Russian Math Olympiad | Best Geometry Problems

[@imetroangola4943]

*Outra solução:*
Usando a lei dos senos no ∆ABC:
sin C/6= sin B/3, onde B=45°-θ e C=135°-3θ= 3(45°-θ)=3B. Daí,
sin C/sin B= 6/3 =>
*sin 3B/sin B= 2 (1)*
Pelas fórmulas trigonométricas, temos:
sin 3B= 3sinB - 4(sinB)^3. Substituindo em (1), obtemos:
3 -4(sinB)^2=2=> (sin B)^2=1/4
sinB=1/2, logo B=30°. Ora, como B=45°-θ, então, *θ=15°.*

[@michaeldoerr5810]

This is a nice geometry problem that I better start practicing in order to sharpen my intuition. I think that if the angle shown was anything other than 45 degrees, it would not be an equilateral triangle. I could be wrong and I may have forgotten that there was a similar problem. And finally a geometry problem that did not require circle theorems!

[@RahulKumar-id5cq]

Could also be solved this way:
From nomenclature mentioned by math booster
AB Sin(45°-§)=AC Sin(180°-(45+3§))
2(Cos§-Sin§)=(Cos3§+Sin3§)
2(Cos§-Sin§)=(Cos§-Sin§)(1+4sin§cos§)
Either Cos§=Sin§ or 2=1+4sin§cos§;
§=45° or 15°
But if §=45° then angle BAC=180° that means ∆BAC become a Straight line that is a special case.
In general perception it doesn't remain a triangle for §=45° so it could not be the value of §.
So §=15° is answer.

[@markwu2939]

From ∠ABD =45°-θ and ∠ACD =135°-3θ, we find that ∠ACD =3∠ABD=3ϕ.
Then we know 6/sin135°=AD/sinϕ and 3/sin45°=AD/sin(3ϕ).
So sin(3ϕ)/sinϕ=2=3-4(sinϕ)², i.e. sinϕ=1/2 and ϕ=30°.
Finally, θ = 45°-ϕ =15°.

[@RahulKumar-id5cq]

Excellent solution by Math booster!

[@holyshit922]

Sine law twice , cosine law once then playing with trigonometric identities solve the problem
From sine rule
x/sin(theta) = 6/sin(135)
x = 6sqrt(2)sin(theta)
y/sin(3theta) = 3/sin(45)
y = 3sqrt(2)sin(3theta)
From cosine law
(x+y)^2 = 6^2+3^2-2*6*3*cos(4theta)
18(4sin^2(theta)+4sin(theta)sin(3theta)+sin^2(3theta)) = 9(5-4cos(4theta))
4(2sin^2(theta)) -4(-2sin(theta)sin(3theta))+2sin^2(3theta) = 5-4cos(4theta)
If we somehow show that triangle ADC is isosceles with base AD then this also would solve the problem
but how to show that length of AC is equal to the length DC

[@ludmilaivanova1603]

it would be much easier if you construct another 45 degrees angle in addition to an existed one and the length of a side will be equal to AD. Two triangles are equal in a straight triangle ADF.

[@haiduy7627]

🎉🎉😊😊❤❤

[@giuseppemalaguti435]

Dal teorema dei seni risulta 2=(cos3θ+sin3θ)/(cosθ-sinθ).?dopo i calcoli risulta θ=45(no) e sin2θ=1/2..θ=15