Japan Math Olympiad Problem | A Very Nice Geometry Challenge 

Construct ∆EAC on AC such that it is Congruent to ∆ABD. ∆EDC is Equilateral and ∆ADE is Isosceles, with ∠DAE = θ+20=50 θ=30°

30 Could use trigometry given diagram. Since AD =DC, then the triangle is an isoceles, hence angle =20 degrees You can assign any length to AD Let AD = 10, then DC = 10 Use the law of cosine to find AC = 18.794 using 10 , 10 and 140 degrees Since BC = BD + DC and BD =AC then BC = 18.794 + 10 = 28.794 Use the law of cosine again using 28.794, 20, andf 18.794 hence theta = 30 degreees.

it was an explicit calculation: 10 print "mathbooster-japan math olympiad problem-a very nice geometry challenge" 20 dim x(1,2),y(1,2):w1=20:w2=180-2*w1:l2=1:l1=l2*sqr(2*(1-cos(rad(w2)))) 30 x(0,0)=0:y(0,0)=0:x(0,1)=l1:y(0,1)=0:x(0,2)=l1*(1-cos(rad(w1)))+l2:y(0,2)=l1*sin(rad(w1)) 40 x(1,0)=x(0,1):y(1,0)=y(0,1):x(1,1)=l1+l2:y(1,1)=0:x(1,2)=x(0,2):y(1,2)=y(0,2) 50 l3=sqr((x(0,0)-x(0,2))^2+(y(0,0)-y(0,2))^2):sw3=l1*sin(rad(w1))/l3 60 if abs(sw3)>1 then stop 70 w3=asn(sw3):print "der gesuchte winkel=";deg(w3):masy=850/y(0,2):masx=1200/x(1,1) 80 if masx run in bbc basic sdl and hit ctrl tab to copy from the results window

You are superb with your visualising the solution and construction to reach there. 🎉

According to the law of sines AD/sinθ=BD/sin(θ+40) and AD/sin20=BD/sin(140) and from it sin(θ)*sin140=sin(θ+40)*sin(20) so θ=30

Eu fiz pela Lei dos Senos e depois pelo truque das cotangentes; no fim, tive que negar uma das hipóteses do truque, o que me deu 30 graus.

The angle is 30°. Looks like I better use this as practice for visualizing the solution.

{20°C+20°A+90°D}=130°CAD {130°CAD180°}= 50°CAD 5^10 5^2^5 1^2^1 2^1 (CAD ➖ 2CAD+1).

AD=a,AC=t...a/sin20=t/sin140...a/sinθ=t/sin(θ+40) ..calcoli ctgθ=csc20-ctg40=√3...θ=30

30 degrees