Challenging Geometry Puzzle: Problem 1580. Share your solution by posting it in the comment box provided.

Audience: Mathematics Education - K-12 Schools, Honors Geometry, and College Level.

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ReplyDeleteDefine points I, M per sketch where I is incenter of ABC

Note that B, D,C, I are concyclic

And ^(DBC)=^(DIC)= 60 and ^(IDC)=30

Angle AEC= ½ arc(ABC)= 120 => ^(DAE)=180-120-30=30

Si triangle ADE is isosceles and AD= 2.DE.cos(30)= 10.sqrt(3)

Let the tangency points on the Excircle D be U,V,W for AB, BC, CA respectively

ReplyDeleteSo < BCD = < DCW = @ (say)

Hence < BCW = 2@ and < BAC = 2@ - 60

Now AD bisects < BAC

So < DAC = 1/2. (2@ - 60) = @ - 30

And < ADC = @ - (@ - 30) = 30

So AED is a 30-30-120 Isosceles Triangle

Hence AD = 2. DE cos 30 = 10 V3

Sumith Peiris

Moratuwa

Sri Lanka