Challenging Geometry Puzzle: Problem 1579. Share your solution by posting it in the comment box provided.
Audience: Mathematics Education - K-12 Schools, Honors Geometry, and College Level.
Gain comprehensive insights! Click below to reveal the complete details.
Click for additional details.
Share your solution by clicking 'Comment' below the post or entering your solution or comment in the 'Enter Comment' field and pressing 'Publish'.
Share your solution by clicking 'Comment' below the post or entering your solution or comment in the 'Enter Comment' field and pressing 'Publish'.
Let GH = h, GF = t, GE = q BE = p and radius = r
ReplyDeleteApply Pythagoras many times
h^2 + 4 = p^2 + q^2 ...... (1) (Triangles GHB & BEG)
h^2 + 64 = t^2 ................(2) (Triangle FGH)
t^2 + r^2 = (r+p)^2 + q^2..(3) (Triangles OFG & OEG)
(2) - (1) t^2 - p^2 - q^2 = 60 ....(4)
(3) - (4) r^2 + p^2 + q^2 = (r+p)^2 + q^ - 60
From which upon simplifying
2rp = 60
So Area S(BCE) = 1/2 rp = 15
Sumith Peiris
Moratuwa
Sri Lanka
Alternate solution using similar triangles
ReplyDeleteLet FB and GE extended meet at P and let BE = p and radius = r
Let < PBE = < ABF = < OFB = @
Since GHBE is concyclic, < EGH = @
Since FOEG is concylic, < AOF = 2@ = < FGE
So < HGF = HGP = @
Hence H is the midpoint of FP and BP = 6
Now Triangles ABF & BEP are similar
So p/6 = 10/2r which yields
pr = 30 and therefore
Area S(BCE) = 1/2 rp = 15
Sumith Peiris
Moratuwa
Sri Lanka