Two concentric circles of radius R and r, with R>r are both intersected by the same secant line. The points of intersection, in order, are A,B,C,D.
Prove AC*CD is constant.
Align the circles so the secant is
parallel to the x-axis, then reflect the secant to create two rectangles. Let the distance from the center of the
circles to the secant be h.<o:p></o:p>
Then from the small circle BC^2 = 4r^2 –
4h^2 and from the large circle AD^2 = 4R^2 – 4h^2.
Subtracting and factoring gives
(AD+BC)*(AD-BC) = 4R^2 – 4r^2.
Now AD = AB+BC+CD and by
symmetry AB=CD so AD=2AB+BC. Then (AD+BC)=2AB+2BC
and (AD-BC)=2AB, so (AB+BC)*(AB) = AC*CD = R^2 – r^2
|
|
Posted by xdog
on 2018-02-15 12:34:16 |
solution
