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Simplify Condition for Chord Length Equals Angle? 
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#1
Jun2412, 05:48 PM

P: 90

Consider the following:
On a circle of radius 1, two points are marked: P1 and P2. Two lines are drawn from the center of the circle: one from the center to P1, the other from the center to P2. The angle between these two lines is [itex]\theta[/itex]. One more line is drawn: from P1 directly to P2. In other words, this third line is a chord on this circle. For the special condition that the length of this chord equals the angle, find a simple expression. i.e. – find a simple expression for [itex]\theta[/itex] given the special condition that chord length = [itex]\theta[/itex] = angle = [itex]\theta[/itex]    So far, all the expressions that I have worked out mix terms of [itex]\theta[/itex] and either sin([itex]\theta[/itex]) or cos([itex]\theta[/itex]); I have not been able to find an expression simply in terms of [itex]\theta[/itex], sin([itex]\theta[/itex]), or cos([itex]\theta[/itex]). For example, following is one of my approaches: Bisect the angle [itex]\theta[/itex], which also divides the chord in half. The chord length is [itex]\theta[/itex]. But this value is also 2 sin([itex]\theta[/itex]/2) Equating these two expressions: 2 sin([itex]\theta[/itex]/2) = [itex]\theta[/itex] or sin([itex]\theta[/itex]/2) = [itex]\theta[/itex]/2 I cannot find a way to simplify this expression further. Any suggestions? 


#2
Jun2412, 07:27 PM

P: 234

You're almost there. The only x that satisfies sin(x)=x is zero, so theta/2=0, so theta=0 is the simple expression you're looking for.
To see geometrically why it's true, use the fact that on the unit circle, the arclength subtended by an angle is equal to the measure of that angle in radians. So you have two paths connecting P1 and P2: one of them is the chord, and the other is the arc along the circle. But you're requiring that their lengths are equal. The chord is a straight line, and the arc is not, so that can't happen unless they both have zero length. 


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