Canada Math Olympiad | A Very Nice Geometry Problem

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[ANTONIOMARTINEZ-zzsp]

This is a really interesting exercise. Thank you for sharing with us. I love this channel!

[harikatragadda]

∆ABC is Isosceles hence, X= BC and θ=40° and ∠ADB =60°
Reflect ∆ADB about AD to form ∆ADE.
∠ADE=60°, implies ∠EDC=60°
Reflect ∆ADE about DE to form ∆EDF.
∠EFD=40°, implies ∠FEC=20°, which in turn implies EF= FC= AE= a
Since DF=c, X= BD+DF+FC=a+b+c

[markwu]

I give another method. Make a line segment DE.The point E is on AC and DE is parallel to AB. Thus, AE=ED=DB=b, because ∠EAD=∠EDA=40°.
Then we take a point F on EC and let ∠FAD=∠AFD=40°. So AD=DF. Here, we find that ∠FDC=∠FCD=20°. So AD=DF=FC=c.
Finally, ∆DEF=∆DBA, that is EF=a. Thus, x=AC=AE+EF+FC=b+a+c.

[ludmilaivanova]

1. DE bisects the angle ADC (120 degrees). Triangles ABD and ADE are equal. AE=a.DE=b.
2. A line sgment from the point D drawn to AC under 20 degrees create a triange ADF angles AFD and DAF=40 degrees.. Triangle ADF is asosceles where AD=DF=c. Thriangle DFC is also isosceles where DF=FC=c.
3. EF =b as sides of an isosceles triange DEF in triangle DEF where anges EDF and EFD=40 degrees.
So, AC=a+b+c.

[michaeldoerr]

Two equally subtended angles result in a+b+c. I shall use that as practice!!!

[solimana-soli]

I got several values ​​for x, but none of them seem to be algebraically easy to handle. We can calculate DC with the bisector theorem and then involve c with Stewart's theorem. We can have further equations using the Cosine law (being an angle equal to 60°), but no expression obtained resolve to x = a+b+c, yet it should be possible...
here are some:
x = ab/a-b
x = ac²/a²-b²

[RealQinnMalloryu]

{80°B+80°A+20c+20A}=200°BACA {200°BACA ➖ 180°}=20°BACA 2^10.2^2^5 1^2^1 2^1 (BACA ➖ 2BACA+1).

[giuseppemalaguti]

θ=40...col teorema dei seni risulta c=(sin20/sin60)x..a=(sin20/sin80)x..b=(sin40sin20/sin80sin60)x...Risulta, dopo i calcoli..(x 'ovviamemte si