Suppose for a positive number \(x\), we have a quadrilateral with sides \(x,2,4-x,2\) in that order. It contains an inscribed circle of radius \(x\) with another inscribed quadrilateral in it that has corresponding parallel edges. Find \(x\).
Solution
Answer: \(\frac45\)
Call the outer quadrilateral \(ABCD\) and the inner one \(EFGH\), with corresponding sides parallel. Since \(EFGH\) is cyclic, opposite angles sum to \(180^\circ\). Parallel sides transfer these angles to \(ABCD\), so it is cyclic too. Its equal opposite sides of length \(2\) make it an isosceles trapezoid.
The inradius is \(x\), so the altitude is \(2x\). The difference of the bases is \((4-x)-x=4-2x\), giving horizontal offset \(2-x\). Thus
\[2x=\sqrt{4-(2-x)^2}=\sqrt{4x-x^2},\qquad 4x^2=4x-x^2,\qquad 5x^2=4x.\]
Since \(x>0\), we obtain \(x=\frac45\).