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 midpointtriangle 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 20200111 08:03:58 