## Sunday, May 21, 2023

### Geometry Problem 1540: Solving for the Length of Chord in a Circle: Analyzing Intersections and Given Values

Geometry Problem 1540. Post your solution in the comment box below.
Level: Mathematics Education, K-12 School, Honors Geometry, College.

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1. Draw the common tangent XCY at C

OG & 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

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

Connect 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)

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