Challenging Geometry Puzzle: Problem 1567. 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 M, N, O, K, G, H per attached sketch

Note that NC=CF=4

And OG=¼(8+12)= 5 and KD=1

OF=OM=GK= r

Triangle CHF similar to GKD ( case AA)

So HF/4= ⅕ => HF= ⅘

And EF= BC+ 2.HF= 9.6

Let K,L,M the midpoints BC,AD, EF respectively and O the circumcenter.

ReplyDeleteKL┴EF. Let BG┴AD , CH┴AD (G,H on AD)

GH=BC=8, BG=KL=CH

TrGAB=TrHDC so AG=DH=(AD-GH)/2=2

AB=AE+BE=AL+BK=6+4=10

From Pythagorean Theorem at GAB we have GB^2=AB^2-AG^2

GB=4sqrt(6)

Also is BE/AB=KM/KL (Thales)→ KM=8sqrt(6)/5 hence OM=KL-KM=2sqrt(6)/5

In right triangle MOE we have EM^2=OE^2-OM^2=576/25 → EM=24/5

Therefore EF=2EM=48/5=9.6

Georgios Kousinioris

Gastouni

Hellas