Arc of a Circle

\overline{AB} is a diameter of a circle of radius 1 unit. \overline{CD} is a chord perpendicular to \overline{AB} that cuts \overline{AB} at E. If the arc CAD is 2/3 of the circumference of the circle, what is the length of the segment \overline{AE}?
Source: NCTM Mathematics Teacher, September 2006

Solution
image
Circumference = 2\pi(r)=2\pi(1)=2\pi. Since measure of major arc CAD=2/3(\mathrm{circumference}), measure of minor arc CBD=1/3(\mathrm{circumference})=2\pi/3. Central angle COD which tends arc CBD therefore measures 2\pi/3 radians or 120^\circ. Triangle COD is isosceles, hence altitude OE is also the angle bisector of central angle COD. Triangle OEC is a 30^\circ\!\textrm{-}60^\circ\!\textrm{-}90^\circ triangle with hypotenuse OC equal 1. Thus, side \overline{OE} has length equal 1/2.
AE=AO+OE=1+1/2=3/2 units

Answer: 3/2 units

Alternative solution
Since chord \overline{AB} is a diameter and chord \overline{CD} is perpendicular to \overline{AB}, arc AD is a mirror image of arc AC in the \overleftrightarrow{AB} line of reflection. Points A,C, and D divide the circumference into three equal arcs each measuring 120^\circ. Hence, ACD is an equilateral triangle where the circumcenter O is also the centroid of the triangle.
By the Concurrency of Medians of a Triangle theorem
AO=(2/3)AE
AE=(3/2)AO=(3/2)(1)=3/2

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About mvtrinh

Retired high school math teacher.
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