to mention if the provlem is solved with pure Geometry , cause sometimes we try to prove the impossible !

Nice solution❤

Nice solution! It is so nice solution that seems to be simple. Congrat

This is a nice solution because this shows that why and how the simplest solution and just staring at you in plain-sight. It just requires right angles to be identified in order to justify "a", a right-angled square, a certain circle theorem that correlates to that square, and then use the circle theorem in order to justify R, and then simplify in terms of a. I almost forgot to include that cos45 degrees is part of the first step of simplifying in terms of "a" and R. And your explanation makes more sense than the comments as always!

I want to have a go at this so I haven't listened to the whole video yet, but... You didn't mention, in the 'givens' preface, whether point O is the center of the circle. Is it?

BP = 2a BO = R√2 OP = BP - BO OP² + a² = R²

Solução linda! Parabéns!

Nice question and nice solition

No hace falta aclarar que O es el centro de la circunferencia?

Hello sir Can yoy tell me why BC = AC x COS 45 Pls reply anybody

Answer: θ ≈ 20.2°

Rabbit out of the hat