Japan Math Olympiad | A Very Nice Geometry Problem
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- Опубліковано 3 жов 2024
- Japan Math Olympiad | A Very Nice Geometry Problem
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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)
Você tinha que provar que primeiro o triângulo ABC é isósceles e depois concluir que os segmentos BO e OP estão no mesmo segmento BP.
Same methos
Your method and solution are so intresting!! Wish you the best 🙂
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!
Nice solution❤
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 !
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?
Yes , O is the center of the circle.
Nice question and nice solition
Solução linda! Parabéns!
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
BP = 2a
BO = R√2
OP = BP - BO
OP² + a² = R²
Answer: θ ≈ 20.2°
θ ≈ 24.2° not θ ≈ 20.2°
Rabbit out of the hat