Geometry Problem 1544. Post your solution in the comment box below.

Level: Mathematics Education, K-12 School, Honors Geometry, College.

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Draw a perpendicular AU from A to OE, U on OE. Let BF cut OE at V.

ReplyDeleteLet OA = OF = OB = a and OU = p.

AOBE is concylic and AO = OB hence OE bisects < AEB (=90).

So Triangles FEV, BEV & FREB are all right isosceles.

< OFA = < OAF = < EBD so Triangles OFE & OBE are congruent ASA.

Now FV = 5 / V2 and AU = UE = 16/V2 = 8V2.

UV = 8V2 - 5/V2 = 11/V2

Now use Pythagoras on Triangles AOU & AEU.

a^2 = p^2 + 128 ...............(1)

a^2 = (p+11/V2)^2 + 25/2.......(2)

(2) - (1); 11V2p + 121/2 = 231/2 so p = 5/V2

OE = (5 + 11 + 5)/V2 = 21/V2

S(AOE ) = (1/2) AU. OE = (1/2). (8V2).(21/V2) = 84

Sumith Peiris

Moratuwa

Sri Lanka

84

ReplyDeleteSince AE is tangent to circle with Center B, Angle AEB = 90 deg,

ReplyDeleteAnd Angle AOB = 90 deg too, hence Quadrilateral AOBE is cyclic.

Angle AEO = Angle ABO = 45 deg.

Draw OH perpendicular AE. We get Triangle OHE as right isosceles triangle. (OH=HE)

Also H is midpoint of chord of AF, hence AH=HF=11/2

And OH = HE = HF + EF = 11/2 + 5 =21/2

Area of Triangle AOE = (1/2)OH.AE = (1/2)*(21/2)*16=84 square units.

Beautiful simple solution Pradyumna

DeleteThank you Sumith !!

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