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.

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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