Let ABCD be an isosceles trapezoid with bases AB and CD and a height 4r.

Inside the trapezoid are three circles of radius r. Two of the circles touch base AB, are separated by a distance r and touch opposite legs. The third circle touches base CD and both legs.

Find the ratio of the bases AB/CD.

This is a bit difficult to describe without a drawing, for me anyway, so I give a rather brief description.

Draw the pic, connect the 3 midpoints of the circles and draw sectory from midpoints to the trapezoids sides where they meet in a right angle. Also extend AD and BC, so these meet at M. The sides of the midpoint-triangle are parallel to AD, BC and AB. From there, you can work out that angle ABC has a tan of 4/3 => half that angle has a tan of 1/2 => AB=7r. For the height of triangle ABM at M we geht, thanks to knowing tan ABC, that it is 14/3 r. From here follows DC/2 = r/2, i.e. DC=r and so the ratio is 1:7.

Posted by JLo on 2020-01-11 08:03:58