We can solve it with only trigonometry without any construction lines at all. Let angle ABC=b sin(3*theta)/11=sin(b)/10 sin(2*theta)/6=sin(b)/x canceling out sin(b),we get sin(3*theta)/sin(2*theta)=x*11/60 (3*sin(theta)-4*sin(theta)^3)/2*sin(theta)*cos(theta)=x*11/60 (3-4*sin(theta)^2)/cos(theta)=x*11/30 4*cos(theta)^2-1=cos(theta)*x*11/30 120*cos(theta)^2-30=cos(theta)*x*11 replacing cos(theta) with (x^2+10^2-5^2)/(2*x*10)=(x^2+75)/(20*x) and simplifying We get x^4+105*x^2-6750=0 (X^2-45)(x^2+150)=0 x=3*sqrt(5)
Boa solução! Se conseguíssemos prova que é possível construir o segmento AE=AC tal que o ângulo BEC=tetra de forma que o ponto E pertença a reta BC, em que E é um ponto a esquerda de B, provaria que BD=AD. O restante seria fácil. Como é uma questão Olímpica, sua solução foi muito plausível! Parabéns! 🎉🎉🎉
the image of a triangle looks strange to me because if we add another Theta to the right of the angle 1 Theta we should get an isosceles triangle and X should be a bisector, median and hight. Am I missing something?
Yes, turns out that triangle ABD is indeed isosceles, but we have no way of knowing this until we find that a = 3, which tells us that angle bisector also bisects the opposite side, which then tells us that triangle is isosceles. But you cannot assume from the start that ABD is isosceles.
@@MarieAnne. Why not? If BC = horizontal line, then BE = DE = 3 and AB = AD. This is an easy question with complicated answers. It is not very challenging working nearly always with pyth. triples or with tan(z) = 1/2 or tan(y) = 3/4...
This easier than it looks bc you have to apply the angle bisector twice. May I ask is it possible to apply the angle busector theorem twice IN THE ABSENCE of 2theta?
We can solve it with only trigonometry without any construction lines at all. Let angle ABC=b
sin(3*theta)/11=sin(b)/10
sin(2*theta)/6=sin(b)/x canceling out sin(b),we get
sin(3*theta)/sin(2*theta)=x*11/60
(3*sin(theta)-4*sin(theta)^3)/2*sin(theta)*cos(theta)=x*11/60
(3-4*sin(theta)^2)/cos(theta)=x*11/30
4*cos(theta)^2-1=cos(theta)*x*11/30
120*cos(theta)^2-30=cos(theta)*x*11
replacing cos(theta) with (x^2+10^2-5^2)/(2*x*10)=(x^2+75)/(20*x) and simplifying
We get x^4+105*x^2-6750=0
(X^2-45)(x^2+150)=0
x=3*sqrt(5)
Cosine rule :
5² = x² + 10² - 2.10.x.cosθ
cosθ = (x²+10²-5²)/10x
cosθ = (x²+75)/20x
Sine rule:
x / sin α = 6 / sin 2θ
10 / sin α = 11/sin 3θ
x/10 = 6/11 . sin 3θ/sin2θ
x.10.11/6 = (3sinθ-4sin³θ)/(2sinθcosθ)
55x/3 = 3/(2cosθ) - 2sin²θ/cosθ
55x/3 = 3/2 . cosθ - 2(1-cos²θ)/cosθ
Replacing cosθ by (x²+75)/20x
we can obtain x = 3√5 cm
( Solved √ )
Excellent
1:10 please explain to me where this pattern comes from, l see it first time
Boa solução! Se conseguíssemos prova que é possível construir o segmento AE=AC tal que o ângulo BEC=tetra de forma que o ponto E pertença a reta BC, em que E é um ponto a esquerda de B, provaria que BD=AD. O restante seria fácil.
Como é uma questão Olímpica, sua solução foi muito plausível! Parabéns! 🎉🎉🎉
@ 4:31 can someone explain AD^2? please. where this come from?
the image of a triangle looks strange to me because if we add another Theta to the right of the angle 1 Theta we should get an isosceles triangle and X should be a bisector, median and hight. Am I missing something?
Yes, turns out that triangle ABD is indeed isosceles, but we have no way of knowing this until we find that a = 3, which tells us that angle bisector also bisects the opposite side, which then tells us that triangle is isosceles. But you cannot assume from the start that ABD is isosceles.
@@MarieAnne. Why not? If BC = horizontal line, then BE = DE = 3 and AB = AD. This is an easy question with complicated answers. It is not very challenging working nearly always with pyth. triples or with tan(z) = 1/2 or tan(y) = 3/4...
cosθ=2/√5..α=ADC..si ottiene ,perciò teorema dei seni,sinα=2sinθ..x=6sin(α-2θ)/sin2θ=15/√5...
This easier than it looks bc you have to apply the angle bisector twice. May I ask is it possible to apply the angle busector theorem twice IN THE ABSENCE of 2theta?
Nice! φ = 30° → sin(3φ) = 1; ∆ ABC → AB = c; BC = BD + CD = 6 + 5; AC = 10; CAD = θ
DAB = 2θ; AD = x = ? BD = BE + DE = (6 - a) + a; DAE = EAB = θ →
10/5 = 2 = AE/a → AE = 2a; AQ = AD + DQ; AP = AE + PE = 2a + (10 - 2a) → PQ = CQ = z →
sin(AQC) = 1; DQ = y → QCD = CAQ = θ → y/5 = z/10 → y = z/2 →
25 = 5z^2/4 → z = 2√5 → y = √5 →
cos(θ) = z/5 = 2√5/5 = (x + y)/10 = (x + √5)/10 = 2√5/5 → x = 3√5
sin(θ) = √5/5 → sin(2θ) = 2sin(θ)cos(θ) = 2(√5/5)(2√5/5) = 4/5 = (a + 5)/10 →
40 = 5(a + 5) = 5a + 25 → 5a = 15 → a = 3 → (6 - a) = 3 → sin(BEA) = 1
fast lane: AB = AD = x → ∆ AEC = pyth. triple: 2(3 - 4 - 5) → tan(θ) = 1/2 → x = 3√5
asnwer=5 isit
asnwer=/3/5 isit
R u a girl