Geometry Problem 1540. Post your solution in the comment box below.

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

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Draw the common tangent XCY at C

ReplyDeleteOG & BC are both perpendicular to AC, so OG//BC

<AFC = B = <XCY = B = <GOC = <GOA = <ACE

Hence AC is a tangent to circle FEC at C

So AE. AF = AC^2

Therefore AC^2 = 3 X 16

AC = V48 = 4V3

Sumith Peiris

Moratuwa

Sri Lanka

https://photos.app.goo.gl/wYT97Ju9ijb1Ww9a6

ReplyDeleteConnect OG and FC

Since AB= 2OC => G is the midpoint of AC and OG is the bisector of angle AOC

We have ^(AFC)=^(AOG)=^(ACD)= u

Triangle AEC similar to ACF ( case AA)

So AE/AC=AC/AF

AC^2=AC.AF= 48 and AC= 4.sqrt(3)

We can use the solution of triangles take ∠BAC = θ , ∠AED = ɣ, AC = x and radius = r;

ReplyDeletethen by property of circle ∠BOC = 2θ

Apply sine rule in ▵OAC which is an isosceles triangle,

sin(2θ)/x=sin(θ)/r

apply sine rule in ▵BAF;

sin(ɣ)/16=1/2r

apply sine rule in ▵DAE;

sin(ɣ)/(r/2)=1/3

∴sin(ɣ)=r/2

∴r=√(48)

and after further calculations θ=60°

∴x=√(48)=4√3

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