Circle in a trapezium

Let $M$ denote the intersection point between the circle and the side $AB$ .
Let $E$ denote the intersection point between the circle and the side $CD$ .
Let $F$ denote the intersection point between the circle and the side $BC$ .

Denote $x$ and $y$ as the distance $EC$ and $MB$ , respectively.

We straight a straight line from $C$ such that it is parallel to $AB$ and it intersects $AB$ at a point $P$ .

Then, $MP = EC = x$ .

Because the circle is tangent to both the sides $AB$ and $CB$ , then $MB = CB = y$ .

Since the $EM = 1$ represents the diameter of the inscribed circle, and it is parallel to $AD$ , then $AD = 1$ .

And $AM = DE = \dfrac12$ can be found be comparing them to the radius of the circle.

We are given that the area of the trapezium is 4, then

$\begin{aligned} 4 &=& \dfrac12 (\text{sum of lengths of the two parallel sides}) \times AD = \dfrac12 (AB + CD) \\ 8 &=& (AM + MB) + (DE + EC) \\ &=& \left(\dfrac12 + y\right) + \left(\dfrac12 + x\right) \\ x + y &=& 7 \qquad \tag 1 \end{aligned}$

And because the triangle $CPB$ is a right triangle, then by Pythagorean theorem ,

$\begin{aligned} (CP)^2 + (PB)^2 & = &(CB)^2 \\ 1^2 + (y-x)^2 & = & (y+x)^2 \\ 4xy & = & 1 \quad \quad \quad \tag 2 \end{aligned}$

By Vieta's formula , $x$ and $y$ satisfy the equation $z^2 - 7z + \dfrac14 = 0$ . Applying the quadratic formula gives

$(x,y) = \left( \dfrac72 \pm 2\sqrt3 ,\dfrac72 \mp 2\sqrt3 \right).$

And so, we can find our ratio $\dfrac{AB}{CD}$ to be $\dfrac{AB}{CD} = \dfrac{AM + MB}{DE + EC} = \dfrac{\frac12 + y}{\frac12 + x} = 7\pm 4\sqrt3$ .

But since $AB > CD$ , then $\dfrac{AB}{CD} > 1$ , so $\dfrac{AB}{CD} = 7-4\sqrt3$ cannot be a solution.

Hence $\dfrac{AB}{CD} = 7 + 4\sqrt3 \approx \boxed{13.93}$ only.

Triangles $\triangle EBF$ and $\triangle GFC$ are similar, therefore $\frac{x}{1/2}=\frac{1/2}{y}$ giving us $x=\frac{1}{4y}$ . The area of the trapezoid can be composed of one rectangle and four triangles and written as

$A=\frac{1}{2}+\frac{1}{2}\times x+\frac{1}{2}\times y=4$

Substituting for $x$ , we get a quadratic equation in $y$

$4y^2-28y+1=0$ with solutions $y_{1}=\frac{1}{2}(7+4\sqrt{3})$ and $y_{2}=\frac{1}{2}(7-4\sqrt{3})$

Interestingly enough $x_{1}=\frac{1}{4y_{1}}=y_{2}$ and $x_{2}=\frac{1}{4y_{2}}=y_{1}$ . Taking the larger one of the two values as $x$ , in accordance with the drawing, we get the final ratio as

$\frac{x+\frac{1}{2}}{y+\frac{1}{2}}$

Quadrilateral POFB is similar to quadrilateral OQCF, so 0.5/x = y/0.5, or xy=1/4. Area = (1+x+y) (0.5) = 4. We have x+ y – 7 = 0, or x + 1/(4x) – 7 = 0.The quadratics eqs. 4x^2 – 28x +1=0, the solution x=0.0359, and y=6.9641. Then AB/CD = (0.5+y)/(0.5+X) = 13.93