In a circle of radius r there are two parallel chords.
The distance between the chords is equal to the average of the lengths of the chords.
Find the relation between the distances from the chords to the center of the circle.
Assuming the chords are vertical in a circle of diameter 1, centred on x=0.5, some values can be assigned quite easily, thanks to Pythagoras:
AK: Distance from Chord 1 to centre
AL: Distance from Chord 2 to centre
B: Distance from origin to Chord 1 (so AK+B=0.5)
B= 0.0 AK=0.5, AL=0
B= 0.1, AK=0.4, AL=0.3
B= 1/(4 + 2 sqrt(2)), AK=sqrt(2)/4, AL=sqrt(2)/4
B= 0.2, AK=0.3, AL=0.4
B= 0.25, AK=0.25, AL=sqrt(3)/4
B=0.5, AK=0, AL=0.5
These values all fall on a circle, also with a radius of 0.5, so the relation between the distances (x,y) from the chords to the center of the circle is x^2+y^2=(1/2)^2
e.g 0.25^2+(sqrt(3)/4)^2 =1/4, 2*(sqrt(2)/4)^2 =1/4
In general: x^2+y^2=r^2
Nore1: Charlie noted that the distance of each chord from the center is half the length of the other chord.
In general, a chord's length is 2*sqrt(r^2-l^2) where l is the distance from the origin to the chord.
In the case of Chord1 this distance is x, which as shown above is sqrt(r^2-y^2), while the length of Chord2 is 2*sqrt(r^2-y^2), given that y is the distance from Chord2 to the origin. Charlie's observation is therefore correct.
Note2: The figure has other interesting properties besides that noted by Charlie. The right triangles formed by connecting an endpoint of the chords are congruent, and the angle between their hypotenuses is always a right angle also. Hence there is likely a more elegant geometric solution than the one given above.
Edited on February 11, 2022, 2:09 am
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Posted by broll
on 2022-02-10 23:13:54 |
Possible Solution
