This Russian Math Olympiad Problem Will Blow Your Mind - Learn the Secret Trick! | 2 Methods
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- Опубликовано: 27 май 2024
- Many Students Failed to Solve This Problem | You should know this trick | 2 Methods
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Rotate ∆ACD clockwise by 140° about C.
This makes an Isosceles ∆ACE with ∠E= 20° and ∠DAE= 80°, making ∆AED Isosceles.
Hence, AE = DE = X+Y, and this in turn makes ∆BAE Isosceles.
Therefore, θ= ∠E = 20°
Very interesting. Method one really nice
I liked the first solution, which I ended up with, after many adventures.😊
In geometry you have to use your *observation and intuition* .
AB=x+y and AC=y . *AB>AC* (cause x+y>y).
In triangle ADC : ∠DAC=60° < 80°=∠ADC => DC *x
I actually understood both methods and I do think that both methods are reasonably intricate. I also think that the secret trick making sure that if the angle can be derived and computed via double angle identity, then you can prove that the triangle has equal sides ans equal angles. And vice versa.
Thank you professor .
The boy who is making this video has really a professor mind 🎉🎉🎉
It can be done using sine law
x/sin(60) = y/sin(80)
x/y = sin(60)/sin(80)
(x+y)/sin(40) = y/sin(theta)
(x+y)/y = sin(40)/sin(theta)
x/y + 1 = sin(40)/sin(theta)
sin(60)/sin(80) + 1 = sin(40)/sin(theta)
(sin(60) + sin(80))/sin(80) = sin(40)/sin(theta)
sin(theta)/sin(40) = sin(80)/(sin(60) + sin(80))
sin(theta) = sin(80)sin(40)/(sin(60) + sin(80))
sin(theta) = sin(80)sin(40)/(2sin(70)cos(10))
sin(theta) = sin(80)sin(40)/(2sin(70)sin(80))
sin(theta) = sin(40)/(2sin(70))
sin(theta) = sin(40)/(2cos(20))
sin(theta) = 2sin(20)cos(20)/(2cos(20))
sin(theta) = sin(20) , but we know that theta must be acute angle so the only one solution is
theta = 20 degrees
Difícil, porém, bela questão. :)
asnwer=65 isit her ㅠ.ㅠ