The radii of the incircle and the circumcircle of a quadrilateral are 7 and 12. Find the distance between the centers of the two circles.
(In reply to
re(2): the right starting point? - kite solution by Jer)
Let ABCD be an isosceles trapezoid with AD parallel to BC and AD<BC; also AB and CD are the non-parallel sides and are congruent.
Since ABCD has an incircle AB+CD=AD+BC, which implies AB=CD=(AD+BC)/2 for an isosceles trapezoid. Let M be the midpoint of AD and N be the midpoint of BC. The center of the incircle will be the midpoint of MN; then MN is a diameter of the incircle. Making substitutions yields AB=AM+BN.
Let E be the base of the altitude from A onto BC. Then EN=AM, which makes BE=BN-AM.
Let r be the inradius. Then MN=AE=2r. ABE is a right triangle; then by Pythagorean Theorem (AM+BN)^2 = (BN-AM)^2 + (2r)^2. This simplifies into r^2 = AM*BN.
Let O be the center of the circumcircle, which will be on MN or its extension. Let x be the length of the offset; then ON=|r-x| and OM=r+x.
Let R be the circumradius. AMO and BNO are right triangles with hypotenuse of length R. Then AM^2 = R^2 - (r+x)^2 and BN = R^2 - (r-x)^2.
Then r^2 = sqrt[R^2 - (r+x)^2] * sqrt[R^2 - (r-x)^2].
Squaring, expanding, and simplifying yields R^4 - 2R^2x^2 - 2R^2r^2 + x^4 - 2x^2r^2 = 0.
Isolate r^2 to get r^2 = [R^4 - 2R^2x^2 + x^4] / [2R^2 + 2x^2]
Finally take the reciprocal of each side and use partial fraction decomposition to get 1/r^2 = 1/(R+x)^2 + 1/(R-x)^2, the same formula as from the kite.
re(3): the right starting point? - isosceles trapezoid
