China Math Olympiad Problem | A Very Nice Geometry Challenge | 2 Methods 

My approach Measures of angles in triangle ADB Angle DBA = 3theta Angle DAB = 90-2theta Angle ADB = 90 - theta From sine rule in ADB (x-y)/sin(3theta) = y/sin(90-theta) (x-y)/y = sin(3theta)/cos(theta) Express sin(3theta)/cos(theta) in terms of tan(theta) From triangle ABC we know that tan(2theta) = y/x x/y - 1 = sin(3theta)/cos(theta) x/y - 1 = 1/tan(2theta) - 1 After comparing these two results we will get quartic equation easy to solve but only tan(theta) = 2 - sqrt(3) will be valid solution so theta = 15

Very nice solution ❤

In △ABC, ∠C = 2θ, ∠B = 90 → ∠A = 90−2θ In △ABC, tan(2θ) = y/x → *x/y* = cot(2θ) = *cos(2θ)/sin(2θ)* In △ABD, ∠A = 90−2θ, ∠B = 3θ → ∠D = 90−θ Using law of sines in △ABD we get: AD/sin(∠B) = AB/sin(∠D) AD/AB = sin(∠B)/sin(∠D) (x−y)/y = sin(3θ)/sin(90−θ) x/y − 1 = sin(3θ)/cosθ cos(2θ)/sin(2θ) − 1 = sin(3θ)/cosθ cos(2θ) − sin(2θ) = sin(3θ) sin(2θ)/cosθ = sin(3θ) * 2 sinθ cosθ / cosθ cos(2θ) − sin(2θ) = 2 sin(3θ) cosθ Using product-to-sum identity, we get: cos(2θ) − sin(2θ) = cos(2θ) − cos(4θ) sin(2θ) = cos(4θ) sin(2θ) = 1 − 2 sin²(2θ) 2 sin²(2θ) + sin(2θ) − 1 = 0 (sin(2θ) + 1) (sin(2θ) − 1) = 0 2θ = −1 or 1/2 Since 2θ is an acute angle (it is a non-right angle of right triangle), then 0 < 2θ < 1 *sin(2θ) = 1/2 → 2θ = 30° → θ = 15°*

Both solutions are nice. The second one is amazing. Congrats

2√2cos(theta) =[tan(theta) -1]/tan(theta)

In The 3rd method you construct a triangle CBE that has an angle 2 theta. ¿When do you conclude that EC contains D ?

Dai teorema dei seni risulta (sin3θ/cosθ)+1=ctg2θ..dopo le semplificazioni(θ=-45non è accettabile)risulta(tgθ)^2-4tgθ+1=0..tgθ=2+√3,θ=75(??)...tgθ=2-√3,θ=15(ok)

Is BD perpendicular to AC?

Thita =15 degree, may be

this was painful to watch

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