Canada Math Olympiad | A Very Nice Geometry Problem

[@ANTONIOMARTINEZ-zz4sp]

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

[@markwu2939]

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.

[@AthSamaras]

Bravo..!!!

[@ludmilaivanova1603]

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.

[@ГалинаСкворцова-ш6с]

Спасибо

[@soli9mana-soli4953]

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

[@michaeldoerr5810]

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

[@hanswust6972]

Beautiful.

[@RealQinnMalloryu4]

{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).

[@batmunkhenkhbaatar9061]

thank you

[@RAG981]

Nice one!

[@giuseppemalaguti435]

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 semplifica)..sin20sin60+sin40sin80+sin20sin80=sin80sin60

[@imetroangola4943]

Proof through trigonometry, it won't work!

[@soli9mana-soli4953]

a me seguendo il tuo metodo viene: sin20sin60+sin20sin40+sin20sin80=sin80sin60

[@mohammadrezaEskandari-r7j]

It was so difficult

[@shrikrishnagokhale3557]

Too good.Thanks