 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Double Secant (Posted on 2018-02-15) 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.

 No Solution Yet Submitted by Jer No Rating Comments: ( Back to comment list | You must be logged in to post comments.) solution | Comment 2 of 3 | 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 Please log in:

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