This Russian Math Olympiad Problem Will Blow Your Mind - Learn the Secret Trick! | 2 Methods

Many Students Failed to Solve This Problem | You should know this trick | 2 Methods
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[@harikatragadda]

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°

[@Irtsak]

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

[@ludmilaivanova1603]

Extend AC by x. AE=y+x. Connect E and D. Triangle DCE is isosceles, Angles E and D in that are =20degrees. Triangles ABD and ADE are equal: AD is common, AB=AE, angle ADB-100 degrees= angle ADE. Therefore, ANgle ABD=ANgle AED = 20.

[@johnbrennan3372]

Very interesting. Method one really nice

[@phungcanhngo]

Thank you professor .

[@aninditaroydebbarman8125]

The boy who is making this video has really a professor mind 🎉🎉🎉

[@michaeldoerr5810]

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.

[@holyshit922]

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

[@professorrogeriocesar]

Difícil, porém, bela questão. :)

[@comdo777]

asnwer=65 isit her ㅠ.ㅠ

[@zdrastvutye]

there is no isosceles triangle, just the sum of angles
and what the description tells:
10 print "mathbooster-this russian math olmypiad problem will blow your mind"
20 w1=80:w2=40:lx=1:ly=sin(rad(w1))/sin(rad(180-w1-w2))
30 swth=ly*sin(rad(w2))/(lx+ly):if abs(swth)>1 then stop
40 wth=deg(asn(swth)):print wth:dim x(1,2),y(1,2):lad=ly*sin(rad(w2))/sin(rad(w1))
50 wbac=180-wth-w2:dlbc=2*(lx+ly)*ly*cos(rad(wbac))
60 lbc=sqr((lx+ly)^2+ly^2-dlbc):x(0,0)=0:y(0,0)=0
70 x(0,1)=lbc-lad*cos(rad(w1))-ly*cos(rad(w2)):y(0,1)=0:x(0,2)=(lx+ly)*cos(rad(wth))
80 y(0,2)=(lx+ly)*sin(rad(wth)):x(1,0)=x(0,1):y(1,0)=0:x(1,1)=lbc:y(1,1)=0
90 x(1,2)=lbc-ly*cos(rad(w2)):y(1,2)=ly*sin(rad(w2)):fe=(x(1,2)-x(0,2))/x(1,2)*100
100 print "der fehler=";fe;"%":masx=1200/lbc:masy=850/y(0,2)
110 if masx
run in bbcbasic sdl and hit ctrl tab to copy from the results window.